Percentage

Name: _____________________

Date: _____________________

Instructions: Answer all questions. Write your answers clearly in the space provided.

Question 1:

The population of a town is 4.2 × 10 6 . If the population increases by 75 per 1000 per annum, then what will be the population after 2 years ?

A. 4633628
B. 4853625
C. 5253495
D. 5853615
Answer: _________
Question 2:

A's marks in Biology are 20 less than 25% of the total marks obtained by him in Biology, Maths and Drawing. If his marks in Drawings be 50, what are his marks in Maths ?

A. 40
B. 45
C. 50
D. Cannot be determined
Answer: _________
Question 3:

If x = 63% of y, then y 2 is approximately what percent of x 2 ?

A. 125%
B. 200%
C. 250%
D. 350%%
Answer: _________
Question 4:

A number is first decreased by 10% and then increased by 10%. The number so obtained is 50 less than the original number. The original number is :

A. 5000
B. 5050
C. 5500
D. 5900
E. 5900
F. 5000
G. 5500
H. 5050
Answer: _________
Question 5:

A city has a population of 300000 out of which 180000 are males. 50% of the population is literate. If 70% of the males are literate, then the percentage of female who are literate is :

A. 20%
B. 25%
C. 35%
D. 45%
Answer: _________
Question 6:

The price of an articles was increased by r%. Later the new price was decreased by r%. If the latest price was Rs. 1, then the original price was :

A. Rs. 1
B. Rs. $$left( {frac{{1 - {r^2}}}{{100}}} ight)$$
C. Rs. $$frac{{sqrt {1 - {r^2}} }}{{100}}$$
D. Rs. $$frac{{10000}}{{left( {10000 - {r^2}} ight)}}$$
Answer: _________
Question 7:

In September 2009, the sales of a product were $$frac{2}{3}$$rd of the that in July 2009. In November 2009, the sales of the product were higher by 5% as compared to September 2009. How much is the percentage of increase on sales in November 2009 with respect to the base figure in July 2009 ?

A. - 20%
B. + 25%
C. - 30%
D. + 40%
Answer: _________
Question 8:

Two numbers are respectively 25% and 60% more than a third number. The ratio of the two numbers is:

A. 20 : 30
B. 20 : 35
C. 25 : 32
D. 21 : 31
Answer: _________
Question 9:

The income of A is 80% of B's income and the expenditure of A is 60% of B's expenditure. If the income of A is equal to 90% of B's expenditure, then by what percentage are the saving of A more than B's savings?

A. 140%
B. 125%
C. 100%
D. 150%
Answer: _________
Question 10:

What is to be added to 15% of 180 so that the sum is equal to 20% of 360?

A. 45
B. 40
C. 60
D. 50
Answer: _________
Question 11:

A sample of 50 litres of glycerine is found to be adulterated to the extent of 20%. How much pure glycerine should be added to it so as to bring down the percentage of impurity to 5% ?

A. 155 litres
B. 150 litres
C. 150.4 litres
D. 140 litres
Answer: _________
Question 12:

A and B are two fixed points 5 cm apart and C is a point an AB such that AC is 3 cm. If the length of AC is increased by 6%, the length CB is decreased by :

A. 6%
B. 7%
C. 8%
D. 9%
Answer: _________
Question 13:

What percentage of the whole week does Ajay spend in office, if his office hours are 9 am to 5 pm from Monday to Friday ?

A. 33.33%
B. 23.81%
C. 25.86%
D. 42.23%
Answer: _________
Question 14:

A's salary was decreased by 50% and subsequently increased by 50%. How much percent does he lose ?

A. 25%
B. 30%
C. 20%
D. No loss
Answer: _________
Question 15:

A man spend $$7frac{1}{2}$$% of his money and after spending 75% of the remaining he had Rs. 370 left. How much money did he have :

A. 1200
B. 1600
C. 1500
D. 1400
Answer: _________
Question 16:

A man spends a part of his monthly income and saves the rest. The ratio of his expenditure to the saving is 61 : 6. If his monthly income is Rs. 8710, the amount of his monthly savings is :

A. Rs. 870
B. Rs. 690
C. Rs. 980
D. Rs. 780
Answer: _________
Question 17:

A number is increased by 15% and then decreased by 25% and the number becomes 22 less than the original number. The original number is :

A. 140
B. 160
C. 120
D. 100
Answer: _________
Question 18:

What is the difference between 0.6 and 0.6% ?

A. 5.94
B. 0.594
C. 60
D. 54
Answer: _________
Question 19:

A number is decreased by 10% and the resulting number is again decreased by 20%. What is the final percentage of decrease ?

A. 25%
B. 26%
C. 27%
D. 28%
Answer: _________
Question 20:

In an office, 40% of the staff is female. 70% of the female staff and 50% of the male staff are married. The percentage of the unmarried staff in the office is :

A. 65%
B. 42%
C. 60%
D. 64%
Answer: _________
Question 21:

If 20% of a = b, then b% of 20 is the same as:

A. 4% of a
B. 5% of a
C. 20% of a
D. None of these
Answer: _________
Question 22:

In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is $$frac{2}{3}$$ of the number of students of 8 years of age which is 48. What is the total number of students in the school?

A. 72
B. 80
C. 120
D. 150
Answer: _________
Question 23:

Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.

A. 2 : 3
B. 1 : 1
C. 3 : 4
D. 4 : 3
Answer: _________
Question 24:

A student multiplied a number by $$frac{3}{5}$$ instead of $$frac{5}{3}$$. What is the percentage error in the calculation?

A. 34%
B. 44%
C. 54%
D. 64%
Answer: _________
Question 25:

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A. 2700
B. 2900
C. 3000
D. 3100
Answer: _________
Question 26:

Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

A. 57%
B. 60%
C. 65%
D. 90%
Answer: _________
Question 27:

Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?

A. Rs. 200
B. Rs. 250
C. Rs. 300
D. None of these
Answer: _________
Question 28:

Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

A. Rs. 15
B. Rs. 15.70
C. Rs. 19.70
D. Rs. 20
Answer: _________
Question 29:

Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.

A. Rs. 6876.10
B. Rs. 6999.20
C. Rs. 6654
D. Rs. 7000
Answer: _________
Question 30:

The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

A. 4.37%
B. 5%
C. 6%
D. 8.75%
Answer: _________
Question 31:

The population of a variety of tiny bush in an experimental field increased by 10% in the first year, increased by 8% in the second year but decreased by 10% in the third year. If the present number of bushes in the experimental field is 26730, then the number of bushes in the beginning was :

A. 25000
B. 27000
C. 28000
D. 24600
Answer: _________
Question 32:

From a container having pure milk, 20% is replaced by water and the process is repeated thrice. At the end of the third operation, the milk is :

A. 40% pure
B. 50% pure
C. 51.2% pure
D. 58.8% pure
Answer: _________
Question 33:

How much $$66frac{2}{3}$$% of Rs. 312 exceeds Rs. 200 ?

A. Rs. 96
B. Rs. 4
C. Rs. 8
D. Rs. 104
Answer: _________
Question 34:

Find the number : 125% of 3060 - 85% of (?) = 408

A. 3890
B. 3940
C. 4015
D. 4020
Answer: _________
Question 35:

David and his wife each receives an 8 percent annual rise. If David receives a raise of Rs. 800 and his wife receives a raise of Rs. 840, what is the difference between their annual incomes after their raises ?

A. Rs. 40
B. Rs. 460
C. Rs. 500
D. Rs. 540
Answer: _________
Question 36:

The price of a car is Rs. 325000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?

A. Rs. 32500
B. Rs. 48750
C. Rs. 76375
D. Rs. 81250
Answer: _________
Question 37:

In an election, a total of 500000 voters participated. A candidate got 255000 votes which was 60% of total valid votes. What was the percentage of invalid votes :

A. 10%
B. 12%
C. 15%
D. $$frac{300}{17}$$%
Answer: _________
Question 38:

If x 80% of y, then what percent of 2x is y ?

A. 40%
B. $$62frac{1}{2}$$%
C. $$66frac{2}{3}$$%
D. 80%
Answer: _________
Question 39:

5 kg of tea and 8 kg of sugar together cost Rs. 172. The price of tea has risen by 20% and that of sugar by 10%. Hence the same quantities of tea and sugar now cost Rs. 199.20. What is the original price of tea per kg ?

A. Rs. 16
B. Rs. 18
C. Rs. 19
D. Rs. 20
Answer: _________
Question 40:

The charges for a five-day trip by a tourist bus for one full ticket and a half-ticket are Rs. 1440 inclusive of boarding charges which are same for a full ticket and a half-ticket. The charges for the same trip for 2 full tickets and one half-ticket inclusive of boarding charges are Rs. 2220. The fare for a half-ticket is 75% of the full ticket. Find the fare and the boarding charges separately for one full ticket.

A. Rs. 580, Rs. 400
B. Rs. 280, Rs. 200
C. Rs. 480, Rs. 300
D. Rs. 380, Rs. 400
Answer: _________
Question 41:

The number that is to be added to 10% of 320 to have the sum as 30% of 230 is :

A. 37
B. 32
C. 23
D. 73
Answer: _________
Question 42:

25% of 120 + 40% of 380 = ? of 637

A. $$frac{2}{7}$$
B. $$frac{1}{7}$$
C. $$frac{4}{7}$$
D. $$frac{3}{7}$$
Answer: _________
Question 43:

A number, on subtracting 15 from it reduces to its 80%. What is 40% of the number ?

A. 75
B. 60
C. 30
D. 90
Answer: _________
Question 44:

If 35% of A's income is equal to 25% of B's income, then the ratio of A's income to B's income is :

A. 7 : 5
B. 5 : 7
C. 4 : 7
D. 4 : 3
Answer: _________
Question 45:

If y% of one hour is, 1 minute 12 seconds, then y is equal to :

A. 2
B. 1
C. $$frac{1}{2}$$
D. $$frac{1}{4}$$
Answer: _________
Question 46:

When the price of an article was reduce by 20%, its sale increased by 80%. What was the net effect on the sale ?

A. 44% increase
B. 44% decrease
C. 66% increase
D. 75% increase
Answer: _________
Question 47:

In an examination, 34% failed in Mathematics and 42% failed in English. If 20% failed in both the subjects, the percentage of students who passed in both subjects was :

A. 54%
B. 50%
C. 44%
D. 56%
Answer: _________
Question 48:

In an election there were only two candidates. One of the candidate secured 40% of votes and is defeated by the other candidate by 298 votes. The total number of votes polled is :

A. 745
B. 1460
C. 1490
D. 1500
Answer: _________
Question 49:

In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as get a mixture of 45% alcohol strength ?

A. 1 : 2
B. 1 : 3
C. 2 : 1
D. 3 : 1
Answer: _________
Question 50:

When 60% of a number is subtracted from another number, the second number reduces to its 52%, the ratio of the first number to the second number is :

A. 6 : 5
B. 5 : 3
C. 5 : 4
D. 4 : 5
Answer: _________
Question 51:

In a factory, producing parts for an automobile, the parts manufactured on the shop floor are required to go through three quality checks, each conducted after a specific part of the processing on the raw materials is completed. Only parts that are not rejected at one stage are put through the subsequent stages of production and testing. If average rejection rates at these testing machines during a month are 10%, 5% and 2% respectively, then what is the effective rejection rate for the whole plant ?

A. 15.20%
B. 16.21%
C. 16.48%
D. 17%
Answer: _________
Question 52:

Depreciation applicable to an equipment is 20%. The value of the equipment 3 years from now will be less by :

A. 45%
B. 48.8%
C. 51.2%
D. 60%
Answer: _________
Question 53:

In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both types of lunch. If 96 people were present, how many did not eat either type of lunch ?

A. 20
B. 24
C. 26
D. 28
Answer: _________
Question 54:

1 litre of water is added to 5 litres of alcohol water solution containing 40% alcohol strength. The strength of alcohol in the new solution will be :

A. $$30\% $$
B. $$33frac{1}{3}\% $$
C. $$33frac{2}{3}\% $$
D. $$33\% $$
Answer: _________
Question 55:

Solve this : 30% of 1225 - 64% of 555 = ?

A. 10.7
B. 12.3
C. 13.4
D. None of these
Answer: _________
Question 56:

One litre of water is evaporated from a 6 litre solution containing 4% sugar. The percentage of sugar in the remaining solution is :

A. $$3frac{1}{3}\% $$
B. $$4\% $$
C. $$4frac{4}{5}\% $$
D. $$5\% $$
Answer: _________
Question 57:

Find the number : 85% of 420 + ?% of 1080 = 735

A. 25
B. 30
C. 35
D. 40
Answer: _________
Question 58:

What is 45% of 25% of $$frac{4}{5}$$ th of 850 ?

A. 67.5
B. 69.5
C. 76.5
D. 83.5
Answer: _________
Question 59:

In an examination it is required to get 36% of the aggregate marks to pass. A student gets 198 marks and is declared failed by 36 marks. What is the maximum aggregate marks a student can get ?

A. 480
B. 550
C. 650
D. Cannot be determined
Answer: _________
Question 60:

It costs Rs. 1 to photocopy a sheet of paper. However, 2% discount is allowed on all photocopies done after first 1000 sheets. How much will it cost to copy 5000 sheets of paper ?

A. Rs. 3920
B. Rs. 3980
C. Rs. 4900
D. Rs. 4920
Answer: _________
Question 61:

A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people, the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people be solicited?

A. 300
B. 250
C. 400
D. 500
Answer: _________
Question 62:

In the Bombay Stock Exchange there are 45% female employees and thus the number of male employees is exceeded by 72. Hence the total no. of employees in the BSE is:

A. 540
B. 720
C. 800
D. 550
Answer: _________
Question 63:

The average weight of a class of students is 67.5 kg. The weight of the class teacher is 25% more than the average weight of the class. The average weight of the class is less than the class teacher by x%. The value of x is:

A. 33.335
B. 25%
C. 20%
D. 22%
Answer: _________
Question 64:

Every day a mango seller sells half his stock, 10% of the stock overnight gets spoiled. If 1983 mangoes rotted over 3 nights then how many did hi start with on the first day ?

A. 25,000
B. 24,000
C. 30,000
D. 32,000
Answer: _________
Question 65:

In a factory there are three types of machine M 1 , M 2 and M 3 which produces 25%, 35% and 40% of the total products respectively. M 1 , M 2 and M 3 produces 2%, 4% and 5% defective products, respectively. what is the percentage of non-defective products ?

A. 89%
B. 97.1%
C. 96.1%
D. 86.1%
Answer: _________
Question 66:

The square of a positive number is 2000% greater than the number itself, then the square of that number is :

A. 1762
B. 1635
C. 441
D. 139
Answer: _________
Question 67:

The cost of a car is 400% greater than the cost of a bike. If there is an increase in the cost of the car is 15% and that of bike 20%. Then the total increase in the cost of the 5 cars and 10 bikes is:

A. 17.5%
B. $$16frac{3}{7}$$ %
C. 18.5%
D. 18.25%
Answer: _________
Question 68:

Connie has a number of gold bars, all of different weights. She gives the 24 lightest bars, which weigh 45% of the total weight, to Brennan. She gives the 13 heaviest bars, which weigh 26% of the total weight, to Maya. She gives the rest of the bars to Blair. How many bars did Blair receive?

A. 14
B. 15
C. 16
D. 17
Answer: _________
Question 69:

At the beginning of a year ,the owner of a jewel shop raised the price of all the jewels in his shop by x% and lowered them by x%. The price of one jewel after this up and down cycle reduced by Rs. 100. The owner carried out the same procedure after a month. After this second up-down cycle,the price of that jewel was Rs. 2304. Find the original price of that jewel(in Rs.)

A. 2500
B. 2550
C. 2600
D. 2650
Answer: _________
Question 70:

From 2000 onwards, till 2003 the price of computers increased every year by 10%. After that due to government subsidy the price of computers decreases every year by 10%. The price of a computer in 2006 will be approx. how much per cent less than the price in 2000 if the same pattern of price is continued :

A. 2
B. 3
C. 4
D. 5
Answer: _________
Question 71:

If the price of a book is first decreased by 25% and then increased by 20%, then the net charges in the price will be :

A. No change
B. 5% increase
C. 5% decrease
D. 10% decrease
Answer: _________
Question 72:

The current birth rate per thousand is 32, whereas the corresponding death rate is 11 per thousand. The net growth rate in terms of population increase in percent is given by :

A. 0.0021%
B. 0.021%
C. 2.1%
D. 21%
Answer: _________
Question 73:

A reduction of 21% in the price of wheat enables a person to buy 10.5 kg more for Rs. 100. What is the reduced price per kg ?

A. Rs. 2
B. Rs. 2.25
C. Rs. 2.30
D. Rs. 2.50
Answer: _________
Question 74:

If X is 90% of Y, then what percent of X is Y ?

A. $$90\%$$
B. $$101frac{1}{9}\% $$
C. $$111frac{1}{9}\% $$
D. $$190\% $$
Answer: _________
Question 75:

30% apples out of 450 are rotten, How many apples are in good condition ?

A. 125
B. 180
C. 240
D. 315
Answer: _________
Question 76:

Find the value : 3.2% of 500 × 2.4% of ? = 288

A. 600
B. 650
C. 700
D. 750
Answer: _________
Question 77:

76% of the students in a school are boys. If the number of girls is 204, then the total number of students is :

A. 760
B. 800
C. 850
D. 900
Answer: _________
Question 78:

If a number is reduced by 40% it becomes two-thirds of another number. What is the ratio of the first number to the second number ?

A. 8 : 9
B. 9 : 8
C. 10 : 9
D. 9 : 10
Answer: _________
Question 79:

In a mixture of milk and water, the proportion of water by weight was 75%. If in the 60 gm mixture 15 gm of water was added, what would be the percentage of water ?

A. 75%
B. 88%
C. 90%
D. 100%
Answer: _________
Question 80:

1.14 expressed as a per cent of 1.9 is:

A. 6%
B. 10%
C. 60%
D. 90%
Answer: _________
Question 81:

In an examination 80% candidates passed in English and 85% candidates passed in Mathematics. If 73% candidates passed in both these subjects, then what per cent of candidates failed in both the subjects?

A. 8
B. 15
C. 27
D. 35
Answer: _________
Question 82:

Half percent, written as a decimal, is

A. 0.2
B. 0.02
C. 0.005
D. 0.05
Answer: _________
Question 83:

If the price of the commodity is increased by 50% by what fraction must its consumption be reduced so as to keep the same expenditure on its consumption?

A. $$frac{1}{4}$$
B. $$frac{1}{3}$$
C. $$frac{1}{2}$$
D. $$frac{2}{3}$$
Answer: _________
Question 84:

The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be

A. 53,900
B. 54,000
C. 54,080
D. 54,900
Answer: _________
Question 85:

A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3cm. if the length of AC is increased by 6%, the length of CB is decreased by

A. 6%
B. 7%
C. 8%
D. 9%
Answer: _________
Question 86:

The cost of an article was Rs.75. The cost was first increased by 20% and later on it was reduced by 20%. The present cost of the article is:

A. Rs. 72
B. Rs. 60
C. Rs. 75
D. Rs. 76
Answer: _________
Question 87:

Each side of a rectangular field diminished by 40%. By how much per cent is the area of the field diminished?

A. 32%
B. 64%
C. 25%
D. 16%
Answer: _________
Question 88:

If the price of a commodity is decreased by 20% and its consumption is increased by 20%, what will be the increase or decrease in expenditure on the commodity?

A. 4% increase
B. 4% decrease
C. 8% increase
D. 8% decrease
Answer: _________
Question 89:

Two numbers A and B such that the sum of 5% of A and 4% of B is $$frac{2}{3}$$rd of the sum of 6% of A and 8% of B. The ratio A : B is -

A. 4 : 3
B. 3 : 4
C. 1 : 1
D. 2 : 3
Answer: _________
Question 90:

If a number is increased by 25% and the resulting number is decreased by 25%, then the percentage increase or decrease finally is :

A. No change
B. Decreased by $$6frac{1}{4}$$%
C. Increased by $$6frac{1}{4}$$%
D. Increased by 6%
Answer: _________
Question 91:

In an examination, 60% of the candidates passed in English and 70% of the candidates passed in Mathematics, but 20% failed in both subjects. If 2500 candidates passed in both the subjects, the number of candidates who appeared at the examination was :

A. 3000
B. 3500
C. 4000
D. 5000
Answer: _________
Question 92:

The value of a property depreciates every year by 10% of its value at the beginning of the year. The present value of the property is Rs. 8100. What was its value 2 years ago ?

A. Rs. 10000
B. Rs. $${left( {frac{{90}}{{11}}} ight)^2} imes 100$$
C. Rs. $${left( {frac{{100}}{{101}}} ight)^2} imes 8100$$
D. Rs. 9801
Answer: _________
Question 93:

If a man receives on one-fourth of his capital 3% interest, on two third 5% and one the remaining 11%, the percentage interest he receives on the whole is :

A. 4.5%
B. 5%
C. 5.5%
D. 5.2%
Answer: _________
Question 94:

The difference between the value of the number increased by 20% and the value of the number decreased by 25% is 36. Find the number ?

A. 7.2
B. 0.8
C. 720
D. 80
Answer: _________
Question 95:

The population of a village has increased annually at the rate of 25%. If at the end of 3 years it is 10000, the population in the beginning of the first year was :

A. 5120
B. 5000
C. 4900
D. 4500
Answer: _________
Question 96:

A saves 20% of his monthly salary. If his monthly expenditure is Rs. 6000, then his monthly savings is :

A. Rs. 1500
B. Rs. 1800
C. Rs. 1200
D. Rs. 4800
Answer: _________
Question 97:

A man spends 75% of his income. His income increased by 20% and he increased his expenditure by 15%. His savings will then be increased by :

A. 33%
B. $$33frac{1}{3}$$%
C. 35%
D. 40%
Answer: _________
Question 98:

Due to fall of 10% in the rate of sugar, 500 gm more sugar can be purchased for Rs. 140. Find the original rate?

A. Rs. 31.11
B. Rs. 29.22
C. Rs. 33.11
D. Rs. 32.22
Answer: _________
Question 99:

Two numbers are respectively 20% and 50% of a third number. What per cent is the first number of second?

A. 10%
B. 20%
C. 30%
D. 40%
Answer: _________
Question 100:

An empty fuel tank of a car was filled with A type petrol. When the tank was half-empty, it was filled with B type petrol. Again when the tank was half-empty, it was filled with A type petrol. When the tank was half-empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank?

A. 33.5%
B. 37.5%
C. 40%
D. 50%
Answer: _________
Question 101:

For an examination it is required to get 36% of maximum marks to pass. A student got 113 marks and failed by 85 marks. The maximum marks for the examination are:

A. 500
B. 550
C. 565
D. 620
Answer: _________
Question 102:

1% of 1% of 25% 1000 is:

A. 0.025
B. 0.0025
C. 0.25
D. 0.000025
Answer: _________
Question 103:

The population of a village increase by 5% annually. If its present population is 4410, then its population 2 years ago was:

A. 4500
B. 4000
C. 3800
D. 3500
Answer: _________
Question 104:

A spider climbed $$frac{{125}}{2}$$ % of the height of the pole in one hour and in the next hour it covered $$frac{{25}}{2}$$ % of remaining height. If pole's height is 192 m, then distance climbed in second hour is:

A. 3m
B. 5m
C. 7m
D. 9m
Answer: _________
Question 105:

What is a percent of b divided by b percent of a?

A. a
B. b
C. 1
D. 10
Answer: _________
Question 106:

In an election 4% of votes cast are invalid. A candidate gets 55% of casted votes and wins the election by 4200 votes. Find the total number of votes casted.

A. 30000
B. 43750
C. 45000
D. 42000
Answer: _________
Question 107:

In a mixture there is 15% of salt. When 30 liter of water is evaporated, salt becomes 20% of mixture. Find the initial quantity of mixture.

A. 120 liters
B. 110 liters
C. 125 liters
D. 115 liters
Answer: _________
Question 108:

The population of a town is 8500. It increases by 20% in the first year and by another 25% in the second year. What would be the population of the town after 2 years ?

A. 10950
B. 11950
C. 12550
D. 12750
Answer: _________
Question 109:

In a survey of a city, it was found that 90 percent of the people in the city own a refrigerator and 15 percent own a washing machine. If everybody owns at least one appliance, what percentage owns both ?

A. 5 percent
B. 8 percent
C. 10 percent
D. None of these
Answer: _________
Question 110:

A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3 cm. If the length of AC is increased by 6%, the length of CB is decreased by :

A. 6%
B. 7%
C. 8%
D. 9%
Answer: _________
Question 111:

In an examination, 34% of the students failed in Mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then the percentage of students who passed in both the subjects was :

A. 44%
B. 50%
C. 54%
D. 56%
Answer: _________
Question 112:

If x% of a is the same as y% of b, then z% of b is :

A. $$frac{{xy}}{z}\% { ext{ of }}a$$
B. $$frac{{yz}}{x}\% { ext{ of }}a$$
C. $$frac{{xz}}{y}\% { ext{ of }}a$$
D. None of these
Answer: _________
Question 113:

In the price of sugar falls by $$2frac{1}{2}\% $$ , a person can buy 9 kg more of sugar for Rs. 1260 than before. If the price had risen by $$12frac{1}{2}\% $$ , how much sugar would he have bought for the same sum ?

A. 288 kg
B. 312 kg
C. 328 kg
D. 336 kg
Answer: _________
Question 114:

At a special sale, 5 tickets can be purchased for the price of 3 tickets. If 5 tickets are purchased at the sale, the amount saved will be what percent of the original price of the 5 tickets ?

A. 20%
B. $$33frac{1}{3}$$ %
C. 40%
D. 60%
Answer: _________
Question 115:

The price of sugar per kg increased from Rs. 16 to Rs. 20. The percentage reduction in the use of sugar so that the expenditure does not increase, should be :

A. 15%
B. 20%
C. 25%
D. 40%
Answer: _________
Question 116:

If 50% of (x - y) = 30% of (x + y), then what percent of x is y ?

A. 25%
B. $$33frac{1}{3}$$%
C. 40%
D. 400%
Answer: _________
Question 117:

In a History examination, the average for the entire class was 80 marks. If 10% of the students scored 95 marks and 20% scored 90 marks, what was the average marks of the remaining students of the class ?

A. 65.5
B. 72.5
C. 75
D. 85
Answer: _________
Question 118:

If (x + 20)% of 250 is 25% more than x% of 220, then 10% of (x + 50) is what percent less than 15% of x?

A. $$16frac{2}{3}$$
B. $$8frac{1}{3}$$
C. $$13frac{1}{3}$$
D. $$33frac{1}{3}$$
Answer: _________
Question 119:

The ratio of expenditure to savings of a woman is 5 : 1. If her income and expenditure are increased by 10% and 20%, respectively, then find the percentage change in her savings.

A. 55%
B. 60%
C. 50%
D. 40%
Answer: _________
Question 120:

A is 80% more than B and C is $$48frac{4}{7}\% $$ xa0less than the sum of A and B. By what percent is C less than A?

A. 30
B. 15
C. 25
D. 20
Answer: _________
Question 121:

Rohit's income is Rs. 32000. If his expenses is 30 percent of total income, then what will be the saving of Rohit?

A. Rs. 18600
B. Rs. 22400
C. Rs. 19200
D. Rs. 24600
Answer: _________
Question 122:

The price of petrol shot up by 5%. Before the hike, the price was Rs. 82 per litre. A man travels 3045 km every month and his car gives a mileage of 15 km per litre. What is the increase in the monthly expenditure (to the nearest Rs.) on the man's travel due to the hike in the petrol prices?

A. 832
B. 859
C. 758
D. 944
Answer: _________
Question 123:

If each edge of a cube is increased by 10% then the percentage increase in its surface area is:

A. 21%
B. 19%
C. 22%
D. 20%
Answer: _________
Question 124:

If the numerator of a fraction is increased by 15% and denominator is decreased by 20%, then the fraction, so obtained, is $$frac{{17}}{{65}}.$$ xa0What is the original fraction?

A. $$frac{{281}}{{1495}}$$
B. $$frac{{278}}{{1495}}$$
C. $$frac{{267}}{{1495}}$$
D. $$frac{{272}}{{1495}}$$
Answer: _________
Question 125:

In a two-candidate election, 10% of the voters did not cast their ballots. 10% of the votes cast were found invalid. The winning candidate received 54% of the valid votes and a 1620-vote majority. Find the number of people on the voter list who have registered to vote.

A. 25000
B. 26000
C. 24500
D. 25500
Answer: _________
Question 126:

If the sum of 40% of a number and 30% of the same number is 70, then the number is:

A. 125
B. 100
C. 150
D. 200
Answer: _________
Question 127:

Two candidates P and Q contested in an election. 70% of the registered voters are P supporters. If 60% of the P supporters and 30% of the Q supporters are expected to vote for candidate P, then what percentage of the registered voters are expected to vote for candidate P?

A. 30%
B. 51%
C. 26%
D. 47%
Answer: _________
Question 128:

The actual area of a rectangle is 60 Cm 2 , but while measuring its length a student decreases it by 20% and the breadth increases by 25%. The percentage error in area, calculated by the student is :

A. 5%
B. 15%
C. 20%
D. No change
Answer: _________
Question 129:

The cost of packaging of the mangoes is 40% the cost of fresh mangoes themselves. The cost of mangoes increased by 30% but the cost of packaging decreased by 50%, then the percentage change of the cost of packed mangoes, if the cost of packed mangoes is equal to the sum of the cost of fresh mangoes and cost of packaging :

A. 14.17%
B. 7.14%
C. 8.87%
D. 6.66%
Answer: _________
Question 130:

220% of a number X is 44. What is 44% of X.

A. 8.8
B. 8.9
C. 6.6
D. 7.7
Answer: _________
Question 131:

The shopkeeper increased the price of a product by 25% so that customer finds difficult to purchase the required amount. But Somehow the customer managed to purchase only 70% of the required amount. What is the net difference in the expenditure on that product ?

A. 55 more
B. 10% more
C. 12.5% less
D. 17.5% less
Answer: _________
Question 132:

A customer asks for the production of x number of goods. The company produces y number of goods daily. Out of which z% are units for sale. The order will be completed in :

A. $$ {frac{x}{{100y imes left( {1 - z} ight)}}} ,{ ext{days}}$$
B. $$ {frac{{100yz}}{x}} ,{ ext{days}}$$
C. $$ {frac{{100x}}{{left( {100 - z} ight)y}}} ,{ ext{days}}$$
D. $$frac{{100}}{{y imes left( {z - 1} ight)}},{ ext{days}}$$
Answer: _________
Question 133:

In the Science City, Kolkata the rate of the ticket is increased by 50% to increased the revenue but simultaneously 20% of the visitor decreased. What is percentage change in the revenue. if it is known that the Science city collects one revenue only from the visitors and it has no other financial supports:

A. +20%
B. -25%
C. +30%
D. -30%
Answer: _________
Question 134:

600 students took the test on Physics and chemistry. 35% students failed in Physics and 45% students failed in chemistry and 40% of those who passed in chemistry also passed in Physics, then how many students failed in both :

A. 16
B. 15
C. 13
D. 12
Answer: _________
Question 135:

An alloy contains the copper and aluminum in the ratio of 7 : 4 While making the weapons from this alloy, 12% of the alloy got destroyed. If there is 12 kg of aluminum in the weapon, then weight of the alloy required is :

A. 14.4 kg
B. 37.5 kg
C. 40 kg
D. 48 kg
Answer: _________
Question 136:

80% of a smaller number is 4 less than 40% of a larger number. The larger number is 85 greater than the smaller one. The sum of these two number is

A. 325
B. 425
C. 235
D. 500
Answer: _________
Question 137:

A number x is mistakenly divided by 10 instead of being multiplied by 10. what is the percentage error in the result?

A. -99%
B. -100%
C. +99%
D. +100%
Answer: _________
Question 138:

5 kg of metal A and 20 kg of metal B are mixed to form an alloy. The percentage of metal A in the alloy is ?

A. 20%
B. 25%
C. 40%
D. None of these
Answer: _________
Question 139:

23% of 8040 + 42% of 545 = ? % of 3000

A. 56.17
B. 63.54
C. 69.27
D. 71.04
Answer: _________
Question 140:

While purchasing one item costing Rs. 400, I had to pay the sales tax at 7% and on another costing Rs. 6400, the sales tax was 9%. What percent of the sales tax I had to pay, taking the two items together on an average ?

A. 8%
B. $$8frac{13}{17}$$%
C. $$8frac{15}{17}$$%
D. $$8frac{1}{2}$$%
Answer: _________
Question 141:

The difference between 54% of a number and 26% of the same number is 22526. What is 66% of that number ?

A. 48372
B. 49124
C. 51218
D. 53097
Answer: _________
Question 142:

If 75% of a number is added to 75, then the result is the number itself. The number is :

A. 50
B. 60
C. 300
D. 400
Answer: _________
Question 143:

40% of 60% of 32% of an amount is Rs. 432. What is the amount ?

A. Rs. 5000
B. Rs. 5600
C. Rs. 6400
D. None of these
Answer: _________
Question 144:

In a test, minimum passing percentage for girls and boys is 35% and 40% respectively. A boy scored 483 marks and failed by 117 marks. What is the minimum passing marks for girls ?

A. 425
B. 500
C. 520
D. 625
Answer: _________
Question 145:

A clothing supplier stores 800 coats in a warehouse, of which 15 percent are full-length coats. If 500 of the short-length coats are removed form the warehouse, then what percent of the remaining costs are full length ?

A. 5.62%
B. 9.37%
C. 35%
D. 40%
Answer: _________
Question 146:

Peter could save 10% of his income. But two years later when his income is increased by 20%, he could save the same amount only as before. By how much percent has his expenditure increased ?

A. 22%
B. $$22frac{2}{9}\% $$
C. $$23frac{1}{3}\% $$
D. 24%
Answer: _________
Question 147:

A papaya tree was planted 2 years ago. It grows at the rate of 20% every year. If at present, the height of three is 540 cm, what was it when the tree was planted ?

A. 324 cm
B. 375 cm
C. 400 cm
D. 432 cm
Answer: _________
Question 148:

The ratio of the number of boys and girls in school at 8 : 12. If 50% of boys and 25% of girls are getting scholarship for their studies, what is the percentage of school students who are not getting any scholarships ?

A. 65%
B. 66%
C. 67%
D. 68%
Answer: _________
Question 149:

In an examination there were 640 boys and 360 girls, 60% of boys and 80% of were successful. The percentage of failure was :

A. 20%
B. 60%
C. 30.5%
D. 32.8%
Answer: _________
Question 150:

In an election between two candidates, 75% of the voters cast their votes, out of which 2% votes were declared invalid. A candidate got 9261 votes which were 75% of the valid votes. The total number of voters enrolled in that election was :

A. 16000
B. 16400
C. 16800
D. 18000
Answer: _________
Question 151:

1 litre of water is added to 5 litres of alcohol-water solution containing 40% alcohol strength. The strength of alcohol in the new solution will be :

A. 30%
B. 33%
C. $$33frac{2}{3}$$%
D. $$33frac{1}{3}$$%
Answer: _________
Question 152:

Each side of a rectangular field is diminished by 40%. By how much percent is the area of the field diminished ?

A. 32%
B. 64%
C. 25%
D. 16%
Answer: _________
Question 153:

There is a ratio of 5 : 4 between two numbers. If 40% of the first number is 12, then what would be 50% of the second number ?

A. 12
B. 24
C. 18
D. Data Inadequate
Answer: _________
Question 154:

The ratio of number of boys and girls in a school 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1 ?

A. 90
B. 120
C. 220
D. 240
Answer: _________
Question 155:

On a certain date, Pakistan has a success rate of 60% against India in all the ODI's played between the two countries. The lost the next 30 ODI's. In a row to India and their success rate comes down to 30%. The total number of ODI's played between the two countries is :

A. 50
B. 45
C. 60
D. 30
Answer: _________
Question 156:

A person who spends $$66frac{2}{3}$$% of his income is able to save Rs. 1200 per month. His monthly expenses (in Rs.) is :

A. 1200
B. 2400
C. 3000
D. 3200
Answer: _________
Question 157:

The cost of an article worth Rs. 100 is increased by 10% first and again increased by 10%. The total increase in rupees is :

A. 20
B. 21
C. 110
D. 121
Answer: _________
Question 158:

Income tax is raised from 4 paise to 5 paise in a rupee but the revenue is increased by 10% only. Find the decrease percent in the amount taxed.

A. 12%
B. 14%
C. 16%
D. None of these
Answer: _________
Question 159:

The population of a town is 189000. It decreases by 8% in the first year and increases by 5% in the second year. What is the population of the town at the end of 2 year ?

A. 182574
B. 185472
C. 191394
D. 193914
Answer: _________
Question 160:

Two vessels contain equal quantities of 40% alcohol. Sachin changed the concentration of the first vessel to 50% by adding extra quantity of pure alcohol. Vivek changed the concentration of the second vessel to 50% replacing a certain quantity of the solution with pure alcohol. By what percentage is the quantity of alcohol added by Sachin more/less than that replaced by Vivek ?

A. $$11frac{1}{9}$$% less
B. $$11frac{1}{9}$$% more
C. $$16frac{2}{3}$$% less
D. $$20$$% more
Answer: _________
Question 161:

Nandini Basu bought an article for Rs. 5844. She gave Rs. 156 to a mechanic to remove its defect. She then sold it for Rs. 5700. What was her loss percent ?

A. 5%
B. 5.5%
C. 2.5%
D. 2.46%
Answer: _________
Question 162:

Rajeev buy goods worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sale tax @ 10%. Find the amount he will have to pay for the goods ?

A. Rs. 6876.10
B. Rs. 6999.20
C. Rs. 6654
D. Rs. 7000
Answer: _________
Question 163:

In a year, a man manages to sell only 65% of the chicken he owns. How many chicken should the man own to sell 47775 chicken in a years ?

A. 55000
B. 68500
C. 73000
D. 82500
Answer: _________
Question 164:

When 125 is subtracted from a number, it reduces to its 37.5 percent. What is 25 percent of that number ?

A. 50
B. 75
C. 125
D. 175
Answer: _________
Question 165:

In an examination it is required to get 40% of the aggregate marks to pass. A student get 261 marks and is declared failed by 4% marks. What are the maximum aggregate marks a student can get ?

A. 700
B. 730
C. 745
D. 765
Answer: _________
Question 166:

If x% of y is the same as $$frac{4}{5}$$ of 80, then the value of xy is :

A. 320
B. 400
C. 640
D. None of these
Answer: _________
Question 167:

The contents of a certain box consist of 14 apples and 23 oranges. How many oranges must be removed from the box so that 70% of the pieces of fruit in the box will be apples ?

A. 6
B. 12
C. 17
D. 36
Answer: _________
Question 168:

If house tax is paid before the due date, one gets a reduction of 12% on the amount of the bill. By paying the tax before the due date, a person got a reduction of Rs. 2,100. The amount (in Rs.) of house tax was:

A. 21,000
B. 17,500
C. 25,000
D. 18,000
Answer: _________
Question 169:

If the price of eraser is reduced by 25%. A person can buy 2 more erasers for a rupee. How many erasers are available for a rupee after reduction?

A. 8
B. 6
C. 4
D. 2
Answer: _________
Question 170:

$$frac{{11}}{5}$$ of a number A is 22% of a number B. The number B is equal to 2.5% of a third number C. If the value of C is 5500, then the sum of 80% of A and 40% of B is:

A. 88
B. 75
C. 48
D. 66
Answer: _________
Question 171:

The base of a triangle is increased by 40%. By what percentage (correct to two decimal places) should its height be increased so that the area increases by 60%?

A. 14.29%
B. 20.01%
C. 15.54%
D. 18.62%
Answer: _________
Question 172:

A class has five sections that have 25, 30, 40, 45 and 60 students, respectively. The pass percentage of these section are 20%, 30%, 35%, 40% and 100% respectively. The pass percentage of the entire class is:

A. 87%
B. 63%
C. 53%
D. 79%
Answer: _________
Question 173:

A crate of fruits contains one spoiled fruit for every 25 fruits. 60% of the spoiled fruits were sold. If the seller had sold 48 spoiled fruits, then the number of fruits in the crate were:

A. 2000
B. 2400
C. 3000
D. 1200
Answer: _________
Question 174:

Rice is now being sold at Rs. 29 per kg. During the last month, its cost was Rs. 25 per kg. By how much percentage should a family reduce its consumption, so as to keep the expenditure the same as before? (correct to nearest integer)

A. 14%
B. 13%
C. 15%
D. 12%
Answer: _________
Question 175:

Some students (only boys and girls) from different schools appeared for an Olympiad exam. 20% of the boys and 15% of the girls failed the exam. The number of boys who passed the exam was 70 more than that of the girls who passed the exam. A total of 90 students failed. Find the number of students that appeared for the exam.

A. 420
B. 400
C. 500
D. 350
Answer: _________
Question 176:

If 25% of half of x is equal to 2.5 times the value of 30% of one fourth of y, then x is what percent more or less than y?

A. 50% more
B. $$33frac{1}{3}\% $$ xa0less
C. $$33frac{1}{3}\% $$ xa0more
D. 50% less
Answer: _________
Question 177:

A fruit seller sells 45% of the oranges that he has along with one more orange to a customer. He then sells 20% of the remaining oranges and 2 more oranges to a second customer. He then sells 90% of the now remaining oranges to a third customer and is still left with 5 oranges. How many oranges did the fruit seller have initially?

A. 100
B. 111
C. 121
D. 120
Answer: _________
Question 178:

A's salary is 50% more than that of B. Then B's salary is less than that of A by :

A. $$50$$%
B. $$33frac{1}{3}$$%
C. $$33frac{1}{4}$$%
D. $$44frac{1}{2}$$%
Answer: _________
Question 179:

If 60% of A = 30% of B, B = 40% of C and C = x% of A, then value of x is :

A. 800%
B. 200%
C. 300%
D. 500%
Answer: _________
Question 180:

In an examination 70% of the candidate passed in English, 80% passed in Mathematics, 10% failed in both subjects. If 144 candidates passed in both, the total number of candidates was :

A. 125
B. 200
C. 240
D. 375
Answer: _________
Question 181:

Raman's salary is increased by 5% this year. If his present salary is Rs. 1806, the last year's salary was :

A. Rs. 1720
B. Rs. 1620
C. Rs. 1520
D. Rs. 1801
Answer: _________
Question 182:

If 80% of A = 50% of B and B = x% of A, then the value of x is :

A. 400
B. 300
C. 160
D. 150
Answer: _________
Question 183:

The percentage of metals in a mine of lead ore is 60%. Now the percentage of silver is $$frac{3}{4}$$% of metals and the rest is lead. If the mass of ore extracted from this mine is 8000 kg, the mass (in kg.) of lead is :

A. 4763
B. 4764
C. 4762
D. 4761
Answer: _________
Question 184:

If 120 is 20% of a number, then 120% of that number will be :

A. 20
B. 120
C. 480
D. 720
Answer: _________
Question 185:

498 is 17% less than a number then the number is :

A. 610
B. 580
C. 600
D. 620
Answer: _________
Question 186:

If radius of a circle is increased by 5%, then the increase in it's area is :

A. 10.25%
B. 5.75%
C. 10%
D. 5%
Answer: _________
Question 187:

If 60% of A's income is equal to 75% of B's income, then B's income is equal to x% of A's income. The value of x is :

A. 70
B. 60
C. 80
D. 90
Answer: _________
Question 188:

A dozen pairs of socks quoted at Rs. 180 are available at discount of 20%. How many pairs of socks can be bought for Rs. 48 ?

A. 3 pairs
B. 4 pairs
C. 2 pairs
D. 5 pairs
Answer: _________
Question 189:

If the population of a town is 64000 and its annual increase is 10%, then its population at the end of 3 years will be :

A. 80000
B. 85184
C. 85000
D. 85100
Answer: _________
Question 190:

In a village three people contested for the post of village Pradhan. Due to their own interest, all the voters voted and no one vote was invalid. The losing candidate got 30% votes. What could be the minimum absolute margin of votes by which the winning candidate led by the nearest rival, if each candidate got an integral percent of votes ?

A. 4
B. 2
C. 1
D. None of these
Answer: _________
Question 191:

I paid Rs. 27.20 as sales tax on a watch worth Rs. 340. Find the rate of sales tax.

A. 8%
B. 9%
C. 10%
D. 12%
Answer: _________
Question 192:

A student scores 55% marks in 8 papers of 100 marks each. He scores 15% of his total marks in English. How much does he score in English ?

A. 44
B. 45
C. 66
D. 77
Answer: _________
Question 193:

A wrist watch of cost price Rs. 1250 was sold by Sharel for Rs. 1500. What was the profit percent ?

A. 21%
B. 24%
C. 25%
D. 20%
Answer: _________
Question 194:

The difference of two numbers is 20% of the larger number. If the smaller number is 20, then the larger number is :

A. 25
B. 45
C. 50
D. 80
Answer: _________
Question 195:

Two candidates fought an election. One of them got 62% of the total votes and won by 432 votes. What is the total number of votes polled ?

A. 1500
B. 1600
C. 1800
D. Cannot be determined
Answer: _________
Question 196:

If x, y, z are three positive integers such that x is greater then y and y is greater than z, then which of the following is definitely true ?

A. x% of y is greater than y% of z
B. y% of x is greater than z% of y
C. z% of x is greater than y% of z
D. All of these
Answer: _________
Question 197:

If the length of a rectangle is increased by 12% and the breadth is decreased by 8%, the net effect on the area is:

A. increase by 3.04%
B. increase by 2.6%
C. decrease by 3.04%
D. decrease by 2.6%
Answer: _________
Question 198:

Raju, Ravi and Ashok contested an election. 5% votes polled were invalid. Raju got 30% of the total votes. Ravi got 32% of the total votes. The winner got 5136 more votes than the person who received the least number of votes. Find the total number of votes polled.

A. 171200
B. 64200
C. 171220
D. 172100
Answer: _________
Question 199:

A certain number of student from school X appeared in an examination and 30% student failed. 150% more students than more from school X, appeared in the same examination from school Y, If 80% of the total number of students who appeared from X and Y passed, then what is the percentage of student who failed from Y?

A. 24
B. 20
C. 16
D. 18
Answer: _________
Question 200:

When an article is sold at 5% discount, then there is a profit of 14%. If the discount is 11%, then what will be the profit?

A. 6.8%
B. 7.2%
C. 7.6%
D. 8.4%
Answer: _________
Question 201:

Three candidates P, Q and R participated in an election. P got 35% more votes than Q, and R got 15% more votes than Q. P over took R by 2,412 votes. If 90% voters voted and no invalid or illegal votes were cast, then what was the number of voters in the voting list?

A. 46,900
B. 42,800
C. 42,210
D. 48,500
Answer: _________
Question 202:

In a manufacturing unit, it was noted that the price of raw material has increased by 25% and the labor cost has gone up from 30% of the cost of raw material to 38% of the cost of the raw material. What percentage of the consumption of raw material be reduced to keep the cost the same as that before the increase?

A. 20.7%
B. 30.2%
C. 24.6%
D. 25.5%
Answer: _________
Question 203:

Ramesh spends 40% of his monthly salary on food, 18% on house rent, 12% on entertainment, and 5% on conveyance. But due to a family function, he has to borrow Rs. 16,000 from a money lender to meet the expenses of Rs. 20,000. His monthly salary is:

A. Rs. 18,000
B. Rs. 16,500
C. Rs. 16,000
D. Rs. 15,000
Answer: _________
Question 204:

The value of a motorcycle depreciates every year by 4%. What will be its value after 2 years, if its present value is Rs. 75,000?

A. Rs. 72,000
B. Rs. 70,120
C. Rs. 69,120
D. Rs. 69,000
Answer: _________
Question 205:

The price of diesel is increased by 26%. A person wants to increase his expenditure by 15% only. By what percentage, correct to one decimal place, should he decrease his consumption?

A. 7.2%
B. 6.5%
C. 8.7%
D. 9.5%
Answer: _________
Question 206:

If X is 20% less then Y, then find the value of $$frac{Y - X}{Y}$$xa0 and $$frac{X}{X - Y}$$ :

A. $$frac{1}{5}$$, - 4
B. 5, $$frac{1}{4}$$
C. $$frac{2}{5}$$, $$-frac{5}{2}$$
D. $$frac{3}{5}$$, $$-frac{3}{5}$$
Answer: _________
Question 207:

The ratio of the number of boys to that of girls in a school is 4 : 1. If 75% of boys and 70% of the girls are scholarship holders, then the percentage of students who do not get scholarship is :

A. 50%
B. 28%
C. 75%
D. 26%
Answer: _________
Question 208:

A man had a certain amount with him. He spent 20% of that to buy any article and 5% of the remaining on transport. Then he gifted Rs. 120. If he is left with Rs. 1400, the amount he spent on transport is :

A. Rs. 76
B. Rs. 61
C. Rs. 95
D. Rs. 80
Answer: _________
Question 209:

The income of a company increases 20% per annum. If its income is Rs. 2664000 in the year 2012. Then its income in the year 2010 was :

A. Rs. 2120000
B. Rs. 1850000
C. Rs. 2820000
D. Rs. 2855000
Answer: _________
Question 210:

An individual pays 30% income tax. On this he has to pay a surcharge of 10%. Thus, the net tax rate, he has pay is :

A. 45%
B. 40%
C. 33%
D. 27%
Answer: _________
Question 211:

An army lost 10% of its men in war. 10% of the remaining died due to disease and 10% of the rest were declared disabled. Thus the strength of the army was reduced to 729000 active men. The original strength of the army was :

A. 1500000
B. 1000000
C. 1200000
D. 1100000
Answer: _________
Question 212:

In an examination, 35% of total students failed in Hindi, 45% failed in English and 20% failed in both. Find the percentage of those students who passed in both the subjects ?

A. 45%
B. 35%
C. 20%
D. 40%
Answer: _________
Question 213:

The price of an article was increased by r%. Later the new price was decreased by r%. If the latest price was Rs. 1, then the original price was :

A. Rs. 1
B. Rs. $$frac{{1 - {r^2}}}{{100}}$$
C. Rs. $$frac{{sqrt {1 - {r^2}} }}{{100}}$$
D. Rs. $$left( {frac{{10000}}{{10000 - {r^2}}}} ight)$$
Answer: _________
Question 214:

The difference of two numbers is 15% of larger sum. The ratio of the larger number to the smaller number is :

A. 23 : 17
B. 11 : 9
C. 17 : 11
D. 23 : 11
Answer: _________
Question 215:

A number if reduced by 25% becomes 225. By what percent should it be increased so that it becomes 375 ?

A. 25%
B. 30%
C. 35%
D. 75%
Answer: _________
Question 216:

A shopkeeper bought 20 kg of rice at Rs. 55 per kg, 25 kg of rice at Rs. 50 per kg, and 35 kg of rice at Rs. 60 per kg. He spent a sum of Rs. 150 on transportation. He mixed all the three types of rice and sold all the stock at Rs. 62.56 per kg. His profit percent in the entire transaction is:

A. 8.8
B. 12.5
C. 10.5
D. 9.2
Answer: _________
Question 217:

The monthly salary of a person was Rs. 75,000. He used to spend on Family Expenses (E), Taxes (T), Charity (C) and rest were his savings. E was 60% of the income, T was 20% of E, and C was 15% of T. When his salary got raised by 40% he maintained the percentage level of E, but T became 30% of E and C became 20% of T. The ratio of the saving of his earlier salary to that of his present salary is:

A. 655 : 644
B. 325 : 337
C. 644 : 655
D. 337 : 325
Answer: _________
Question 218:

If 49% of X = Y, then Y% of 50 is:

A. 40% of Y
B. 24.5% of Y
C. 50% of X
D. 24.5% of X
Answer: _________
Question 219:

In a class, if 60% of the students are boys and the number of girls is 36, then the number of boys is:

A. 65
B. 54
C. 60
D. 58
Answer: _________
Question 220:

When the price of an item was reduced by 20%, its sale increased by x%. If there is an increase of 25% in receipt of the revenue, then the value of x is:

A. 54.35
B. 57.75
C. 56.25
D. 55.35
Answer: _________
Question 221:

A student multiplied a number with $$frac{3}{4}$$ instead of $$frac{4}{3}.$$ What is the error percentage?

A. 59.67%
B. 43.75%
C. 67.45%
D. 39.34%
Answer: _________
Question 222:

A person saves $$33frac{1}{3}\% $$ xa0of his income. If the saving increases by 22% and the expenditure increases by 10%, then the percentage increase in his income is:

A. 14%
B. 16%
C. 18%
D. 22%
Answer: _________
Question 223:

The sum of weights of A and B is 80 kg. 50% of A's weight is $$frac{5}{6}$$ times the weight of B. Find the difference between their weights.

A. 20 kg
B. 10 kg
C. 25 kg
D. 15 kg
Answer: _________
Question 224:

In an election, candidate X got 70% of the overall valid votes. If 20% of the overall votes were declared invalid and the total numbers of votes is 64000, then find the number of valid votes polled in favour of the candidate.

A. 35840
B. 45000
C. 40000
D. 35800
Answer: _________
Question 225:

In an election between two candidates, 65% of the voters cast their votes, out of which 3% of the votes were decided to be invalid. A candidate got 81965 votes which are 65% of the total valid votes. What is the total number of votes enrolled in that election?

A. 200000
B. 190000
C. 215000
D. 185000
Answer: _________
Question 226:

A monthly return railway ticket cost 25 percent more than a single ticket. A week's extension can be had for the former by paying 5 percent of the monthly ticket's cost. If the money paid for the monthly ticket (with extension) is Rs. 84, the price of the single ticket is :

A. Rs. 48
B. Rs. 64
C. Rs. 72
D. Rs. 80
Answer: _________
Question 227:

Fresh grapes contain 80 percent water while dry grapes contain 10 percent water. If the weight of dry grapes is 250 kg what was its total weight when it was fresh ?

A. 1000 kg
B. 1100 kg
C. 1125 kg
D. 1225 kg
Answer: _________
Question 228:

When income tax is 3 paise in a rupee, a person's net income is Rs. 237650. What will it be when the income tax is raised to 7 paise ?

A. Rs. 233000
B. Rs. 231650
C. Rs. 227850
D. None of these
Answer: _________
Question 229:

In a certain month a base ball team that played 60 games had won 30% of its games played. After a phenomenal winning streak this team raised its average to 50%. How many games must the team have won in a row to attain this average ?

A. 12
B. 20
C. 24
D. 30
Answer: _________
Question 230:

The numbers are respectively $$12frac{1}{2}\% $$xa0 and $$25\% $$ more than a third number. The as a percentage of the second number is :

A. 50%
B. 60%
C. 75%
D. 90%
Answer: _________
Question 231:

Two successive price increases of 10% each on an article are equivalent to a single price increase of :

A. 19%
B. 20%
C. 21%
D. 22%
Answer: _________
Question 232:

In a market research project, 20% opted for Nirma detergent whereas 60% opted for Surf Blue detergent. The remaining individuals were not certain. If the difference between those who opted for Surf Blue and those who were uncertain was 720, how many respondents were covered in the survey ?

A. 1440
B. 1800
C. 3600
D. Data inadequate
Answer: _________
Question 233:

In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examinations ?

A. 30000
B. 35000
C. 37000
D. None of these
Answer: _________
Question 234:

the price of sugar increases by 32%. A family reduces its consumption so that the expenditure of the sugar is up only by 10%. If the total consumption of the sugar before the price rise was 10 kg per month, then the consumption of sugar per month at present (in kg) is :

A. $$8frac{1}{3}$$
B. $$8frac{1}{2}$$
C. $$8frac{3}{4}$$
D. 9
Answer: _________
Question 235:

On a test consisting of 250 questions, Jassi answered 40% of the first 125 questions correctly. What percent of the other 125 questions does she need to answer correctly for her grade on the entire exam to be 60% ?

A. 60%
B. 75%
C. 80%
D. Cannot be determined
E. 60%
F. 75%
G. 80%
H. Cannot be determined
Answer: _________
Question 236:

The schedule working hour of a labour in a week if 48 hours and he gets Rs. 480 for that. Over time rate is 25% more than the the basic salary rate. In a week a labour gets Rs. 605, how many hours altogether he works in that week.

A. 49 hours
B. 52 hours
C. 55 hours
D. 58 hours
Answer: _________
Question 237:

In an election 4% of the votes caste become invalid. Winner gets 55% of casted votes and wins the election by a margin of 4800 votes. Find the total number of votes casted.

A. 45000
B. 48000
C. 50000
D. 52000
Answer: _________
Question 238:

A reduction of 10% in the price of cloth enables a man to buy 6 meters of cloth more for Rs. 2160. Find the reduced price and also the original price of cloth per meter.

A. Rs. 36, Rs. 40
B. Rs. 40, Rs. 36
C. Rs. 36, Rs. 44
D. Rs. 44, Rs. 36
Answer: _________
Question 239:

A gardener increased the rectangular garden by increasing its length by 40% and decreasing its width by 20%. The area of the new garden:

A. Has increased by 20%
B. Has increased by 12%
C. Has increased by 8%
D. Is exactly the same as the old area
Answer: _________
Question 240:

If A exceeds B by 40%, B is less than C by 20%, then A : C is

A. 28 : 25
B. 26 : 25
C. 3 : 2
D. 3 : 1
Answer: _________
Question 241:

A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A. 45%
B. $$45frac{5}{{11}}\% $$
C. $$54frac{6}{{11}}\% $$
D. 55%
Answer: _________
Question 242:

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

A. 39, 30
B. 41, 32
C. 42, 33
D. 43, 34
Answer: _________
Question 243:

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

A. 588 apples
B. 600 apples
C. 672 apples
D. 700 apples
Answer: _________
Question 244:

What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?

A. 1%
B. 14%
C. 20%
D. 21%
Answer: _________
Question 245:

If A = x % of y and B = y % of x , then which of the following is true?

A. A is smaller than B
B. A is greater than B
C. Relationship between A and B cannot be determined
D. If x is smaller than y , then A is greater than B
Answer: _________
Question 246:

Out of two numbers, 40% of the greater number is equal to 60% of the smaller. If the sum of the numbers is 150, then the greater number is :

A. 70
B. 80
C. 90
D. 60
Answer: _________
Question 247:

If the total monthly income of 16 persons is Rs. 80800 and the income of one of them is 120% of the average income, them his income is :

A. Rs. 5050
B. Rs. 6060
C. Rs. 6160
D. Rs. 6600
Answer: _________
Question 248:

If three-fifth of sixty percent of a number is 36, the number is :

A. 100
B. 80
C. 75
D. 90
Answer: _________
Question 249:

X has twice as much money as that of Y and Y has 50% more money than that of Z. If the average money of all of them is Rs. 110, then the money, which X has is :

A. Rs. 55
B. Rs. 60
C. Rs. 90
D. Rs. 180
Answer: _________
Question 250:

The sum of two numbers is 520. If the bigger number is decreased by 4% and the smaller number is increased by 12% then the numbers obtained are equal. The smaller number is :

A. 280
B. 240
C. 210
D. 300
Answer: _________
Question 251:

Tulsiram's salary is 20% more then that of Kashyap. If Tulsiram saves Rs. 720 which is 4% of his salary, then Kashyap's salary is :

A. Rs. 15000
B. Rs. 12000
C. Rs. 10000
D. Rs. 22000
Answer: _________
Question 252:

The ratio of the number of boys to that of girls in a village is 3 : 2. If 30% of boys and 70% of girls appeared in an examination, the ratio of the number of students, appeared in the examination to that not appeared in the same examination is :

A. 1 : 1
B. 27 : 23
C. 9 : 14
D. 23 : 27
Answer: _________
Question 253:

The value of a machine is Rs. 6250. It decreases by 10% during the first year, 20% during the second year and 30% during the third year, What will be the value of the machine after 3 years ?

A. Rs. 2650
B. Rs. 3050
C. Rs. 3150
D. Rs. 3510
Answer: _________
Question 254:

A person gave 20% of his income to his elder son, 30% of remaining to the younger son and 10% of the balance, he donate to a trust. He is left with Rs. 10080. His income was :

A. Rs. 50000
B. Rs. 40000
C. Rs. 30000
D. Rs. 20000
Answer: _________
Question 255:

The weight of an empty bucket is 25% of the weight of the bucket when filled with some liquid. Some of the liquid has been removed. Then, the bucket, along with the remaining liquid, weighed three-fifths of the original weight. What percentage of the liquid has been removed ?

A. 40%
B. $$62frac{1}{2}$$
C. $$56frac{2}{3}$$%
D. $$53frac{1}{3}$$%
Answer: _________
Question 256:

The ratio 5 : 4 expressed as a percent equals :

A. 12.5%
B. 40%
C. 80%
D. 125%
E. 125%
F. 80%
G. 405
H. 12.5%
Answer: _________
Question 257:

Solve (550% of 250) ÷ 275 = (?)

A. 15
B. 1.5
C. 0.5
D. None of these
Answer: _________
Question 258:

270 candidates appeared for an examination, of which 252 passed. The pass percentage is :

A. 80%
B. $$83frac{1}{2}$$%
C. $$90frac{1}{3}$$%
D. $$93frac{1}{3}$$%
Answer: _________
Question 259:

What will come in the place of (?) in the expression below : x% of y is y% of (?)

A. x
B. 100x
C. $$frac{x}{100}$$
D. $$frac{y}{100}$$
Answer: _________
Question 260:

In the expression xy 2 , the value of both variables x and y are decreased by 20%. By this, the value of the expression is decreased by :

A. 40%
B. 48.8%
C. 51.2%
D. 80%
Answer: _________
Question 261:

The value of a machine depreciates at the rate of 12 percent per annum. It was purchased three years ago. Its present value is Rs. 29644.032. What was the purchase price of the machine ?

A. Rs. 38900
B. Rs. 39000
C. Rs. 43500
D. Rs. 48700
Answer: _________
Question 262:

From 5 litres of a 20% solution of alcohol in water, 2 litres of solution is taken out and 2 litres of water is added to it. Find the strength of alcohol in the new solution.

A. 10%
B. 12%
C. 15%
D. 18%
Answer: _________
Question 263:

A man ordered a length of rope by telephone from his nearest hardware shop. But when a worker in the shop brought the rope, he found that the man on the telephone had miswritten the order by interchange feet and inches. As a result of this, the length of rope received was only 30% of the length he had ordered. The length of the rope which the man ordered was between :

A. 6 ft and 7 inches
B. 7$$frac{1}{2}$$ ft and 9 ft
C. 9 ft and 2 inches
D. 10$$frac{1}{2}$$ ft and 12 ft
Answer: _________
Question 264:

60% of 264 is the same as :

A. 10% of 44
B. 15% of 1056
C. 30% of 132
D. None of these
Answer: _________
Question 265:

The number which exceeds 16% of it by 42 is :

A. 50
B. 52
C. 58
D. 60
Answer: _________
Question 266:

Solve this : (23.6% of 1254) - (16.6% of 834) = ?

A. 153.5
B. 155.5
C. 157.5
D. 159.5
Answer: _________
Question 267:

One-fourth of sixty percent of a number is equal to two-fifth of twenty percent of another number. What is the respective ratio of the first number to the second number ?

A. 4 : 7
B. 5 : 9
C. 8 : 13
D. Cannot be determined
Answer: _________
Question 268:

In an examination, 96% of students passed and 500 students failed. How many students did appear at the examination ?

A. 14000
B. 12500
C. 12000
D. 13500
Answer: _________
Question 269:

The price of a certain item is increased by 15%. If consumer wants to keeps his expenditure on the item same as before, how much percent must he reduce his consumption of that item ?

A. $$10frac{20}{{23}}\% $$
B. $$13frac{1}{{23}}\% $$
C. $$16frac{2}{{3}}\% $$
D. $$15\% $$
Answer: _________
Question 270:

The difference between 38% of a number and 24% of the same number is Rs. 135.10. What is 40% of that number ?

A. 370
B. 378
C. 386
D. 394
Answer: _________
Question 271:

Aman's expense is 30% more than Vimal's and Vimal's expense is 10% less than Raman's. If the sum of their expenses is Rs. 6447, then what would be Aman's expense ?

A. Rs. 1890
B. Rs. 2100
C. Rs. 2200
D. Rs. 2457
Answer: _________
Question 272:

If A is 150 percent of B, then B is what percent of (A + B) ?

A. $$33frac{1}{3}\% $$
B. $$40\% $$
C. $$66frac{2}{3}\% $$
D. $$75\% $$
Answer: _________
Question 273:

From the salary of an officer, 10% is deducted as house rent, 20% of the rest, he spends on conveyance, 20% of the rest he pays as income tax and 10% of the balance, he spends on clothes. Then, he is left with Rs. 15552. Find his total salary.

A. Rs. 25000
B. Rs. 30000
C. Rs. 35000
D. Rs. 40000
Answer: _________
Question 274:

In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects. If 2500 candidates appeared at the examination, how many passed in either subject but not in both ?

A. 325
B. 1175
C. 2125
D. None of these
Answer: _________
Question 275:

14% of 14 + 28% of 28 + 92% of 96 - 15% of 85 = ?

A. 8.37
B. 85.37
C. 89.37
D. None of these
Answer: _________
Question 276:

What percent of 88 is 33 ?

A. 34.5%
B. 35.5%
C. 36.5%
D. 37.5%
Answer: _________
Question 277:

40% of 60% of $$frac{3}{5}$$th of a number is 504. What is 25% of $$frac{2}{5}$$th of that number ?

A. 175
B. 180
C. 350
D. 360
Answer: _________
Question 278:

Nupur invests Rs. 89856, which is 26% of her annual income, in mutual funds. What is her monthly income ?

A. Rs. 2398.50
B. Rs. 28800
C. Rs. 28990
D. Rs. 33606.25
Answer: _________
Question 279:

Two numbers A and B are such that the sum of 5% of A and 4% of B is two-thirds of the sum of 6% of A and 8% of B. Find the ratio of A : B ?

A. 2 : 3
B. 1 : 1
C. 3 : 4
D. 4 : 3
Answer: _________
Question 280:

In a college election between two candidates, one candidate got 55% of the total valid votes. 15% of the votes were invalid. If the total votes were 15200, what is the number of valid votes the other candidate got ?

A. 5814
B. 6840
C. 7106
D. 8360
Answer: _________
Question 281:

If x% of y is equal to z, what percent of z is x ?

A. $$frac{{{y^2}}}{{100}}$$ %
B. $$frac{y}{{{{100}^2}}}$$ %
C. $$frac{{100}}{y}$$ %
D. $$frac{{{{100}^2}}}{y}$$ %
Answer: _________
Question 282:

A part of Rs. 9600 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from both portions is the same, what is the total income from the two investments ?

A. Rs. 320
B. Rs. 380
C. Rs. 410
D. Rs. 440
Answer: _________
Question 283:

Of the 50 researchers in a work group, 40% will be assigned to Team A and the remaining 60% to Team B. However, 70% of the researchers prefer team A and 30% prefer Team B. What is the least possible number of researchers who will not be assigned to the team they prefer ?

A. 15
B. 20
C. 30
D. 35
Answer: _________
Question 284:

A fraction in reduced form is such that when it is squared and then its numerator is reduced by $$33frac{1}{3}$$ % and denominator is reduced to 20%, its result is twice the original fraction. The sum of numerator and denominator is:

A. 8
B. 13
C. 17
D. 15
Answer: _________
Question 285:

A student appeared in the Mock CAT. The test paper contained 3 sections namely QA, DI and VA. The percentage marks in all VA was equal to the average of percentage marks in all the 3 sections. Coincidentally, if we reverse the digit of the percentage marks of QA we get the percentage marks of DI. The percentage marks in VA scored by student could be:

A. 81
B. 48
C. 66
D. 69
Answer: _________
Question 286:

A shopkeeper first raises the price of Jewellery by x% then he decreases the new price by x%. After such up down cycle, the price of a Jewellery decreased by Rs. 21025. After a second up down cycle the Jewellery was sold for Rs. 484416. What was the original price of the jewellery.

A. Rs. 5,26,000
B. Rs. 6,00,625
C. Rs. 5,25,625
D. Rs. 5,00,000
Answer: _________
Question 287:

A company has 12 machines of equal efficiency in its factory. The annual manufacturing expenses are Rs. 24,000 and the establishment charges are Rs. 10,000. The annual output of the company is Rs. 48,000. The annual output and manufacturing costs are directly proportional to the no. of machines while the share holders get the 10% profit, which is directly proportional to the annual output of the company. If 8.33% of machines remained close throughout the year. Then the percentage decrease in the amount of share holders is :

A. 14.28%
B. 11.11%
C. 16.66%
D. 8.33%
Answer: _________
Question 288:

Every month a man consumes 25 kg rice and 9 kg wheat. The price of rice is 20% of the price of wheat and thus he spends total Rs. 350 on the rice and wheat per month. If the price of wheat is increased by 20% then what is the percentage reduction of rice consumption for the same expenditure of Rs. 350? Given that the price of rice and consumption of wheat is constant :

A. 40%
B. 25%
C. 36%
D. 24%
Answer: _________
Question 289:

The price of raw materials has gone up by 15%, labor cost has also increased from 25% of the cost of raw material to 30% of the cost of raw material. By how much percentage should there be reduction in the usage of raw materials so as to keep the cost same?

A. 28%
B. 17%
C. 27%
D. 24%
Answer: _________
Question 290:

A sales executive gets 20% bonus of the total sales value and 10% commission besides the bonus on the net profit after charging such commission. If the total sales value be Rs. 10 lakh per annum and the total profit of the company be Rs. 1.32 lakh, then his total earning per annum will be, given that he is not entitled to receive any fixed salary from the company :

A. 2.3 lakh
B. 2.32 lakh
C. 2.12 lakh
D. 3.2 lakh
Answer: _________
Question 291:

A shepherd had n goats in the year 2000. In 2001 the no. of goats increased by 40%. In 2002 the no. of goats declined to 70%. In 2003 the no. of goats grew up 30%. In 2004, he sold 10% goats and then he had only 34,398 goats. The percentage increase of the no. of goats in this duration was :

A. 16.66%
B. 14.66%
C. 11.33%
D. 20%
Answer: _________
Question 292:

In an office in Singapore there are 60% female employees. 50 % of all the male employees are computer literate. If there are total 62% employees computer literate out of total 1600 employees, then the no. of female employees who are computer literate ?

A. 690
B. 674
C. 672
D. 960
Answer: _________
Question 293:

The price of a car depreciates in the first year by 25% in the second year by 20% in third year by 15% and so on. The final price of the car after 3 years, if the present cost of the car is Rs. 10,00,000 :

A. 7,80,000
B. 1,70,000
C. 6,90,000
D. 5,10,000
Answer: _________
Question 294:

In an election, 30% of the voters voted for candidate A whereas 60% of the remaining voted for candidate B. The remaining voters did not vote. If the difference between those who voted for candidate A and those who did not vote was 1200, how many individuals were eligible for casting vote in that election ?

A. 10000
B. 45000
C. 60000
D. 72000
Answer: _________
Question 295:

x% of x is the same as 10% of

A. $$frac{{x}}{{10}}$$
B. $$frac{{{x^2}}}{{10}}$$
C. $$frac{{{x^3}}}{{10}}$$
D. None of these
Answer: _________
Question 296:

In a certain organisation, 40% employees are matriculates, 50% of the remaining are graduates and the remaining 180 are post-graduates. What is the number of graduate employees ?

A. 180
B. 240
C. 300
D. 360
Answer: _________
Question 297:

In an examination in which full marks were 800, A gets 20% more than B, B gets 20% more than C, and C gets 15% less than D. If A got 576, what percentage of full marks did D get (approximately) ?

A. 45.73
B. 51.27
C. 58.82
D. 61.71
Answer: _________
Question 298:

The price of an article is reduced by 25%. In order to retain the original price, the present price has to be increased by :

A. 20%
B. 25%
C. 33$$frac{1}{3}$$%
D. 50%
Answer: _________
Question 299:

A person's salary is decreased by steps of 20%, 15% and 10%. Approximately by what percent should the reduced salary be increases so as to get back the original salary ?

A. 39%
B. 44%
C. 56%
D. 63%
Answer: _________
Question 300:

A building worth Rs. 133100 is constructed on land worth Rs. 72900. After how many years will the value of both be the same if land appreciates at 10% p.a. and building depreciates at 10% p.a. ?

A. $$1frac{1}{2}$$
B. 2
C. $$2frac{1}{2}$$
D. 3
Answer: _________
Question 301:

A bag contains 600 coins of 25p denomination and 1200 coins of 50p denomination. If 12% of 25p coins and 24% of 50p are removed, the percentage of money removed from the bag is nearly :

A. 15.6%
B. 17.8%
C. 21.6%
D. 30%
Answer: _________
Question 302:

Mr. More spent 20% of his monthly income on food and 15% on children's education. 40% of the remaining he spent on entertainment and transport together and 30% on medical. He is left with an amount of Rs. 8775 after all these expenditures. What is Mr. More's monthly income ?

A. Rs. 35000
B. Rs. 38000
C. Rs. 40000
D. Rs. 42000
Answer: _________
Question 303:

If the price of erasers goes down by 25%, a man can buy 2 more erasers for a rupee. How many erasers are available for a rupee ?

A. 2
B. 4
C. 6
D. 8
Answer: _________
Question 304:

Find the : (550% of 250) ÷ 275 = ?

A. 15
B. 0.5
C. 1.5
D. 25
Answer: _________
Question 305:

By how much percent is four-fifths of 70 lesser than five-sevenths of 112 ?

A. 24%
B. 30%
C. 36%
D. 42%
Answer: _________
Question 306:

The firm uses the following function to calculate the production out (PO) : PO = 5.3C 2 L 15 , where C = capital invested and L = labour employed. If the capital invested (C) increases by 20 percent, the change in PO will be :

A. 20% decrease
B. 32% increase
C. 44% increase
D. 56% increase
Answer: _________
Question 307:

Find the 92.5% of 550 = ?

A. 506.45
B. 508.75
C. 518.55
D. 521.65
Answer: _________
Question 308:

40% of 265 + 35% of 180 = 50% of ?

A. 84.5
B. 169
C. 253.5
D. 338
Answer: _________
Question 309:

If a exceeds b by x%, then which one of the following equations is correct ?

A. $$a - b = frac{x}{100}$$
B. $$b = a + 100x$$
C. $$a = frac{bx}{100 + x}$$
D. $$a = b + frac{bx}{100}$$
Answer: _________
Question 310:

10% of 5 and 5% of 10 add up to

A. 0.10
B. 0.25
C. 1.0
D. 2.5
Answer: _________
Question 311:

In an examination, 30% and 35% students respectively failed in History and Geography while 27% students failed in both subjects. If the number of students passing the examination is 248, find the total number of students who appeared in the examination.

A. 380
B. 400
C. 425
D. 725
Answer: _________
Question 312:

In an examination, the percentage of students qualified to the number of students appeared from school A is 70%. In school B, the number of students appeared is 20% more than the students appeared from school A and the number of students qualified from school B is 50% more than the students qualified from school A. What is the percentage of students qualified to the number of students appeared from school B ?

A. 30%
B. 70%
C. 78.5%
D. 87.5%
Answer: _________
Question 313:

In an examination, 65% of the students passed in Mathematics, 48% passed in Physics and 30% passed in both. How much percent of students failed in both the subjects ?

A. 17%
B. 43%
C. 13%
D. 47%
Answer: _________
Question 314:

At an election there were two candidate got 38% of votes and lost by 7200 votes. The total numbers of valid votes were :

A. 13000
B. 13800
C. 16200
D. 30000
Answer: _________
Question 315:

In 2 kg mixture of copper and aluminium, 30% is copper. How much aluminium powder should be added to the mixture so that the quantity of copper becomes 20% ?

A. 900 gms
B. 800 gms
C. 1000 gms
D. 1200 gms
Answer: _________
Question 316:

If 20% of (A + B) = 50% of B, then value of $$frac{2A - B}{2A + B}$$ xa0 is :

A. $$frac{1}{2}$$
B. $$frac{1}{3}$$
C. $$frac{1}{41}$$
D. 1
Answer: _________
Question 317:

Ticket for all but 100 seats in a 10000 seat stadium were sold. Of the ticket sold, 20% were sold at half price and the remaining tickets were sold at the full price of Rs. 20. The total revenue from the ticket sales, (in Rs.) was :

A. 158400
B. 178200
C. 180000
D. 198000
Answer: _________
Question 318:

The average of marks obtained by 100 candidates in a certain examination is 30. If the average marks of passed candidates is 35 and that of the failed candidates is 10, what is the number of candidates who passed the examinations ?

A. 60
B. 70
C. 80
D. 90
Answer: _________
Question 319:

If 90% of A = 30% of B and B = 2x% of A, then the value of x is :

A. 450
B. 400
C. 300
D. 150
Answer: _________
Question 320:

18% of which number is equal to 12% of 75 ?

A. 50
B. 100
C. 2
D. $$frac{3}{2}$$
Answer: _________
Question 321:

Out of his total income, Mr. Kapur spends 20% on house rent and 70% of the rest on house hold expenses. If he saves Rs. 1800 what is his total income (in Rs.) ?

A. Rs. 7800
B. Rs. 7000
C. Rs. 8000
D. Rs. 7500
Answer: _________
Question 322:

Two numbers are in the ratio 2 : 3. If 20% of the smaller number added to 20, is equal to the sum of 10% of the larger number and 25, the the smaller number is :

A. 100
B. 160
C. 180
D. 200
Answer: _________
Question 323:

x * 12 = 75% of 336 Find x.

A. 27
B. 25
C. 21
D. 19
Answer: _________
Question 324:

A shop sells floor tiles at Rs. 48 per square meter. A contractor employs a machine that polishes the tiles that damages 10% of the total number of tiles which cannot be used any more. Calculate the amount that needs to be paid by contractor to tile shop owner, if the hall is of a square shape and has a perimeter of 400 meters?

A. Rs. 4,00,000
B. Rs. 5,00,000
C. Rs. 5,28,000
D. Rs. 3,65,000
Answer: _________
Question 325:

125% of 860 + 75% of 480 = ?

A. 1415
B. 1385
C. 1435
D. None of these
Answer: _________
Question 326:

A person subscribing to sky cable for a year pack Rs. 1785. If the monthly subscription is Rs. 175, how much discount does a yearly subscriber get?

A. 11%
B. 13%
C. 15%
D. 18%
Answer: _________
Question 327:

In a metro train there are 600 passengers out of which 34% are females. Fare of each male is Rs. 20 and each female's fare is 25% less than each male. What is the total revenue generated by all the passengers together?

A. Rs. 10880
B. Rs. 10980
C. Rs. 10740
D. Rs. 10680
Answer: _________
Question 328:

In a test, minimum passing percentage for girls and boys are 45% and 60% respectively. A boy scored 767 marks and failed by 313 marks. What are the minimum passing marks for girls?

A. 910
B. 920
C. 840
D. 810
Answer: _________
Question 329:

50 minutes is what percentage of a day (approx.)-

A. 3.5%
B. 4%
C. 5%
D. 6%
Answer: _________
Question 330:

In an examination 36% are pass marks. If an examine gets 17 marks and fails by 10 marks, what are the maximum marks?

A. 90
B. 75
C. 60
D. 55
Answer: _________
Question 331:

Find $$frac{{100}}{3}$$ % of 600

A. 100
B. 150
C. 200
D. 225
Answer: _________
Question 332:

If 75% of the students in a school are boys and the number of girls is 420, the number of boys is:

A. 1176
B. 1350
C. 1260
D. 1125
Answer: _________
Question 333:

If the numerator of a fraction is increased by 60% and the denominator is increased by 40%, then resultant fraction is $$frac{{16}}{{63}}.$$xa0The original fraction is:

A. $$frac{2}{9}$$
B. $$frac{5}{9}$$
C. $$frac{2}{{11}}$$
D. $$frac{4}{9}$$
Answer: _________
Question 334:

In an examination, B obtained 20%, more marks than those obtained by A, and A obtained 10% less marks than those obtained by C. D obtained 20% marks than those obtained by C. By what percentage are the marks obtained by D more than those obtained by A?

A. $$43frac{1}{3}\% $$
B. $$23frac{1}{3}\% $$
C. $$13frac{1}{3}\% $$
D. $$33frac{1}{3}\% $$
Answer: _________
Question 335:

Kavita's attendance in her school for the academic session 2018-2019 was 216 days. On computing her attendance, it was observed that her attendance was 90%. The total working days of the school were:

A. 250
B. 194
C. 240
D. 195
Answer: _________
Question 336:

In an examination, 92% of the students passed and 480 students failed. If so, how many students appeared in the examination?

A. 5800
B. 6200
C. 6000
D. 5000
Answer: _________
Question 337:

Sachin scored 120 runs, which included 6 boundaries and 4 sixes. What percentage of his total score did he make by running between the wickets?

A. $$46frac{4}{9}\% $$
B. $$33frac{1}{3}\% $$
C. 60%
D. 45%
Answer: _________
Question 338:

Renu saves 20% of her income. If her expenditure increase by 20% and income increase by 29%, then her saving increase by:

A. 55%
B. 65%
C. 54%
D. 60%
Answer: _________
Question 339:

If the radius of a cylinder is decreased by 20% and the height is increased by 20% to form a new cylinder, then the volume will be decreased by:

A. 23.2%
B. 22.3%
C. 32.2%
D. 20.5%
Answer: _________
Question 340:

The sum of the number of male and female students in an institute is 100. If the number of male students is x, then the number of female students becomes x% of the total number of students. Find the number of male students.

A. 50
B. 65
C. 45
D. 60
Answer: _________
Question 341:

If 91% of A is 39% of B, and B is x% of A, then the value of x is:

A. $$frac{{500}}{3}$$
B. $$frac{{400}}{3}$$
C. $$frac{{700}}{3}$$
D. $$frac{{200}}{3}$$
Answer: _________
Question 342:

The reduction of 20% in the price of rice enables a person to obtain 50 kg more for Rs. 450. Find the original price of rice per kg.

A. Rs. 1
B. Rs. 2
C. Rs. 1.25
D. Rs. 2.25
Answer: _________
Question 343:

The percentage increase in the surface area of a cube when each side is doubled is :

A. 200%
B. 300%
C. 150%
D. 50%
Answer: _________
Question 344:

In an examination, 35% of the candidates failed in Mathematics and 25% in English. If 10% failed in both Mathematics and English, then how much percent of candidates passed in both the subjects ?

A. 50%
B. 55%
C. 57%
D. 60%
Answer: _________
Question 345:

The population of a town increases every year by 4%. If population was 50000 in starting, then after two years will be :

A. 53900
B. 54000
C. 54080
D. 54900
Answer: _________
Question 346:

In a town, the population was 8000. In one year, male population increased by 10% and female population increased by 8% but the total population increased by 9%. The number of males in the town was :

A. 4000
B. 4500
C. 5000
D. 6000
Answer: _________
Question 347:

The population of a village is 25000. One-fifth are females and the rest are males, 5% of males and 40% of females are uneducated. What percentage on the whole are educated ?

A. 75%
B. 88%
C. 55%
D. 85%
Answer: _________
Question 348:

In a factory, the production of cycles rose to 48400 from 40000 in 2 years. The rate of growth per annum is ?

A. 10.5%
B. 9%
C. 8%
D. 10%
Answer: _________
Question 349:

A student scored 32% marks in science subjects out of 300. How much should he score in language papers out of 200 if he is to get overall 46% marks ?

A. 72%
B. 67%
C. 66%
D. 60%
Answer: _________
Question 350:

Present population of a village is 67600, It has been increasing annually at the rate of 4%. What was the population of the village two years ago ?

A. 62500
B. 63000
C. 64756
D. 65200
Answer: _________
Question 351:

In a college, 40% of the students were allotted group A, 75% of the remaining were given group B and the remaining 12 students were given group C. Then the number of students who applied for the group is :

A. 100
B. 60
C. 80
D. 92
Answer: _________
Question 352:

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had :

A. 588 apples
B. 600 apples
C. 672 apples
D. 700 apples
Answer: _________
Question 353:

When expressed as a fraction 64% would mean :

A. $$frac{16}{25}$$
B. $$frac{9}{64}$$
C. $$frac{8}{81}$$
D. $$frac{12}{121}$$
Answer: _________
Question 354:

Solve this : (12% of 555) + (15% of 666) = ?

A. 166.5
B. 167.5
C. 168.5
D. None of these
Answer: _________
Question 355:

Solve this : 12% of 5000 = ?

A. 600
B. 620
C. 680
D. 720
Answer: _________
Question 356:

Krishna's present salary is Rs. 3500. It will increase by 10% next year. What will be Krishna's salary after the increment ?

A. Rs. 3850
B. Rs. 3950
C. Rs. 4000
D. Rs. 3900
Answer: _________
Question 357:

Solve this : 32% of 825 + 25% of 1440 = 1025 - (?)

A. 456
B. 206
C. 223
D. 401
Answer: _________
Question 358:

Find the : 38% of 341 = ?

A. 120.68
B. 129.58
C. 135.78
D. 136.28
Answer: _________
Question 359:

If 0.03 is X% of 0.3, then the value of X is :

A. 3
B. 10
C. 30
D. Cannot be determined
Answer: _________
Question 360:

What is 28% of 36% of $$frac{5}{7}$$th of 5000 ?

A. 360
B. 375
C. 420
D. 480
Answer: _________
Question 361:

If x% of y is y% of (?)

A. x
B. 100x
C. $$frac{x}{100}$$
D. $$frac{y}{100}$$
Answer: _________
Question 362:

Solve this : $$frac{4}{3}$$ of 25% $$frac{18}{19}$$ of 57 = ?

A. 36
B. 8
C. 18
D. 12
Answer: _________
Question 363:

A litre of water evaporates from 6L of sea water containing 4% salt. Find the percentage of salt in the remaining solution.

A. $$5frac{1}{2}\% $$
B. $$3frac{1}{2}\% $$
C. 3%
D. $$4frac{4}{5}\% $$
Answer: _________
Question 364:

Two discount of 8% and 12% are equal to a single discount of:

A. 20%
B. 19.04%
C. 22.96%
D. 22%
Answer: _________
Question 365:

In a library 60% of the books are in Hindi, 60% of the remaining books are in English rest of the books are in Urdu. If there are 3600 books in English, then total no. of books in Urdu are:

A. 2400
B. 2500
C. 3000
D. 3200
Answer: _________
Question 366:

In Sabarmati Express, there as many wagons as there are the no. of seats in each wagon and not more than one passenger can have the same berth (seat). If the middlemost compartment carrying 25 passengers is filled with 71.428% of its capacity, then find the maximum no. of passengers in the train that can be accommodated if it has minimum 20% seats always vacant.

A. 500 seats
B. 786 seats
C. 980 seats
D. 1060 seats
Answer: _________
Question 367:

The population of a village is 5000 and it increases at the rate of 2% every year. After 2 years, the population will be:

A. 5116
B. 5202
C. 5200
D. 5204
Answer: _________
Question 368:

In a class, the no. of boys is more than the no. of girls by 12% of the total strength. The ratio of boys and girls is:

A. 15 : 11
B. 11 : 14
C. 14 : 11
D. 8 : 11
Answer: _________
Question 369:

In an office there were initially N employees. The HR manager first hired P% employees then after a month Q% employees left the office, the value of (P - Q) is:

A. PQ
B. $$frac{{{ ext{PQ}}}}{{100}}$$
C. $$frac{{ ext{P}}}{{ ext{Q}}}$$
D. $$frac{{ ext{Q}}}{{ ext{P}}}$$
Answer: _________
Question 370:

The amount of work in a leather factory is increased by 50%. By what percent is it necessary to increase the number of workers to complete the new amount of work in previously planned time, if the productivity of the new labour is 25% more.

A. 60%
B. 66.66%
C. 40%
D. 33.33%
Answer: _________
Question 371:

A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the expouser of the coloured cubes to the outside is minimum. The percentage of exposed area that is coloured is:

A. 25.9%
B. 44.44%
C. 35%
D. 32%
Answer: _________
Question 372:

78% of 750 + 34% of x = 30% of 2630. Find x.

A. 960
B. 600
C. 800
D. 750
Answer: _________
Question 373:

Kamal has 160 toffees. He gave 5% toffees to Ravi, 15% toffees to Anita and one-fourth of the toffees to Gagan. How many toffees are left with Kamal after the distribution ?

A. 78
B. 69
C. 88
D. 79
Answer: _________
Question 374:

If a is 60% of b, then what percent of 4a is 5b ?

A. $$frac{{25}}{{12}}$$ %
B. $$148$$ %
C. $$frac{{625}}{{3}}$$ %
D. $$240$$ %
Answer: _________
Question 375:

Solve this : (0.85% of 405) + (2.25% of 550) = ?

A. 13.8175
B. 14.7125
C. 15.7150
D. 16.7175
Answer: _________
Question 376:

An alloy of gold and silver weight 50 g. It contains 80% gold. How much gold should be added to the alloy so that percentage of gold is increased to 90 ?

A. 30 g
B. 40 g
C. 50 g
D. 60 g
Answer: _________
Question 377:

One-eighth of a number is 41.5. What will 69% of that number be ?

A. 219.12
B. 225.76
C. 229.08
D. 232.4
Answer: _________
Question 378:

A company pays rent of Rs. 25000 per month for office space to its owner. But if the company pays the annual rent at the beginning of the year the owner gives a discount of 5% on the total annual rent. What is the annual amount the company pays to the owner after the discount ?

A. Rs. 275000
B. Rs. 285000
C. Rs. 295000
D. Rs. 300000
Answer: _________
Question 379:

Solve this : (0.56% of 225) × (3.25% of 430) = ?

A. 15.3195
B. 15.6175
C. 17.3075
D. 17.6085
Answer: _________
Question 380:

An interval of 3 hours 40 minutes is wrongly estimated as 3 hours 45.5 minutes. The error percentage is :

A. 2.5%
B. 5%
C. 5.2%
D. 5.5%
Answer: _________
Question 381:

A number increased by $$37frac{1}{2}$$ % gives 33. The number is :

A. 22
B. 24
C. 25
D. 27
Answer: _________
Question 382:

If A's salary is 25% more than B's salary, then B's salary is how much lower than A's salary?

A. $$33frac{1}{3}\% $$
B. 25%
C. 20%
D. $$16frac{2}{3}\% $$
Answer: _________
Question 383:

Population of a town increase 2.5% annually but is decreased by 0.5% every year due to migration. What will be the percentage increase in 2 years?

A. 5%
B. 4.04%
C. 4%
D. 3.96%
Answer: _________
Question 384:

In an election between two candidates, the winner got 65% of the total votes cast and won the election by a majority of 2748 votes. What is the total number of votes cast if no vote is declared invalid?

A. 8580
B. 8720
C. 9000
D. 9160
Answer: _________
Question 385:

Narayan spends 30% of his income on education and 50% of the remaining on food. He gives Rs. 1000 as monthly rent and now has Rs. 1800 left with him. What is his monthly income?

A. Rs. 8000
B. Rs. 7000
C. Rs. 9000
D. Rs. 6000
Answer: _________
Question 386:

P is 6 times greater than Q then by what per cent is Q smaller than P?

A. 84%
B. 85.5%
C. 80%
D. 83.33%
Answer: _________
Question 387:

If two numbers are respectively 30% and 40% more than a third number, what percent is the first of the second?

A. $$92frac{6}{7}\% $$
B. $$84frac{4}{5}\% $$
C. 80%
D. 75%
Answer: _________
Question 388:

The population of a city is 35000. On an increase of 6% in the number of men and an increase of 4% in the number of women, the population would become 36760. What was the number of women initially?

A. 18000
B. 19000
C. 17000
D. 20000
Answer: _________
Question 389:

The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid.

A. 77%
B. 75%
C. 88%
D. 98%
Answer: _________
Question 390:

The price of rice falls by 20%. How much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously?

A. 5 kg
B. 15 kg
C. 25 kg
D. 30 kg
Answer: _________
Question 391:

If decreasing 110 by x% gives the same result as increasing 50 by x%, then x% of 650 is what percentage (correct to the nearest integer) more than (x - 10)% of 780?

A. 17%
B. 12%
C. 18%
D. 14%
Answer: _________
Question 392:

If 25% of 400 + 35% of 1260 + 27% of 1800 = 1020 + x, then the value of x lies between:

A. 6 to 10
B. 0 to 5
C. 11 to 15
D. 16 to 20
Answer: _________
Question 393:

The monthly salary of a person was Rs. 50,000. He used to spend on family expenses (E), Taxes (T), Charity (C), and the rest were his saving. E was 60% of the income, T was 20% of E, and C was 15% of T. When his salary got raised by 40%, he maintained the percentage level of E, but T becomes 30% of E and C becomes 20% of T. The difference between the two savings (in Rs.) is:

A. 128
B. 220
C. 130
D. 250
Answer: _________
Question 394:

Raju spends 10 percent and 20 percent of his income on transport and food respectively. He spends 30 percent of the remaining income on clothing. He saves rest of his income. If his saving is Rs. 26460, then what will be total expenditure on food and clothing together?

A. Rs. 26420
B. Rs. 22140
C. Rs. 24480
D. Rs. 23440
Answer: _________
Question 395:

The monthly salaries of A and B are the same. A, B and C donate 10%, 8% and 9% respectively, of their monthly salaries to a charitable trust. The difference between the donations of A and B is Rs. 400. The total donation by A and B is Rs. 900 more than that of C. What is the monthly salary of C?

A. Rs. 25,000
B. Rs. 30,000
C. Rs. 27,000
D. Rs. 36,000
Answer: _________
Question 396:

Price of rice is decreased by 25 percent and therefore a person can purchase 30 kg more rice in the same expenditure. If expenditure is Rs. 5400, then what was the original price of rice per kg?

A. Rs. 90 kg
B. Rs. 60 kg
C. Rs. 50 kg
D. Rs. 75 kg
Answer: _________
Question 397:

If a positive number 'k' when multiplied by 30% of itself gives a number which is 170% more than the number 'k' then the number 'k' is equal to:

A. 9
B. 5
C. 6
D. 7
Answer: _________
Question 398:

If 40% of a number is less than its 60% by 30, then the 20% of that number is:

A. 60
B. 40
C. 50
D. 30
Answer: _________
Question 399:

If A is 40% less than B and C is 40% of the sum of A and B, then by what percentage is B greater than C?

A. 60
B. $$56frac{1}{4}$$
C. $$40frac{1}{8}$$
D. 36
Answer: _________
Question 400:

If A's salary is 30% more than B's salary, then by what percentage is B's salary less than that of A? (correct to one decimal place)

A. 17.5%
B. 23.1%
C. 25%
D. 19.7%
Answer: _________
Question 401:

Rishu saves x% of her income. If her income increases by 26% and the expenditure increases by 20%. then her savings increase by 50%. What is the value of x?

A. 25
B. 30
C. 20
D. 10
Answer: _________
Question 402:

If the price of petrol is increased by 28%, by what percentage should the consumption be decreased by the consumer, if the expenditure on petrol remains unchanged? (Correct to 2 decimal places)

A. 12.35%
B. 21.88%
C. 20.25%
D. 25.75%
Answer: _________
Question 403:

Salary of Mohit is 60% more than Vijay. Salary of Vijay is how much percent less than Mohit?

A. 45%
B. 42.5%
C. 47.5%
D. 37.5%
Answer: _________
Question 404:

In an election contested between two candidates, 15% of the total voters did not cast their votes and 100 votes got disqualified. The candidate who won the election won it by securing 45% of the total votes and won by a margin of 400 votes. Find the total number of voters?

A. 6,000
B. 3,600
C. 10,000
D. 3,500
Answer: _________
Question 405:

What percent of Rs. 2650 is Rs. 1987.50 ?

A. 60%
B. 75%
C. 80%
D. 90%
Answer: _________
Question 406:

The price of a commodity which was Rs. 250 three years ago is Rs. 2000 now. The annual rate of increase in the price is ?

A. $$100\%$$
B. $$200\%$$
C. $$266frac{2}{3}\%$$
D. None of these
Answer: _________
Question 407:

Solve this : (7.9% of 134) - (3.4% of 79) = ?

A. 7.3
B. 7.8
C. 8.1
D. 8.6
Answer: _________
Question 408:

Solve this : 36% of 365 + (?)% of 56.2 = 156.69

A. 30
B. 35
C. 40
D. 45
Answer: _________
Question 409:

Twenty-five percent of Reena's yearly income is equal to seventy-five percent of Anubhab monthly income . If Anubhab yearly income is Rs. 240000, What is the Reena's monthly income ?

A. Rs. 60000
B. Rs. 12000
C. Rs. 5200
D. Cannot be determined
Answer: _________
Question 410:

What will be the answer : 140% of 56 + 56% of 140 = ?

A. 78.4
B. 87.4
C. 156.6
D. 158.6
Answer: _________
Question 411:

A bakery opened with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 60% of the remaining rolls were sold between noon and closing time. How many dozen rolls were left unsold ?

A. 6
B. 8
C. 10
D. 12
Answer: _________
Question 412:

Find the missing value : 35568 ÷ ? of 650 = 456

A. 12
B. 14
C. 16
D. 18
Answer: _________
Question 413:

The value of which of the following fractions is less than twenty percent ?

A. $$frac{5}{61}$$
B. $$frac{2}{3}$$
C. $$frac{2}{5}$$
D. $$frac{1}{4}$$
Answer: _________
Question 414:

Twelve percent of Kaushal's monthly salary is equal to sixteen percent of Nandini's monthly salary. Sanal's monthly salary is half that of Nandini's. If Sonal's annual salary is Rs. 1.08 lacs, what is Kaushal's monthly salary ?

A. Rs. 18000
B. Rs. 20000
C. Rs. 24000
D. Rs. 26000
Answer: _________
Question 415:

30% of a number when subtracted from 91, gives the number itself. Find the number.

A. 60
B. 65
C. 75
D. 70
Answer: _________
Question 416:

The population of village is 1,00,000. The rate of increase is 10% per annum. Find the population at the start of the third year?

A. 1,33,100
B. 1,21,000
C. 1,18,800
D. 1,20,000
Answer: _________
Question 417:

In the recent, climate conference in New York, out of 700 men, 500 women, 800 children present inside the building premises, 20% of the men, 40% of the women and 10% of the children were Indians. Find the percentage of people who were not Indian?

A. 77%
B. 73%
C. 79%
D. 83%
Answer: _________
Question 418:

Out of the total production of iron from hematite, an ore of Iron, 20% of the ore gets wasted, and out of the remaining iron, only 25% is pure iron. If the pure iron obtained in a year from a mine of hematite was 80,000 kg, then the quantity of hematite mined from that mine in the year is

A. 5,00,000 kg
B. 4,00,000 kg
C. 4,50,000 kg
D. None of these
Answer: _________
Question 419:

A man buys a truck for Rs. 2,50,000. The annual repair cost comes to 2% of the price of purchase. Besides, he has to pay an annual tax of Rs. 2000. At what monthly rent must he rent out the truck to get a return of 15% on his net invests of the first year?

A. Rs. 3359
B. Rs. 2500
C. Rs. 4000
D. Rs. 3212.5
Answer: _________
Question 420:

Ram spends 30% of his salary on house rent, 30% of the rest he spends on his children's education and 24% of the total salary he spends on clothes. After his expenditure, he is left with Rs. 2500. What is Ram's salary?

A. Rs. 11,494.25
B. Rs. 20,000
C. Rs. 10,000
D. Rs. 15,000
Answer: _________
Question 421:

A report consists of 20 sheets each of 55 lines and each such line consist of 65 characters. This report is reduced onto sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in number of sheets is closer to

A. 20%
B. 5%
C. 30%
D. 35%
Answer: _________
Question 422:

The price of Maruti car rises by 30 percent while the sales of the car come down by 20%. What is the percentage change in the total revenue?

A. - 4%
B. - 2%
C. + 4%
D. + 2%
Answer: _________
Question 423:

1.14 expressed as a percent of 1.9 is :

A. 6%
B. 10%
C. 60%
D. 90%
Answer: _________
Question 424:

42% of a number is 892.5. What is 73% of that number ?

A. 1466.25
B. 1508.75
C. 1551.25
D. 1636.25
Answer: _________
Question 425:

A toy merchant announces 25% rebate in prices of balls. If one needs to have a rebate of Rs. 40, then how many balls each costing Rs. 32, he should purchase ?

A. 5
B. 6
C. 7
D. 10
Answer: _________
Question 426:

Solve this : 85% of 485.5 = 50% of ?

A. 675.75
B. 735.65
C. 825.35
D. 915.5
Answer: _________
Question 427:

Vinay decided to donate 5% of his salary. On the day of donation he changed his mind and donated Rs. 1687.50, which was 75% of what he had decided earlier. How much is Vinay's salary ?

A. Rs. 33750
B. Rs. 37500
C. Rs. 45000
D. Cannot be determined
Answer: _________
Question 428:

64% of a number is 2592. What is 88% of that number ?

A. 3202
B. 3458
C. 3564
D. 3826
Answer: _________
Question 429:

Rajan got 76 percent marks and Sonia got 480 marks in a test. The maximum marks of the test equal to the marks obtained by Rajan and Sonia together. How many marks did Rajan score in the test ?

A. 1450
B. 1520
C. 1540
D. 2000
Answer: _________
Question 430:

30% of 28% of 480 is the same as

A. 15% of 56% of 240
B. 60% of 28% of 240
C. 60% of 56% of 240
D. None of these
Answer: _________
Question 431:

To meet a government requirement, a bottler must test 5 percent of its spring water and 10 percent of its sparkling water for purity. If a customer ordered 120 cases of spring water and 80 cases of sparkling water, then what percent of all the cases must the bottler test before he can send it out ?

A. 6.5%
B. 7.0%
C. 7.5%
D. 8.0%
Answer: _________
Question 432:

A 14.4 kg gas cylinder runs for 104 hours when the smaller burner on the gas stove is fully opened while it runs for 80 hours when the larger burner on the gas stove is fully opened. Which of these value is the closest to the percentage difference in the usage of gas per hour, of the smaller burner over the larger burner ?

A. 23.07%
B. 26.23%
C. 30%
D. 32.23%
Answer: _________
Question 433:

When water is changed into ice, its volume increases by 9%. If ice change into water, the percentage decrease in volume is :

A. $$8frac{{28}}{{109}}\% $$
B. 9%
C. 10%
D. 18%
Answer: _________
Question 434:

A tree increases annually by $$frac{1}{8}$$ of its height. By how much will it increase after $$2frac{1}{2}$$ years if it stands today 8 m high ?

A. 10.75 m
B. 11.85 m
C. 12.25 m
D. 15.60 m
Answer: _________
Question 435:

How many litres of a 30% alcohol solution should be added to 40 litres of a 60% alcohol solution to prepare a 50% solution ?

A. 20
B. 24
C. 30
D. 32
Answer: _________
Question 436:

Solve [180% of (?)] ÷ 2 = 504

A. 400
B. 480
C. 560
D. 600
Answer: _________
Question 437:

The marked price of an article is Rs. 2400. The shopkeeper gives successive discounts of x% and 15% to the customer. If the customer pays Rs. 1876.80 for the article, find the value of x :

A. 9%
B. 8%
C. 12%
D. 11%
Answer: _________
Question 438:

5 out of 2250 parts of earth is sulphur. What is the percentage of sulphur in earth ?

A. $$frac{11}{50}$$%
B. $$frac{2}{9}$$%
C. $$frac{1}{45}$$%
D. $$frac{2}{45}$$%
Answer: _________
Question 439:

In a class of 65 students and 4 teachers, each student got sweet that are 20% of the total number of students and each teacher got sweets that are 40% of the total number of students. How many sweets are there ?

A. 104
B. 845
C. 949
D. 897
Answer: _________
Question 440:

If 35% of a number is 175, then what percent of 175 is that number ?

A. 35%
B. 65%
C. 280%
D. None of these
Answer: _________
Question 441:

A number reduced by 25% becomes 225. What percent should it be increase so that it becomes 390 ?

A. 25%
B. 30%
C. 35%
D. 45%
Answer: _________
Question 442:

In an examination it is required to get 296 of the total maximum aggregate marks to pass. A student gets 259 marks and is decided failed. The difference of marks obtained by the student and that required to pass is 5%. What are the maximum aggregate marks a students can get ?

A. 690
B. 740
C. 780
D. Cannot be determined
Answer: _________
Question 443:

45% of 300 + $$sqrt {?} $$ = 56% of 750 - 10% of 250

A. 60
B. 130
C. 260
D. 67600
Answer: _________
Question 444:

The enrolment of students in a school increases from 560 to 581. What is the percent increase in the enrolment ?

A. 2.75%
B. 3.25%
C. 3.72%
D. 3.75%
Answer: _________
Question 445:

If the average of number, its 75% and its 25% is 240, then the number is :

A. 280
B. 320
C. 360
D. 400
Answer: _________
Question 446:

A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets ?

A. $$45$$%
B. $$45frac{5}{11}$$%
C. $$54frac{6}{11}$$%
D. $$55$$%
Answer: _________
Question 447:

0.01 is what percent of 0.1 ?

A. $$frac{1}{100}$$%
B. $$frac{1}{10}$$%
C. 10%
D. 100%
Answer: _________
Question 448:

10% of the voters did not cast their vote in an election between two candidates. 10% of the votes polled were found invalid. The successful candidate got 54% of the valid votes and won by a majority of 1620 votes. The number of voters enrolled on the voters list was :

A. 25000
B. 33000
C. 35000
D. 40000
Answer: _________
Question 449:

Subtracting 6% of x from x is equivalent to multiplying x by how much ?

A. 0.094
B. 0.94
C. 9.4
D. 94
Answer: _________
Question 450:

605 sweets are distributed equally among children in such a way that the number of sweets received by each child is 20% of the total number of children. how many sweets did each child receive ?

A. 11
B. 24
C. 45
D. Cannot determined
Answer: _________
Question 451:

Asha's monthly income is 60% of Deepak's monthly income and 120% of Maya's income. What is Maya's monthly income if Deepak's monthly income is Rs. 78000 ?

A. Rs. 36000
B. Rs. 39000
C. Rs. 42000
D. Cannot be determined
Answer: _________
Question 452:

By how much percent must a motorist increase his speed in order of reduce by 20%, the time taken to cover a certain distance ?

A. 20%
B. 25%
C. 30%
D. 35%
Answer: _________
Question 453:

In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?

A. 360
B. 280
C. 320
D. 330
Answer: _________
Question 454:

After three successive equal percentage rise in the salary the sum of 100 rupees turned into 140 rupees and 49 paise. Find the percentage rise in the salary.

A. 12%
B. 22%
C. 66%
D. 82%
Answer: _________
Question 455:

A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6:7:8:9:10. In all papers together, the candidate obtained 60% of the total marks then, the number of papers in which he got more than 50% marks is

A. 1
B. 3
C. 4
D. 5
Answer: _________
Question 456:

The length, breadth and height of a room are in ratio 3:2:1. If breadth and height are halved while the length is doubled, then the total area of the four walls of the room will

A. remain the same
B. decrease by 13.64%
C. decrease by 15%
D. decrease by 18.75%
Answer: _________
Question 457:

One bacterium splits into eight bacteria of the next generation. But due to environment, only 50% of one generation can produced the next generation. If the seventh generation number is 4096 million, what is the number in first generation?

A. 1 million
B. 2 million
C. 4 million
D. 8 million
Answer: _________
Question 458:

The rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994 is Rs. 20 per kg. The inflation rates of the years 1994 and 1995 are expected to be 8% each. The expected price of sugar on January 1, 1996 would be

A. Rs. 23.60
B. Rs. 24
C. Rs. 24.20
D. Rs. 24.60
Answer: _________
Question 459:

In an examination, questions were asked in five sections. Out of the total students, 5% candidates cleared the cut-off in all the sections and 5% cleared none. Of the rest, 25% cleared only one section and 20% cleared four sections. If 24.5% of the entire candidates cleared two sections and 300 candidates cleared three sections. Find out how many candidates appeared at the examination?

A. 1000
B. 1200
C. 1500
D. 2000
Answer: _________
Question 460:

A clock is set right at 12 noon on Monday. It losses $$frac{1}{2}$$ % on the correct time in the first week but gains $$frac{1}{4}$$ % on the true time during the second week. The time shown on Monday after two weeks will be

A. 12 : 25 : 12
B. 11 : 34 : 48
C. 12 : 50 : 24
D. 12 : 24 : 16
Answer: _________
Question 461:

If a 36 inches long strip cloth shrinks to 33 inches after being washed, how many inches long will the same strip remain after washing if it were 48 inches long?

A. 47 inches
B. 44 inches
C. 45 inches
D. 46 inches
Answer: _________
Question 462:

(X% of Y) + (Y% of X) is equal to:

A. X% of Y
B. 20% of XY
C. 2% of XY
D. 2% of 100 XY
Answer: _________
Question 463:

2 is what percent of 50 ?

A. 2%
B. 2.5%
C. 4%
D. 5%
Answer: _________
Question 464:

15% of 45% of a number is 105.3. What is 24% of that number.

A. 385.5
B. 374.4
C. 390
D. 375
Answer: _________
Question 465:

If 60% of A = $$frac{3}{4}$$ of B, then A : B is :

A. 9 : 20
B. 20 : 9
C. 4 : 5
D. 5 : 4
Answer: _________
Question 466:

In an examination, 93% of students passed and 259 failed. The total number of students appearing at the examination was :

A. 3700
B. 3850
C. 3950
D. 4200
Answer: _________
Question 467:

When 75 added to 75% of a number, the answer is the number. Find 40% of that number.

A. 100
B. 80
C. 120
D. 160
Answer: _________
Question 468:

If 15% of (A + B) = 25% of (A - B), then what percent of B equal to A ?

A. 10%
B. 60%
C. 200%
D. 400%
Answer: _________
Question 469:

When 60 is subtracted from 60% of a number, the result is 60. The number is :

A. 120
B. 150
C. 180
D. 200
Answer: _________
Question 470:

In an examination A got 25% marks more than B, B got 10% less than C and C got 25% more than D. If D got 320 marks out of 500, the marks obtained by A were :

A. 405
B. 450
C. 360
D. 400
Answer: _________
Question 471:

A reduction of 10% in the price of an apple enable a man to buy 10 apples more for Rs. 54. The reduced price of apples per dozen is :

A. Rs. 6.48
B. Rs. 12.96
C. Rs. 10.80
D. Rs. 14.40
Answer: _________
Question 472:

There are 600 boys in a hostel. Each plays either hockey or football or both. If 75% play hockey and 45% play football, how many play both
?

A. 48
B. 60
C. 80
D. 120
Answer: _________
Question 473:

A man spends 80% of his income. With increase in the cost of living, his expenditure increases by $$37frac{1}{2}$$% and his income increases by $$16frac{2}{3}$$%. His present savings are :

A. $$5frac{3}{7}$$%
B. $$5frac{5}{7}$$%
C. $$6frac{1}{2}$$%
D. $$6frac{2}{3}$$%
Answer: _________
Question 474:

At the college entrance examination, each candidate is admitted or rejected according to whether he has passed or failed the tests. Of the candidates who are really capable, 80% pass the tests and of the incapable, 25% pass the test. Given that 40% of the candidates are really capable, the proportion of capable college students is about:

A. 68%
B. 70%
C. 73%
D. 75%
Answer: _________
Question 475:

What percent of 7.2 kg is 18 gms ?

A. 0.025%
B. 0.25%
C. 2.5%
D. 25%
Answer: _________
Question 476:

? % of 450 + 46% of 285 = 257.1

A. 31
B. 28
C. 32
D. 34
Answer: _________
Question 477:

Anand has drawn an angle of measure 45° 27' when he was asked to draw an angle of 45° . The percentage error in his drawing is ?

A. 0.5%
B. 1%
C. 1.5%
D. 2.0%
Answer: _________
Question 478:

Solve this : 15% of 578 + 22.5% of 644 = ?

A. 231.4
B. 231.6
C. 231.8
D. 233.6
Answer: _________
Question 479:

Solve : 105.27% of 1200.11 + 11.80% of 2360.85 = 21.99% of (?) + 1420.99

A. 500
B. 240
C. 310
D. 550
Answer: _________
Question 480:

31% of employees pay tax in the year 2008. Non-tax paying employees are 20700. The total number of employees is :

A. 31160
B. 64750
C. 30000
D. 66775
Answer: _________
Question 481:

If 30% of A is added to 40% of B, the answer is 80% of B. What percentage of A is B ?

A. 30%
B. 40%
C. 70%
D. 75%
Answer: _________
Question 482:

A village lost 12% of its goats in a flood and 5% of remainder died from diseases. If the number left now is 8360. What was the original number before the flood ?

A. 1000
B. 10000
C. 100000
D. 8360
Answer: _________
Question 483:

If 'basis points' are defined so that 1 percent is equal to 100 basis points, then by how many basis points is 82.5 percent greater than 62.5 percent.

A. 0.2
B. 20
C. 200
D. 2000
Answer: _________
Question 484:

0.01 is what percent of 0.1 ?

A. 10%
B. $$frac{1}{10}$$%
C. 100%
D. $$frac{1}{100}$$%
Answer: _________
Question 485:

One-fifth of half of a number is 20. Then 20% of that number is :

A. 80
B. 60
C. 20
D. 40
Answer: _________
Question 486:

In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is :

A. 1 : 2
B. 3 : 4
C. 1 : 4
D. 3 : 5
Answer: _________
Question 487:

If P% of P is 36, then P is equal to :

A. 3600
B. 600
C. 60
D. 15
Answer: _________
Question 488:

If A is equal to 20% of B and B is equal to 25% of C. Then what percentage of C equal to A ?

A. 10%
B. 15%
C. 5%
D. 20%
Answer: _________
Question 489:

The marked price of an article is Rs. 5000 but due to festive offer a certain percent of discount is declared. Mr. x availed this opportunity and bought the article at reduced price. He then sold it at Rs. 5000 and thereby made a profit of $$11frac{1}{9}$$%. The percentage of discount allowed was ?

A. 10%
B. $$3frac{1}{3}$$%
C. $$7frac{1}{2}$$%
D. $$11frac{1}{9}$$%
Answer: _________
Question 490:

Heinz produces tomato puree by boiling tomato juice. Tomato puree has 20% water whereas tomato juice has 90% water.How many litres of tomato puree will be obtained from 20 litres of tomato juice ?

A. 2 litres
B. 3 litres
C. 2.5 litres
D. 5 litres
Answer: _________
Question 491:

What is the percentage change in the result when we add 50 to a certain number x, instead of subtracting 50 from the same number x?

A. 50%
B. 75%
C. 100%
D. 300%
Answer: _________
Question 492:

In a school, there are 100 students. 60% of the students are boys, 40% of whom play hockey and the girls don't play hockey, 75% of girls play badminton. There are only two games to be played. The number of student who don't play any game is:

A. 10%
B. 20%
C. 36%
D. 46%
Answer: _________
Question 493:

A book consist of 30 pages, 25 line on each page and 35 characters on each line. If this content is written in another note book consisting 30 lines and 28 characters per line then the required no. of pages will how much percent greater than previous pages?

A. 4.16%
B. 5%
C. 6.66%
D. 7%
Answer: _________
Question 494:

A fraction in reduced form is such that when it is squared and then its numerator is increased by 25% and the denominator is reduced t0 80% it results in $$frac{5}{8}$$ of original fraction. The product of the numerator and denominator is :

A. 6
B. 12
C. 10
D. 7
Answer: _________
Question 495:

80% of a number added to 80 gives the result as the number itself, then the number is :

A. 200
B. 300
C. 400
D. 480
Answer: _________
Question 496:

Reena goes to a shop to buy a radio costing Rs. 2568. The rate of sales tax is 7% and the final value is rounded off to the next higher integer. She tells the shopkeeper to reduce the price of the radio so that she has to pay Rs. 2568 inclusive of sales tax. Find the reduction needed in the price of the radio.

A. Rs. 180
B. Rs. 210
C. Rs. 168
D. Rs. 170
Answer: _________
Question 497:

Australia scored a total of X runs in 50 overs. India tied the scores in 20% less overs. If India's average run rate had been 33.33% higher the scores would have been tied 10 overs earlier. Find how many runs were scored by Australia?

A. 250
B. 240
C. 200
D. 190
Answer: _________
Question 498:

In 2000, the market shares of the toilet soaps Margo, Palmolive and dove were 40%, 30% and 30% respectively. Starting from the next year, a new soap enters into the market each year and gets 10% of the market share. The existing soap share the remaining market share in the same ratio as they did in the previous year. What percent of the total market share will mango have in 2002?

A. 28%
B. 32%
C. 32.4%
D. 34%
Answer: _________
Question 499:

In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?

A. 30000
B. 35000
C. 37000
D. 39000
Answer: _________
Question 500:

In an election between two candidates, 85% of the voters cast their votes, out which 4% of the votes were declared invalid. A candidate got 6936 votes which were 85% of the valid votes. Find the total number of voters enrolled in that election.

A. 10,700
B. 10,500
C. 10,800
D. 10,000
Answer: _________
Question 501:

In an election between two candidates, 5% of the registered voters did not cast their vote. 10% of the votes were found to be either invalid or of NOTA. The winning candidate received 60% votes in his favour and won the election by 17271 votes. Find the number of registered voters.

A. 90252
B. 100000
C. 101000
D. 102500
Answer: _________
Question 502:

ln an examination, the number of students who passed and the number of students who failed were in the ratio 25 : 4. If one more students had appeared and passed and the number of failed students was 3 less than earlier, the ratio of passed students to failed students would have become 22 : 3. What is the difference between the number of students who, initially, passed the examination and the number of students who failed the examination?

A. 132
B. 174
C. 126
D. 150
Answer: _________
Question 503:

Anuja owns $$66frac{2}{3}\% $$ xa0of a property. If 30% of the property that she owns is worth Rs. 1,25,000, then 45% of the value (in Rs.) of the property is:

A. 2,70,000
B. 2,81,250
C. 2,25,000
D. 2,62,500
Answer: _________
Question 504:

Ankita's weight is 20% less than that of her grandmother. The grandmother weights 26 kg less than grandmother's husband, whose weight is 81 kg. If Ankita's brother is 8 kg heavier than Ankita, then what is the weight (in kg) of Ankita's brother?

A. 60
B. 19
C. 36
D. 52
Answer: _________
Question 505:

If each side of a square is decreased by 17%, then by what percentage does its area decrease?

A. 30.79%
B. 31.11%
C. 25%
D. 44.31%
Answer: _________
Question 506:

The volume of the water in two tanks, A and B, is in the ratio of 6 : 5. The volume of water in tank A is increased by 30%. By what percentage should the volume of water in tank B be increased so that both the tanks have the same volume of water?

A. 56%
B. 18%
C. 15%
D. 30%
Answer: _________
Question 507:

The monthly salary of a person was Rs. 1,60,000. He used to spend on three heads Personal and family expenses (P), Taxes (T) and Education loan (E). The rest were his savings. P was 50% of the income, E was 20% of P and T was 15% of E. When his salary got raised by 30%, he maintained the percentage level of P, but E became 30% of P and T became 20% of E. The sum of the two savings (in Rs.) is:

A. 2,11,680
B. 1,28,160
C. 1,18,000
D. 1,62,810
Answer: _________
Question 508:

The number of students in a class is 45, out of which $$33frac{1}{3}\% $$ xa0are boys and the rest are girls. The average score of girls is Science is $$66frac{2}{3}\% $$ xa0more than that of boys. If the average score of all the students is 78, then the average score of girls is:

A. 54
B. 65
C. 78
D. 90
Answer: _________
Question 509:

A saves 35% of his income. If his income increase by 20.1% and his expenditure increase by 20%, then by what percentage do his saving increase or decrease? (correct to one decimal place)

A. 20.3% of increase
B. 18.5% of decrease
C. 19.75% of decrease
D. 21.9% of increase
Answer: _________
Question 510:

A man spends 75% of his income, if his income increases by 28% and his expenditure increases by 20%, then what is the increase or decrease percentage in his savings?

A. 52% increase
B. 13% decrease
C. 13% increase
D. 52% decrease
Answer: _________
Question 511:

A, B and C spend 80%, 85% and 75% of their incomes, respectively. If their savings are in the ratio 8 : 9 : 20 and the difference between the incomes of A and C is Rs. 18,000, then the income of B is:

A. Rs. 24,000
B. Rs. 36,000
C. Rs. 27,000
D. Rs. 30,000
Answer: _________
Question 512:

A man started off a business with a certain capital amount. In the first year, he earned 60% profit and donated 50% of the total capital (initial amount + profit). He followed the same procedure with the remaining capital after the second and the third year. If at the end of the three years, he is left with Rs. 15,360, what was the initial amount (in Rs.) with which the man started his business?

A. 20,000
B. 30,000
C. 25,000
D. 32,000
Answer: _________
Question 513:

A hotel is giving a discount of 12% on the booking of 2 or more rooms. Additionally, the hotel is offering a 5% discount only on payment using any card of SBI. Rakesh booked 2 rooms in the hotel for a day at the rate of Rs. 1,500 per room per day. While checking out, he paid the bill using SBI Silver Card. How much amount did he have to pay?

A. Rs. 2,498
B. Rs. 1,254
C. Rs. 2,508
D. Rs. 2,618
Answer: _________
Question 514:

Rita's income is 15% less than Richa's income. By what percent is Richa's income more than Rita's income?

A. $$15frac{{11}}{{17}}\% $$
B. $$17frac{{11}}{{17}}\% $$
C. $$16frac{{11}}{{17}}\% $$
D. $$14frac{{11}}{{17}}\% $$
Answer: _________
Question 515:

Three years ago, Raman's salary was Rs. 45,000. His salary is increased by 10 percent, A percent and 20 percent in first, second and third year respectively. Raman's present salary is Rs. 83160. What is the value of A?

A. 40
B. 30
C. 50
D. 54
Answer: _________
Question 516:

Rama spent $$frac{5}{8}$$ of her weekly salary on rent and $$frac{1}{3}$$ of the remaining on food, remaining Rs. 40 available for other expenses. Rama's weekly salary (in Rs.) is:

A. 160
B. 170
C. 140
D. 150
Answer: _________
Question 517:

In an election between two candidates, the defeated candidate secured 42% of the valid votes polled and lost the election by 7,68,400 votes. If 82,560 votes were declared invalid and 20% people did NOT cast their vote, then the invalid votes were what percentage (rounded off to 1 decimal place) of the votes which people did NOT cast?

A. 10.6%
B. 9.8%
C. 12.9%
D. 6.8%
Answer: _________
Question 518:

A number first increased by 40% and then decreased by 25% again increased by 15% and then decreased by 20%, What is the net increase/decrease percent in the number?

A. 7.2% decrease
B. 6.4% increase
C. 3.4% increase
D. 3.4% decrease
Answer: _________
Question 519:

$$frac{{25\% ,{ ext{of}}left( {50\% ,{ ext{of }}30\% ,{ ext{of }}150}
ight)}}{{40\% ,{ ext{of }}2250}}$$ xa0 xa0 is equal to:

A. 0.625%
B. 0.225%
C. 0.825%
D. 0.25%
Answer: _________
Question 520:

In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?

A. 7600
B. 8000
C. 8400
D. 8800
Answer: _________
Question 521:

Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get ?

A. 45%
B. 57%
C. 60%
D. 65%
Answer: _________
Question 522:

Fresh grapes contain 80% while dry grapes contain 10% water. If the weight of dry grapes is 250 kg, what was its total weight when it was fresh?

A. 1000 kg
B. 1125 kg
C. 1225 kg
D. 1100 kg
Answer: _________
Question 523:

A population of variety of tiny bush in an experiment field increased by 10% in the first year, increased by 8% in the second year but decreased by 10% in third year. If the present number of bushes in the experiment field is 26730, then the number of variety of bushes in beginning was:

A. 35000
B. 27000
C. 25000
D. 36000
Answer: _________
Question 524:

If a% of x is equal to b% of y, then of c% of y is what % of x ?

A. c %
B. $$frac{{{ ext{ac}}}}{{ ext{b}}}$$ %
C. $$frac{{{ ext{bc}}}}{{ ext{a}}}$$ %
D. abc %
Answer: _________
Question 525:

Mr. X salary increased by 20%. On the increase, the tax rate is 10% higher. The percentage increase in tax liability is:

A. 20%
B. 22%
C. 23%
D. 24%
Answer: _________
Question 526:

The total emoluments of A and B are equal. However, A gets 65% of his basic salary as allowances and B gets 80% of his basic salary as allowances. What is the ratio of the basic salaries of and B?

A. 16 : 13
B. 5 : 7
C. 12 : 11
D. 7 : 9
Answer: _________
Question 527:

Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A traveled uniformly with average speed of 4 km/hr. The other man traveled with varying speed as follows: In the first hour his speed 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When / where will they meet each other?

A. 7 hours after starting
B. 10 hours after starting
C. 35 km from A
D. Mid-way between A and B
Answer: _________
Question 528:

In company there are 75% skilled workers and reaming are unskilled. 80% of skilled workers and 20% of unskilled workers are permanent. If number of temporary workers is 126, then what is the number of total workers ?

A. 480
B. 510
C. 360
D. 377
Answer: _________
Question 529:

Population of a district is 2,96,000 out of which 1,66,000 are male. 50% of the population is literate. If 70% males are literate, then the number of woman who are literate, is

A. 32,200
B. 31,800
C. 66,400
D. 48,000
Answer: _________
Question 530:

What is 20% of 25% of 300 ?

A. 150
B. 60
C. 45
D. 15
Answer: _________
Question 531:

0.001 is equivalent to :

A. 10%
B. 1%
C. 0.01%
D. 0.1%
Answer: _________
Question 532:

What percentage of 3.6 kg is 72 gms. ?

A. 32%
B. 22%
C. 12%
D. 2%
Answer: _________
Question 533:

The tax imposed on an article is decreased by 10% and its consumption is increased by 10%. Find the percentage change in revenue from it.

A. 10% increase
B. 2% decrease
C. 1% decrease
D. 11% increase
Answer: _________
Question 534:

The length of a rectangle is increased by 10% and breadth decreased by 10% Then the area of the new rectangle is :

A. Neither decreased nor increased
B. Increased by 1%
C. Decreased by 1%
D. Decreased 10%
Answer: _________
Question 535:

The price of an article was first increased by 10% and then again by 20%. If the last increased price was Rs. 33, the original price was :

A. Rs. 30
B. Rs. 27.50
C. Rs. 26.50
D. Rs. 25
Answer: _________
Question 536:

The price of a commodity rises from Rs. 6 per kg to Rs. 7.50 per kg. If the expenditure cannot increase, the percentage of reduction in consumption is :

A. 15%
B. 20%
C. 25%
D. 30%
Answer: _________
Question 537:

What % of a day is 30 minutes ?

A. 2.83
B. 2.083
C. 2.09
D. 2.075
Answer: _________
Question 538:

A scored 72% in a paper with a maximum marks of 900 and 80% in another paper with a maximum marks of 700. If the result is based on the combined percentage of two papers, the combined percentage is :

A. 75.5%
B. 76%
C. 76.5%
D. 77%
Answer: _________
Question 539:

In an examination a candidate must secure 40% marks to pass. A candidate, who gets 220 marks, fails by 20 marks. Find the maximum marks for the examination ?

A. 1200
B. 300
C. 600
D. 450
Answer: _________
Question 540:

If x% of 500 = y% of 300 and x% of y% of 200 = 60, then x = ?

A. $$10sqrt 2 $$
B. $$20sqrt 2 $$
C. $$15sqrt 2 $$
D. $$30sqrt 2 $$
Answer: _________
Question 541:

In a city, 35% of the population is composed of migrants, 20% of whom are from rural areas. Of the local population, 48% is female while this figure for rural and urban migrants is 30% and 40% respectively. What percentage of the total population comprises of females ?

A. 42.75%
B. 44.50%
C. 48%
D. None of these
Answer: _________
Question 542:

A person speeds 75% of his income. If his income increase by 20% and expenses increase by 15%, his saving will increase by :

A. $$17frac{1}{2}$$%
B. $$20$$%
C. $$33frac{1}{2}$$%
D. $$35$$%
Answer: _________
Question 543:

The population of a town is 1771561. If it had been increasing at 10% per annum, its population 6 years ago was :

A. 1000000
B. 1100000
C. 1210000
D. 1331000
Answer: _________
Question 544:

In some quantity of ghee, 60% is pure ghee and 40% is vanaspati. If 10 kg pure ghee is added, then the strength of vanaspati ghee becomes 20%. The original quantity was :

A. 10 kg
B. 15 kg
C. 20 kg
D. 25 kg
Answer: _________
Question 545:

In a graduate class of 200, 40% are women and $$frac{1}{5}$$ become lecturers. If the number of men who become lecturers is twice that of women, calculate approximate percentage of men who became lecturers.

A. 16%
B. 18%
C. 205
D. 27%
Answer: _________
Question 546:

Nagaraj could save 10% of his income. But 2 years later, when his income increased by 20%, he could save the same amount only as before. By how much percentage has his expenditure increased ?

A. $$22frac{2}{9}\% $$
B. $$23frac{1}{3}\% $$
C. $$24frac{2}{9}\% $$
D. $$25frac{2}{9}\% $$
Answer: _________
Question 547:

6 c.c. of a 20% solution of alcohol in water is mixed with 4 c.c. of a 60% solution of alcohol in water. The alcoholic strength of the mixture is?

A. 20%
B. 26%
C. 36%
D. 40%
Answer: _________
Question 548:

A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ?

A. 8%
B. 9%
C. 10%
D. 11%
Answer: _________
Question 549:

If Rs. 2800 is $$frac{2}{7}$$ percent of the value of a house, the worth of the house (in Rs.) is :

A. 800000
B. 980000
C. 1000000
D. 1200000
Answer: _________
Question 550:

Two persons contested an election of Parliament. The winning candidate secured 57% of the total votes polled and won by a majority of 42000 votes. The number of total votes polled is :

A. 500000
B. 600000
C. 300000
D. 400000
Answer: _________
Question 551:

In a factory 60% of the workers are above 30 years and of these 75% are male and the rest are females. If there are 1350 male workers above 30 years, the total number of workers in the factory is :

A. 3000
B. 2000
C. 1800
D. 1500
Answer: _________
Question 552:

In the last financial year, a car company sold 41800 cars. In this year, the target is sale 51300 cars. By what percent must the sale be increased ?

A. $$11frac{9}{22}$$%
B. $$8frac{9}{22}$$%
C. $$8frac{11}{23}$$%
D. $$22frac{8}{11}$$%
Answer: _________
Question 553:

One third of a number is 96. What will 67% of that number be ?

A. 192.96
B. 181.44
C. 169.92
D. 204.48
Answer: _________
Question 554:

If the duty on an article is reduced by 40% of his present rate by how much percent must its consumption increase in order that the revenue remains unaltered ?

A. 60%
B. $$62frac{1}{3}$$%
C. 72%
D. $$66frac{2}{3}$$%
Answer: _________
Question 555:

If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction becomes $$frac{5}{2}$$. The original fraction is :

A. $$frac{24}{19}$$
B. $$frac{3}{18}$$
C. $$frac{95}{48}$$
D. $$frac{48}{95}$$
Answer: _________
Question 556:

A line of length 1.5 metres was measured as 1.55 metres by mistakes. What will be the value of error percent ?

A. 0.05%
B. $$3frac{7}{31}$$%
C. $$3frac{1}{3}$$%
D. 0.08%
Answer: _________
Question 557:

A reduction of 20% in the price of wheat enables Bhuvnesh to buy 5 kg more wheat for Rs. 320. The original rate (in rupees per kg) of wheat was :

A. 16
B. 18
C. 20
D. 21
Answer: _________
Question 558:

If the height of a cylinder is increased by 15% and the radius of its base is decreased by 10% then the percentage change in its curved surface area is :

A. 2.5% increased
B. 3.5% increased
C. 2.5% decreased
D. 3.5% decreased
Answer: _________
Question 559:

Two numbers are respectively 20% and 50% of the third number. What percent is the first number of the second ?

A. 10%
B. 20%
C. 30%
D. 40%
Answer: _________
Question 560:

25% of the candidates who appeared in an examination failed and only 450 students qualify the exam. The number of students who appeared in the examination was :

A. 700
B. 600
C. 550
D. 500
Answer: _________
Question 561:

In an assembly election, a candidate got 55% of the total valid votes. 2% of the total votes were declared invalid. If the total number of voters is 104000, then the number of valid votes polled in favour of the candidate is :

A. 56506
B. 56650
C. 56560
D. 56056
Answer: _________
Question 562:

75 gm of sugar solution has 30% sugar in it. Then the quantity of sugar that should be added to the solution to make the quantity of the sugar 70% in the solution is :

A. 125 gm
B. 100 gm
C. 120 gm
D. 130 gm
Answer: _________
Question 563:

In an examination, 1100 boys and 900 girls appeared, 50% of the boys and 40% of the girls passed the examination. The percentage of candidates who failed :

A. 45%
B. 45.5%
C. 50%
D. 54.5%
Answer: _________
Question 564:

Christy donated 10% of his income to an orphanage and deposited 20% of the remainder in his bank. If he has now Rs. 7200 left, what is his income :

A. Rs. 10000
B. Rs. 8000
C. Rs. 9000
D. Rs. 8500
Answer: _________
Question 565:

The average marks obtained in a class of 50 students is 70%. The average of first 25 is 60% and that of 24 is 80%. What is the mark obtained by the last student ?

A. 90%
B. 60%
C. 80%
D. 70%
Answer: _________
Question 566:

If 50% of (P - Q) = 30% of (P + Q) and Q = x% of P, then the value of x is :

A. 30
B. 25
C. 20
D. 50
Answer: _________
Question 567:

If x% of a is the same as y% of b, then z% of b will be :

A. $$frac{yz}{x}$$% of a
B. $$frac{zx}{y}$$% of a
C. $$frac{xy}{z}$$% of a
D. $$frac{y}{z}$$% of a
Answer: _________
Question 568:

A candidate who gets 20% marks in a examination fails by 30 marks but another candidate who gets 32% gets 42 marks more than the pass marks. Then the percentage of pass marks is :

A. 52%
B. 50%
C. 33%
D. 25%
Answer: _________
Question 569:

A man gives 50% of his money to his son and 30% to his daughter. 80% of the rest is donate to a trust. If he is left with 16000 now, how much money did he have in the beginning ?

A. Rs. 400000
B. Rs. 40000
C. Rs. 90000
D. Rs. 80000
Answer: _________
Question 570:

An empty fuel tank of a car was filled with A type petrol. When the tank was half-empty, it was filled with B type petrol. Again when the tank was half-empty, it was filled with A type petrol. When the tank was half-empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank ?

A. 33.5%
B. 37.5%
C. 40%
D. 50%
E. 37.5%
F. 35.5%
G. 30%
H. 60%
Answer: _________
Question 571:

If a = b × $$frac{{d}}{{c}}$$
b, c and d are increased by 10%, then by how much does a increase ?

A. 10%
B. 11%
C. 20%
D. 21%
Answer: _________
Question 572:

Last year, the population of a town was x and if it increases at the same rate, next year it will be y. the present population of the town is

A. $$frac{{x + y}}{2}$$
B. $$frac{{y - x}}{2}$$
C. $$frac{{2xy}}{{x + y}}$$
D. $$sqrt {xy} $$
Answer: _________
Question 573:

Vicky's salary is 75% more than Ashu's. Vicky got a raise of 40% on his salary while Ashu got a raise of 25% on his salary. By what percent is Vicky's salary more than Ashu's?

A. 96%
B. 51.1%
C. 90%
D. 52.1%
Answer: _________
Question 574:

An ore contains 25% of an alloy that has 90% iron. Other than this, in the remaining 75% of the ore, there is no iron. How many kilograms of the ore are needed to obtain 60 kg of pure iron?

A. 250 kg
B. 275 kg
C. 300 kg
D. 266.66 kg
Answer: _________
Question 575:

Ms. Pooja invests 13% of her monthly salary, i.e., Rs. 8554 in Mediclaim Policies. Later she invests 23% of her monthly salary on Child Education Policies
also she invests another 8% of her monthly salary on Mutual Funds. What is the total annual amount invested by Ms. Pooja ?

A. Rs. 28952
B. Rs. 43428
C. Rs. 173712
D. Rs. 347424
Answer: _________
Question 576:

The monthly expenses of a person are $$66frac{2}{3}\% $$ xa0more than her monthly savings. If her monthly income increases by 44% and her monthly expenses increase by 60%, then there is an increase of Rs. 1,040 in her monthly savings. What is the initial expenditure (in Rs.)?

A. 10,000
B. 13,000
C. 12,000
D. 9,000
Answer: _________
Question 577:

$$frac{5}{9}$$ part of the population in a village are males. If 30% of the males are married, the percentage of unmarried females in the total population is :

A. 20%
B. $$27frac{7}{9}$$%
C. 40%
D. 70%
Answer: _________
Question 578:

1100 boys and 700 girls are examined in a test
42% of the boys and 30% of the girls pass. The percentage of the total who failed is :

A. $$58\% $$
B. $$62frac{2}{3}\% $$
C. $$64\% $$
D. $$78\% $$
Answer: _________
Question 579:

The price of the sugar rise by 25%. If a family wants to keep their expenses on sugar the same as earlier, the family will have to decrease its consumption of sugar by

A. 25%
B. 20%
C. 80%
D. 75%
Answer: _________
Question 580:

In a village, each of the 60% of families has a cow
each of the 30% of families has a buffalo and each of the 15% of families has both a cow and buffalo. In all there are 96 families in the village. How many families do not have a cow or a buffalo ?

A. 20
B. 24
C. 26
D. 28
Answer: _________

Answer Key

1: B
Solution: Percentage annual increase : $$eqalign{
& = left( {frac{{75}}{{1000}} imes 100}
ight)\% cr
& = 7frac{1}{2}\% cr
& = frac{{15}}{2}\% cr} $$ Population after 2 years : $$eqalign{
& = 4200000{left( {1 + frac{{15}}{{2 imes 100}}}
ight)^2} cr
& = left( {4200000 imes frac{{43}}{{40}} imes frac{{43}}{{40}}}
ight) cr
& = 4853625 cr} $$
2: D
Solution: Let B + M + D = $$x$$ Then, B = 25 of $$x$$ - 20 $$eqalign{
& = left( {frac{{25x}}{{100}} - 20}
ight) cr
& = left( {frac{x}{4} - 20}
ight) cr} $$ And D = 50 ∴ $$frac{x}{4}$$ - 20 + M + 50 = $$x$$ or M = $$left( {frac{{3x}}{{4}} - 30}
ight)$$ So, marks in Maths cannot be determined.
3: C
Solution: x = 63% of y ⇒ x = $$frac{63y}{100}$$ ⇒ y = $$frac{100x}{63}$$ ∴ Required percentage : $$eqalign{
& = left( {frac{{{y^2}}}{{{x^2}}} imes 100}
ight)\% cr
& = left[ {{{left( {frac{{100x}}{{63}}}
ight)}^2} imes frac{1}{{{x^2}}} imes 100}
ight]\% cr
& = left( {frac{{10000}}{{3969}} imes 100}
ight)\% cr
& = 251.96\% approx 250\% cr} $$
4: A, F
Solution: Let the original number be $$x$$ Final number obtained : = 110% of (90% of $$x$$) = $$left( {frac{{110}}{{100}} imes frac{{90}}{{100}} imes x}
ight)$$ = $$frac{{99x}}{{100}}$$ $$eqalign{
& herefore x - frac{{99x}}{{100}} = 50 cr
& Rightarrow frac{x}{{100}} = 50 cr
& Rightarrow x = 50 imes 100 cr
& Rightarrow x = 5000 cr} $$
5: A
Solution: Number of males = 180000 Number of females : = (300000 - 180000) = 120000 Number of literates : = 50% of 300000 = 150000 Number of literate males : = 70% of 180000 = 126000 Number of literate females : = (150000 - 126000) = 24000 ∴ Required percentage : = $$left( {frac{{24000}}{{120000}} imes 100}
ight)\% $$ = 20%
6: D
Solution: Let the original price be Rs. $$x$$ ∴ (100 - r)% of (100 + r)% of $$x$$ = 1 $$eqalign{
& Rightarrow frac{{left( {100 - r}
ight)}}{{100}} imes frac{{left( {100 + r}
ight)}}{{100}} imes x cr
& Rightarrow x = frac{{100 imes 100}}{{left( {100 - r}
ight)left( {100 + r}
ight)}} cr
& Rightarrow x = frac{{10000}}{{left( {10000 - {r^2}}
ight)}} cr} $$
7: C
Solution: Let the total sales in July 2009 be Rs. $$x$$ Then, sales in September 2009 = Rs. $$frac{2x}{3}$$ Sales in November 2009 : = 105% of Rs. $$frac{2x}{3}$$ = Rs. $$left( {frac{{105}}{{100}} imes frac{{2x}}{3}}
ight)$$ = Rs. $$frac{7x}{10}$$ Decrease in sales : = Rs. $$left( {x - frac{{7x}}{{10}}}
ight)$$ .
= Rs. $$frac{3x}{10}$$ ∴ Decrease % : = $$left( {frac{{3x}}{{10}} imes frac{1}{x} imes 100}
ight)\% $$ = 30%
8: C
Solution: [x08egin{array}{*{20}{c}}
{ ext{I}}&{}&{{ ext{II}}}&{{ ext{III}}} \
{125}&:&{160}&{100} \
{25}&:&{32}&{}
end{array}]
9: A
Solution: Income of A is 80% of B $$eqalign{
& A = B imes frac{{80}}{{100}} cr
& frac{A}{B} = frac{{4x}}{{5x}} cr} $$ Expenditure of A is 60% of expenditure of B $$eqalign{
& A = B imes frac{{60}}{{100}} cr
& frac{A}{B} = frac{{3y}}{{5y}} cr} $$ Income of A is 90% of income of B $$eqalign{
& 4x = frac{{90}}{{100}} imes 5y cr
& 8x = 9y cr
& frac{x}{y} = frac{9}{8} cr} $$ Saving ratio of A and B = (4x - 3y) : (5x - 5y) xa0 [∴ 8x = 9y] = (4 × 9 - 3 × 8) : (5 × 9 - 5 × 8) = (36 - 24) : (45 - 40) = 12 : 5 More = 12 - 5 = 7 More% $$ = frac{7}{5} imes 100 = 140\% $$ Alternate solution [x08egin{array}{*{20}{c}}
{}&{{ ext{A}},,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{B}}} \
{{ ext{Income}} o }&{4 imes 5 imes 9 = 180,,,,,,,,,,5 imes 5 imes 9 = 225} \
{{ ext{Expenditure}} o }&{3 imes 4 imes 10 = 120,,,,,,,,,,5 imes 4 imes 10 = 200} \
{{ ext{Saving}} o }&{overline {underline {,,,,,,,,,,,,,,,,60,,,,,,,,,,,,,,,,,:,,,,,,,,,,,,,,,,,25,,,,,,,,,,,,,,,,} } }
end{array}] Ratio of A saving to B saving = 60 : 25 = 12 : 5 More = 12 - 5 = 7 More% $$ = frac{7}{5} imes 100 = 140\% $$
10: A
Solution: $$eqalign{
& { ext{Let be the number is }}x cr
& 180 imes frac{{15}}{{100}} + x = frac{{360 imes 20}}{{100}} cr
& 27 + x = 72 cr
& x = 72 - 27 cr
& x = 45 cr} $$
11: B
Solution: Give the amount of solution = 50litres 20% of impurity in a given solution i.e. 40 liters of glycerine and 10 liters of impurities Now to keep the impurity 10 liters and we added glycerine to a solution to bring down impurities level 5%. i.e. 10 is 5% of 200. So we need to added glycerine 'x' amount of glycerine to make solution of 200liters ∴ x = 200 - (40 + 10) = 150 litres of glycerine
12: D
Solution: After increment of 6% new length of AC = 3 + $$frac{3 × 6}{100}$$ = 3.18cm Required % decrease = $$frac{0.18}{2}$$ × 100 = 9%
13: B
Solution: Time spend by Ajoy in a day = 8 hrs Time spend by Ajay in a week = 8 × 5 = 40 hrs Percentage time spend in a week : = $$frac{40}{24 × 7}$$ × 100 = 23.81%
14: A
Solution: Let A salary's in starting is = 100 Now after decreased by 50% A salary become = 50 Again 50% of salary increase i.e. 50% of 50 = 25 Current salary is = 50 + 25 = 75 Actual decrease in salary of A = 100 - 75 = 25 Actual % lose = $$frac{25}{100}$$ × 100 = 25%
15: B
Solution: Let the 100unit is the salary of the person After spending 7.5% of his salary remain salary 92.5
Now again he spended 75% of 92.5 and remain 25% which have value Rs. 370 According to the question, 92.5 × 25% unit = 370 92.5 × $$frac{1}{4}$$ unit = 370 1 unit = $$frac{370 × 4}{92.5}$$ 100 unit = $$frac{370 × 4 × 100}{92.5}$$ = Rs. 1600
16: D
Solution: Expense + Saving = Income 61 xa0 xa0 + xa0 xa0 6 xa0 xa0 = xa0 xa0 67 67 units = Rs. 8710 1 unit = $$frac{{8710}}{{67}}$$ xa0= 130 6 unit = Rs. 780
17: B
Solution: Let the number is = 100x Now after 15% of increase = 100x + 15% of 100x = 115x Now 25% decrease = 115x - 25% of 115x = 115x - 28.75x = 86.25x Actual decreased = 100x - 86.25x = 13.75x According to the question, 13.75x = 22 x = $$frac{22}{13.75}$$ 100x = $$frac{22}{13.75}$$ xa0 × 100 = 160 ∴ Original number = 160
18: B
Solution: 0.6% = $$frac{0.6}{100}$$ = 0.006 So, difference = 0.6 - 0.006 = 0.594
19: D
Solution: Find decreased profit = 10 + 20 - $$frac{10 × 20}{100}$$ = 30 - 2 = 28%
20: B
Solution: Let the total staff = 100 40% of female, i.e 40 female and 60 male 70% of female staff is married i.e. 70% of 40 female = 28 Number of unmarried female = 40 - 28 = 12 50% of male staff is married i.e 50% of 60 male = 30 Number of unmarried male = 60 - 30 = 30 Total number of unmarried staff = 12 + 30 = 42 % of unmarried staff = $$frac{42}{100}$$ × 100 xa0 = 42%
21: A
Solution: $$eqalign{
& 20\% ,{ ext{of}},a = b Rightarrow frac{{20}}{{100}}a = b cr
& herefore b\% ,{ ext{of}},20 = {frac{b}{{100}} imes 20} cr
& = {frac{{20}}{{100}}a imes frac{1}{{100}} imes 20} cr
& = frac{4}{{100}}a = 4\% ,,{ ext{of }},a cr} $$
22: N/A
Solution: Let the number of students be x. Then, Number of students above 8 years of age $$eqalign{
& = left( {100 - 20}
ight)\% ,{ ext{of}},,x = 80\% ,,{ ext{of}},,x cr
& herefore 80\% ,{ ext{of}},,x = 48 + frac{2}{3},{ ext{of}},,48 cr
& Rightarrow frac{{80}}{{100}}x = 80 cr
& Rightarrow x = 100 cr} $$
23: D
Solution: $$eqalign{
& 5\% ,{ ext{of}},A + 4\% ,{ ext{of}},B = frac{2}{3}left( {6\% ,{ ext{of}},A + ,8,{ ext{of}},B}
ight) cr
& Rightarrow frac{5}{{100}}A + frac{4}{{100}}B = frac{2}{3}left( {frac{6}{{100}}A + frac{8}{{100}}B}
ight) cr
& Rightarrow frac{1}{{20}}A + frac{1}{{25}}B = frac{1}{{25}}A + frac{4}{{75}}B cr
& Rightarrow left( {frac{1}{{20}} - frac{1}{{25}}}
ight)A = left( {frac{4}{{75}} - frac{1}{{25}}}
ight)B cr
& Rightarrow frac{1}{{100}}A = frac{1}{{75}}B cr
& frac{A}{B} = frac{{100}}{{75}} = frac{4}{3} cr
& herefore { ext{Required}},{ ext{ratio}} = 4:3 cr} $$
24: D
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{number}},{ ext{be}},x cr
& { ext{Then,}},{ ext{error}} = frac{5}{3}x - frac{3}{5}x = frac{{16}}{{15}}x. cr
& { ext{Error}}\% = left( {frac{{16x}}{{15}} imes frac{3}{{5x}} imes 100}
ight)\% cr
& ,,,,,,,,,,,,,,,,,,,,, = 64\% cr} $$
25: A
Solution: Number of valid votes = 80% of 7500 = 6000 ∴ Valid votes polled by other candidate = 45% of 6000 = $$ {frac{{45}}{{100}} imes 6000} $$ = 2700
26: A
Solution: Total number of votes polled = (1136 + 7636 + 11628) = 20400 ∴ Required percentage = $$left( {frac{{11628}}{{20400}} imes 100}
ight)\% $$ = $$57\% $$
27: B
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{sum}},{ ext{paid}},{ ext{to}},{ ext{y}},{ ext{per}},{ ext{week}},{ ext{be}},Rs.,z cr
& { ext{Then}},,z + 120\% ,of,z = 550 cr
& Rightarrow z + frac{{120}}{{100}}z = 550 cr
& Rightarrow frac{{11}}{5}z = 550 cr
& Rightarrow z = {frac{{550 imes 5}}{{11}}} = 250 cr} $$
28: C
Solution: Let the amount taxable purchases be Rs. x $$eqalign{
& { ext{Then}},6\% ,,{ ext{of}},x = frac{{30}}{{100}} cr
& Rightarrow x = {frac{{30}}{{100}} imes frac{{100}}{6}} = 5 cr
& herefore { ext{Cost of tax free items}} cr
& = { ext{Rs}}{ ext{.}}left[ {25 - left( {5 + 0.30}
ight)}
ight] cr
& = { ext{Rs}}{ ext{.}},19.70 cr} $$
29: A
Solution: $$eqalign{
& { ext{Rebate}} = 6\% ,{ ext{of Rs}}.6650 cr
& = { ext{Rs}}{ ext{.}}left( {frac{6}{{100}} imes 6650}
ight) cr
& = { ext{Rs}}{ ext{.}},399 cr
& { ext{Sales tax}} = {kern 1pt} 10\% ,{ ext{of Rs}}{ ext{.}}left( {6650 - 399}
ight) cr
& = { ext{Rs}}{ ext{.}}left( {frac{{10}}{{100}} imes 6251}
ight) cr
& = { ext{Rs}}.625.10 cr
& herefore { ext{Final}}{kern 1pt} { ext{amount}} cr
& = { ext{Rs}}{ ext{.}}left( {6251 + 625.10}
ight) cr
& = { ext{Rs}}{ ext{.}},6876.10 cr} $$
30: B
Solution: $$eqalign{
& { ext{Increase}},{ ext{in}},{ ext{10}},{ ext{years}} cr
& = {262500 - 175000} cr
& = 87500 cr
& { ext{Increase}}\% cr
& = left( {frac{{87500}}{{175000}} imes 100}
ight)\% = 50\% cr
& herefore { ext{Required}},{ ext{average}} cr
& = {frac{{50}}{{10}}} \% = 5\% cr} $$
31: A
Solution: Number of bushes in the beginning : $$eqalign{
& = frac{{26730}}{{left( {1 + frac{{10}}{{100}}}
ight)left( {1 + frac{8}{{100}}}
ight)left( {1 - frac{{10}}{{100}}}
ight)}} cr
& = left( {26730 imes frac{{10}}{{11}} imes frac{{25}}{{27}} imes frac{{10}}{9}}
ight) cr
& = 25000 cr} $$
32: C
Solution: Let total quantity of original milk = 1000 gm Milk after first operation = 80% of 1000 = 800 gm Milk after second operation = 80% of 800 = 640 gm Milk after third operation = 80% of 640 = 512 gm ∴ Strength of final mixture = 51.2%
33: C
Solution: $$66frac{2}{3}$$% of Rs. 312/- exceeds Rs. 200 by Rs. $$x$$ According to the question, required difference $$eqalign{
& x = { ext{Rs}}{ ext{. }}left( {312 imes frac{{200}}{3}\% - 200}
ight) cr
& x = { ext{Rs}}{ ext{. }}left( {312 imes frac{{200}}{{3 imes 100}} - 200}
ight) cr
& x = { ext{Rs}}{ ext{. }}left( {312 imes frac{{200}}{{300}} - 200}
ight) cr
& x = { ext{Rs}}{ ext{. }}left( {208 - 200}
ight) cr
& x = { ext{Rs}}{ ext{. 8}} cr} $$
34: D
Solution: Let 125% of 3060 - 85% of x = 408 Then, $$eqalign{
& 85\% { ext{ of }}x = left( {frac{{125}}{{100}} imes 3060}
ight) - 408 cr
& Rightarrow frac{{85}}{{100}}x = 3825 - 408 cr
& Rightarrow frac{{17x}}{{20}} = 3417 cr
& Rightarrow x = left( {frac{{3417 imes 20}}{{17}}}
ight) cr
& Rightarrow x = 4020 cr} $$
35: D
Solution: Let David's annual income be Rs. x and his wife's annual income be Rs. y Then, 8% of x = 800 ⇒ $$frac{8}{100}$$x = 800 ⇒ x = $$left( {frac{{800 imes 100}}{{8}}}
ight)$$ ⇒ x = 10000 And, 8% of y = 840 ⇒ $$frac{8}{100}$$y = 840 ⇒ y = $$left( {frac{{840 imes 100}}{{8}}}
ight)$$ ⇒ y = 10500 ∴ Required difference : = [(10500 + 840) - (10000 + 800)] = Rs. (11340 - 10800) = Rs. 540
36: C
Solution: Amount paid to car owner = 90% of 85% of Rs. 325000 = Rs. $$left( {frac{90}{{100}} imes frac{85}{100} imes 325000}
ight)$$ = Rs. 248625 ∴ Required difference : = Rs. (325000 - 248625) = Rs. 76375
37: C
Solution: Let the number of valid votes be x Then, 60% of x = 255000 ⇒ x = $$left( {frac{{255000 imes 100}}{{60}}}
ight)$$ ⇒ x = 425000 Number of invalid votes : = (500000 - 425000) = 75000 ∴ Required percentage : = $$left( {frac{{75000 imes 100}}{{500000}}}
ight)$$xa0 xa0 % = 15%
38: B
Solution: x = 80% of y ⇒ x = $$frac{80}{100}$$y ⇒ $$frac{y}{x}$$ = $$frac{5}{4}$$ ⇒ $$frac{y}{2x}$$ = $$frac{5}{8}$$ ∴ Required percentage : $$eqalign{
& = left( {frac{y}{{2x}} imes 100}
ight)\% cr
& = left( {frac{5}{8} imes 100}
ight)\% cr
& = 62frac{1}{2}\% cr} $$
39: D
Solution: Let the original price of tea be Rs. x per kg and that of sugar be Rs. y per kg Then, 5x + 8y = 172 ⇒ 15x + 24y = 516.....(i) And, 120% of 5x + 110% of 8y = 199.20 ⇒ 600x + 880y = 19920 ⇒ 15x + 22y = 498 .....(ii) Subtracting (ii) from (i), we get 2y = 18 or y = 9 Putting y = 9 in (i), we get: x = 20
40: C
Solution: Let the fare for a full ticket be Rs. x and the boarding charges be Rs. y per ticket. Then, fare for a half-ticket : = 75% of Rs. x = Rs. $$frac{3x}{4}$$ $$eqalign{
& herefore left( {x + y}
ight) + left( {frac{3}{4}x + y}
ight) = 1440 cr
& Rightarrow frac{{7x}}{4} + 2y = 1440 cr
& Rightarrow 7x + 8y = 5760 ..... (i) cr
& { ext{And, 2}}left( {x + y}
ight) + left( {frac{3}{4}x + y}
ight) = 2220 cr
& Rightarrow frac{{11x}}{4} + 3y = 2220 cr
& Rightarrow 11x + 12y = 8880 ..... (ii) cr} $$ Solving (i) and (ii), we have, x = 480, y = 300
41: A
Solution: Let the number be x According to the question, x + 320 × $$frac{10}{100}$$ = 230 × $$frac{30}{100}$$ x + 32 = 69 x = 37 Hence, required number = 37
42: A
Solution: 120 × $$frac{25}{100}$$ + 380 × $$frac{40}{100}$$ = x × 637 ⇒ 30 + 152 = x × 637 ⇒ $$frac{182}{637}$$ = x ⇒ x = $$frac{2}{7}$$ ⇒ Required answer = $$frac{2}{7}$$
43: C
Solution: Let the number = x According to the question, x - 15 = $$frac{80x}{100}$$ x - 15 = $$frac{4x}{5}$$ 5x - 75 = 4x x = 75 Required answer = 75 × $$frac{40}{100}$$ = 30
44: B
Solution: A 35% = B 25% $$frac{A}{B}$$ = $$frac{5}{7}$$ A : B = 5 : 7
45: A
Solution: 1 hour = 60 minutes 1 min + 12 sec = 1 + $$frac{12}{60}$$ = $$frac{6}{5}$$ min According to the question, 60 × $$frac{y}{100}$$ = $$frac{6}{5}$$ y = 2
46: A
Solution: Let conside 100 item sold on Rs.100 ∴ Total Amount = 100 × 100 = Rs. 10000 Now After 20% discount price become Rs.80 Total Number of Sale increased after discount is 80%. i.e. 100 + 80% of 100 = 180(Item sold) ∴ Total Amount = 180 × 80 = Rs. 14400 Increased in amount = 14400 - 10000 = 4400 % Increase in amount = $$frac{{4400 imes 100}}{{10000}} $$ = 44% (increase)
47: C
Solution: Failed students in Mathematics = 34% Failed students in English = 42% Student failed in both subject math and english = 20% Student only fail in Math = 34 - 20 = 14% Student only fail in English = 42 - 20 = 22% Percentage of passed students in both subjects : = 100 - [ student fail in maths + student fail in english + student fail in both subject] = [100 - (14 + 20 + 22)] = 44%
48: C
Solution: Let the total number of the voter is 100x and the winner get 60x votes and loser get 40x votes. ∴ According to question 60x - 40x = 298 ⇒ x = $$frac{298}{20}$$ ∴ Total Number of voter = 100 × $$frac{298}{20}$$ xa0 = 1490
49: B
Solution: By using Alligation rule, Hence, required ratio = 1 : 3
50: D
Solution: Let the first and second number is a and b respectively, b - $$frac{60a}{100}$$ = $$frac{52b}{100}$$ b - $$frac{52b}{100}$$ = $$frac{60a}{100}$$ $$frac{48b}{100}$$ = $$frac{60a}{100}$$ 4b = 5a $$frac{a}{b}$$ = $$frac{4}{5}$$ ⇒ a : b = 4 : 5
51: B
Solution: Let the number of parts before the quality checks be 100 Then, number of parts passed after quality checks = (100 - 2)% of (100 - 5)% of (100 - 10)% of 100 = 98% of 95 of 90% of 100 = $$left( {frac{{98}}{{100}} imes frac{{95}}{{100}} imes frac{{90}}{{100}} imes 100}
ight)$$ = $$left( {frac{{8379}}{{100}}}
ight)$$ = 83.79 ∴ Effective rejection rate : = (100 - 83.79)% = 16.21%
52: B
Solution: Let the present value be Rs. 100 Value after 3 years : $$eqalign{
& = { ext{Rs}}{ ext{. }}left[ {100 imes {{left( {1 - frac{{20}}{{100}}}
ight)}^3}}
ight] cr
& = { ext{Rs}}{ ext{. }}left( {100 imes frac{4}{5} imes frac{4}{5} imes frac{4}{5}}
ight) cr
& = { ext{Rs}}{ ext{. 51}}{ ext{.20}} cr} $$ ∴ Reduction in value : = (100 - 51.20)% = 48.8%
53: B
Solution: $$eqalign{
& nleft( A
ight) = left( {frac{{60}}{{100}} imes 96}
ight) = frac{{288}}{5} cr
& nleft( B
ight) = left( {frac{{30}}{{100}} imes 96}
ight) = frac{{144}}{5} cr
& nleft( {A cap B}
ight) = left( {frac{{15}}{{100}} imes 96}
ight) = frac{{72}}{5} cr
& herefore nleft( {A cup B}
ight): cr
& = nleft( A
ight) + nleft( B
ight) - nleft( {A cap B}
ight) cr
& = frac{{288}}{5} + frac{{144}}{5} - frac{{72}}{5} cr
& = frac{{360}}{5} cr
& = 72 cr} $$ So, people who had either or both types of lunch = 72 Hence, people who had neither type of lunch = (96 - 72) = 24
54: B
Solution: Quantity of alcohol : = $$left( {frac{{40}}{{100}} imes 5}
ight)$$ xa0 litres = 2 litres ∴ New strength : $$eqalign{
& = left( {frac{2}{6} imes 100}
ight)\% cr
& = 33frac{1}{3}\% cr} $$
55: B
Solution: Given, 30% of 1225 - 64% of 555 = (?) $$eqalign{
& = left( {frac{{30}}{{100}} imes 1225}
ight) - left( {frac{{64}}{{100}} imes 555}
ight) cr
& = left( {367.5 - 355.2}
ight) cr
& = 12.3 cr} $$
56: C
Solution: Quantity of sugar $$eqalign{
& = left( {frac{4}{{100}} imes 6}
ight)kg cr
& = 0.24,,kg cr} $$ ∴ New percentage : $$eqalign{
& = left( {frac{{0.24}}{5} imes 100}
ight)\% cr
& = 4frac{4}{5}\% cr} $$
57: C
Solution: Given, (85% of 420) + (?% of 1080) = 735 $$ Rightarrow left( {frac{{85 imes 420}}{{100}}}
ight) imes left( {frac{{(?) imes 1080}}{{40}}}
ight)$$ xa0 xa0 xa0 $$= 735$$ $$eqalign{
& Rightarrow 357 + left( {frac{{(?) imes 1080}}{{100}}}
ight) = 735 cr
& Rightarrow left( {frac{{(?) imes 1080}}{{100}}}
ight) = left( {735 - 357}
ight) cr
& Rightarrow left( {frac{{(?) imes 1080}}{{100}}}
ight) = 378 cr
& Rightarrow (?) = frac{{378 imes 100}}{{1080}} cr
& Rightarrow (?) = 35 cr} $$
58: C
Solution: Given expression : $$eqalign{
& = left( {frac{{45}}{{100}} imes frac{{25}}{{100}} imes frac{4}{5} imes 850}
ight) cr
& = frac{{153}}{2} cr
& = 76.5 cr} $$
59: C
Solution: Let the maximum marks be x Then, 36% of x = 198 + 36 ⇒ $$frac{36x}{100}$$ = 234 ⇒ x = $$frac{234 × 100}{36}$$ ⇒ x = 650
60: D
Solution: Total cost : = Rs. [1 × 1000 + (100 - 2)% of 1 × 4000] = Rs. (1000 + 0.98 × 4000) = Rs. (1000 + 3920) = Rs. 4920
61: A
Solution: Let the number of people be x who has been asked for the donations
People already solicited = 60% of x = 0.6x Remaining people = 40% of x = 0.4x Amount collected from the people solicited, = 600 × 0.6x = 360x 360x = 75% of the amount collected Remaining amount = 25% = 120x Thus, Average donations from remaining people, $$eqalign{
& = frac{{120{ ext{x}}}}{{0.4{ ext{x}}}} cr
& = 300 cr} $$
62: B
Solution: Female employees = 45% Male employees = 55% Difference = 55 - 45 = 10% → 10% = 72 → 1% = $$frac{{72}}{{10}}$$ → 100 = 7.2 × 100 = 720
63: B
Solution: 100 == 25% ↑ ==> 125 == x%↓ ==> 100 $${ ext{x}}\% = frac{{25 imes 100}}{{100}} = 25\% $$ We don't need the actual value to calculate the answer
64: B
Solution: Day 1 2 3 Initial mangoes x (let) 0.45x 0.2025x Sales 0.5x 0.225x 0.10125x Over night remaining 0.5x 0.225x 0.10125x Rotten 0.05x 0.0225x 0.010125x Next day stock 0.45x 0.2025x Thus, Total mangoes rotten amount is ⇒ 0.05x + 0.0225x + 0.010125x = 0.082625x = 1983x = 24,000
65: C
Solution: Non-defective products M 1 = 25 × 0.98 = 24.5% Non-defective products M 2 = 35 × 0.96 = 33.6% Non-defective products M 3 = 40 × 0.95 = 38% Percentage of non-defective products = 24.5 + 33.6 + 38 = 96.1%
66: C
Solution: $$eqalign{
& { ext{Let the positive number be x}}. cr
& { ext{According}},{ ext{to}},{ ext{the}},{ ext{question}}, cr
& x imes x = x + frac{{left( {x imes 2000}
ight)}}{{100}} cr
& {x^2} = x + 20x cr
& {x^2} - 21x = 0 cr
& { ext{Either}}, cr
& x = 0,,,or cr
& x = 21 cr
& { ext{21}},{ ext{is}},{ ext{the}},{ ext{possible}},{ ext{value}}{ ext{.}} cr
& { ext{Then}},{ ext{square}},{ ext{of}},21,{ ext{is}},441 cr} $$
67: B
Solution: Let the bike's initial cost be x and then car's initial cost be 5x
After the increase,
Bike price = 1.2x Car price = 5.75x Initial total cost of 5 cars and 10 bikes, = 25x + 10x = 35x New cost, = 28.75x + 12x = 40.75x Change in cost = (40.75x - 35x) = 5.75x % change = $$frac{{5.75{ ext{x}} imes 100}}{{35}} = 16frac{3}{7}\% $$
68: B
Solution: The average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya (heavy). Let the total weight of all the bars be X. The weight of the bars given to Brennan, = 45% of X = 0.45X The weight of the bars given to Maya, = 26% of X = 0.26X The weight of the bars given to Claire = rest = 29% of X = 0.29X The average weight of the bars given to Brennan, $$ = frac{{{ ext{Weight}}}}{{{ ext{Number}},{ ext{of}},{ ext{bars}}}} = frac{{0.45{ ext{X}}}}{{24}}$$ The average weight of the bars given to Maya = Weight / number of bars = $$frac{{0.26{ ext{X}}}}{{13}}$$ Similarly, if the number of bars given to Blair = B, then the average weight of the bars given to Blair = $$frac{{0.29{ ext{X}}}}{{ ext{B}}}$$ As, the average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya So , Option (B) is the right answer
69: A
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{original}},{ ext{price}} = y, cr
& { ext{After}},{ ext{first}},{ ext{change,}},{ ext{it}},{ ext{becomes}}, cr
& y imes left( {1 + {frac{x}{{100}}} }
ight) cr
& { ext{After}},{ ext{second}},{ ext{change,}},{ ext{it}},{ ext{becomes}} cr
& y imes left( {1 + {frac{x}{{100}}} }
ight)left( {1 - {frac{x}{{100}}} }
ight) cr
& = yleft( {1 - {{left( {frac{x}{{100}}}
ight)}^2}}
ight) cr
& { ext{Thus}}, cr
& {x^2} imes y = {10^6} - - - - left( 1
ight) cr
& {x^2} = frac{{{{10}^6}}}{y} cr
& { ext{Now}}, cr
& y{left( {1 - {frac{{{{10}^6}}}{{10000y}}} }
ight)^2} cr
& = 2304left( {{ ext{similar}},{ ext{to}},{ ext{above}}}
ight) cr
& y{left( {1 - frac{{100}}{y}}
ight)^2} = 2304 cr
& y = 2500 cr} $$
70: B
Solution: Year Value 2000 100 (Let) 2001 110 2002 121 2003 133.1 2004 119.79 2005 107.811 2006 97.0299 $$eqalign{
& { ext{Now}}, cr
& frac{{ {left( {100 - 97.0299}
ight) imes 100} }}{{100}} approx 3 cr} $$
71: D
Solution: Let the original price be Rs. 100 New final price : = 120% of (75% of Rs. 100) = Rs. $$left( {frac{{120}}{{100}} imes frac{{75}}{{100}} imes 100}
ight)$$ = Rs. 90 ∴ Decrease = 10%
72: C
Solution: Net growth on 1000 $$= (32 - 11)$$ $$= 21$$ Net growth on 100 $$eqalign{
& = left( {frac{{21}}{{1000}} imes 100}
ight)\% cr
& = 2.1\% cr} $$
73: A
Solution: Let the original price = Rs. $$x$$ per kg Reduced price = Rs. $$left( {frac{{79x}}{{100}}}
ight)$$ per kg $$eqalign{
& herefore frac{{100}}{{79x}} - frac{{100}}{x} = 10.5 cr
& Rightarrow frac{{10000}}{{79x}} - frac{{100}}{x} = 10.5 cr
& Rightarrow 10000 - 7900 = 10.5 imes 79x cr
& Rightarrow x = frac{{2100}}{{10.5 imes 79}} cr} $$ ∴ Reduced price : $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {frac{{79}}{{100}} imes frac{{2100}}{{10.5 imes 79}}}
ight){ ext{ per kg}} cr
& = { ext{Rs}}{ ext{. 2 per kg}} cr} $$
74: C
Solution: $$eqalign{
& X = frac{{90}}{{100}}Y cr
& Rightarrow X = frac{9}{{10}}Y cr
& Rightarrow Y = frac{{10}}{9}X cr
& Rightarrow frac{Y}{X} = frac{{10}}{9} cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{Y}{X} imes 100}
ight)\% cr
& = left( {frac{{10}}{9} imes 100}
ight)\% cr
& = 111frac{1}{9}\% cr} $$
75: D
Solution: Number of rotten apples : $$eqalign{
& = 30\% { ext{ of }}450 cr
& = left( {frac{{30}}{{100}} imes 450}
ight) cr
& = 135 cr} $$ ∴ Number of good apples = 450 - 135 = 315
76: D
Solution: Let 3.2% of 500 × 2.4% of x = 288 Then, $$ Rightarrow left( {frac{{32}}{{10}} imes frac{1}{{100}} imes 500}
ight) imes $$ xa0 xa0 $$left( {frac{{24}}{{10}} imes frac{1}{{100}} imes x}
ight)$$ xa0 $$ = 288$$ $$eqalign{
& Rightarrow 16 imes frac{{3x}}{{125}} = 288 cr
& Rightarrow x = frac{{288 imes 125}}{{16 imes 3}} cr
& Rightarrow x = 750 cr} $$
77: C
Solution: Let the total number of students be x Then, $$eqalign{
& left( {100 - 76}
ight)\% { ext{ of }}x = 204 cr
& Rightarrow 24\% { ext{ of }}x = 204 cr
& Rightarrow frac{{24}}{{100}}x = 204 cr
& Rightarrow x = left( {frac{{204 imes 100}}{{24}}}
ight) cr
& Rightarrow x = 850 cr} $$
78: C
Solution: Let the numbers be $$x$$ and $$y$$ respectively Then, 60% of $$x$$ = $$frac{{2}}{{3}}$$ $$y$$ ⇒ $$frac{{60}}{{100}}$$ $$x$$ = $$frac{{2}}{{3}}$$ $$y$$ ⇒ $$frac{{3}}{{5}}$$ $$x$$ = $$frac{{2}}{{3}}$$ $$y$$ ⇒ $$frac{{x}}{{y}}$$ = $$frac{{2}}{{3}}$$ × $$frac{{5}}{{3}}$$ ⇒ $$frac{{x}}{{y}}$$ = $$frac{{10}}{{9}}$$ ∴ x : y = 10 : 9
79: N/A
Solution: Weight of water in 60 gm mixture : = 75% of 60 gm = $$left( {frac{{75}}{{100}} imes 60}
ight)$$ xa0 gm = 45 gm Weight of water in 75 gm mixture : = (45 + 15) gm = 60 gm ∴ Required percentage : = $$left( {frac{{60}}{{75}} imes 100}
ight)$$ xa0 % = 80%
80: C
Solution: $$eqalign{
& { ext{Required}},{ ext{Percentage}} cr
& = frac{{ {1.14 imes 100} }}{{1.9}} cr
& = 60\% cr} $$
81: A
Solution: Students passed in English = 80% Students passed in Math's = 85% Students passed in both subjects = 73% Then, number of students passed in at least one subject = (80+85)-73 = 92%. [The percentage of students passed in English and Maths individually, have already included the percentage of students passed in both subjects. So, We are subtracting percentage of students who have passed in both subjects to find out percentage of students at least passed in one subject.] Thus, students failed in both subjects = 100-92 = 8%.
82: C
Solution: $$eqalign{
& { ext{As}},{ ext{we}},{ ext{know}},,1\% = frac{1}{{100}} cr
& { ext{Hence}},, cr
& frac{1}{2}\% = {frac{1}{2} imes frac{1}{{100}}} cr
& ,,,,,,,,,,, = frac{1}{{200}} cr
& ,,,,,,,,,,, = 0.005 cr} $$
83: B
Solution: Let the initial price of the commodity be 100. After 50% increase in price, It will become, 100 ------50% increase----> 150. Now, we have to reduce the consumption to keep expenditure 100. Increase in price= 150 - 100 = 50 We have to reduce the consumption, $$eqalign{
& = frac{{50}}{{150}} imes 100 cr
& = frac{1}{3}{ ext{ or 33}}{ ext{.33}}\% cr} $$ Other Method: Here, we use, Final product constant graphic. 100 ==50% up== 150===33.33% down===>100 Consumption Reduce = 33.33% = $$frac{1}{3}$$
84: C
Solution: Here we can use the compound interest based formula, $$eqalign{
& { ext{Population after }}n{ ext{ years}} cr
& = P imes {left[ {1 + {frac{r}{{100}}} }
ight]^n} cr
& { ext{Population after 2 years}} cr
& = 50000 imes {left[ {1 + {frac{4}{{100}}} }
ight]^2} cr} $$ Population after 2 years = 54080 Alternatively, we can use, net percentage change graphic as well, 50,000------4%↑---→ 52,000---- 4%↑---→ 54,080. Then, population after 2 years= 54,080. In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100.
85: D
Solution: As A and B are fixed, C is any point on AB, so if AC is increases then CB decreases. A ________3 cm_________ C _____2 cm____ B Then, solution can be visualized as, Increase in AC 6% = $$frac{{106 imes 3}}{{100}} = 3.18,{ ext{cm}}{ ext{.}}$$ Decrease in CB = 0.18 cm % decrease = $$frac{{0.18}}{2} imes 100 = 9\% $$ Alternatively, AC = 3 Cm. BC = 2 Cm. Increase in AC by 6%, then New, AC = 3 + 6% of 3 = 3 + 0.18 = 3.18 cm. 0.18 cm increase in AC means 0.18 cm decrease in BC as already mentioned AB as the fixed point. So, % decrease in BC, $$eqalign{
& = frac{{{ ext{Actual}},{ ext{Decrease}},{ ext{in}},{ ext{BC}}}}{{{ ext{Original BC}}}} imes 100 cr
& = frac{{0.18}}{2} imes 100 = 9\% cr} $$
86: A
Solution: Initial Cost = Rs. 75 After 20% increase in the cost, it becomes, (75 + 20% of 75) = Rs. 90 Now, Cost is decreased by 20%, So cost will become, (90 - 20% of 90) = Rs. 72 So, present cost is Rs. 72
87: B
Solution: Let the Original length of the rectangle be 20 unit and breadth be 10 unit. Then Original Area = length *breadth = 20*10 = 200 Square unit. 40% decrease in each side, then Length = (20 - 40% of 20) = 12 unit. Breadth = (10 - 40% of 10) = 6 unit. Now, Area = 12 × 6 = 72 Square unit. Decrease in area = 200 - 72 = 128 square unit. % Decrease in Area = $$frac{{128}}{{200}} imes 100 = 64\% $$ Mind Calculation Method: Let the original area be 100 square unit. 100 === 40%↓(decrease in length) ==⇒ 60 === 40%↓ === (decrease in breadth) ===> 36 Diminished in area = 100 - 36 = 64%
88: B
Solution: Let the initial expenditure on the commodity be Rs. 100. Now, the price decreases by 20%, Current Price = (100 - 20% of 100) = Rs. 80. Same time due to decrements in price 20% consumption has been increased. So, Current expenses on commodity = (80 + 20% of 80)= Rs. 96. Here, the initial expenditure was Rs. 100 which became 96 at the end, it means there is 4% decrements in the expenditure of the commodity. Mind Calculation Method: 100 === 20%↓(Decrements in Price) ===> 80 === 20%↑(Increment in Consumption) ===> 96. Thus, there is a decrements of 4%
89: A
Solution: According to the question, $$eqalign{
& frac{5}{{100}}A + frac{4}{{100}}B = frac{2}{3}left[ {frac{{6A}}{{100}} + frac{{8B}}{{100}}}
ight] cr
& Rightarrow 5A + 4B = frac{2}{3}left( {6A + 8B}
ight) cr
& Rightarrow 15A + 12B = 12A + 16B cr
& Rightarrow 3A = 4B cr
& Rightarrow frac{A}{B} = frac{4}{3} cr
& Rightarrow A:B = 4:3 cr} $$
90: B
Solution: Required % decrements : $$eqalign{
& = frac{{{x^2}}}{{100}}\% cr
& = frac{{{{left( {25}
ight)}^2}}}{{100}} cr
& = 6frac{1}{4}\% cr} $$
91: D
Solution: Candidates failed in English = (100 - 60)% = 40% Candidate failed in Mathematics = (100 - 70)% = 30% 20% Candidate fail in Both. Candidate only fail in English = 40 - 20 = 20% Candidate only fail in Mathematics = 30 - 20 = 10% Students passed in both subjects = 100 - (20% Candidate fail in Both + Candidate only fail in English + Candidate only fail in Mathematics) = 100 - (20 + 20 + 10) = 50% 50% of students = 2500 Total students = $$frac{{2500}}{{50}} imes 100$$ = 5000
92: A
Solution: 10% = $$frac{1}{10}$$ [x08egin{gathered}
{ ext{Initial }},,,,,,{ ext{Final}} hfill \
,,,{ ext{10}},,,,,,,,,,,,,,,9 hfill \
,,,10,,,,,,,,,,,,,,,9 hfill \
overline {,,100,,,,,,,,,,,,,81,,,} hfill \
end{gathered} ] ⇒ 81 units = Rs. 8100 ⇒ 1 unit = Rs. 100 ⇒ 100 units = Rs. 10000 ⇒ Value of property 2 years ago ⇒ Rs. 10000
93: B
Solution: Let the total capital = 1200 According to the question, one-fourth of 1200 = 300, two third of 1200 = 800, remain 100 Interest on 300 = 3% of 300 = 9 Interest on 800 = 5% of 800 = 40 Interest on 100 = 11% of 100 = 11 Total interest = (9 + 40 + 11) = 60 Required percentage : = $$frac{60}{1200}$$ × 100 = 5%
94: D
Solution: Let the number = x According to the question, ⇒ x × $$frac{120}{100}$$ - x × $$frac{75}{100}$$ = 36 ⇒ 120x - 75x = 3600 ⇒ 45x = 3600 ⇒ x = $$frac{3600}{45}$$ ⇒ x = 80 Hence, required number = 80
95: A
Solution: 25% = $$frac{1}{4}$$ [x08egin{gathered}
{ ext{Initial }},,,,,,{ ext{Final}} hfill \
,,,4,,,,,,,,,,,,,,,5 hfill \
,,,4,,,,,,,,,,,,,,,5 hfill \
,,,4,,,,,,,,,,,,,,,5 hfill \
overline {,,,,,,64,,,,,,,,,,125,,,} hfill \
end{gathered} ] ⇒ 125 units = 10000 ⇒ 1 unit xa0 xa0 xa0 = 80 ⇒ 64 units = 5120 ⇒ Population at the beginning of 1 st year = 5120
96: A
Solution: Let the salary = 100 units Savings = 20% Savings = 100 × $$frac{20}{100}$$ = 20 units Expenditure = (100 - 20) = 80 units According to the question, 80 units = Rs. 6000 1 unit xa0 xa0 = Rs. 75 Savings = 75 × 20 = Rs. 1500
97: C
Solution: Let the income of the man = Rs. 100 ∴Intial Expenditure = Rs.75 Now new income become = 100 + 20% of 100 = Rs. 120 New Expenditure = 75 + 15% of 75 = Rs.86.25 Intial Saving = 100 - 75 = Rs. 25 New Saving = 120 - 86.25 = Rs. 33.75 Required percentage increase : = $$frac{(33.75 - 25)}{25}$$ xa0 × 100 = 35%
98: A
Solution: Money spent originally = Rs. 140
Less Money to be spent now = 10% of 140 = Rs. 14 Rs. 14 now yield 500 gm sugar So, Present rate of sugar = Rs. 28 per kg. If the present value is Rs. 90, the original value = Rs. 100 If the present value is Rs. 28 the original value $$ = { ext{Rs}}{ ext{. }}frac{{100}}{{90}} imes 28 = { ext{Rs}}{ ext{. }}31.11$$
99: D
Solution: $$eqalign{
& { ext{Let third number is x}}. cr
& { ext{Then}},{ ext{first}},{ ext{no}}{ ext{.}} cr
& 20\% ,{ ext{of}},x = frac{{20x}}{{100}} cr
& { ext{Second}},{ ext{number}} cr
& = 50\% ,{ ext{of}},x = frac{{50x}}{{100}} cr
& { ext{Percent of first no of second no,}} cr
& = {frac{{ {frac{{20x}}{{100}}} }}{{ {frac{{50x}}{{100}}} }}} imes 100 cr
& = frac{{ {2 imes 100} }}{{20}} cr
& = 40\% cr} $$
100: B
Solution: Let the capacity of the tank be 100 litres Initially: A type petrol = 100 litres After first operation: A type petrol = $$frac{{100}}{2}$$ = 50 litres B type petrol = 50 litres After second operation:
A type petrol = $$frac{{50}}{2}$$ + 50 = 75 litres B type petrol = $$frac{{50}}{2}$$ = 25 litres After third operation: A type petrol = $$frac{{75}}{2}$$ = 37.5 litres B type petrol = $$frac{{25}}{2}$$ + 50 = 62.5 litres Required percentage = 37.5%
101: B
Solution: 36% marks = 113 + 85 ⇒ 36% marks = 198 So, ⇒ 1% marks = $$frac{{198}}{{36}}$$ = 5.5 ⇒ 100% marks = 5.5 × 100 = 550
102: A
Solution: $$eqalign{
& 1\% ,{ ext{of}},1\% ,{ ext{of}},25\% ,1000 cr
& = 1\% ,{ ext{of}},1\% ,{ ext{of}},, {frac{{ {25 imes 1000} }}{{100}}} cr
& = 1\% ,{ ext{of}},1\% ,{ ext{of}},250 cr
& = 1\% ,{ ext{of}},, {frac{{ {1 imes 200} }}{{100}}} cr
& = 1\% ,{ ext{of}},,2.5 cr
& = frac{{2.5}}{{100}} cr
& = 0.025 cr} $$
103: B
Solution: Using compounding formula $${ ext{a}} + { ext{b}} + frac{{{ ext{ab}}}}{{100}}$$ we have $$eqalign{
& 5 + 5 + frac{{25}}{{100}} cr
& = 10.25\% cr} $$ so if the population 2 yrs ago be x then $$eqalign{
& { ext{x}} + frac{{10.25x}}{{100}} = 4410 cr
& { ext{or}},,110.25{ ext{x}} = 441000 cr
& herefore { ext{x}} = 4000 cr} $$
104: D
Solution: $$eqalign{
& { ext{Height}},{ ext{of}},{ ext{the}},{ ext{Pole}} = 192,m. cr
& { ext{Spider}},{ ext{covered}},{ ext{in}},{ ext{first}},{ ext{hour}}, cr
& = frac{{125}}{2}\% ,of,192 cr
& = frac{{ {125 imes 192} }}{{ {2 imes 100} }} cr
& = 120,m cr
& { ext{Remaining Pole}} = 192 - 120 = 72,m cr
& { ext{Spider}},{ ext{covered}},{ ext{in}},{ ext{second}},{ ext{hour}}, cr
& = frac{{25}}{2}\% ,{ ext{of}},{ ext{ remaining}},{ ext{ height}} cr
& = frac{{ {25 imes 72} }}{{ {2 imes 100} }} cr
& = 9,m cr} $$
105: C
Solution: $$eqalign{
& a,{ ext{percent}},{ ext{of}},b cr
& = {frac{a}{{100}}} imes b = {frac{{ab}}{{100}}} cr
& b,{ ext{percent}},{ ext{of}},a cr
& = {frac{b}{{100}}} imes a = {frac{{ab}}{{100}}} cr
& { ext{on}},{ ext{division}},{ ext{we}},{ ext{get}},1 cr} $$
106: B
Solution: Winner gets 55% of votes
As 4% votes were declared invalid so 96% would be the valid votes
So,
Winner gets 55% of 96% valid votes
Winner gets % valid votes = $$frac{{55 imes 96}}{{100}}$$ xa0 = 52.8% votes
Loser gets = 96 - 52.8 = 43.2% votes
Difference = 9.6% Now, 9.6% = 4200 So, 1% = $$frac{{4200}}{{9.66}}$$ Thus, 100% Votes = $$frac{{4200 imes 100}}{{9.66}}$$ xa0 = 43750 Hence, Total Voters = 43,750 Alternatively, Let total number of voters were X. Invalid Votes = 4% Valid Votes = 96% Total Valid Votes = 96% of X = $$frac{{96{ ext{x}}}}{{100}}$$xa0 = 0.96X Winner gets 55% of Valid Votes, = $$frac{{0.96 imes 55}}{{100}}$$ xa0 = 0.528X votes Loser gets = (0.96X - 0.528X) = 0.432X Difference = 0. 528 - 0.432 = 0. 096X 0.096X = 4200 ∴ X = 43750
107: A
Solution: Let Mixture was X initially Initial Salt = 15% of X = 0.15X Initial Water = 85% of X = 0.85X 30 liters of water evaporated So, Final water = (0.85X - 30) Now, salt rises to 20% of the mixture, it means as water dried up salt rises Now, Total Mixture = (0.85X -30 + 0.15X) Final salt = 20% of(0.85X - 30 + 0.15X) As, Initial Amount of salt = Final amount of salt , we have 0.15X = 20% of(0.85X - 30 + 0.15X) 0.15X = 0.20X - 6 Or, 0.5X = 6 Thus, X = 120 Initial amount of the mixture was 120 liters
108: D
Solution: Population after 2 years : $$eqalign{
& = 8500 imes left( {1 + frac{{20}}{{100}}}
ight)left( {1 + frac{{25}}{{100}}}
ight) cr
& = left( {850 imes frac{6}{5} imes frac{5}{4}}
ight) cr
& = 12750 cr} $$
109: A
Solution: $$eqalign{
& nleft( A
ight) = 90 cr
& nleft( B
ight) = 15 cr
& nleft( {A cup B}
ight) = 100 cr
& { ext{So, }} cr
& { ext{n}}left( {A cap B}
ight) cr
& = nleft( A
ight) + nleft( B
ight) - nleft( {A cup B}
ight) cr
& = 90 + 15 - 100 cr
& = 5 cr} $$ ∴ Percentage of people who own both = 5%
110: D
Solution: Original length of AB = 5 cm Original length of AC = 3 cm Original length of CB = (5 - 3) cm = 2 cm New length AC : = 106% of 3 cm = $$left( {frac{{106}}{{100}} imes 3}
ight)$$xa0 cm = 3.18 cm New length of CB : = (5 - 3.18) cm = 1.82 cm Decrease in length of CB : = (2 - 1.82) cm = 0.18 cm ∴ Decrease % : = $$left( {frac{{0.18}}{2} imes 100}
ight)\% $$ = 9%
111: D
Solution: $$eqalign{
& nleft( A
ight) = 34 cr
& nleft( B
ight) = 42 cr
& nleft( {A cap B}
ight) = 20 cr
& { ext{So, }} cr
& { ext{n}}left( {A cup B}
ight) cr
& = nleft( A
ight) + nleft( B
ight) - nleft( {A cap B}
ight) cr
& = 34 + 42 - 20 cr
& = 56 cr} $$ ∴ Percentage failed in either or both the subjects = 56%
112: C
Solution: $$eqalign{
& x\% { ext{ of }}a = y\% { ext{ of }}a cr
& Rightarrow frac{x}{{100}}a = frac{y}{{100}}b cr
& Rightarrow b = left( {frac{x}{y}}
ight)a cr
& herefore z\% { ext{ of }}b: cr
& = left( {z\% { ext{ of }}frac{x}{y}}
ight)a cr
& = left( {frac{{xz}}{{y imes 100}}}
ight)a cr
& = left( {frac{{xz}}{y}}
ight)\% { ext{ of }}a cr} $$
113: B
Solution: Let the original price of sugar be Rs. $$x$$ per kg Then, reduced price : $$eqalign{
& 97frac{1}{2}\% { ext{ of Rs}}{ ext{. }}x cr
& = { ext{Rs}}{ ext{. }}left( {frac{{195}}{2} imes frac{1}{{100}} imes x}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{39x}}{{40}} cr
& herefore frac{{1260}}{{left( {frac{{39x}}{{40}}}
ight)}} - frac{{1260}}{x} = 9 cr
& Rightarrow frac{{16800}}{{13x}} - frac{{1260}}{x} = 9 cr
& Rightarrow 13x = frac{{420}}{9} cr
& Rightarrow x = frac{{140}}{{39}} cr} $$ Increased price : $$eqalign{
& 112frac{1}{2}\% { ext{ of Rs}}{ ext{. }}frac{{140}}{{39}} cr
& = { ext{Rs}}{ ext{. }}left( {frac{{225}}{2} imes frac{1}{{100}} imes frac{{140}}{{39}}}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{105}}{{26}} cr} $$ ∴ Quiantity of sugar bought for Rs. 1260 : $$eqalign{
& = left( {1260 imes frac{{26}}{{105}}}
ight){ ext{ kg}} cr
& = { ext{312 kg}} cr} $$
114: C
Solution: Let the original price of each ticket be Rs. 100 Then, original price of 5 tickets = Rs. 500 Sale price of 5 tickets = Rs. 300 Amount saved : = Rs. (500 - 300) = Rs. 200 ∴ Required percentage : = $$left( {frac{{200}}{{500}} imes 100}
ight)$$ xa0% = 40%
115: B
Solution: Let original consumption = 100 kg and new consumption = $$x$$ kg So, ⇔ 100 × 16 = $$x$$ × 20 ⇔ $$x$$ = 80 ∴ Reduction in consumption = 20%
116: A
Solution: $$eqalign{
& 50\% { ext{ of }}left( {x - y}
ight) = 30\% { ext{ of }}left( {x + y}
ight) cr
& Rightarrow 5left( {x - y}
ight) = 3left( {x + y}
ight) cr
& Rightarrow 5x - 5y = 3x + 3y cr
& Rightarrow 2x = 8y cr
& Rightarrow y = frac{x}{4} cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{y}{x} imes 100}
ight)\% cr
& = left( {frac{x}{4} imes frac{1}{x} imes 100}
ight)\% cr
& = 25\% cr} $$
117: C
Solution: Let the number of students in the class be 100 and let the required average be $$x$$ Then, ⇒ (10 × 95) + (20 × 90) + (70 × $$x$$) = (100 × 80) ⇒ 70$$x$$ = 8000 - (950 + 1800) ⇒ 70$$x$$ = 5250 ⇒ $$x$$ = 75
118: A
Solution: $$eqalign{
& frac{{250 imes left( {x + 20}
ight)}}{{100}} = frac{{220 imes x}}{{100}} imes frac{{125}}{{100}} cr
& 25x + 500 = 22 imes x imes frac{5}{4} cr
& 100x + 2000 = 110x cr
& 10x = 2000 cr
& x = 200 cr
& frac{{left( {x + 50}
ight) imes 10}}{{100}} = frac{{250 imes 10}}{{100}} = 25 cr
& frac{{x imes 15}}{{100}} = frac{{200 imes 15}}{{100}} = 30 cr
& { ext{Less}}\% = frac{5}{{30}} imes 100 = 16frac{2}{3}\% cr} $$
119: D
Solution: [x08egin{array}{*{20}{c}}
{{ ext{Income}}}&{{ ext{Expenditure}}}&{{ ext{Saving}}} \
x08egin{gathered}
600 hfill \
,,,,,,,,,,{ downarrow ^{ + 10\% }} hfill \
660 hfill \
end{gathered} &x08egin{gathered}
500 hfill \
,,,,,,,,,{ downarrow ^{ + 20\% }} hfill \
600 hfill \
end{gathered} &x08egin{gathered}
100 hfill \
hfill \
60 hfill \
end{gathered}
end{array}] Decrease in saving = $$frac{{40}}{{100}}$$ × 100 = 40%
120: D
Solution: $$eqalign{
& {x08f{Given:}} cr
& { ext{A is }}80\% { ext{ more than B}} cr
& { ext{C is }}48frac{4}{7}\% { ext{ less than A}} + { ext{B}} cr
& {x08f{Formula}},{x08f{used:}} cr
& 80\% o frac{4}{5} cr
& {x08f{Calculation:}} cr
& Rightarrow { ext{A}} = left( {1 + frac{4}{5}}
ight){ ext{B}} = frac{{9{ ext{B}}}}{5} cr
& Rightarrow { ext{C}} = left( {100\% - 48frac{4}{7}\% }
ight)left( {{ ext{A}} + { ext{B}}}
ight) cr
& Rightarrow { ext{C}} = 51frac{3}{7}\% imes left( {frac{{9B}}{5} + { ext{B}}}
ight) cr
& Rightarrow { ext{C}} = frac{{3.6}}{7} imes frac{{14{ ext{B}}}}{5} cr
& Rightarrow { ext{C}} = frac{{7.2{ ext{B}}}}{5} cr
& { ext{Difference between A and C}} = frac{{1.8{ ext{B}}}}{5} cr
& Rightarrow { ext{Percentage}} = frac{{frac{{1.8{ ext{B}}}}{5}}}{{frac{{9{ ext{B}}}}{5}}} imes 100 cr
& = 0.2 imes 100 cr
& = 20\% cr
& herefore { ext{The required percentage}} = 20\% cr
& cr
& {x08f{Alternate}},{x08f{solution:}} cr
& { ext{Let the value of B be 100}} cr
& Rightarrow { ext{A}} = 100 imes 180\% = 180 cr
& Rightarrow left( {{ ext{A}} + { ext{B}}}
ight) = 280 cr
& { ext{According to the question}} cr
& { ext{C}} = 51frac{3}{7}\% { ext{ of }}280 cr
& Rightarrow { ext{C}} = frac{{360}}{{700}} imes 280 = 144 cr
& { ext{So, required }}\% = frac{{180 - 144}}{{180}} imes 100\% = 20\% cr
& herefore { ext{The required }}\% { ext{ is }}20\% cr} $$
121: B
Solution: Rohit income = 32000 Cost = 30% Saving = 70% Saving = 32000 × $$frac{{70}}{{100}}$$ = 22400
122: A
Solution: Before hike price → Rs. 82 per litre 5% → $$frac{{82 imes 5}}{{100}}$$ xa0= Rs. 4.1 per litre Distance = 3045 km Petrol use = $$frac{{3045}}{{15}}$$ xa0= 203 Increase expenditure = 203 × 4.1 = 832.3 = Rs. 832
123: A
Solution: $${l_1} o {l_2}$$ [10xrightarrow[{ + 10 uparrow }]{}{l_2}] $$eqalign{
& {S_1} o {S_2} cr
& 6 imes {10^2} o 6 imes {11^2} cr
& 100 o 121 cr
& = frac{{21}}{{100}} imes 100 cr
& = 21\% cr} $$
124: D
Solution: $$eqalign{
& { ext{Let fraction}} = frac{x}{y} cr
& frac{{xleft( {115}
ight)}}{{y imes 80}} = frac{{17}}{{65}} cr
& frac{x}{y} = frac{{17}}{{65}} imes frac{{16}}{{23}} = frac{{272}}{{1495}} cr} $$
125: A
Solution: Given 10% of the voters did not cast their vote and 10% of the polled vote were found invalid The winner candidate got 54% of the valid vote and beat the opposition by 1620 votes Concept used: The percentage is calculated as based on 100 i.e. 100 is the base 40% means 40 out of 100 Calculation: Let, total enrolled voter be $$x$$ 10% did not cast vote means casted or polled vote = $$frac{{9x}}{{10}}$$ 10% vote is invalid That means valid vote = $$frac{{90}}{{100}} imes frac{{9x}}{{10}}$$ $$ Rightarrow frac{{81x}}{{100}}$$ The winner candidate got 54% of the polled vote means loosed got (100 - 54) = 46% vote Winner candidate got total $$left( {frac{{54}}{{100}} imes frac{{81x}}{{100}}}
ight)$$ xa0 vote And the looser candidate got $$left( {frac{{46}}{{100}} imes frac{{81x}}{{100}}}
ight)$$ xa0 vote Accordingly, $$eqalign{
& left( {frac{{54}}{{100}} imes frac{{81x}}{{100}}}
ight) - left( {frac{{46}}{{100}} imes frac{{81x}}{{100}}}
ight) = 1620 cr
& Rightarrow left( {frac{{81x}}{{100}}}
ight) imes left( {frac{{54 - 46}}{{100}}}
ight) = 1620 cr
& Rightarrow frac{{left( {81 imes 8}
ight)x}}{{10000}} = 1620 cr
& Rightarrow x = frac{{1620 imes 10000}}{{81 imes 8}} cr
& Rightarrow x = 25000 cr} $$ ∴ Total 25000 voters registered in the voter list.
126: B
Solution: Let, number = x 40% of x + 30% of x = 70 70% of x = 70 x = 100
127: B
Solution: Given: Registered voters for P supporters = 70% 60% of the P supporters and 30% of the Q supporters are expected to vote for candidate P. Concept used: Let total voters be 1000 Then 700 are P supporters and 300 are Q supporters Total votes of P ⇒ 60% of 700 + 30% of 300 ⇒ 420 + 90 ⇒ 510 Percentage of P voters ⇒ $$510 imes frac{{100}}{{1000}}$$ ⇒ 51% Hence, the required value is 51%
128: D
Solution: 100 === 25% ↑===> 125 ===> 20% ↓===> 100
So there is no change in the area of rectangle Alternatively : Let area be 1 Area of rectangle, = l × b = 0.8 × 1.25 = 1
129: B
Solution: Cost of fresh mangoes + Cost of packaging = Total cost Let initial Cost of fresh, mangoes = 100 Then, packaging cost = 40 Thus, Initial total cost = 100 + 40 = 140 After increasing in cost of fresh mangoes 30%, Cost of fresh mangoes = 130 And cost of packing go down by 50 % so, Cost of packing = 20 Now Total cost = 130 + 20 = 150 Increased cost = 150 - 140 = 10 % increased = $$ = frac{{10 imes 100}}{{140}} = 7.14\% $$
130: A
Solution: Given, 220% of X = 44 Or, X = 20 Thus, 44% of 20 $$ = frac{{44 imes 20}}{{100}} = 8.8$$
131: C
Solution: Let initially the quantity and rate be 100 each Quantity × rate = Expenditure 100 × 100 = 10000 Now, Increase in price is 25% and new quantity is 70% of original Quantity × rate = Expenditure 70 × 125 = 8750 Decreased expenditure, = 10000 - 8750 = 1250 % decrease $$ = frac{{1250 imes 100}}{{10000}} = 12.5\% $$
132: C
Solution: $$eqalign{
& { ext{Daily}},{ ext{supply}}, cr
& = left( {100 - z}
ight)\% ,{ ext{of}},y cr
& = frac{{ {left( {100 - z}
ight)y} }}{{100}} cr
& { ext{Thus,}},{ ext{required}},{ ext{number}},{ ext{of}},{ ext{days}} cr
& = {frac{{ {100x} }}{{left( {100 - z}
ight)y}}} cr} $$
133: A
Solution: Let the initial revenue be 100 100 === 50% ↑ (Ticket up) ===> 150 === 20% ↓ (Visitors down) ===> 120 There is 20% increase in the revenue
134: D
Solution: Result Physics Chemistry Failed 35% 45% Passed 65% 55% Now, 55% of 600 passed in Chemistry = $$frac{{55 imes 600}}{{100}}$$ xa0 = 330 From question, 40% of those who passed in Chemistry(330 Students) also passed in physics = $$frac{{40 imes 330}}{{100}}$$ = 132, so, 132 students passed in both Therefore % who passed in both = $$frac{{132 imes 100}}{{600}} = 22\% $$ Passed in both subject = 22% of total students
% of students who passed in either Physics or Chemistry or both,
= (65 + 55) - 22 = 98%
Thus, percentage of students who failed in both subjects = 2%
Number of students who failed = 2% of 600 = 12
135: B
Solution: Copper : Aluminum = 7 : 4
Let Copper and Aluminum in the weapon be 7x and 4x respectively
Given,
Aluminum in weapon = 12 kg So, → 4x = 12 → x = 3 Copper = 7x = 7 × 3 = 21 Kg. Total alloy in the weapon = 12 + 21 = 33 kg But 12% alloy get destroyed in making the weapon, i.e. 88% alloy is used in the weapon, so, → 88 % alloy = 33 kg → 100 % alloy = 37.5 kg
136: C
Solution: Let the smaller number be x and larger number be y According to the question, 80% of x + 4 = 40% of y 4y - 8x = 40 y - 2x = 10 --------- (1) y - x = 85 ----------- (2) By using (1) and (2), we get x = 75 y = 160 x + y = 235
137: A
Solution: Actual result = 10x By mistake it has been divided by 10 = $$frac{x}{{10}}$$ %Change = $$ {frac{{10x - left( {frac{{10}}{x}}
ight)}}{{10x}}} $$ xa0 × 100 = 99 = -99% Since, actual value is greater than the wrong value. Alternatively, Let, x = 10 Actual result = 10 × 10 = 100 Wrong result = $$frac{{10}}{{10}}$$ = 1 Change = (1 - 100) = -99 % Change = - 99%
138: A
Solution: Weight of metal A = 5kg Total weight of the alloy : = (5 + 20) kg = 25 kg ∴ Required percentage : $$eqalign{
& = left( {frac{5}{{25}} imes 100}
ight)\% cr
& = 20\% cr} $$
139: C
Solution: Let 23% of 8040 + 42% of 545 = x% of 3000 Then, $$left( {frac{{23}}{{100}} imes 8040}
ight) + left( {frac{{42}}{{100}} imes 545}
ight)$$ xa0 xa0 xa0 = $$left( {frac{x}{{100}} imes 3000}
ight)$$ $$eqalign{
& Rightarrow 30x = 1849.2 + 228.9 cr
& Rightarrow 30x = 2078.1 cr
& Rightarrow x = frac{{2078.1}}{{30}} cr
& Rightarrow x = 69.27 cr} $$
140: C
Solution: Total sales tax paid = 7% of Rs. 400 + 9% of Rs. 6400 = Rs. $$left( {frac{7}{{100}} imes 400 + frac{9}{{100}} imes 6400}
ight)$$ = Rs. (28 + 576) = Rs. 604 Total cost o the items : = Rs. (400 + 6400) = Rs. 6800 ∴ Required percentage $$eqalign{
& = left( {frac{{604}}{{6800}} imes 100}
ight)\% cr
& = 8frac{{15}}{{17}}\% cr} $$
141: D
Solution: Let the number be = $$x$$ Then, ⇒ 54% of $$x$$ - 26% of $$x$$ = 22526 ⇒ $$frac{{54}}{{100}}$$ $$x$$ - $$frac{{26}}{{100}}$$ $$x$$ = 22526 ⇒ $$frac{{28}}{{100}}$$ $$x$$ = 22526 ⇒ $$x$$ = $$left( {frac{{22526 × 100}}{28}}
ight)$$ ⇒ $$x$$ = 80450 ∴ 66% of 80450 = $$left( {frac{{66}}{{100}} imes 80450}
ight)$$ = 53097
142: C
Solution: Let the number be x Then, 75% of x + 75 = x ⇔ x - $${frac{{75}}{100}}$$ x = 75 ⇔ x - $${frac{{3}}{4}}$$ x = 75 ⇔ $${frac{{x}}{4}}$$ = 75 ⇔ x = 300
143: D
Solution: Let the required amount be Rs. x Then, 40% of 60% of 32% of x = 432 ⇒ $$frac{{40}}{{100}} imes frac{{60}}{{100}} imes frac{{32}}{{100}}$$ × $$x$$ = 432 ⇒ x = $$frac{{432 imes 100 imes 100 imes 100}}{{40 imes 60 imes 32}}$$ ⇒ x = 5625
144: N/A
Solution: Let the maximum marks be x Then, 40% of x = 483 + 117 ⇒ $$frac{{40x}}{{100}}$$ = 600 ⇒ x = $$left( {frac{{600 imes 100}}{{40}}}
ight)$$ ⇒ x = 1500 ∴ Minimum passing marks for girls : = 35% of 1500 = $$left( {frac{{35}}{{100}} imes 1500}
ight)$$ = 525
145: D
Solution: Number of short length coats before removal : = (100 - 15)% of 800 = 85% of 800 = $$left( {frac{{85}}{{100}} imes 800}
ight)$$ = 680 Number of short length coats after removal : = (680 - 500) = 180 Total number of coats after removal : = (800 - 500) = 300 Number of full length coats removal : = (300 - 180) = 120 ∴ Required percentage : = $$left( {frac{{120}}{{300}} imes 100}
ight)$$ xa0 % = 40%
146: B
Solution: Let original income - Rs. 100 Then, saving = Rs. 10 and expenditure = Rs. 90 New income = Rs. 120 New saving = Rs. 10 New expenditure = Rs. (120 - 10) = Rs. 110 Increase in expenditure = Rs. (110 - 90) = Rs. 20 ∴ Income % : $$eqalign{
& = left( {frac{{20}}{{90}} imes 100}
ight)\% cr
& = 22frac{2}{9}\% cr} $$
147: B
Solution: Height of the tree 2 years ago : $$eqalign{
& = frac{{540}}{{{{left( {1 + frac{{20}}{{100}}}
ight)}^2}}}cm cr
& = left( {540 imes frac{5}{6} imes frac{5}{6}}
ight)cm cr
& = 375,,cm cr} $$
148: A
Solution: Let the number of boys is = 8x Number of girls = 12x Boys are not getting scholarship = 8x × $$frac{1}{2}$$ = 4x Girls are not getting scholarship = 12x × $$frac{3}{4}$$ = 9x % of students not getting scholarship = $$frac{(4x + 9x)}{20x}$$ xa0 × 100 = 65%
149: D
Solution: The number of failure boys = $$frac{640 × 40}{100}$$ = 256 The number of failure girls = $$frac{360 × 20}{100}$$ = 72 Percentage of failure students = $$frac{(72 + 256)}{640 + 360}$$ = 32.8%
150: C
Solution: Total valid votes got candidates = $$frac{9261}{75}$$ × 100 = 12348 Let total number of votes = x Total votes polled = x × $$frac{75}{100}$$ = $$frac{75x}{100}$$ Valid votes = $$frac{75x}{100}$$ × $$frac{98}{100}$$ According to the question, ⇔ $$frac{75x}{100}$$ × $$frac{98}{100}$$ = 12348 ⇔ x = 16800 Hence, total votes = 16800
151: D
Solution: 40% = $$frac{{2,, o ,,{ ext{Alcohol}}}}{{5,, o ,,{ ext{Mixture}}}}$$ Water : Alcohol 3 xa0 : xa0 2 Required percentage = $$frac{2}{(5 + 1)}$$ × 100 = $$frac{2}{6}$$ × 100 = $$frac{1}{3}$$ × 100 = $$33frac{1}{3}$$%
152: B
Solution: By using successive formula Net decrease in area $$eqalign{
& = 40 + 40 - frac{{40 imes 40}}{{100}} cr
& = 64\% cr} $$
153: A
Solution: 1 st number : 2 nd number 5 xa0 xa0 : xa0 xa0 4 40% = $$frac{2}{5}$$ 2 units = 12 5 units = 30 So, 1 st number = 30 2 nd number = $$frac{30}{5}$$ × 4 = 24 50% of 2 nd = 24 × $$frac{1}{2}$$ = 12
154: B
Solution: Number of a boy in the school = 720 × $$frac{7}{12}$$ xa0 = 420 Number of a girl in the school = 720 × $$frac{5}{12}$$ xa0 = 300 To maintain 1 : 1 we will have to add number of girls = 420 - 300 = 120
155: C
Solution: Let a total of 5x match is played between Pakistan and India. Pakistan won 3x matches and India won 2x matches According to the question, But now India won 30 matches in a row and hence India's success percentage increase to 70%. Now, Total match = 5x + 30 (2x + 30) = (5x + 30) × $$frac{70}{100}$$ 20x + 300 = 35x + 210 15x = 90 x = 6 So, total match : = 5x + 30 = 5 × 6 + 30 = 60 matches
156: B
Solution: $$66frac{2}{3}$$% = $$frac{2}{3}$$ Let the income of the person = 3 units Expenditure = 2 units Savings = (3 - 2) = 1 unit According to the question, 1 unit = Rs. 1200 2 units = 2 × 1200 = Rs. 2400
157: B
Solution: % increase = 10 + 10 + $$frac{10 × 10}{100}$$ xa0= 21% Total increase = $$frac{100 × 21}{100}$$ xa0= Rs. 21
158: A
Solution: Let the original taxed amount be Rs. x and new taxed amount be Rs. y Let the original revenue be Rs. 100 Then, 4% of x = 100 or x = $$left( {frac{{100 imes 100}}{{4}}}
ight)$$ x = Rs. 2500 New revenue = 110% of Rs. 100 = Rs. 110 Then, 5% of y = 110 or y = $$left( {frac{{110 imes 100}}{{5}}}
ight)$$ y = Rs. 2200 Decrease in taxed amount : = Rs. (2500 - 2200) = Rs. 300 ∴ Decrease % : $$=$$$${ ext{ }}left( {frac{{300}}{{2500}} imes 100}
ight)$$ $$= 12$$ %
159: A
Solution: Population at the end of 2 years = 189000 $$left( {1 - frac{8}{{100}}}
ight)$$ $$left( {1 + frac{5}{{100}}}
ight)$$ = $$left( {189000 imes frac{{23}}{{25}} imes frac{{21}}{{20}}}
ight)$$ = 182574
160: D
Solution: Let each vessel contain 100 litres of 40% alcohol Suppose Sachine added x litres of pure alcohol. Then, $$eqalign{
& Leftrightarrow frac{{40 + x}}{{100 + x}} = frac{{50}}{{100}} cr
& Leftrightarrow frac{{40 + x}}{{100 + x}} = frac{1}{2} cr
& Leftrightarrow 80 + 2x = 100 + x cr
& Leftrightarrow x = 20 cr} $$ Suppose Vivek replaced y litres. Then, alcohol in y litres = 40% of y = $$frac{2y}{5}$$ $$eqalign{
& herefore frac{{40 - frac{{2y}}{5} + y}}{{100}} = frac{{50}}{{100}} cr
& Rightarrow frac{{40 - frac{{2y}}{5} + y}}{{100}} = frac{1}{2} cr
& Rightarrow 80 + frac{{6y}}{{25}} = 100 cr
& Rightarrow y = frac{{20 imes 5}}{6} cr
& Rightarrow y = frac{{50}}{3} cr} $$ Required percentage : $$eqalign{
& = left[ {frac{{left( {20 - frac{{50}}{3}}
ight)}}{{left( {frac{{50}}{3}}
ight)}} imes 100}
ight]\% cr
& = left( {frac{{10}}{3} imes frac{3}{{50}} imes 100}
ight)\% cr
& = 20\% cr} $$
161: A
Solution: Actual cost price of an article : = Rs. (5844 + 156) = Rs. 6000 Selling price of that article = Rs. 5700 Loss = Rs. (6000 - 5700) = Rs. 300 Loss percent : = $$frac{300}{6000}$$ × 100 = 5%
162: A
Solution: Rebate = 6% of Rs. 6650 $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {frac{6}{{100}} imes 6650}
ight) cr
& = { ext{Rs}}{ ext{. 399}} cr} $$ Sales tax = 10 % of (6650 - 399) $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {frac{{10}}{{100}} imes 6251}
ight) cr
& = { ext{Rs}}{ ext{. 625}}{ ext{.10}} cr} $$ ∴ Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
163: N/A
Solution: Let the required number of chicken be x Then, 65% of x = 47775 $$eqalign{
& Rightarrow frac{{65}}{{100}}x = 47775 cr
& Rightarrow x = frac{{47775 imes 100}}{{65}} cr
& Rightarrow x = 73500 cr} $$
164: A
Solution: Let the number be x Then, $$eqalign{
& Rightarrow x = 125 = 37frac{1}{2}\% { ext{ of }}x cr
& Rightarrow x - 125 = frac{{75}}{2} imes frac{1}{{100}} imes x cr
& Rightarrow x - 125 = frac{{3x}}{8} cr
& Rightarrow x - frac{{3x}}{8} = 125 cr
& Rightarrow frac{{5x}}{8} = 125 cr
& Rightarrow x = left( {frac{{125 imes 8}}{5}}
ight) cr
& Rightarrow x = 200 cr
& herefore 25\% { ext{ of 200}} cr
& = left( {frac{{25}}{{100}} imes 200}
ight) cr
& = 50 cr} $$
165: N/A
Solution: Let the maximum marks be x. Then, $$eqalign{
& left( {40 - 4}
ight)\% { ext{ of }}x = 261 cr
& Rightarrow 36\% { ext{ of }}x = 261 cr
& Rightarrow frac{{36}}{{100}}x = 261 cr
& Rightarrow x = left( {frac{{261 imes 100}}{{36}}}
ight) cr
& Rightarrow x = 725 cr} $$
166: D
Solution: $$frac{x}{100}$$ × y = $$frac{4}{5}$$ × 80 ⇒ xy = 64 × 100 ⇒ xy = 6400
167: C
Solution: Total number of fruits = (14 + 23) = 37 Let x oranges be removed. Then, 70% of (37 - x) = 14 ⇒ 7 (37 - x) = 140 ⇒ 37 - x = 20 ⇒ x = 17
168: B
Solution: [x08egin{array}{*{20}{c}}
{D = }&{12\% }& o &{2100} \
{}&{1\% }& o &{175} \
{}&{100\% }& o &{17500}
end{array}]
169: A
Solution: According to the question, 1 unit → 2 3 units → 2 × 3 = 6 ereaser 4 units → 2 × 4 = 8 ereaser
170: D
Solution: $$eqalign{
& frac{{11}}{5}A = frac{{22}}{{100}}B cr
& frac{A}{B} = frac{1}{{10}} cr
& B = frac{{25}}{{1000}}C cr
& frac{B}{C} = frac{1}{{40}} cr} $$ [x08egin{array}{*{20}{c}}
{A,,,:,,,B,,,:,,,C} \
{,1,,,,,,,,,,10,,,,,,,,,,10} \
{,1,,,,,,,,,,,,1,,,,,,,,,,,40} \
{overline {,,1,,:,,40,,:,,400,} }
end{array}] $$eqalign{
& 400 o 5500 cr
& 400 o frac{{55}}{4} cr
& A o frac{{55}}{4} cr
& B o frac{{550}}{4} cr
& { ext{According to the question,}} cr
& left( {frac{{80}}{{100}} imes frac{{55}}{4}}
ight) + left( {frac{{40}}{{100}} imes frac{{550}}{4}}
ight) cr
& = 11 + 55 cr
& = 66 cr} $$
171: A
Solution: $$eqalign{
& { ext{Base}} - 40\% = frac{2}{5} cr
& { ext{Area}} - 60\% = frac{3}{5} cr
& { ext{Area}} o { ext{5:8}} cr
& { ext{Base}} o { ext{5:7}} cr
& { ext{Height}} o frac{5}{5}:frac{8}{7} cr
& underbrace {7,,:,,8}_1 cr
& frac{1}{7} imes 100 = 14.29\% cr} $$
172: C
Solution: Number of students of all sections = 25, 30, 40, 45, 60 $$eqalign{
& 25 imes frac{1}{5} = 5 cr
& 30 imes frac{3}{{10}} = 9 cr
& 40 imes frac{7}{{20}} = 14 cr
& 45 imes frac{2}{5} = 18 cr
& 60 imes frac{{100}}{{100}} = 60 cr} $$ 106 → Passed student Total student = 200 $${ ext{Pass}}\% = frac{{106}}{{200}} imes 100 = 53\% $$
173: A
Solution: Let the number of Crate = x Spoiled Fruit = x $$eqalign{
& { ext{Now, }}frac{{x imes 60}}{{100}} = 48 cr
& x = 80 cr} $$ The number of Fruits = 80 × 25 = 2000
174: A
Solution: Given: New selling price = Rs. 29 per kg Old selling price = Rs. 25 per kg Expenditure = constant Formula used: (i) Expenditure = Price × Quantity (ii) Percentage difference = $$frac{{{ ext{New price}} - { ext{Old price}}}}{{{ ext{Old price}}}} imes 100$$ Calculations: Let expenditure, initial selling price and initial quantity be C, P 1 and Q 1 respectively. C = P 1 × Q 1 ⇒ C = 25 × Q 1 ⇒ Q 1 = $$frac{{ ext{C}}}{{25}}$$ Let final selling price and final quantity be P 2 and Q 2 respectively. C = P 2 × Q 2 ⇒ C = 29 × Q 2 ⇒ Q 2 = $$frac{{ ext{C}}}{{29}}$$ Percentage difference $$ = frac{{{{ ext{Q}}_2} - {{ ext{Q}}_1}}}{{{{ ext{Q}}_1}}} imes 100$$ $$eqalign{
& Rightarrow left[ {frac{{frac{{ ext{C}}}{{29}} - frac{{ ext{C}}}{{25}}}}{{frac{{ ext{C}}}{{25}}}}}
ight] imes 100 cr
& Rightarrow left[ {frac{{frac{{25{ ext{C}} - 29{ ext{C}}}}{{25 imes 29}}}}{{frac{{ ext{C}}}{{25}}}}}
ight] imes 100 cr
& Rightarrow frac{{ - left( {frac{{4{ ext{C}}}}{{725}}}
ight)}}{{frac{{ ext{C}}}{{25}}}} imes 100 cr
& Rightarrow frac{{ - left( {4{ ext{C}} imes { ext{25}}}
ight)}}{{725{ ext{C}}}} imes 100 cr
& Rightarrow - 13.79\% approx - 14\% cr} $$ ∴ By 14% should a family reduce its consumption, so as to keep the expenditure the same as before. Alternate solution [x08egin{array}{*{20}{c}}
{}&{{ ext{Price}}}&{{ ext{Consumption}}} \
{{ ext{Old}}}&{25}&{{mathbf{29}}} \
{{ ext{New}}}&{29}&{{mathbf{25}}} \
{{ ext{Expenditure}}}&{25 imes 29}&{29 imes 25}
end{array}] Thus, family should reduce their consumption by 4 kg % Reduction $$ = frac{4}{{29}} imes 100 = 13.79 approx 14\% $$ ∴ The family should reduce the consumption by 14%.
175: C
Solution: Let number of boys = B and number of girls = G According to the question, B × 80% - G × 85% = 70 ⇒ 80B - 85G = 7000 . . . . . . (i) Again, B × 20% + G × 15% = 90 ⇒ 20B + 15G = 9000 ⇒ 80B + 60G = 36000 . . . . . . (ii) From equation (i) - (ii) 145G = 29000 ⇒ G = 200 20B + 15 × 200 = 9000 ⇒ 20B = 6000 ⇒ B = 300 ∴ Total students = 300 + 200 = 500
176: A
Solution: $$eqalign{
& y imes frac{1}{4} imes frac{{30}}{{100}} imes 2.5 = x imes frac{1}{4} imes frac{1}{2} cr
& frac{x}{y} = frac{3}{2} cr
& x:y = underbrace {3,:,2}_{ + 1} cr
& \% = frac{1}{2} imes 100 = 50\% cr} $$
177: D
Solution: $$eqalign{
& { ext{Let total numbe of oranges}} = x cr
& left{ {left( {frac{{x imes 55}}{{100}} - 1}
ight) imes frac{{80}}{{100}} - 2}
ight} imes frac{{10}}{{100}} = 5 cr
& left( {frac{{x imes 55}}{{100}} - 1}
ight) imes frac{{80}}{{100}} = 50 + 2 cr
& left( {frac{{x imes 55}}{{100}} - 1}
ight) = frac{{52 imes 10}}{8} cr
& frac{{x imes 55}}{{100}} - 1 = 65 cr
& frac{{x imes 55}}{{100}} = 66 cr
& x = frac{{66 imes 100}}{{55}} cr
& x = 120 cr} $$
178: B
Solution: Let salary of B = 100 ∴ Salary of A = 100 + 50% of 100 = 150 B salary is lesser then A = 150 - 100 = 50 Required % = $$frac{50}{150}$$ × 100 = $$33frac{1}{3}$$%
179: D
Solution: According to the question, A : B : C 1 : 2 : 5 ( 60A = 30B) $$frac{A}{B}$$ = $$frac{1}{2}$$ C = 5 A = 1 Required answer = $$frac{5}{1}$$ × 100 = 500%
180: C
Solution: Failed candidates in English = (100 - 70) = 30% Failed candidates in Mathematics = (100 - 80) = 20% Candidates who fail in both subject = 10%
Candidates who only fail in English = 30 - 10 = 20% Candidates who only fail in Mathematics = 20 - 10 = 10% Percentage of passed students in both subject = 100 - (Candidates who only fail in English + Candidates who only fail in Mathematics + Candidates who fail in both subject) = 100 - (20 + 10 + 10) = 60% According to the question, 60% of students = 144 Total students : = $$frac{144}{60}$$ × 100 = 240
181: A
Solution: Required last year salary = $$frac{1806}{100 + 5}$$ xa0 × 100 = Rs. 1720
182: C
Solution: 80% of A = 50% of B ⇒ $$frac{80}{100}$$A = $$frac{50}{100},$$B ⇒ 8A = 5B ⇒ A = $$frac{5}{8}$$B Put value of A in given equation, B = x% of A ⇒ B = $$frac{x}{100}$$ × $$frac{5}{8},$$B ⇒ x = $$frac{100 × 8}{5}$$ ⇒ x = 160
183: B
Solution: Mass of lead ore = 8000 kg ⇒ Mass of metal in lead ore = 60% of 8000 = 4800 kg ⇒ Mass of silver in metal = $$frac{3}{4}$$% of 4800 = 36 kg ⇒ Mass of lead in ore = 4800 - 36 = 4764 kg
184: D
Solution: Let the number = x According to the question, x × $$frac{20}{100}$$ = 120 x = 600 Required answer = 600 × $$frac{120}{100}$$ = 720 Alternate 20% represents → 120 1% → $$frac{120}{20}$$ So, 120% = $$frac{120}{20}$$ × 120 = 720
185: C
Solution: Let the number = x According to the question, ⇒ x - $$frac{17x}{100}$$ = 498 ⇒ 100x - 17x = 49800 ⇒ 83x = 49800 ⇒ x = $$frac{49800}{83}$$ = 600
186: A
Solution: Quicker approach $$ uparrow $$ xa0 in A = a + b + $$frac{ab}{100}$$ Here a = b = 5% $$ uparrow $$ xa0 in A : $$eqalign{
& = left( {5 + 5 + frac{{5 imes 5}}{{100}}}
ight)\% cr
& = 10.25\% cr} $$
187: C
Solution: According to the question, ⇒ 60 × $$frac{A}{100}$$ = 75 × $$frac{B}{100}$$ ⇒ 4A = 5B ⇒ B = $$frac{4}{5}$$A ⇒ A × $$frac{x}{100}$$ = B (given) ⇒ A × $$frac{x}{100}$$ = $$frac{4}{5}$$A ⇒ x = 80
188: B
Solution: Price after discount = $$frac{180 × 80}{100}$$ = Rs. 144 Price of 1 pair of socks = Rs. $$frac{144}{12}$$ = Rs. 12 Required number of pairs = $$frac{48}{12}$$ = 4 pairs
189: B
Solution: Present population = 64000 1 st year = 66400 2 nd year = 6400 + 640 3 rd year = 6400 + 2 × 640 + 64 Total population after 3 years = 64000 + 3 × 6400 + 3 × 640 + 64 = 85184 Alternate: Population after n years $$eqalign{
& Rightarrow { ext{p}}' = { ext{p}}{left( {1 pm frac{{ ext{R}}}{{100}}}
ight)^{ pm { ext{n}}}} cr
& { ext{p}}' = 64000{left( {1 pm frac{{10}}{{100}}}
ight)^3} cr
& ,,,,,,,, = 85184 cr} $$
190: B
Solution: Let total votes in village = 100 Losing candidate got votes = 30% of 100 = 30 votes Winner's and nearer rival's votes = 100 - $$frac{13}{100}$$ × 100 = 70 ∴ If both get equal votes then, it should be, 35 ∴ Minimum difference between winner and nearer rival = (36 - 34) = 2
191: A
Solution: Rate of sales tax $$eqalign{
& = left( {frac{{27.20}}{{340}} imes 100}
ight)\% cr
& = 8\% cr} $$
192: C
Solution: Total marks obtained by the student = 55% of 800 = $$frac{55}{100}$$ × 800 = 440 ∴ Marks scored in English = 15% of 440 = $$frac{15}{100}$$ × 440 = 66
193: D
Solution: Cost price of wrist watch = Rs. 1250 Sale price of wrist watch = Rs. 1500 Profit percent : $$eqalign{
& = left( {frac{{1500 - 1250}}{{1250}}}
ight) imes 100 cr
& = frac{{250}}{{1250}} imes 100 cr
& = frac{{25000}}{{1250}} cr
& = 20\% cr} $$
194: A
Solution: Let the larger number be x Then, x - 20 = $$frac{20}{100}$$x ⇔ x - $$frac{1}{5}$$x = 20 ⇔ $$frac{4}{5}$$x = 20 ⇔ x = 20 × $$frac{5}{4}$$ ⇔ x = 25
195: C
Solution: Let the total number of votes polled be x Then, votes polled by other candidate = (100 - 62)% of x = 38% of x ∴ 62% of x - 38% of x = 432 ⇒ $$frac{24x}{100}$$ = 432 ⇒ x = $$frac{432 × 100}{24}$$ ⇒ 1800
196: D
Solution: (a) x% of y = $$frac{xy}{100}$$ and y of z = $$frac{yz}{100}$$ x > y, y < z ⇒ xy > yz ⇒ $$frac{xy}{100}$$ > $$frac{yz}{100}$$ ⇒ x% of y > y% of z (b) y% of x = $$frac{xy}{100}$$ and z% of y = $$frac{yz}{100}$$ As proved above, y% of x > z% of y (c) z% of x = $$frac{xy}{100}$$ and y% of z = $$frac{yz}{100}$$ x > y ⇒ xz > yz ⇒ $$frac{xz}{100}$$ > $$frac{yz}{100}$$ ⇒ z% of x > y% of z
197: A
Solution: $$eqalign{
& 12\% o frac{{ + 3}}{{25}} cr
& 8\% o frac{{ - 2}}{{25}} cr
& { ext{Length}} o 25:28 cr
& underline {{ ext{Breadth}} o 25:23} cr} $$ $${ ext{Increase }}\% = frac{{19}}{{625}} imes 100 = 3.04\% $$
198: A
Solution: Let total vote = 100 Invalid vote = 5 Raju got = 30 Ravi got = 32 Ashok got = 95 - (30 + 32) = 33 33 - 30 = 3 unit → 5136 1 unit → $$frac{{5136}}{3}$$ 100 unit → 171200
199: C
Solution: Calculation: 30% of the students from school X failed Let the number of students from school X be 100 ⇒ Number of students who failed = 30 ⇒ Number of students who passed = (100 - 30) = 70 According to the question, 150% more students than school X, appeared in the examination from school Y Number of students from school Y = $$100 + left( {100 imes frac{{150}}{{100}}}
ight)$$ ⇒ Number of students from school Y = 250 Again, according to the question, 80% of the total number of students from X and Y passed Total students from school X and Y = 100 + 250 = 350 ⇒ Total number of students who passed = $$350 imes frac{{80}}{{100}}$$ ⇒ Total number of students who passed = 280 Now, Number of students who passed from school Y = 280 - 70 ⇒ Number of students who passed from school Y = 210 Number of students who failed from school Y = 250 - 210 ⇒ Number of students who failed from school Y = 40 Percentage of students who failed from Y = $$frac{{40}}{{250}} imes 100$$ ⇒ Percentage of students who failed from Y = 16% ∴ 16% of students failed from school Y.
200: A
Solution: Given: For Discount = 5%, Profit = 14% Formula used: Selling price = Marked price - Discount Cost price = $$frac{{100}}{{100 + { ext{Profit}}}} imes { ext{Selling price}}$$ Profit % = $$frac{{{ ext{Selling price}} - { ext{Cost price}}}}{{{ ext{Cost price}}}} imes 100$$ Calculation: Let the Marked price of the article = Rs. 100 Discount = 5% ⇒ Selling price = Rs. 95 Profit = 14% ⇒ Cost price = $${ ext{Rs}}{ ext{. }}left( {frac{{100}}{{100 + 14}}}
ight) imes 95$$ Cost price = $${ ext{Rs}}{ ext{. }}frac{{250}}{3}$$ xa0= Rs. 83.3 For the Discount of 11%, Selling price = Rs. 89 Then Profit % = $$frac{{89 - 83.3}}{{83.3}} imes 100$$ Profit = 6.84 ≈ 6.8 Hence, the correct answer is option A.
201: A
Solution: $$35\% = frac{7}{{20}},,15\% = frac{3}{{20}}$$ $$eqalign{
& Rightarrow 90\% = 42210 cr
& Rightarrow 100\% = frac{{42210}}{{90}} imes 100 = 46900 cr} $$
202: C
Solution: Raw material → 25% increase
203: C
Solution: Let monthly salary = 100x Total expenditure = (40 + 18 + 12 + 5)x = 75x 25x = 20,000 - 16,000 x $$ = frac{{4000}}{{25}} = 160$$ Hence monthly salary = 100x = 100 × 160 = 16,000
204: C
Solution: $$4\% o frac{{ - 1}}{{25}}$$
205: C
Solution: Given: Percentage increase in the price of diesel = 26% Percentage increase in total expenditure = 15% Concept used: P × C = E Where P is Price, C is Consumption and E is Expenditure Calculation: Let initial price be P 1 , consumption be C 1 , and expenditure be E 1 P 1 × C 1 = E 1 ⇒ $${{ ext{C}}_1} = frac{{{{ ext{E}}_1}}}{{{{ ext{P}}_1}}}$$ Let new price be P 2 , new quantity consumed be C 2 , and new expenditure be E 2 P 2 = P 1 + 26% of P 1 ⇒ P 2 = 1.26P 1 E 2 = E 1 + 15% of E 1 ⇒ E 2 = 1.15E 1 As, P 2 × C 2 = E 2 ⇒ 1.26P 1 × C 2 = 1.15E 1 ⇒ C 2 = $$frac{{{ ext{1}}{ ext{.15}}{{ ext{E}}_1}}}{{{ ext{1}}{ ext{.26}}{{ ext{P}}_1}}}$$ ⇒ C 2 = 0.9126 × $$frac{{{{ ext{E}}_1}}}{{{{ ext{P}}_1}}}$$ ⇒ C 2 = 0.9126C 1 Decrease in consumption = C 1 - C 2 ⇒ Decrease in consumption = C 1 - 0.9126C 1 = 0.0874C 1 Percentage decrease in consumption = $$frac{{{ ext{Decrease in consumption}}}}{{{ ext{Initial consumption}}}} imes 100$$ ⇒ Percentage decrease in consumption = $$frac{{0.0874{{ ext{C}}_1}}}{{{{ ext{C}}_1}}} imes 100$$ ∴ The percentage decrease in consumption is 8.7% (correct to 1 decimal place)
206: A
Solution: 20% = $$frac{1}{5}$$ = $$frac{4}{5}$$ Xxa0 : xa0 Y 4 xa0 : xa0 5 Let X = 4a Y = 5a Hence, ⇔ $$frac{Y - X}{Y}$$ = $$frac{5a - 4a}{5a}$$ = $$frac{a}{5a}$$ = $$frac{1}{5}$$ ⇔ $$frac{X}{X - Y}$$ = $$frac{4a}{4a - 5a}$$ = $$frac{4a}{- a}$$ = - 4 Hence required answer = $$frac{1}{5}$$, - 4
207: D
Solution: Let the number of boys = 400 Let the number of girls = 100 Total number of students who do not get scholarship = 400 × $$frac{25}{100}$$ + 100 × $$frac{30}{100}$$ = 100 + 30 = 130 Required percentage = $$frac{130}{500}$$ × 100 = 26%
208: D
Solution: Let the total amount = x According to the question, x × $$frac{80}{100}$$ × $$frac{95}{100}$$ = (120 + 1400) x × $$frac{4}{5}$$ × $$frac{19}{20}$$ = 1520 x = $$frac{1520 × 100}{76}$$ x = 2000 Total amount = 2000 Amount spent on transport = 2000 × $$frac{80}{100}$$ × $$frac{5}{100}$$ = Rs. 80
209: B
Solution: Let the income in 2010 be P ⇒ R = 20% ⇒ Income of year 2012 ⇒ Rs. 2664000 $$eqalign{
& Rightarrow { ext{Income of 2012}} = P{left[ {1 + frac{R}{{100}}}
ight]^2} cr
& Rightarrow 2664000 = P{left[ {1 + frac{{20}}{{100}}}
ight]^2} cr
& Rightarrow 2664000 = P imes frac{6}{5} imes frac{6}{5} cr
& Rightarrow { ext{Income in 2010 = 1850000}} cr} $$
210: C
Solution: Net tax rate = 30 + $$frac{30 × 10}{100}$$ = 33%
211: B
Solution: Initial number of soldiers in the army = x According to the question, $$eqalign{
& Rightarrow { ext{x}} imes frac{{90}}{{100}} imes frac{{90}}{{100}} imes frac{{90}}{{100}} = 729000 cr
& Rightarrow x = frac{{729000 imes 1000}}{{9 imes 9 imes 9}} cr
& Rightarrow x = 1000000 cr} $$ Alternate: 10% → $$frac{1}{10}$$ 10 → 9 in war 10 → 9 in disease 10 → 9 disabled 100 → 729 729 → 729000 1 → 1000 10000 → 1000 × 1000 = 1000000
212: D
Solution: Failed students in Hindi = 35% Failed students in English = 45% Student failed in both subject hindi and english = 20% Student only fail in hind = 35 - 20 = 15% Student only fail in English = 45 - 20 = 25% Percentage of passed students in both subjects : = 100 - [ student fail in hindi + student fail in english + student fail in both subject] = [100 - (15 + 25 + 20)] = 40%
213: D
Solution: r% = $$frac{r}{100}$$ Initial Price Final 100 (100 + r) 100 (100 - r) 10000 (100 + r) (100 - r) According to the question, (100 + r) (100 - r) units = Rs. 1 (10000 - r 2 ) units = Rs. 1 1 unit = $$left( {frac{{1}}{{10000 - {r^2}}}}
ight)$$ Original price = $$left( {frac{{10000}}{{10000 - {r^2}}}}
ight)$$
214: A
Solution: Let the number are a and b where a > b According to the question, (a - b) = $$frac{15}{100}$$ (a + b) (a - b) = $$frac{3}{20}$$ (a + b) 20a - 20b = 3a + 3b 17a = 23b $$frac{a}{b}$$ = $$frac{23}{17}$$ Required ratio = 23 : 17
215: A
Solution: Let the number = x According to the question, $$frac{x × (100 - 25)}{100}$$ xa0 = 225 x = $$frac{225 × 100}{75}$$ x = 300 Required percentage : = $$frac{(375 - 300)}{300}$$ xa0 × 100 = 25%
216: A
Solution: Rs. 55 per kg → 20 kg Rice Rs. 50 per kg → 25 kg Rice Rs. 60 per kg → 35 kg Rice 55 × 20 = Rs. 1100 50 × 25 = Rs. 1250 60 × 35 = $$underline {,{ ext{Rs}}{ ext{. 2100}},} $$ = Rs. 4450 + Rs. 150 Total Rice 80 kg → Rs. 4600 1 kg → Rs. 57.50 62.56 57.50 $$overline {,,,5.06,} $$ $$frac{{5.06}}{{57.5}}$$ xa0× 100 = 8.8%
217: A
Solution: Let total income = 100 E = 60 T = 12 C = 1.8 Saving = 100 - 73.8 = 26.2 After increase income = 140 E = 84 T = 25.2 C = 5.04 Saving = 140 - (114.24) = 25.76 Ratio = 26.2 : 25.76 = 2620 : 2576 = 655 : 644
218: D
Solution: $$eqalign{
& frac{{X imes 49}}{{100}} = Y cr
& frac{X}{Y} = frac{{100}}{{49}} cr
& 50 imes Y\% cr
& = frac{{2 imes 50 imes 49\% }}{2} cr
& = frac{{100 imes 49\% }}{2} cr
& = X imes 24.5\% cr} $$
219: B
Solution: 40% → 36 20% → 18 60% → 54
220: C
Solution: $$20\% = frac{1}{5}$$ [x08egin{array}{*{20}{c}}
{}&{{ ext{Price}}}&{{ ext{Total}}}&{{ ext{Sales}}}&{} \
{{ ext{Old}} o }&5&{100}&{frac{{100}}{5}}&{ = 20} \
{{ ext{New}} o }&4&{125}&{frac{{125}}{4}}&{ ,,,,,,, = 31.25}
end{array}] $$eqalign{
& { ext{Increase in sale}} = 31.25 - 20 = 11.25 cr
& { ext{Increase }}\% = frac{{11.25}}{{20}} imes 100 = 56.25\% cr} $$
221: B
Solution: $$eqalign{
& frac{3}{4},,frac{4}{3} cr
& { ext{LCM of 4, 3}} = 12 cr
& { ext{First number}} cr
& = frac{3}{4} imes 12 = 9 cr
& { ext{Second number}} cr
& = frac{4}{3} imes 12 = 16 cr
& { ext{Error percentage}} cr
& = frac{7}{{16}} imes 100 = 43frac{3}{4}\% = 43.75\% cr} $$
222: A
Solution: $$eqalign{
& 33frac{1}{3}\% = frac{1}{3} cr
& { ext{Let income}} = 300 cr
& { ext{Saving}} = 100 cr
& { ext{Expenditure}} = 300 - 100 = 200 cr} $$ [x08egin{array}{*{20}{c}}
{}&{{ ext{Income}}}& = &{{ ext{Expenditure}}}& + &{{ ext{Saving}}} \
{{ ext{Old}} o }&{300}& = &{200}& + &{100} \
{}&{}&{}&{,,,,,,,,,,{ downarrow ^{ + 10\% }}}&{}&{,,,,,,,,,,{ downarrow ^{ + 22\% }}} \
{{ ext{New}} o }&x& = &{220}& + &{122}
end{array}] $$eqalign{
& x = 220 + 122 = 342 cr
& { ext{Increase}} = 342 - 300 = 42 cr
& { ext{Increase }}\% = frac{{42}}{{300}} imes 100 = 14\% cr} $$
223: A
Solution: $$eqalign{
& { ext{A}} + { ext{B}} = 80 cr
& { ext{A}} imes frac{1}{2} = frac{5}{6}{ ext{B}} cr
& frac{{ ext{A}}}{{ ext{B}}} = frac{5}{3} cr} $$ $$eqalign{
& 8,{ ext{unit}} o 80{ ext{ kg}} cr
& 1,{ ext{unit}} o 10{ ext{ kg}} cr
& 2,{ ext{unit}} o x08oxed{20{ ext{ kg}}} cr} $$
224: A
Solution: Total vote = 100 Valid vote = 100 - 20 = 80 X = 70% of 80 = 56 100 unit → 64000 1 unit → 640 56 unit → 56 × 640 = 35840
225: A
Solution: $$eqalign{
& { ext{Valid}} cr
& 65\% o 81965 cr
& 1\% o 1261 cr
& 100\% o 126100 cr
& 65\% { ext{ of }}97\% o 126100 cr
& 65 imes frac{{97}}{{100}} o 126100 cr
& 65mu o 130000 cr
& 100mu o 200000 cr} $$
226: B
Solution: Let the cost of a single ticket be Rs. $$x$$ The, cost of monthly return ticket : $$eqalign{
& = 125\% ,,{ ext{of}},,{ ext{Rs}}.x cr
& = { ext{Rs}}{ ext{. }}left( {frac{{125}}{{100}}x}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{5x}}{4} cr} $$ Cost of monthly return ticket with extension : $$eqalign{
& = 105\% ,,{ ext{of}},,{ ext{Rs}}.frac{{5x}}{4} cr
& = { ext{Rs}}{ ext{. }}left( {frac{{105}}{{100}} imes frac{{5x}}{4}}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{21x}}{{16}} cr} $$ $$eqalign{
& herefore frac{{21x}}{{16}} = 84 cr
& Rightarrow x = left( {frac{{84 imes 16}}{{21}}}
ight) cr
& Rightarrow x = 64 cr} $$
227: C
Solution: Quantity of pulp in fresh grapes = Quantity of pulp in dry grapes = (100 - 10)% of 250 kg = 90% of 250 kg = 225 kg Let the total weight of fresh grapes be $$x$$ kg Then, ⇒ (100 - 80)% of $$x$$ = 225 ⇒ 20% of $$x$$ = 225 ⇒ $$frac{x}{5}$$ = 225 ⇒ $$x$$ = 225 × 5 ⇒ $$x$$ = 1125 Hence, total weight of fresh grapes = 1125 kg
228: C
Solution: Let the person's gross income be Rs. $$x$$ Then, ⇒ (100 - 3)% of $$x$$ = 237650 ⇒ 97% of $$x$$ = 237650 ⇒ $$frac{97x}{100}$$ = 237650 ⇒ $$x$$ = $$left( {frac{{237650 imes 100}}{{97}}}
ight)$$ ⇒ $$x$$ = 245000 ∴ When income tax is raised to 7%, we have : Net income : = (100 - 7)% of Rs. 245000 = 93% of Rs. 245000 = Rs. $$left( {frac{{93}}{{100}} imes 245000}
ight)$$ = Rs. 227850
229: C
Solution: Number of games already won = 30% of 60 = 18 Let the required number of games be $$x$$ Then, $$eqalign{
& Rightarrow frac{{18 + x}}{{60 + x}} imes 100 = 50 cr
& Rightarrow frac{{18 + x}}{{60 + x}} = frac{1}{2} cr
& Rightarrow 36 + 2x = 60 + x cr
& Rightarrow x = 24 cr} $$
230: D
Solution: Let the third number be $$x$$ Then, first number : = $$112frac{1}{2}\% $$xa0 of $$x$$ = $$frac{9x}{8}$$ Second number : = 125% of $$x$$ = $$frac{5x}{4}$$ ∴ Required percentage : = $$left( {frac{{9x}}{8} imes frac{{4}}{{5x}} imes 100}
ight)$$ xa0 % = 90%
231: C
Solution: Let the original price be Rs. $$x$$ New final price : = 110% of (110% of Rs. 100) = Rs. $$left( {frac{{110}}{{100}} imes frac{{110}}{{100}} imes 100}
ight)$$ = Rs. 121 ∴ Total increase : = (121 - 100)% = 21%
232: B
Solution: Let the total number of respondents be $$x$$ Percentage of uncertain individuals : = [100 - (20 + 60)]% = 20% ∴ 60% of $$x$$ - 20% of $$x$$ = 720 ⇒ 40% of $$x$$ = 720 ⇒ $$frac{{40x}}{{100}}$$ = 720 ⇒ $$x$$ = $$left( {frac{{720 imes 100}}{{40}}}
ight)$$ ⇒ $$x$$ = 1800
233: A
Solution: Let the total number of applicants be $$x$$ Number of eligible candidates = 95 % of $$x$$ Eligible candidates of other categories = 15% of (95 of $$x$$) = $$left( {frac{{15}}{{100}} imes frac{{95}}{{100}} imes x}
ight)$$ = $$frac{{57x}}{{400}}$$ ∴ $$frac{{57x}}{{400}}$$ = 4275 ⇒ $$x$$ = $$left( {frac{{4275 imes 400}}{{57}}}
ight)$$ ⇒ $$x$$ = 30000
234: A
Solution: Let the original price of sugar be Rs. 10 per kg Then, original expenditure = Rs. (10 × 10) = Rs. 100 New expenditure = 110% of Rs. 100 = Rs. 110 New price of sugar = 132% of Rs. 10 = Rs. 13.20 New consumption : = $$frac{110}{13.20}$$ kg = $$frac{25}{3}$$ kg = $$8frac{1}{3}$$ kg
235: C, G
Solution: Number of questions answered correctly : = 40% of 125 = 50 For 60% grade, number of questions to be answered correctly : = 60% of 250 = 150 Remaining number of question to be answered correctly : = 150 - 50 = 100 ∴ Required percentage : = $$left( {frac{{100}}{{125}} imes 100}
ight)\% $$ = 80%
236: D
Solution: Schedule working hours in week = 48
Total pay in a week for schedule working hours = Rs. 480 Pay per hour for schedule working hours = $$frac{{480}}{{48}}$$ = Rs. 10 Pay per hour for over time = 10 + 25% of 10 = Rs. 12.5 Total pay in that particular week = Rs. 605 Extra pay = 605 - 480 = 125 So, total over time = $$frac{{125}}{{12.5}}$$ xa0= 10 hours Thus, total work hour altogether in that week = 48 + 10 = 58 hours
237: C
Solution: Winner gets 55% of votes. As 4% votes were declared invalid so 96% would be the valid votes So, Winner gets 55% of 96% valid votes Winner gets % valid votes = $$frac{{55 imes 96}}{{100}}$$ xa0 = 52.8% votes Loser gets = 96 - 52.8 = 43.2% votes Difference = 9.6% Now, 9.6% = 4800 So, 1% = $$frac{{4800}}{{9.66}}$$ Thus, 100% Votes = $$frac{{4200 imes 100}}{{9.66}}$$ xa0 = 50000 Hence, Total Voters = 50000 Alternatively, Let total number of voters were X Invalid Votes = 4% Valid Votes = 96% Total Valid Votes = 96% of X = $$frac{{96{ ext{x}}}}{{100}}$$ = 0.96X Winner gets 55% of Valid Votes, $$ = frac{{0.96 imes 55}}{{100}} = 0.528{ ext{x}}$$ xa0 xa0 xa0votes Loser gets = (0.96X - 0.528X) = 0.432X Difference = 0. 528 - 0.432 = 0.096X 0.096X = 4800 X = 50000
238: A
Solution: Money spent originally = Rs. 2160 Less money to be spent for now for the same length of cloth, = 10% of 2160 = Rs. 216 It means Rs. 216 enables a man to buy 6 meters of cloth So, reduced price = $$frac{{216}}{6}$$ = Rs. 36 per meter And the original price = $$frac{{100 imes 36}}{{90}}$$ xa0 = Rs. 40 per meter
239: B
Solution: Let original area of the garden was 100 square unit. Increase or decrease in area can be easily determined by this graphic: 100 == 40% up length ==> 140 == 20% down width ==> 112 (Final Area) So, there is 12% increase in area of the garden.
240: A
Solution: Let B = 100 Then, A = 100 + 40% of 100 = 140 Let C = X X - 20% of X = 100 0.8X = 100 X = $$frac{{100}}{{0.8}}$$ = 125 ∴ A : C = $$frac{{140}}{{125}}$$ = 28 : 25
241: B
Solution: Number of runs made by running = 110 - (3 x 4 + 8 x 6) = 110 - (60) = 50 ∴ Required percentage $$eqalign{
& = left( {frac{{50}}{{110}} imes 100}
ight)\% cr
& = 45frac{5}{{11}}\% cr} $$
242: C
Solution: Let their marks be (x + 9) and x Then, $$eqalign{
& x + 9 = frac{{56}}{{100}}left( {x + 9 + x}
ight) cr
& Rightarrow 25left( {x + 9}
ight) = 14left( {2x + 9}
ight) cr
& Rightarrow 3x = 99 cr
& Rightarrow x = 33 cr} $$ So, their marks are 42 and 33
243: D
Solution: Suppose originally he had x apples Then, $$eqalign{
& left( {100 - 40}
ight)\% ,{ ext{of}},x = 420 cr
& Rightarrow frac{{60}}{{100}} imes x = 420 cr
& Rightarrow x = {frac{{420 imes 100}}{{60}}} = 700 cr} $$
244: C
Solution: Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69. Number of such number = 14 $$ = left( {frac{{14}}{{70}} imes 100}
ight)\% = 20\% $$
245: N/A
Solution: $$eqalign{
& x\% ,{ ext{of}},y = {frac{x}{{100}} imes y} cr
& ,,,,,,,,,,,,,,,,,,,, = {frac{y}{{100}} imes x} cr
& ,,,,,,,,,,,,,,,,,,,, = y\% ,{ ext{of}},x cr
& herefore A = B cr} $$
246: C
Solution: Let the greater and smaller number is a and b respectively According to the question, Case (i) a × $$frac{40}{100}$$ = b × $$frac{60}{100}$$ 2a = 3b a = $$frac{3}{2}$$b ..... (i) Case (ii) a + b = 150 ..... (ii) From equation (i) and (ii) $$frac{3}{2}$$b + b = 150 5b = 300 b = 60 Value of b = 60 put in equation (i) a = $$frac{3}{2}$$ × 60 = 90 Hence, greater number = 90
247: B
Solution: Average income = Rs. $$frac{80800}{16}$$ = Rs. 5050 Hence, required income = Rs. 5050 × $$frac{120}{100}$$ = Rs. 6060
248: A
Solution: Let the number be x According to the question, x × $$frac{60}{100}$$ × $$frac{3}{5}$$ = 36 ⇒ x = $$frac{36 × 25}{9}$$ ⇒ x = 100
249: D
Solution: 50% = $$frac{1}{2}$$ Let Z has 2 units of money According to the question, X xa0 : xa0 Y xa0 : xa0 Z 6 xa0 : xa0 3 xa0 : xa0 2 $$frac{(6 + 3 + 2)}{3}$$ xa0 units = Rs. 110 ⇒ 11 units = Rs. 330 1 unit = Rs. 30 6 units = 30 × 6 = Rs. 180 Hence, X has Rs. 180
250: B
Solution: Let the bigger number is a and the smaller number is (520 - a) According to the question, a × $$frac{(100 - 4)}{100}$$ xa0 = (520 - a) × $$frac{(100 + 12)}{100}$$ $$frac{96a}{100}$$ = (520 - a) $$frac{112}{100}$$ 96a = (520 - a) 112 13a = 3640 a = 280 Hence, bigger number = 280 Smaller number = (520 - 280) = 240
251: A
Solution: Salary of Tulsiram = Rs. $$frac{720}{4}$$ × 100 = Rs. 18000 Salary of Kashyap = 18000 × $$frac{100}{120}$$ = Rs. 15000
252: D
Solution: According to the question, Let the total number of students = 100 Ratio of Boys and Girls =$$frac{3}{2}$$ Total number of Boy = $$frac{3}{5} imes 100 = 60$$ Total number of Girl = $$frac{2}{5} imes 100 = 40$$ 30% of boy appeared in exam = 30% of 60 = 18 70% of girl appeared in exam = 70% of 40 = 28 Total number of student appeared in exam = 18 + 28 = 46 Students not appeared in exam = 100 - 46 = 54 $$ herefore frac{{{ ext{Ratio of students appeared in exam}}}}{{{ ext{Not appeared in exam}}}}$$ = $$frac{46}{54}$$ = $$frac{23}{27}$$ = 23 : 27
253: C
Solution: Current value of machine $$eqalign{
& = 6250 imes frac{{90}}{{100}} imes frac{{80}}{{100}} imes frac{{70}}{{100}} cr
& = { ext{Rs}}{ ext{. 3150}} cr} $$
254: D
Solution: Let Person income is 100x 20% his income give to elder son, i.e. 20x Remaining income 100x -20x = 80x Now 30% of his remaining income given to his younger son i.e. 30% of 80x = 24x Remain saving is 80x - 24x = 56x Now 10% of his remaining income he donated to a trust i.e. 10% of 56x = 5.6x Remain income = 56x - 5.6x = 50.4x According to question 50.4x = 10080 x = $$frac{{10080}}{{50.4}}$$ = 200 Person income = 100 x 200 = Rs. 20000
255: D
Solution: Let the weight of the bucket when it is full, be 1 kg. Then, weight of empty bucket = 25% of 1 kg = $$frac{1}{4}$$ kg Weight of liquid in the bucket = 1 - $$frac{1}{4}$$ kg = $$frac{3}{4}$$ kg On removing the liquid, weight of (bucket + liquid) = $$frac{3}{5}$$ kg Weight of liquid in the bucket = $$frac{3}{5}$$ - $$frac{1}{4}$$ kg = $$frac{7}{20}$$ kg Weight of liquid removed = $$frac{3}{4}$$ - $$frac{7}{20}$$ = $$frac{8}{20}$$ = $$frac{2}{5}$$ kg Hence, required percentage = $$frac{2}{5}$$ × $$frac{4}{3}$$ × 100% = $$frac{160}{3}$$% = $$53frac{1}{3}$$%
256: D, E
Solution: 5 : 4 = $$frac{5}{4}$$ = $$frac{5}{4}$$ × 100% = 125%
257: D
Solution: Given (550% of 250) ÷ 275 = (?) ⇒ (?) = $$frac{550 × 250}{100}$$ xa0 ÷ 275 ⇒ (?) = (55 × 25) ÷ 275 ⇒ (?) = $$frac{55 × 25}{275}$$ ⇒ (?) = 5
258: D
Solution: Pass percentage : $$eqalign{
& = left( {frac{{252}}{{270}} imes 100}
ight)\% cr
& = frac{{280}}{3}\% cr
& = 93frac{1}{3}\% cr} $$
259: A
Solution: x% of y = $$frac{xy}{100}$$..... (i) And y% of x = $$frac{xy}{100}$$ ..... (ii) From (i) and (ii) x% of y = y% of x
260: B
Solution: Let X and Y denote the new values of x and y respectively. Then, X = 80% of x = $$frac{4x}{5}$$ Y = 80% of y = $$frac{4y}{5}$$ $$eqalign{
& herefore X{Y^2} = frac{{4x}}{5} imes {left( {frac{{4y}}{5}}
ight)^2} cr
& ,,,,,,,,,,,,,,,,,, = frac{{4x}}{5} imes frac{{16{y^2}}}{5} cr
& ,,,,,,,,,,,,,,,,,, = frac{{64}}{{125}}x{y^2} cr} $$ Decrease in the value : $$eqalign{
& = left( {x{y^2} - frac{{64}}{{125}}x{y^2}}
ight) cr
& = frac{{61}}{{125}}x{y^2} cr} $$ ∴ Decrease % $$eqalign{
& = frac{{61}}{{125}}x{y^2} cr
& = left( {frac{{61x{y^2}}}{{125}} imes frac{1}{{x{y^2}}} imes 100}
ight)\% cr
& = 48.8\% cr} $$
261: C
Solution: Purchase price : $$eqalign{
& = { ext{Rs}}{ ext{.}}left[ {frac{{29644.032}}{{{{left( {1 - frac{{12}}{{100}}}
ight)}^3}}}}
ight] cr
& = { ext{Rs}}{ ext{.}}left( {29644.032 imes frac{{25}}{{22}} imes frac{{25}}{{22}} imes frac{{25}}{{22}}}
ight) cr
& = { ext{Rs}}.43500 cr} $$
262: B
Solution: Quantity of alcohol in 5 litres solution = 20% of 5 litres = 1 litre Quantity of alcohol removed = 20% of 2 litres = 400 ml Quantity of alcohol in new solution : = (1000 - 400)ml = 600 ml ∴ Strength of alcohol in new solution $$eqalign{
& = left( {frac{{600}}{{5000}} imes 100}
ight)\% cr
& = 12\% cr} $$
263: C
Solution: Suppose the man ordered x feet y inches of the rope. Since x and y represent inches in the miswritten and actual order respectively, so each one of x and y is less than 12. [ $$x08ecause $$1 feet = 12 inches] Actual order = x feet y inches = (12x + y) inches Miswritten order = y feet x inches = (12y + x) inches ∴ (12y + x) = 30% of (12x + y) = $$frac{3}{10}$$ (12x + y) ⇒ 10 (12y + x) = 3 (12x + y) ⇒ 26x = 117y ⇒ $$frac{x}{y}$$ = $$frac{117}{26}$$ = $$frac{9}{2}$$ Since x < 12, y < 12, so x = 9, y = 2 Hence, the man ordered 9 feet 2 inches of rope.
264: B
Solution: 60% of 264 $$eqalign{
& = left( {frac{{60}}{{100}} imes 264}
ight) cr
& = 158.40 cr} $$ 10% of 44 $$eqalign{
& = left( {frac{{10}}{{100}} imes 44}
ight) cr
& = 4.40 cr} $$ 15% of 1056 $$eqalign{
& = left( {frac{{15}}{{100}} imes 1056}
ight) cr
& = 158.40 cr} $$ 30% of 132 $$eqalign{
& = left( {frac{{30}}{{100}} imes 132}
ight) cr
& = 39.60 cr} $$ ∴ 60% of 264 = 15% of 1056
265: A
Solution: Let the number be $$x$$ Then, $$eqalign{
& x - 16\% ,,{ ext{of}},x = 42 cr
& Leftrightarrow x - frac{{16}}{{100}}x = 42 cr
& Leftrightarrow x - frac{4}{{25x}} = 42 cr
& Leftrightarrow frac{{21}}{{25}}x = 42 cr
& Leftrightarrow x = left( {frac{{42 imes 25}}{{21}}}
ight) cr
& Leftrightarrow x = 50 cr} $$
266: C
Solution: $$left( {23.6\% { ext{ of 1254}}}
ight) - left( {16.6\% { ext{ of 834}}}
ight)$$ $$ = left( {frac{{236}}{{10}} imes frac{1}{{100}} imes 1254}
ight) - $$ xa0 xa0 $$left( {frac{{166}}{{10}} imes frac{1}{{100}} imes 834}
ight)$$ $$eqalign{
& = frac{1}{{1000}}left( {236 imes 1254 - 166 imes 834}
ight) cr
& = frac{{12}}{{1000}}left( {24662 - 11537}
ight) cr
& = left( {frac{{12 imes 13125}}{{1000}}}
ight) cr
& = 157.5 cr} $$
267: N/A
Solution: Let the first number be x and the second number be y $$frac{1}{4}{ ext{ of 60% of }}x = frac{2}{5}{ ext{ of 20% of }}y$$ $$eqalign{
& Rightarrow frac{1}{4} imes frac{{60}}{{100}} imes x = frac{2}{5} imes frac{{20}}{{100}} imes y cr
& Rightarrow frac{{3x}}{{20}} = frac{{2y}}{{25}} cr
& Rightarrow frac{x}{y} = frac{2}{{25}} imes frac{{20}}{3} cr
& Rightarrow frac{x}{y} = frac{8}{{15}} cr} $$
268: B
Solution: Let number of students who appeared at the examination be $$x$$ Passed percentage = 96% Failed percentage = 4% According to given information we get Unsuccessful students = 4% ∴ 4% of $$x$$ = 500 ⇒ $$left( {frac{{x imes 4}}{{100}}}
ight)$$xa0 = 500 ⇒ $$x$$ = $$left( {frac{{500 imes 100}}{{4}}}
ight)$$ ⇒ $$x$$ = 12500
269: B
Solution: Reduction in consumption : $$eqalign{
& = left[ {frac{R}{{left( {100 + R}
ight)}} imes 100}
ight]\% cr
& = left( {frac{{15}}{{115}} imes 100}
ight)\% cr
& = frac{{300}}{{23}}\% cr
& = 13frac{1}{{23}}\% cr} $$
270: C
Solution: Let the number be x Then, $$eqalign{
& 38\% { ext{ of }}x - 24\% ,,{ ext{of }}x = 135.10 cr
& Rightarrow frac{{38}}{{100}}x - frac{{24}}{{100}}x = 135.10 cr
& Rightarrow frac{{14}}{{100}}x = 135.10 cr
& Rightarrow x = left( {frac{{135.10 imes 100}}{{14}}}
ight) cr
& Rightarrow x = 965 cr
& herefore 40\% { ext{ of 965 :}} cr
& = left( {frac{{40}}{{100}} imes 965}
ight) cr
& = 386 cr} $$
271: D
Solution: Let Raman's expense be Rs. $$x$$ Then, Vimal's expenses : = 90% of $$x$$ $$eqalign{
& = { ext{ Rs}}{ ext{. }}left( {frac{{90}}{{100}} imes x}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{9}{{10}}x{ ext{ }} cr} $$ Aman's expense : $$eqalign{
& = { ext{130% of Rs}}{ ext{. }}left( {frac{{9x}}{{10}}}
ight) cr
& = { ext{ Rs}}{ ext{. }}left( {frac{{130}}{{100}} imes frac{{9x}}{{10}}}
ight) cr
& = { ext{ Rs}}{ ext{. }}frac{{117x}}{{100}} cr
& herefore frac{{117x}}{{100}} + frac{{9x}}{{10}} + x = 6447 cr
& Rightarrow frac{{117x + 90x + 100x}}{{100}} = 6447 cr
& Rightarrow 307x = 644700 cr
& Rightarrow x = frac{{644700}}{{307}} cr
& Rightarrow x = 2100 cr
& { ext{Hence, Aman's expenses :}} cr
& = { ext{Rs}}{ ext{. }}left( {frac{{117 imes 2100}}{{100}}}
ight) cr
& = { ext{Rs}}{ ext{. 2457}} cr} $$
272: B
Solution: $$eqalign{
& A = 150\% { ext{ of }}B cr
& Rightarrow A = frac{{150}}{{100}}B cr
& Rightarrow frac{A}{B} = frac{3}{2} cr
& Rightarrow frac{A}{B} + 1 = frac{3}{2} + 1 cr
& Rightarrow frac{{A + B}}{B} = frac{5}{2} cr
& Rightarrow frac{B}{{A + B}} = frac{2}{5} cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{B}{{A + B}} imes 100}
ight)\% cr
& = left( {frac{2}{5} imes 100}
ight)\% cr
& = 40\% cr} $$
273: B
Solution: Let the total salary be Rs. $$x$$ Then, (100 -10)% of (100 - 20)% of (100 - 20)% of (100 - 10)% of $$x$$ = 15552 $$ Leftrightarrow left( {frac{{90}}{{100}} imes frac{{80}}{{100}} imes frac{{80}}{{100}} imes frac{{90}}{{100}} imes x}
ight)$$ xa0 xa0 xa0 $$ = 15552$$ $$eqalign{
& Leftrightarrow x = left( {frac{{15552 imes 10000}}{{64 imes 81}}}
ight) cr
& Leftrightarrow x = 30000 cr} $$
274: B
Solution: Failed in 1 st subject = $$frac{35}{100}$$ × 2500 = 875 Failed in 2 nd subject = $$frac{42}{100}$$ × 2500 = 1050 Failed in both = $$frac{15}{100}$$ × 2500 = 375 Failed in 1 st subject only = (875 - 375) = 500 Failed in 2 nd subject only = (1050 - 375) = 675 ∴ Passed in 2 nd only + Passed in 1 st only = (675 + 500) = 1175
275: B
Solution: It is given that, (14% of 14) + (28% of 28) + (92% of 96) - (15% of 85) = ? $$left( ?
ight) = left( {frac{{14 imes 14}}{{100}}}
ight) + left( {frac{{28 imes 28}}{{100}}}
ight) + $$ xa0 xa0 xa0 $$left( {frac{{92 imes 96}}{{100}}}
ight) - left( {frac{{15 imes 85}}{{100}}}
ight)$$ $$eqalign{
& left( ?
ight) = left( {1.96 + 7.84 + 88.32 - 12.75}
ight) cr
& left( ?
ight) = left( {98.12 - 12.75}
ight) cr
& left( ?
ight) = 85.37 cr} $$
276: D
Solution: Required percentage : $$eqalign{
& = left( {frac{{33}}{{88}} imes 100}
ight)\% cr
& = frac{{75}}{2}\% cr
& = 37.5\% cr} $$
277: C
Solution: Let the number be x Then, 40% of 60% of $$frac{3}{5}$$ of x = 504 ⇒ $$frac{40}{100}$$ × $$frac{60}{100}$$ × $$frac{3}{5}$$ × x = 504 ⇒ $$frac{18}{125}$$x = 504 ⇒ x = $$frac{504 × 125}{18}$$ ⇒ x = 3500 ∴ 25% of $$frac{2}{5}$$ of 3500 : = $$frac{25}{100}$$ × $$frac{2}{5}$$ × 3500 = 350
278: B
Solution: Let Nupur's annual salary be Rs. x Then, 26% of x = 89856 ⇒ $$frac{26x}{100}$$ = 89856 ⇒ x = $$frac{89856 × 100}{26}$$ ⇒ x = 345600 ∴ Nupur's monthly income : = Rs. $$frac{345600}{12}$$ = Rs. 28800
279: D
Solution: 5% of A + 4% of B = $$frac{2}{3}$$ (6% of A + 8% of B) $$ Leftrightarrow frac{5}{{100}}A + frac{4}{{100}}B = $$ xa0 xa0 $$frac{2}{3}left( {frac{6}{{100}}A + frac{8}{{100}}B}
ight)$$ $$eqalign{
& Leftrightarrow frac{1}{{20}}A + frac{1}{{25}}B = frac{1}{{25}}A + frac{4}{{75}}B cr
& Leftrightarrow left( {frac{1}{{20}} - frac{1}{{25}}}
ight)A = left( {frac{4}{{75}} - frac{1}{{25}}}
ight)B cr
& Leftrightarrow frac{1}{{100}}A = frac{1}{{75}}B cr
& Leftrightarrow frac{A}{B} = frac{{100}}{{75}} cr
& Leftrightarrow frac{A}{B} = frac{4}{3} cr} $$ ∴ A : B = 4 : 3
280: A
Solution: Number of valid votes = (100 - 15)% of 15200 = 85% of 15200 = $$frac{85}{100}$$ × 15200 = 12920 Valid votes polled by other candidate : = (100 - 55)% of 12920 = $$frac{45}{100}$$ × 12920 = 5814
281: D
Solution: x% of y = z ⇒ $$frac{x}{100}$$y = z ⇒ $$frac{x}{z}$$ = $$frac{100}{y}$$ ∴ Required percentage $$eqalign{
& = left( {frac{x}{z} imes 100}
ight)\% cr
& = left( {frac{{100}}{y} imes 100}
ight)\% cr
& = left( {frac{{{{100}^2}}}{y}}
ight)\% cr} $$
282: N/A
Solution: Let the sum invested at 5% be Rs. x and that invested at 3% be Rs. (9600 - x) Then, 5% of x = 3% of (9600 - x) ⇒ 5x = 3 (9600 - x) ⇒ 8x = 28800 ⇒ x = 3600 Hence, total income : = 5% of x + 3% of (9600 - x) = Rs. (5% of 3600 + 3% of 6000) = Rs. (180 + 180) = Rs. 360
283: A
Solution: Number of researchers who prefer Team A = 70% of 50 = 35 Number of researchers who prefer Team B = (50 - 35) = 15 Number of researchers assigned to Team A = 40% of 50 = 20 Number of researchers assigned to Team B = (50 - 20) = 30 To find the least possible number of researchers who will not be assigned to the team they prefer, we assume that the maximum number of researchers get the team they prefer. So,Number of researchers who are assigned to the team they prefer = 20 (Team A) + 15 (Team B) = 35 ∴ Required number = (50 - 35) = 15
284: A
Solution: Let the fraction be $$frac{{ ext{x}}}{{ ext{y}}}$$
When fraction is squared its numerator is reduced by $$33frac{1}{3}$$ and denominator is reduced by 20% $$eqalign{
& { ext{According}},{ ext{to}},{ ext{question,}} cr
& {left( {frac{x}{y}}
ight)^2} imes frac{{33left( {frac{1}{3}}
ight)\% }}{{20\% }} = 2left( {frac{x}{y}}
ight) cr
& { ext{Or}},,{left( {frac{x}{y}}
ight)^2} imes frac{{left( {frac{2}{3}}
ight)}}{{left( {frac{1}{5}}
ight)}} = 2left( {frac{x}{y}}
ight) cr
& { ext{Or}},,frac{x}{y} = frac{3}{5} cr
& { ext{Sum of numerator and denominator is}} cr
& left( {x + y}
ight) = 3 + 5 cr
& ,,,,,,,,,,,,,,,,,,, = 8 cr} $$
285: C
Solution: Let the percentage marks in QA =(10a + b)% Let the percentage marks in DI = (10b + a)% Let the percentage marks in VA = x% Now, according to the question, we have, $$eqalign{
& frac{{left( {10{ ext{a}} + { ext{b}}}
ight) + { ext{x}} + left( {10{ ext{b}} + { ext{a}}}
ight)}}{3} = { ext{x}} cr
& 11{ ext{a}} + 11{ ext{b}} + { ext{x}} = 3{ ext{x}} cr
& { ext{or, x}} = frac{{11left( {{ ext{a}} + { ext{b}}}
ight)}}{2} cr} $$ Clearly, we can see that the percentage of the VA section will be a multiple of 11 So, required answer will be 66
286: C
Solution: Let the initial price = Rs. 10000p Price after first increment = 10000p + 100xp Price after first decrement = 10000p + 100xp - (100px + px 2 ) = 10000p - px 2 Now, total decrement, px 2 = 21025 . . . . . (1) Price after second increment, = 10000p - px 2 + 100xp - $$frac{{{ ext{p}}{{ ext{x}}^3}}}{{100}}$$ Price after second increment, = 10000p - p 2 + 100xp - $$frac{{{{ ext{p}}^3}}}{{100}}$$ - 100xp + $$frac{{{ ext{p}}{{ ext{x}}^3}}}{{100}}$$ - px 2 + $$frac{{{ ext{p}}{{ ext{x}}^4}}}{{10000}}$$ = 10000p - 2px 2 + $$frac{{{ ext{p}}{{ ext{x}}^2}}}{{10000}}$$ = 484416 . . . . . . (2) On solving equation (1) and (2), We get x = 20 Substituting back we get, p = 5,25,625
287: A
Solution: No. of Machines Output Manuf. cost Est. cost Total cost Profit 12 48, 000 24, 000 10, 000 34, 000 14, 000 11 44, 000 22, 000 10, 000 32, 000 12, 000 Profit, = Output - total cost
= 44000 - 32000
= Rs. 12000
Initial value of share holders,
= 14000 $$ imes frac{{10}}{{100}}$$
= Rs. 1400
New value of share holders,
= 12000 $$ imes frac{{10}}{{100}}$$
= Rs. 1200 Decrease in Share holder value = 1400 - 1200 = 200 percentage decrease in the value of shareholders is : $$eqalign{
& = frac{{200 imes 100}}{{1400}} cr
& = 14.28\% cr} $$
288: C
Solution: Let the price of wheat is x per kg. Then price of wheat will be 5x per kg.
Expenditure on rice = 25 × x = 25x
Expenditure of wheat = 9 × 5x = 45x
Total cost,
25x + 45x = 350
70x = 350
x = 5
Hence, price of Rice = Rs. 5 per kg. Price of wheat = 25 per kg. Now, price of wheat = 25 ---- 20% ↑----> Rs. 30 per kg. Let the new amount of rice is N kg, then N*5 + 9*30 = 350 N = 16 kg. % decrease in the amount of rice $$eqalign{
& = frac{{left( {25 - 16}
ight) imes 100}}{{25}} cr
& = 36\% cr} $$
289: B
Solution: Let the initial cost of raw material be 100. So, initial labor cost was 25 and net cost was 125
Now, 15% increment in raw materials cost and labor cost has gone up to 30% from 25 % Raw material cost = 115 And Labor cost = (115 × 30%) = 34.5 So, New net cost, = 115 + 34.5 = 149.5 Difference of labor cost = 149.5 - 125 = 24.5 % reduction = $$frac{{24.5 imes 100}}{{149.5}}$$ xa0xa0= 17%(approx.)
290: C
Solution: $$eqalign{
& { ext{His bonus}}, cr
& = frac{{ {20 imes 1000000} }}{{100}} cr
& = 2, ext{lakh} cr
& { ext{Total}},{ ext{profit}} = { ext{Net}},{ ext{profit}} + frac{{ {10 imes { ext{net}},{ ext{profit}}} }}{{100}} cr
& 1.32, ext{lakh} = { ext{Net}},{ ext{profit}} imes left[ {1 + {frac{{10}}{{100}}} }
ight] cr
& { ext{Net}},{ ext{profit}} = frac{{132000}}{{1.1}} = 120000 cr
& { ext{Commission}}, cr
& = left( {{ ext{Total}},{ ext{profit}} - { ext{Net}},{ ext{profit}}}
ight) cr
& = 132000 - 120000 cr
& = 12000 cr
& { ext{Hence}}, { ext{his}},{ ext{total}},{ ext{earnings}} cr
& = 2, ext{lakh} + 12000 cr
& = Rs.,212000 cr} $$
291: B
Solution: There is no need of the number of goats given i.e. 34,398. Initially, let there be 100 goats. Then 100 == 40% ↑==> 140 == 30%↓(declined to 70%) ==> 98 == 30%↑ ==> 127.4 == 10%↓(sold) ==> 114.66 Hence, % increase = 14.66% [As 100 becomes 114.66]
292: C
Solution: Total employees = 1600 Female employees, 60% of 1600 $$ = frac{{60 imes 1600}}{{100}} = 960$$ Then male employees = 640 50% of male are computer literate, = 320 male computer literate 62% of total employees are computer literate, $$ = frac{{62 imes 1600}}{{100}} = 992$$ xa0 xa0 computer literate Thus, Female computer literate = 992 - 320 = 672 Alternatively : Let 60% employees are female and 40% are male Then, 20% of male are computer literate and 42% are female computer literate Female computer literate $$ = frac{{1600 imes 42}}{{100}} = 672$$
293: D
Solution: Price after third depreciation, 100 ==25%↓ ==> 75 == 20%↓==>60 == 15% ↓ ==> 51 The price will be, = Rs. 5,10,000 Alternatively : 1000000 × 0.75 × 0.80 × 0.85 = Rs. 5,10,000
294: C
Solution: Let the number of persons eligible to vote be x Then, votes who voted for A = 30% of x Votes who voted for B = 60% of (70% of x) = $$left( {frac{{60}}{{100}} imes frac{{70}}{{100}} imes 100}
ight)$$ xa0 % of x = 42% of x Voters who did not vote = [100 - (30 + 42)]% of x = 28% of x ∴ 30% of x - 28% of x = 1200 ⇒ 2% of x = 1200 ⇒ x = $$left( {frac{{1200 × 100}}{{2}}}
ight)$$ ⇒ x = 60000
295: B
Solution: Let x% of x = 10% of y Then, $$eqalign{
& Rightarrow frac{x}{{100}} imes x = frac{{10}}{{100}} imes y cr
& Rightarrow y = frac{{{x^2}}}{{100}} imes 10 cr
& Rightarrow y = frac{{{x^2}}}{{10}} cr} $$
296: A
Solution: Let the total number of employees be x Then, (100 - 50)% of (100 - 40)% of x = 180 50% of 60% of x = 180 $$eqalign{
& Rightarrow left( {frac{{50}}{{100}} imes frac{{60}}{{100}} imes x}
ight) = 180 cr
& Rightarrow x = left( {frac{{180 imes 10}}{3}}
ight) cr
& Rightarrow x = 600 cr} $$ ∴ Number of graduate employees : $$=$$ 50% of 60% of x $$eqalign{
& = left( {frac{{50}}{{100}} imes frac{{60}}{{100}} imes 600}
ight) cr
& = 180 cr} $$
297: C
Solution: $$eqalign{
& A = frac{{120}}{{100}}B, cr
& B = frac{{120}}{{100}}C, & cr
& C = frac{{85}}{{100}}D cr
& herefore B = frac{5}{6}A cr
& C = frac{5}{6}B, & cr
& D = frac{{20}}{{17}}C cr
& B = frac{5}{6} imes 576 = 480 cr
& C = frac{5}{6} imes 480 = 400 cr
& D = frac{{20}}{{17}} imes 400 = frac{{8000}}{{17}} cr} $$ So, required percentage : $$eqalign{
& = left( {frac{{8000}}{{17}} imes frac{1}{{800}} imes 100}
ight)\% cr
& = 58.82\% cr} $$
298: C
Solution: Let the original price = Rs. 100 New price = Rs. 75 Increase on 75 = 25 Increase on 100 : $$eqalign{
& = left( {frac{{25}}{{75}} imes 100}
ight)\% cr
& = frac{{100}}{3}\% cr
& = 33frac{1}{3}\% cr} $$
299: D
Solution: Let original salary = Rs. 100 New salary : 90% of 85% of 80% of Rs. 100 $$eqalign{
& = { ext{Rs}}{ ext{.}}left( {frac{{90}}{{100}} imes frac{{85}}{{100}} imes frac{{80}}{{100}} imes 100}
ight) cr
& = { ext{Rs}}{ ext{.}}frac{{306}}{5} cr} $$ Increase on $$frac{{306}}{5}$$ : = $$left( {100 - frac{{306}}{5}}
ight)$$ = $$frac{194}{5}$$ Increase on 100 : $$eqalign{
& = left( {frac{{194}}{5} imes frac{5}{{306}} imes 100}
ight)\% cr
& = frac{{9700}}{{153}}\% . cr
& = 63.39 approx 63\% cr} $$
300: D
Solution: Let the required time be n years Then, $$ Leftrightarrow 72900 imes {left( {1 + frac{{10}}{{100}}}
ight)^n} = $$ xa0 xa0 xa0$$133100 , imes $$ xa0$${left( {1 - frac{{10}}{{100}}}
ight)^n}$$ $$eqalign{
& Leftrightarrow {left( {frac{{11}}{{10}}}
ight)^n} imes {left( {frac{{10}}{9}}
ight)^n} = frac{{133100}}{{72900}} cr
& Leftrightarrow {left( {frac{{11}}{9}}
ight)^n} = frac{{1331}}{{729}} cr
& Leftrightarrow {left( {frac{{11}}{9}}
ight)^n} = {left( {frac{{11}}{9}}
ight)^3} cr
& Leftrightarrow n = 3 cr} $$
301: C
Solution: Total money : $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {600 imes frac{{25}}{{100}} + 1200 imes frac{{50}}{{100}}}
ight) cr
& = { ext{Rs}}{ ext{. 750}} cr} $$ 25 paise coins removed : $$eqalign{
& = left( {frac{{12}}{{100}} imes 600}
ight) cr
& = 72 cr} $$ 50 paise coins removed : $$eqalign{
& = left( {frac{{24}}{{100}} imes 1200}
ight) cr
& = 288 cr} $$ Money removed : $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {72 imes frac{{25}}{{100}} + 288 imes frac{{50}}{{100}}}
ight) cr
& = { ext{Rs}}{ ext{. 162}} cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{{162}}{{750}} imes 100}
ight)\% cr
& = 21.6\% cr} $$
302: N/A
Solution: Let Mr. More's monthly income be Rs. $$x$$ Then, [100 - (40 + 30)]% of [100 - 20 + 15]% of $$x$$ = 8775 ⇒ 30% of 65% of $$x$$ = 8775 ⇒ $$left( {frac{{30}}{{100}} imes frac{{65}}{{100}} imes x}
ight)$$ xa0 = 8775 ⇒ $$frac{39}{200}$$ $$x$$ = 8775 ⇒ $$x$$ = $$left( {frac{{8775 imes 200}}{{39}}}
ight)$$ ⇒ $$x$$ = 45000
303: C
Solution: Let the original price of an eraser be Rs. x Reduced price = 75% of Rs. x = Rs. $$frac{3x}{4}$$ $$eqalign{
& herefore frac{1}{{left( {frac{{3x}}{4}}
ight)}} - x = 2 cr
& Rightarrow frac{4}{{3x}} - frac{1}{x} = 2 cr
& Rightarrow frac{1}{{3x}} = 2 cr
& Rightarrow 6x = 1 cr
& Rightarrow x = frac{1}{6} cr} $$ Hence, number of erasers available for a rupee = 6
304: N/A
Solution: = (550% of 250) ÷ 275 = $$frac{550}{100}$$ × 250 × $$frac{1}{275}$$ = $$frac{55 × 25}{275}$$ = 5
305: B
Solution: $$frac{4}{5}$$ × 70 = 56 and $$frac{5}{7}$$ × 112 = 80 ∴ Required percentage : $$eqalign{
& = left( {frac{{80 - 56}}{{80}} imes 100}
ight)\% cr
& = left( {frac{{24}}{{80}} imes 100}
ight)\% cr
& = 30\% cr} $$
306: C
Solution: PO initial = 5.3C 2 L 15 PO new = 5.3 (1.2C) 2 L 15 = 1.44 (5.3C 2 L 15 ) = 1.44PO initial Increase in PO = 0.44PO initial ∴ Increase % = (0.44 × 100)% = 44%
307: B
Solution: = 92.5% of 550 = $$frac{925}{10}$$ × $$frac{1}{100}$$ × 550 = 508.75
308: D
Solution: Let 40% of 265 + 35% of 180 = 50% of x Then, $$left( {frac{40}{{100}} imes 265}
ight)$$ xa0 + $$left( {frac{35}{{100}} imes 180}
ight)$$ xa0 = $$left( {frac{50}{{100}} imes x}
ight)$$ ⇒ 106 + 63 = $$frac{x}{2}$$ ⇒ x = 169 × 2 ⇒ x = 338
309: D
Solution: A exceeds b by x% ⇒ a = b + x% of b ⇒ a = b +$$frac{bx}{100}$$
310: C
Solution: = 10% of 5 + 5% of 10 = $$left( {frac{{10}}{{100}} imes 5}
ight)$$ xa0 + $$left( {frac{{5}}{{100}} imes 10}
ight)$$ = 0.5 + 0.5 = 1.0
311: B
Solution: Percentage of failed candidates = (30 + 35 - 27)% = 38% Percentage of passed candidates = (100 - 38)% = 62% Let the total number of students appeared be x Then, 62% of x = 248 ⇒ x = $$frac{248 × 100}{62}$$ ⇒ x = 400
312: D
Solution: Let number of students appeared from school A = 100 Then, number of students qualified from school A = 70 Number of students appeared from school B = 120 Number of students qualified from school B : = $$frac{150}{100}$$ × 70 = 105 ∴ Required percentage : $$eqalign{
& = left( {frac{{105}}{{120}} imes 100}
ight)\% cr
& = 87.5\% cr} $$
313: A
Solution: Pass students in Mathematics = 65% Pass students in Physics = 48% Student pass in both subject Mathematics and Physics = 30% Student only pass in Mathematics = 65 - 30 = 35% Student only pass in Physics = 48 - 30 = 18% Percentage of Failed students in both subjects : = 100 - [ student pass only in Mathematics + student pass only in Physics + student pass in both subject] = [100 - (35 + 18 + 30)] = 17%
314: D
Solution: Let the total number of votes = 100x losser candidate get 38% of vote i.e. = 38x and winner will get = 100x - 38x = 62x According to the question, 62x - 38x = 7200 ⇒ 24x = 7200 ⇒ x = 300 Total votes = 100x = 100 × 300 = 30000
315: C
Solution: According to the question, Mixture of copper and aluminium = 2000 gms 30% copper = $$frac{30}{100}$$ × 2000 = 600 gms Aluminium in mixture = 2000 - 600 = 1400gm Now 'x' weight of mixture have and 600 gm copper become 20% of its total weight. i.e. x = $$frac{{600}}{{20}} imes 100 $$ xa0 = 3000gm Total amount of Aluminium in mixture = 3000 - 600 = 2400 Additional aluminium powder added = 2400 - 1400 = 1000gm
316: A
Solution: $$frac{20}{100}$$ (A + B) = $$frac{50}{100}$$ (B) 2A + 2B = 5B 2A = 3B A = $$frac{3}{2}$$ put value of A in given equation $$frac{2A - B}{2A + B}$$ = $$frac{3B - B}{3B + B}$$ = $$frac{2B}{4B}$$ = $$frac{1}{2}$$
317: B
Solution: Total seats = 10000 Ticket sold = (10000 - 100) = 9900 According to the question, Total revenue = 9900 × $$frac{20}{100}$$ × 10 + 9900 × $$frac{80}{100}$$ × 20 = 9900 × 2 + 9900 × 16 = 9900 (2 + 16) = Rs. 178200
318: C
Solution: By mixture and allegation method. 4 : 1 = 5 units 5 units = 100 1 unit = 20 Pass candidate : = 4 units = 4 × 20 = 80
319: D
Solution: 90% of A = 30% of B 90A = 30B ⇒ B = 3A ..... (i) B = $$frac{2x}{100}$$ × A 3A = $$frac{2x}{100}$$ × A ⇒ x = 150
320: A
Solution: Let the number = x According to the question, ⇒ x × $$frac{18}{100}$$ = $$frac{12}{100}$$ × 75 ⇒ 18x = 12 × 75 ⇒ x = $$frac{12 × 75}{18}$$ ⇒ x = 50 Hence, required number = 50
321: D
Solution: Let Mr.kapur income is 100x House rent 20%, i.e. 20x Remaining income 100x - 20x = 80x Now 70% of his remaining income spend in house hold expenses i.e. 70% of 80x = 56x Remain saving is 80x - 56x = 24x According to question 24x = 1800 x = $$frac{{1800}}{{24}} = 75$$ Mr.kapur income = 100 × 75 = 7500
322: A
Solution: Let the numbers are 2x and 3x respectively According to the question, 2x × $$frac{20}{100}$$ + 20 = 3x × $$frac{10}{100}$$ + 25 $$frac{2}{5}$$x + 20 = $$frac{3}{10}$$ + 25 $$frac{3}{10}$$x - $$frac{2}{5}$$x = - 5 3x - 4x = - 50 x = 50 Hence, required smaller number = 2x = 2 × 50 = 100
323: C
Solution: $$eqalign{
& 12x = frac{{ {75 imes 336} }}{{100}} cr
& x = frac{{ {75 imes 336} }}{{ {100 imes 12} }} cr
& x = 21 cr} $$
324: C
Solution: Let the side of the square shaped hall be X meter
Then, Perimeter = 4X 4X = 400 X = 100m Area of the hall = 100 × 100 = 10000 sq. meter. Now, The cost on total tiles = 10000 × 48 = Rs. 480000 But, 10% damage has occurred on tiles which will also be included in cost i.e Total cost = 480000 + 10% of 480000 = 480000 + 48000
Total cost = Rs. 5,28,000
325: C
Solution: $$eqalign{
& { {frac{{ {125 imes 860} }}{{100}}} + {frac{{75 imes 480}}{{100}}} } cr
& = 1075 + 360 cr
& = 1435 cr} $$
326: C
Solution: Yearly subscription rate = Rs. 1785 Charge for 12 month as rate Rs. 175 per month = 12 × 175 = Rs. 2100 Discount = 2100 - 1785 = Rs. 315 % discount = $$frac{{315 imes 100}}{{2100}}$$ xa0 = 15%
327: B
Solution: Total Passengers = 600 No. of females = $$frac{{600 imes 34}}{{100}}$$ xa0= 204 No. of male passengers = 600 - 204 = 396
Fare of each male = Rs. 20 Fare of female, 15% less, so, = $$frac{{20 imes 75}}{{100}}$$ xa0= Rs. 15 each Total revenue generated by male = 396 × 20 = Rs. 7920
Total revenue generated by female = 204 × 15 = 3060
Total Revenue = 7920 + 3060 = Rs. 10980
328: D
Solution: Minimum pass percentage for Boys = 60%
A boy gets 767 and failed by 313, it means
767 + 313 = 60%
1080 = 60%
60% = 1080 Or, 1% = $$frac{{1080}}{{60}}$$ Or, 45% = $$frac{{1080 imes 45}}{{60}}$$ xa0 = 810 So minimum passing marks for girls = 810
329: A
Solution: 50 = x% of 24 × 60 x = 3.472% x = 3.5% approx.
330: B
Solution: Pass mark = (17 + 10) = 27 Let maximum marks be x Then 36% of x = 27 Or, $$frac{{36{ ext{x}}}}{{100}}$$ = 27 Or, 36x = 2700 Hence, x = 75
331: C
Solution: $$eqalign{
& frac{{100}}{3}\% ,,{ ext{of}},,600 cr
& = frac{{100 imes 600}}{{3 imes 100}} cr
& = 200 cr} $$
332: C
Solution: No. of boys = 75% No. Girls = 25% = 420 Now comparing, 25% = 420 1% = $$frac{{420}}{{25}}$$ So, 75% = $$frac{{420 imes 75}}{{25}}$$ xa0= 1260 No. of boys = 1260
333: A
Solution: $$eqalign{
& { ext{Let numerator}} = 5x cr
& { ext{denominator}} = 5y cr
& { ext{So as per question}} cr
& frac{{5x + 3x}}{{5y + 2y}} = frac{{16}}{{23}} o frac{x}{y} = frac{2}{9} cr
& { ext{So the original fraction}},frac{{5x}}{{5y}} = frac{2}{9} cr} $$
334: D
Solution: Let C marks = 100 More = 120 - 90 = 30 More % $$ = frac{{30}}{{90}} imes 100 = 33frac{1}{3}\% $$
335: C
Solution: $$eqalign{
& 90\% o 216 cr
& 10\% o 24 cr
& 100\% o x08oxed{240} cr} $$
336: C
Solution: $$eqalign{
& 8\% o 480 cr
& 1\% o 60 cr
& 100\% o x08oxed{6000} cr} $$
337: C
Solution: Run by boundaries = 6 × 4 + 4 × 6 = 48 By running = 120 - 48 = 72 % = $$frac{{72}}{{120}}$$ × 100 = 60%
338: B
Solution: [x08egin{array}{*{20}{c}}
{{ ext{Income}}}&{{ ext{Expenditure}}}& = &{{ ext{Saving}}} \
x08egin{gathered}
100 hfill \
,, downarrow hfill \
129 hfill \
end{gathered} &x08egin{gathered}
80 hfill \
, downarrow hfill \
96 hfill \
end{gathered} &{}&{left. x08egin{gathered}
20 hfill \
, downarrow hfill \
33 hfill \
end{gathered}
ight)13}
end{array}] $$frac{{13}}{{20}} imes 100 = 65\% $$ Alternate Solution : First, let's assume Renu's initial income is Rs. 100 (we can pick any number, 100 makes the math easier). If she saves 20%, her initial savings are 20% of Rs. 100 = Rs. 20 . Her initial expenditure is then Rs. 100 (income) - Rs. 20 (savings) = Rs. 80 . Now, her expenditure increases by 20%. So, the increase in expenditure is 20% of Rs. 80 = Rs. 16 . Her new expenditure is Rs. 80 + Rs. 16 = Rs. 96 . Her income also increases, by 29%. So, the increase in income is 29% of Rs. 100 = Rs. 29 . Her new income is Rs. 100 + Rs. 29 = Rs. 129 . Her new savings are now Rs. 129 (new income) - Rs. 96 (new expenditure) = Rs. 33 . To find the percentage increase in savings, we compare the new savings to the old savings. The increase in savings is Rs. 33 (new savings) - Rs. 20 (old savings) = Rs. 13 . The percentage increase in savings is (Increase in savings / Original savings) * 100 = (Rs. 13 / Rs. 20) * 100 = 65% . Therefore, her saving increases by 65% . The correct answer is Option B: 65% .
339: A
Solution: Radius = 5 : 4 Height = 5 : 6 Volume = 5 2 × 5 : 4 2 × 6 Decrease % = $$frac{{29}}{{125}}$$ × 100 = 2.23%
340: A
Solution: x% of 100 = (100 - x) x = 100 - x 2x = 100 x = 50
341: C
Solution: $$eqalign{
& 91\% { ext{ of }}A = 39\% { ext{ of }}B cr
& frac{{91}}{{100}} imes A = frac{{39}}{{100}} imes B cr
& B = frac{{91}}{{39}}A cr
& B = x\% { ext{ of }}A cr
& frac{{91}}{{39}} imes A = frac{x}{{100}} imes A cr
& x = frac{{91}}{{39}} imes 100 = frac{{700}}{3} cr} $$
342: D
Solution: $$eqalign{
& { ext{Reduction of money}} = 450 imes frac{{20}}{{100}} = 90 cr
& { ext{After reduction price}} = frac{{90}}{{50}} = { ext{Rs}}{ ext{. }}1.8{ ext{ per kg}} cr
& { ext{Increase in price }}20\% = frac{1}{5} cr
& { ext{4 unit}} = frac{9}{5} cr
& { ext{1 unit}} = frac{9}{{5 imes 4}} cr
& { ext{5 unit}} = frac{9}{{5 imes 4}} imes 5 = frac{9}{4} = { ext{Rs}}{ ext{. }}2.25 cr} $$
343: B
Solution: We know that ⇒ Total surface area of a cube ⇒ 6a 2 ⇒ If each side is doubled i.e. 2a ⇒ Total surface area of a cube ⇒ 6 × (2a) 2 ⇒ 24a 2 Increase in surface area = 24a 2 - 6a 2 = 18a 2 ∴ % Surface area of cube will increase : $$frac{{18{a^2}}}{{6{a^2}}} imes 100 $$ = 300%
344: A
Solution: Failed students in Mathematics = 35% Failed students in English = 25% Student failed in both subject Mathematics and english = 10% Student only fail in Mathematics = 35 - 10 = 25% Student only fail in English = 25 - 10 = 15% Percentage of passed students in both subjects : = 100 - [ student fail in Mathematics + student fail in english + student fail in both subject] = [100 - (25 + 15 + 10)] = 50%
345: C
Solution: 4% = $$frac{1}{25}$$ [x08egin{gathered}
{ ext{Initial }},,,,,,{ ext{Final}} hfill \
,,,25,,,,,,,,,,,,,,,26 hfill \
,,,25,,,,,,,,,,,,,,,26 hfill \
overline {,,,,,625,,,,,,,,,,,676,,,} hfill \
end{gathered} ] According to the question, 625 units = 50000 1 unit xa0 xa0 = $$frac{50000}{625}$$ = 80 676 units = 80 × 676 = 54080 Hence, population after two years = 54080
346: A
Solution: Note : In such type of questions use allegation method to save your valuable time. Required number of males = $$frac{8000}{(1 + 1)}$$ xa0× 1 = 4000
347: B
Solution: Number of females : = 25000 × $$frac{1}{5}$$ = 5000 Number of males : = 25000 - 5000 = 20000 Number of educated females : = 5000 × $$frac{60}{100}$$ = 3000 Number of educated males = 20000 × $$frac{95}{100}$$ = 19000 Total educated population = 22000 Percentage of educated population = $$frac{22000}{25000}$$ xa0× 100 = 88%
348: D
Solution: The production of cycles rose to 48400 from 40000 in 2 years ⇒ Present production = 40000 ⇒ After two years = 48000 ⇒ Time = 2 years ⇒ Rate of increment = ? According to the question, Production after 2 years = Present production $${left( {1 + frac{{ ext{R}}}{{100}}}
ight)^t}$$ $$eqalign{
& Rightarrow 48400 = 40000{left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow frac{{484}}{{400}} = {left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow 1 + frac{{ ext{R}}}{{100}} = frac{{22}}{{20}} cr
& Rightarrow frac{{ ext{R}}}{{100}} = frac{1}{{10}} cr
& Rightarrow { ext{R}} = 10\% cr} $$ ⇔ Rate of increment = 10%
349: B
Solution: Total marks = (300 + 200) × $$frac{46}{100}$$ = 230 Marks obtained by the students in science = 300 × $$frac{32}{100}$$ = 96 Required marks in language paper = (230 - 96) = 134 Required percentage = $$frac{134}{200}$$ × 100 = 67% Hence, required percentage = 67%
350: A
Solution: Let the initial population = x According to the question, $$eqalign{
& x imes frac{{104}}{{100}} imes frac{{104}}{{100}} = 67600 cr
& x = frac{{67600 imes 100 imes 100}}{{104 imes 104}} cr
& ,,,,,, = 62500 cr} $$ Hence required population = 62500
351: C
Solution: Let total number of student apply in college 100x 40% student alloted in group A , i.e. 40x Remaining Student 100x - 40x = 60x Now 75% of remaining student got alloted in Group B i.e. 75% of 60x = 45x Remain student is 60x - 45x = 15x (which alloted in Group C) According to question 15x = 12 x = $$ frac{12}{15} $$ Total Number of Student applied in college = 100x = 100 x $$ frac{12}{15} $$ = 80
352: D
Solution: Required apples : = $$frac{420}{(100 - 40)}$$ xa0× 100 = $$frac{420}{60}$$ × 100 = 700
353: A
Solution: 64% = $$frac{64}{100}$$ = $$frac{16}{25}$$
354: A
Solution: Given 12% of 555 + 15% of 666 $$eqalign{
& Leftrightarrow (?) = left( {frac{{12}}{{100}} imes 555}
ight) + left( {frac{{15}}{{100}} imes 666}
ight) cr
& ,,,,,,,,,,,,,,,, = left( {66.6 + 99.9}
ight) cr
& ,,,,,,,,,,,,,,,, = 166.5 cr} $$
355: A
Solution: 12% of 5000 = $$frac{12}{100}$$ × 5000 = 600
356: A
Solution: ∵ Krishna's present salary = Rs. 3500 Salary increase by 10% Increased salary of Krishna : = $$frac{3500 × 110}{100}$$ = Rs. 3850
357: D
Solution: Given 32% of 825 + 25% of 1440 = 1025 - (?) (?) = [1025 - (32% of 825) - (25% of 1440)] $$ Rightarrow (?) = 1025 - left( {frac{{32 imes 825}}{{100}}}
ight)$$ xa0 xa0 $$ - left( {frac{{25 imes 1440}}{{100}}}
ight)$$ $$eqalign{
& Rightarrow (?) = left( {1025 - 264 - 360}
ight) cr
& Rightarrow (?) = 401 cr} $$
358: B
Solution: = 38% of 341 = $$frac{38}{100}$$ × 341 = $$frac{12958}{100}$$ = 129.58
359: B
Solution: X% of 0.3 = 0.03 ⇒ $$frac{X}{100}$$ × 0.3 = 0.03 ⇒ X = $$frac{0.03 × 100}{0.3}$$ ⇒ X = 10
360: A
Solution: Given expression : $$eqalign{
& = left( {frac{{28}}{{100}} imes frac{{36}}{{100}} imes frac{5}{7} imes 5000}
ight) cr
& = 360 cr} $$
361: A
Solution: x% of y $$eqalign{
& = left( {frac{x}{{100}} imes y}
ight) cr
& = left( {frac{y}{{100}} imes x}
ight) cr} $$ = y% of x
362: C
Solution: Given, $$frac{4}{3}$$ of 25% $$frac{18}{19}$$ of 57 = ? $$eqalign{
& ? = left[ {left( {frac{{25}}{{100}} imes frac{4}{3}}
ight) imes left( {frac{{18}}{{19}} imes 57}
ight)}
ight] cr
& ? = frac{1}{4} imes frac{4}{3} imes frac{{18}}{{19}} imes 57 cr
& ? = 18 cr} $$
363: D
Solution: $$eqalign{
& { ext{Quantity of salt in 6L of sea water,}} cr
& = frac{{ {6 imes 4} }}{{100}} = 0.24 cr
& { ext{Percentage of salt in 5L of sea water,}} cr
& = frac{{ {0.24 imes 100} }}{5} = 4frac{4}{5}\% cr} $$
364: B
Solution: After first discount,
100 ---- 8%↓ ----> 92 After second discount, 92 ---- 12%↓ ----> 80.96 Single discount = 100 - 80.96 = 19.04%
365: A
Solution: Let there are X books in the library. Number of Hindi books = 60% of X = $$frac{{60{ ext{X}}}}{{100}}$$ xa0= 0.6X Remaining Books = X - 0.6X = 0.4X Number English books = 40% of reaming books = 60% of 0.4X = 0.24X. Urdu Books = X-0.6X -0.24X = 0.16X Given, 0.24X = 3600 X $$ = frac{{3600}}{{0.24}} = 15000$$ Urdu Books = 0.16X = 0.16 × 15000 = 2400
366: C
Solution: Total number of passenger in each compartment = $$frac{{ {25 imes 7} }}{5}$$xa0 = $$35$$ Total berth = 35 2 = 1225 Maximum available capacity $$eqalign{
& = frac{{ {1225 imes 80} }}{{100}} cr
& = 980,{ ext{seats}} cr} $$
367: B
Solution: $$eqalign{
& { ext{Population after two years}}, cr
& = 5000 imes {left[ {1 + {frac{2}{{100}}} }
ight]^2} cr
& = 5202 cr
& { ext{Alternatively}}, cr
& 5000 = = 2\% uparrow Rightarrow 5100 = = 2\% uparrow Rightarrow 5202 cr} $$
368: C
Solution: Let the no. of total student in the class = 100 and number of boy = X
and 12% of the 100 is 12 Number of girl is x - 12 total number of student is x + (x - 12) = 100 therefore x = 56. Then, No of boys = 56
No. of girls = 44
Boys : Girls = 56 : 44 = 14 : 11
369: B
Solution: $$eqalign{
& frac{{ ext{P}}}{{100 + { ext{P}}}} = frac{{ ext{Q}}}{{100}} cr
& { ext{or}},,100left( {{ ext{P}} - { ext{Q}}}
ight) = { ext{PQ}} cr
& { ext{or}},,left( {{ ext{P}} - { ext{Q}}}
ight) = frac{{{ ext{PQ}}}}{{100}} cr} $$
370: C
Solution: Men × Time = Work
100 × 1 = 100 unit work
150 × 1 = 150 unit work
Extra man power = 50 But since, new workers are $$frac{5}{4}$$ time as efficient as existing workers Thus, Actual no. of workers = $$frac{{50}}{{frac{5}{4}}}$$ = 40 workers % required = $$frac{{40 imes 100}}{{100}} = 40\% $$
371: B
Solution: Total number of cubes = 160 + 56 = 216 Thus, side of cube = 6 unit No. of cubes without exposure = (6 - 2) 3 = 64 Thus 64 cubes will be inside of a big cube Now, rest cubes = 160 - 64 = 96 No. of cubes with one face outside = 6 × (4 × 4) = 96 Required % = $$frac{{90 imes 100}}{{216}} = 44.44\% $$
372: B
Solution: $$eqalign{
& 78\% ,{ ext{of}},,750 + 34\% ,{ ext{of}},x = 30\% ,{ ext{of}},,2630 cr
& or,, {frac{{ {78 imes 750} }}{{100}}} + frac{{34x}}{{100}} = left( {30 imes 2630}
ight) imes 100 cr
& or,,78 imes 750 + 34x = 30 imes 2630 cr
& or,,34x = 78900 - 58500 cr
& or,,x = frac{{20400}}{{34}} cr
& { ext{Hence}},,x = 600 cr} $$
373: C
Solution: Number of toffees distributed : = 5% of 160 + 15% of 160 + $$frac{{1}}{{4}}$$ of 160 = $$ left( {frac{5}{{100}} imes 160}
ight) + left( {frac{{15}}{{100}} imes 160}
ight) + $$ xa0 xa0 xa0$$left( {160 imes frac{1}{4}}
ight)$$ = 8 + 24 + 40 = 72 ∴ Number of toffees left behind : = 160 - 72 = 88
374: C
Solution: $$eqalign{
& a = 60\% { ext{ of }}b cr
& Rightarrow a = frac{{60}}{{100}}b cr
& Rightarrow b = frac{5}{3}a cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{{5b}}{{4a}} imes 100}
ight)\% cr
& = left( {5 imes frac{5}{3}a imes frac{1}{{4a}} imes 100}
ight)\% cr
& = left( {frac{{625}}{3}}
ight)\% cr} $$
375: N/A
Solution: $$left( {0.85\% { ext{ of }}405}
ight) + left( {2.25\% { ext{ of }}550}
ight)$$ $$ = left( {frac{{85}}{{100}} imes frac{1}{{100}} imes 405}
ight) + $$ xa0 xa0 $$left( {frac{{225}}{{100}} imes frac{1}{{100}} imes 550}
ight)$$ $$eqalign{
& = frac{{225}}{{10000}}left( {153 + 550}
ight) cr
& = left( {frac{{225 imes 703}}{{10000}}}
ight) cr
& = 15.8175 cr} $$
376: C
Solution: Quantity of gold in the alloy : = 80% of 50 g = $$left( {frac{{80}}{{100}} imes 50}
ight)$$ xa0g = 40 g Let x g of gold be added Then, $$eqalign{
& Rightarrow frac{{40 + x}}{{50 + x}} = frac{{90}}{{100}} cr
& Rightarrow frac{{40 + x}}{{50 + x}} = frac{9}{{10}} cr
& Rightarrow 400 + 10x = 450 + 9x cr
& Rightarrow x = 50 cr} $$
377: C
Solution: Let the number be x Then, $$eqalign{
& Rightarrow frac{1}{8}x = 41.5 cr
& Rightarrow x = 41.5 imes 8 cr
& Rightarrow x = 332 cr
& herefore 69\% ,,{ ext{of}},,332 cr
& = left( {frac{{69}}{{100}} imes 332}
ight) cr
& = 229.08 cr} $$
378: B
Solution: Total annual rent : = Rs. (25000 × 12) = Rs. 300000 Discount : = 5% of Rs. 300000 = Rs. $$left( {frac{5}{{100}} imes 300000}
ight)$$ = Rs. 15000 ∴ Annual rent paid after discount : = Rs. 300000 - 15000 = Rs. 285000
379: D
Solution: $$left( {0.56\% ,,{ ext{of}},,255}
ight) imes $$ xa0 $$left( {3.25\% ,,{ ext{of}},,430}
ight)$$ $$ = left( {frac{{56}}{{100}} imes frac{1}{{100}} imes 225}
ight) imes $$ xa0 xa0 $$left( {frac{{325}}{{100}} imes frac{1}{{100}} imes 430}
ight)$$ $$eqalign{
& = left( {frac{{126}}{{100}} imes frac{{13975}}{{1000}}}
ight) cr
& = 1.26 imes 13.975 cr
& = 17.6085 cr} $$
380: A
Solution: Correct time : = 3 hrs 40 min = (3 × 60 + 40) min = 220 min Error = 5.5 min Error % : = $$left( {frac{{5.5}}{{220}} imes 100}
ight)$$xa0 % = $$frac{5}{2}$$ % = 2.5%
381: B
Solution: Let the number be x Then, $$eqalign{
& left( {100 + 37frac{1}{2}}
ight)\% ,,{ ext{of}},,x = 33 cr
& Rightarrow 137frac{1}{2}\% ,,{ ext{of}},,x = 33 cr
& Rightarrow frac{{275}}{2} imes frac{1}{{100}} imes x = 33 cr
& Rightarrow x = left( {frac{{33 imes 2 imes 100}}{{275}}}
ight) cr
& Rightarrow x = 24 cr} $$
382: C
Solution: Let B's Salary is Rs. 100. Then,
A's Salary = (100 + 25% of 100) = Rs. 125 Difference between A's Salary and B's Salary = 125 - 100 = Rs. 25 % Difference (lower) = $$frac{{25}}{{125}} imes 100 = 20\% $$ Mind Calculation Method: 100(B salary) === 25%↑ ===> 125(A salary) === 20%↓ ===> 100 (B salary) B's salary is 20% lower than A's
383: B
Solution: Net percentage increase in Population = (2.5 - 0.5) = 2% each year. Let the Original Population of the town be 100. Population of Town after 1 year = (100 + 2% of 100) = 102. Population of the town after 2nd year = (102 + 2% of 102 ) = 104.04 Now, % increase in population = $$frac{{4.04}}{{100}} imes 100 = 4.04\% $$ Mind Calculation Method: 100 == 2% Up(1 st year) ==> 102 == 2%Up(2 nd year) ==> 104.04 % population increase in 2 years = 4.04%.
384: D
Solution: Winner gets 65% of valid votes and loser gets 35% of votes Difference between this two = 2748 (65-35)% = 2748 30% = 2748 Total number of voters, 100% = $$frac{{2748 imes 100}}{{30}}$$
= 9160
385: A
Solution: Narayan's saving and rent = 1000 + 1800 = Rs. 2800 Let his monthly income be Rs. 100 30% of his income he spent on education i.e. Rs. 30 Remaining = 100 - 30 = 70 50% of remaining on food = $$frac{{70 imes 50}}{{100}} = { ext{Rs}}{ ext{. 35}}$$ Now, that 35 must be equal to his saving and rent i.e. 35 = 2800 then, 1 = $$frac{{2800}}{{35}}$$ 100 = $$frac{{2800 imes 100}}{{35}} = { ext{Rs}}{ ext{. 8000}}$$ So, his income = Rs. 8000
386: D
Solution: Let Q = 10. Then, P = 60. Q is 50 less than P. Q, % less than P = $$frac{{50}}{{60}} imes 100 = 83.33\% $$ Alternative Method 10 (Q) == (6 times greater) ==> 60(P) == x%↓(Less than Q) ==> 10 (Q) Now, $${ ext{x}} = frac{{50 imes 100}}{{60}} = 83.33\% $$
387: A
Solution: Let the third number be 100. Then, 1 st number = 130 2 nd number = 140 % 1 st number to the 2 nd number $$eqalign{
& = frac{{130 imes 100}}{{140}} cr
& = frac{{650}}{7} cr
& = 92frac{6}{7}\% cr} $$
388: C
Solution: $$eqalign{
& { ext{Let number of men in the population be }}x cr
& { ext{Number of women}} = left( {35000 - x}
ight) cr
& { ext{Increase in the number of men}} cr
& = 6\% ,of,x = frac{{6x}}{{100}} cr
& { ext{Increase in the number of women}} cr
& = left( {3500 - x}
ight) imes frac{4}{{100}} cr
& { ext{Increase in whole population}} cr
& = 36760 - 35000 = 1760 cr
& { ext{Now}}, cr
& frac{{6x}}{{100}} + left[ {left( {35000 - x}
ight) imes frac{4}{{100}}}
ight] = 1760 cr
& left[ {left( {6x - 4x}
ight) + 35000 imes frac{4}{{100}}}
ight] = 1760 cr
& 2x + 35000 imes 4 = 1760 imes 100 cr
& 2x = 176000 - 35000 imes 4 cr
& x = 18000 cr
& { ext{Number}},{ ext{of}},{ ext{men}} = 18000 cr
& { ext{Number}},{ ext{of}},{ ext{women}} cr
& = 35000 - 18000 cr
& = 17000 cr} $$
389: D
Solution: Let each side of the cuboid be 10 unit initially.
Initial Volume of the cuboid, = length * breadth * height = 10 × 10 × 10 = 1000 cubic unit.
After increment dimensions become,
Length = (10 + 10% of 10) = 11 unit.
Breadth = (10 + 20% of 10) = 12 unit. Height = (10 + 50% of 10) = 15 unit. Now, present volume = 11 × 12 × 15 = 1980 cubic unit. Increase in volume = 1980 - 1000 = 980 cubic unit. % increase in volume = $$frac{{980}}{{1000}} imes 100 = 98\% $$ Mind Calculation Method: 100 == 50%↑(height effects) ==> 150 == 20%↑(breadth) ==> 180 == 10%↑(length effects) ==> 198 Change in volume = 98% [We can take net percentage change in any order]
390: C
Solution: Let Rs. 100 be spend on rice initially for 20 kg. As the price falls by 20%, new price for 20 kg rice, = (100 - 20% of 100) = 80 New price of rice = $$frac{{80}}{{20}}$$ = Rs. 4 per kg. Rice can bought now at = $$frac{{100}}{{4}}$$ = 25 kg.
391: D
Solution: $$eqalign{
& 110 imes frac{{left( {100 - x}
ight)}}{{100}} = 50 imes frac{{left( {100 + x}
ight)}}{{100}} cr
& 11left( {100 - x}
ight) = 5left( {100 + x}
ight) cr
& 1100 - 11x = 500 + 5x cr
& 16x = 600 cr
& x = frac{{800}}{{16}} = frac{{75}}{2} = 37frac{1}{2}\% cr
& { ext{Now,}} cr
& { ext{650}} imes frac{{75}}{{200}} o 780 imes frac{{55}}{{200}} cr
& frac{{25}}{2} o 11 cr
& 25 o 22 cr
& = frac{3}{{22}} imes 100 = 14\% cr} $$
392: A
Solution: 400 × 25% + 1260 × 35% + 180 × 27% = 1020 + x 100 + 441 + 486 = 1020 + x 1027 = 1020 + x x = 7
393: B
Solution: Income = Rs. 50,000 [x08egin{array}{*{20}{c}}
{{ ext{Income}}}& = &{50,000}&{70,000} \
{}&{}&{,,,,,, downarrow 60\% }&{,,,,,, downarrow 60\% } \
{{ ext{Expenses}}}& = &{30,000}&{42,000} \
{}&{}&{,,,,,, downarrow 20\% }&{,,,,,, downarrow 30\% } \
{{ ext{Tax}}}& = &{6,000}&{12,600} \
{}&{}&{,,,,,, downarrow 15\% }&{,,,,,, downarrow 20\% } \
{{ ext{Charity}}}& = &{900}&{2,520} \
{{ ext{Saving}}}& = &{overline {underline {,,13,100,,} } }&{overline {underline {,,12,880,,} } }
end{array}] Difference the savings = Rs. 13,100 - Rs. 12,880
= Rs. 220
394: B
Solution: $$eqalign{
& { ext{Let, total income is }}100 cr
& { ext{Transport costs}} = 100 imes frac{{10}}{{100}} = 10 cr
& { ext{Food costs}} = 100 imes frac{{20}}{{100}} = 20 cr
& { ext{Remain}} = 100 - left( {20 + 10}
ight) = 70 cr
& { ext{Clothes costs}} = 70 imes frac{{30}}{{100}} = 21 cr
& { ext{Saving}} = 100 - left( {10 + 20 + 21}
ight) = 49 cr
& { ext{Cost on food and clothes}} = 20 + 21 = 41 cr
& 49 o 26460 cr
& 1 o 540 cr
& herefore 41 o 540 imes 41 = { ext{Rs}}{ ext{. }}22,140 cr} $$
395: B
Solution: Let the salary of A and B be 100x each and C be 100y Money donate by A = 100x × $$frac{{10}}{{100}}$$ = 10x Money donate by B = 100x × $$frac{8}{{100}}$$ = 8x Money donate by C = 100y × $$frac{9}{{100}}$$ = 9y According to the question ⇒ 10x - 8x = 400 ⇒ 2x = 400 ⇒ x = 200 Total donation of A and B = 10x + 8x = 18x ⇒ 18 × 200 = 3600 The total donation by A and B is Rs. 900 more than that of C ⇒ 3600 - 9y = 900 ⇒ 9y = 2700 ⇒ y = 300 Monthly salary of C = 100y = 100 × 300 = 30000 ∴ Monthly salary of C is Rs. 30000
396: B
Solution: Price × Quantity = Expenditure, if expenditure is fix. then, ⇒ 90 kg Rice → Rs. 5400 ⇒ 1 kg Rice → Rs. 60
397: A
Solution: $$eqalign{
& k imes left( {k imes 30\% }
ight) = k imes 270\% cr
& Rightarrow k imes k imes frac{{30}}{{100}} = k imes frac{{270}}{{100}} cr
& Rightarrow k = 9 cr} $$
398: D
Solution: $$eqalign{
& { ext{Let the number be }}x cr
& frac{{x imes 40}}{{100}} = frac{{x imes 60}}{{100}} - 30 cr
& 40x = 60x - 3000 cr
& 20x = 3000 cr
& x = 150 cr
& frac{{x imes 20}}{{100}} = frac{{150 imes 20}}{{100}} = 30 cr} $$
399: B
Solution: $$eqalign{
& { ext{C}} = left( {{ ext{A}} + { ext{B}}}
ight) imes 40\% cr
& = frac{{160 imes 40}}{{100}} cr
& = 64 cr
& { ext{B is more than C}} cr
& = frac{{36}}{{64}} imes 100 cr
& = 56frac{1}{4}\% cr} $$
400: B
Solution: Given: Salary of A is 30% more than salary of B Calculation: Let the salary of B be 100x Salary of A = 100x + 100x × 30% ⇒ 130x Percentage difference = $$frac{{130{ ext{x}} - 100{ ext{x}}}}{{130{ ext{x}}}} imes 100$$ = $$frac{{30{ ext{x}}}}{{130{ ext{x}}}} imes 100$$ = 23.07% ≈ 23.1% ∴ B's salary is 23.1% less than that of A.
401: C
Solution: [x08egin{array}{*{20}{c}}
{{ ext{Income}}}&{{ ext{Saving}}}&{{ ext{Expenditure}}} \
{ Rightarrow 100}&x&{100 - x} \
{ Rightarrow 126}&{1.5x}&{120 - 1.2x}
end{array}] $$eqalign{
& { ext{Now, Income }} - { ext{ Saving}} = { ext{Expenditure}} cr
& 126 - 1.5x = 120 - 1.2x cr
& 6 = 0.3x cr
& x = 20 cr} $$
402: B
Solution: $$eqalign{
& = frac{{28}}{{128}} imes 100 cr
& = 21.875\% cr
& = 21.88\% cr} $$
403: D
Solution: Given: Salary of Mohit is 60% more than Vijay Formula used: Percentage = $$frac{{{ ext{Difference}}}}{{{ ext{Larger value}}}} imes 100$$ Calculations: Let the salary of Vijay = Rs. 100 ⇒ So, the salary of Mohit = 100 + 60% of 100 = 100 + 60 = Rs. 160 According to the formula, $$eqalign{
& Rightarrow { ext{Percentage}} = frac{{{ ext{Difference}}}}{{{ ext{Larger value}}}} imes 100 cr
& = frac{{160 - 100}}{{160}} imes 100 cr
& = frac{{60}}{{160}} imes 100 cr
& = 37.5\% cr} $$ Hence, the salary of Vijay and less than Mohit by 37.5%
404: A
Solution: Let total voters be 100%x Total voters cast = Total votes % - % votes who did not cast vote ⇒ 100%x - 15% ⇒ 85%x Votes got disqualified = 100 Thus, total votes cast = 85%x - 100 Winner got 45% of the total votes cast: Loser got = 45% - 400 According to the question, 85%x - 100 = 45%x + 45%x - 400 ⇒ 300 = 90%x - 85%x ⇒ 300 = 5%x ⇒ x = 6000 ∴ The total number of voters is 6000
405: B
Solution: Required percentage : $$eqalign{
& = left( {frac{{1987.50}}{{2650}} imes 100}
ight)\% cr
& = left( {frac{{19875}}{{265}} imes frac{1}{{100}} imes 100}
ight)\% cr
& = 75\% cr} $$
406: A
Solution: $$eqalign{
& Rightarrow 250{left( {1 + frac{R}{{100}}}
ight)^3} = 2000 cr
& Rightarrow {left( {1 + frac{R}{{100}}}
ight)^3} = frac{{2000}}{{250}} cr
& Rightarrow {left( {1 + frac{R}{{100}}}
ight)^3} = 8 = {left( 2
ight)^3} cr
& Rightarrow left( {1 + frac{R}{{100}}}
ight) = 2 cr
& Rightarrow frac{R}{{100}} = 1 cr
& Rightarrow R = 100 \% cr} $$
407: N/A
Solution: (7.9% of 134) - (3.4% of 79) $$ = left( {frac{{79}}{{10}} imes frac{1}{{100}} imes 134}
ight) - $$ xa0 xa0 $$left( {frac{{34}}{{10}} imes frac{1}{{100}} imes 79}
ight)$$ $$eqalign{
& = frac{{79}}{{1000}} imes left( {134 - 34}
ight) cr
& = frac{{79}}{{1000}} imes 100 cr
& = frac{{79}}{{10}} cr
& = 7.9 cr} $$
408: D
Solution: Let 36% of 365 + x% of 56.2 = 156.69 Then, $$ Rightarrow left( {frac{{36}}{{100}} imes 365}
ight) + left( {frac{x}{{100}} imes 56.2}
ight)$$ xa0 xa0 xa0 $$ = 156.69$$ $$eqalign{
& Rightarrow frac{{281}}{{500}}x = 156.69 - 131.4 cr
& Rightarrow frac{{281}}{{500}}x = 25.29 cr
& Rightarrow x = frac{{25.29 imes 500}}{{281}} cr
& Rightarrow x = 45 cr} $$
409: N/A
Solution: Let Reena's yearly income be Rs. x Anubhab monthly income = Rs. $$frac{240000}{12}$$ = Rs. 20000 Then, ⇒ 25% of x = 75% of 20000 ⇒ $$frac{25}{100}$$x = $$frac{75}{100}$$ × 20000 ⇒ $$frac{x}{4}$$ = 15000 ⇒ x = 60000 ∴ Reena's monthly income : = Rs. $$frac{60000}{12}$$ = Rs. 5000
410: N/A
Solution: = 140% of 56 + 56% of 140 $$eqalign{
& = left( {frac{{140}}{{100}} imes 56}
ight) + left( {frac{{56}}{{100}} imes 140}
ight) cr
& = 78.4 + 78.4 cr
& = 156.8 cr} $$
411: B
Solution: Number of rolls sold by noon = $$frac{1}{2}$$ of 40 dozen = 20 dozen Number of rolls sold between noon and closing time : = 60% of 20 dozen = $$left( {frac{{60}}{{100}} imes 20}
ight)$$ xa0 dozen = 12 dozen Number of rolls left unsold : = [40 - (20 + 12)] dozen = 8 dozen
412: A
Solution: Let 35568 ÷ x% of 650 = 456 Then, $$eqalign{
& Rightarrow 35568 div left( {frac{x}{{100}} imes 650}
ight) = 456 cr
& Rightarrow frac{{13x}}{2} imes 456 = 35568 cr
& Rightarrow 2964x = 35568 cr
& Rightarrow x = frac{{35568}}{{2964}} cr
& Rightarrow x = 12 cr} $$
413: N/A
Solution: $$eqalign{
& frac{5}{6} = left( {frac{5}{6} imes 100}
ight)\% cr
& ,,,,,, = 83frac{1}{3}\% cr
& frac{2}{3} = left( {frac{2}{3} imes 100}
ight)\% cr
& ,,,,,, = 66frac{2}{3}\% cr
& frac{2}{5} = left( {frac{2}{5} imes 100}
ight)\% cr
& ,,,,,, = 40\% cr
& frac{1}{4} = left( {frac{1}{4} imes 100}
ight)\% cr
& ,,,,,, = 25\% cr
& frac{2}{{11}} = left( {frac{2}{{11}} imes 100}
ight)\% cr
& ,,,,,,,,, = 18frac{2}{{11}}\% < 20\% cr} $$
414: C
Solution: Sonal's annual salary = Rs. 108000 ∴ Sonal's monthly salary = Rs. $$left( {frac{{108000}}{{12}}}
ight)$$ = Rs. 9000 Nandini's monthely salary = Rs. (9000 × 2) = Rs. 18000 Let Kaushal's monthly salary be Rs. x Then, 12% of x = 16% of 18000 ⇒ $$frac{{12}}{{100}}$$x = $$left( {frac{{16}}{{100}} imes 18000}
ight)$$ ⇒ $$frac{{12}}{{100}}$$x = 2880 ⇒ x = $$left( {frac{{2880 × 100}}{{12}}}
ight)$$ ⇒ x = 24000
415: D
Solution: Let the number be x
According to the question, 91 - $$frac{{30{ ext{x}}}}{{100}}$$ = x 9100 - 30x = 100x Or, 9100 = 130x Or, x = $$frac{{9100}}{{130}}$$ Hence, x = 70
416: B
Solution: 100000 == 10%↑(1 st year) ==> 110000 == 10%↑(2 nd year) ==> 121000 Population at starting of 3 rd year = 121000 [By using net percentage change method, we have saved our self from cumber some formulas on population which is based on compound interest].
417: C
Solution: Number of Indians men present there = $$frac{{700 imes 20}}{{100}} = 140$$ Indian women = $$frac{{500 imes 40}}{{100}} = 200$$
Indian children = $$frac{{800 imes 10}}{{100}} = 80$$ Total member present in climate conference = 700 + 500 + 800 = 2000 Total Indian = 200 + 140 + 80 = 420 Hence, % of Indian present there = $$frac{{420 imes 100}}{{2000}} = 21\% $$ % of people who were not Indian = 100 - 21 = 79%
418: B
Solution: Let 100 kg of hematite be obtained then 20% of it get wasted that means 80 kg of ore remains. Pure iron = 25% of remaining ore = $$frac{{80 imes 25}}{{100}} = 20,{ ext{kg}}$$ 20 kg pure Iron is obtained from 100 of hematite 1 kg pure Iron is obtained from = $$frac{{100}}{{20}}$$ hematite Then, 80000 kg pure Iron is obtained from =$$,frac{{100}}{{20}} imes 80000 = ,$$ xa0 xa0$$400000,{ ext{kg}}$$ xa0 hematite.
419: D
Solution: The total cost for the year = 250000 + 2% of 2500000 + 2000 = Rs. 257000
For getting return of 15% he must earn = $$frac{{257000 imes 15}}{{100}}$$ = Rs. 38550 per year
Then, Monthly Rent = $$frac{{38550}}{{12}}$$ = Rs. 3212.5
420: C
Solution: Let Ram's salary be x
He spends on rent = 30% of x = $$frac{{30{ ext{x}}}}{{100}}$$
He spends on education = 30% from rest of the salary = $$frac{{30 imes 70{ ext{x}}}}{{100 imes 100}} = frac{{21{ ext{x}}}}{{100}}$$
He Spends on clothes = 24% of total salary = $$frac{{24{ ext{x}}}}{{100}}$$
Saving = 2500 Salary of ram = x $$eqalign{
& frac{{30{ ext{x}}}}{{100}} + frac{{21{ ext{x}}}}{{100}} + frac{{24{ ext{x}}}}{{100}} + 2500 = { ext{x}} cr
& { ext{or,}},frac{{75x}}{{100}} = { ext{x}} - 2500 cr
& { ext{or,}},75{ ext{x}} = 100{ ext{x}} - 250000 cr
& { ext{or, }}100{ ext{x}} - 75{ ext{x}} = 250000 cr
& { ext{or, x}} = frac{{250000}}{{25}} cr
& { ext{or, x}} = 10000 cr
& { ext{Ram's}},{ ext{salary = Rs}}{ ext{.}},10000 cr} $$
421: A
Solution: Let x be the page required when report is retyped Now, we can use work equivalence method 20 × 55 × 65 = 70 × 65 × x Or, x = $$frac{{20 imes 55 imes 65}}{{70 imes 65}}$$ Or, x = 15.70 = 16 pages. Hence, % reduction in pages = $$left( {20 - 16}
ight) imes frac{{100}}{{20}}$$ xa0 $$, = ,$$ $$20\% $$
422: C
Solution: Let initial price of Maruti Car be Rs. 100 As price increases 30%, price of car will become, (100 + 30% of 100) = Rs. 130 Due to increase in price, sales is down by 20%. It means, it is going make 20% less revenue as expected after increment of price So, New revenue = (130 - 20% of 130) = Rs. 104 The initial revenue was Rs. 100 which becomes Rs. 104 at the end. It means there is 4% increment in the total revenue Mind Calculation Method: 100 == 30%↑(price effect) ==> 130 == 20%↓(sales effects) ==> 104 Hence, 4% rises
423: C
Solution: Required percentage : $$eqalign{
& = left( {frac{{1.14}}{{1.9}} imes 100}
ight)\% cr
& = left( {frac{{114}}{{190}} imes 100}
ight)\% cr
& = 60\% cr} $$
424: C
Solution: Let the number be x Then, $$eqalign{
& 42\% ,,{ ext{of}},,x = 892.50 cr
& Rightarrow frac{{42x}}{{100}} = 892.50 cr
& Rightarrow x = left( {frac{{892.5 imes 100}}{{42}}}
ight) cr
& Rightarrow x = 2125 cr
& herefore 73\% ,,{ ext{of}},,2125 cr
& = left( {frac{{73}}{{100}} imes 2125}
ight) cr
& = 1551.25 cr} $$
425: A
Solution: Rebate on one ball : = 25% of Rs. 32 = Rs. $$left( {frac{{25}}{{100}} imes 32}
ight)$$ = Rs. 8 ∴ Required number of balls : = $$frac{40}{8}$$ = 5
426: C
Solution: Lets 85% of 485.5 = 50% of x Then, $$eqalign{
& Rightarrow left( {frac{{85}}{{100}} imes 485.5}
ight) = frac{{50x}}{{100}} cr
& Rightarrow frac{x}{2} = 412.675 cr
& Rightarrow x = 825.35 cr} $$
427: C
Solution: Let Vinay's salary be Rs. $$x$$ Then, 75% of 5% of $$x$$ = 1687.50 ⇒ $$frac{75}{100}$$ × $$frac{5}{100}$$ × $$x$$ = 1687.50 ⇒ $$x$$ = $$left( {frac{{1687.50 imes 100 imes 100}}{{75 imes 5}}}
ight)$$ ⇒ $$x$$ = 45000
428: C
Solution: Let the number be x Then, 64% of x = 2592 ⇒ $$frac{64x}{100}$$ = 2592 ⇒ x = $$left( {frac{{2592 imes 100}}{{64}}}
ight)$$ ⇒ x = 4050 ∴ 88% of 4050 : = $$left( {frac{{88}}{{100}} imes 4050}
ight)$$ = 3564
429: B
Solution: Let the maximum marks be $$x$$ Then, 76% of $$x$$ + 480 = x ⇒ $$x$$ - 76% of $$x$$ = 480 ⇒ 24% of $$x$$ = 480 ⇒ $$frac{24x}{100}$$ = 480 ⇒ $$x$$ = $$left( {frac{{480 imes 100}}{{24}}}
ight)$$ ⇒ $$x$$ = 2000 ∴ Marks score by Rajan : = 76% of 2000 = $$left( {frac{{76}}{{100}} imes 2000}
ight)$$ = 1520
430: D
Solution: Clearly, 60% of 28% of 240 : $$eqalign{
& = left( {frac{{60}}{{100}} imes frac{{28}}{{100}} imes 240}
ight) cr
& = left( {frac{{30}}{{100}} imes frac{{28}}{{100}} imes 2 imes 240}
ight) cr
& = left( {frac{{30}}{{100}} imes frac{{28}}{{100}} imes 480}
ight) cr
& = 30\% ,,{ ext{of}},,285\% ,,{ ext{of}},,480 cr} $$
431: B
Solution: Number of cases required to be tested : $$eqalign{
& = 5\% ,,{ ext{of}},,120 + 10\% ,,{ ext{of}},,80 cr
& = left( {frac{5}{{100}} imes 120}
ight) + left( {frac{{10}}{{100}} imes 80}
ight) cr
& = 6 + 8 cr
& = 14 cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{{14}}{{120 + 80}} imes 100}
ight)\% cr
& = left( {frac{{14}}{{200}} imes 100}
ight)\% cr
& = 7\% cr} $$
432: A
Solution: Consumption of gas in the smaller burner in 1 hour : $$eqalign{
& = left( {frac{{14.4}}{{104}}}
ight){ ext{kg}} cr
& = frac{9}{{65}}{ ext{kg}} cr} $$ Consumption of gas in the larger burner in 1 hour : $$eqalign{
& = left( {frac{{14.4}}{{80}}}
ight){ ext{kg}} cr
& = frac{9}{{50}}{ ext{kg}} cr} $$ Difference in consumption : $$eqalign{
& = left( {frac{9}{{50}} - frac{9}{{65}}}
ight){ ext{kg}} cr
& = frac{{27}}{{650}}{ ext{kg}} cr} $$ Required percentage difference : $$eqalign{
& { ext{ = }}left( {frac{{27}}{{650}} imes frac{{50}}{9} imes 100}
ight)\% cr
& = left( {frac{{300}}{{13}}}
ight)\% cr
& = 23.07\% cr} $$
433: A
Solution: Let V denote the volume of 1 c.c. of water Then, volume of ice formed from it : = 109% of V = $$frac{109}{100}$$V When this ice change into water, decrease in volume : = $$frac{109}{100}$$V - V = $$frac{9}{100}$$V ∴ Decrease % : $$eqalign{
& = left( {frac{{9V}}{{100}} imes frac{{100}}{{109V}} imes 100}
ight)\% cr
& = frac{{900}}{{109}}\% cr
& = 8frac{{28}}{{109}}\% cr} $$
434: A
Solution: Percentage annual increase $$eqalign{
& = left( {frac{1}{8} imes 100}
ight)\% cr
& = frac{{25}}{2}\% cr} $$ Height after $$2frac{1}{2}$$ years $$eqalign{
& = left[ {8{{left( {1 + frac{{25}}{{2 imes 100}}}
ight)}^2}left( {1 + frac{{25}}{{4 imes 100}}}
ight)}
ight]m cr
& = left( {8 imes frac{9}{8} imes frac{9}{8} imes frac{{17}}{{16}}}
ight)m cr
& = left( {frac{{1377}}{{128}}}
ight)m cr
& = 10.75,,m cr} $$
435: A
Solution: Let x litres of 30% alcohol solution be added. Then, 30% of x + 60% of 40 = 50% of (x + 40) ⇒ 30x + 60 × 40 = 50 (x + 40) ⇒ 30x + 2400 = 50x + 2000 ⇒ 20x = 400 ⇒ x = 20
436: C
Solution: Give (180% of ?) ÷ 2 = 504 ⇒ ? × $$frac{180}{100}$$ × 2 = 504 ⇒ ? × $$frac{180}{100}$$ = 504 × 2 ⇒ ? × $$frac{180}{100}$$ = 1008 ⇒ ? = $$frac{1008 × 100}{180}$$ xa0 = 560
437: B
Solution: Marked price of an article = Rs. 2400 According to the question, 2400 × (100 - x)% of 85% = 1876.80 ⇒ 2400 × $$frac{100 - x}{100}$$ × $$frac{85}{100}$$ = 1876.80 ⇒ (100 - x) = $$frac{1876.80 × 100 × 100}{2400 × 85}$$ ⇒ (100 - x) = $$frac{18768000}{204000}$$ ⇒ (100 - x) = 92 ⇒ x = 100 - 92 ⇒ x = 8%
438: B
Solution: Required percentage : $$eqalign{
& = left( {frac{5}{{2250}} imes 100}
ight)\% cr
& = frac{2}{9}\% cr} $$
439: C
Solution: Number of sweets obtained by each student : = 20% of 65 = 13 Number of sweets obtained by each teacher : = 40% of 65 = 26 ∴ Total number of sweets = 65 × 13 + 4 × 26 = 845 + 104 = 949
440: D
Solution: Let the number be x Then, 35% of x = 175 ⇔ $$frac{35}{100}$$ × x = 175 ⇔ $$frac{175 × 100}{35}$$ xa0 = 500 Now, let y% of 175 = 500 Then, $$frac{y}{100}$$ × 175 = 500 ⇔ y = $$frac{500 × 100}{175}$$ ⇔ y = $$frac{2000}{7}$$ ⇔ y = $$285frac{5}{7}$$
441: B
Solution: Let the number be x Then, x - 25% of x = 225 ⇒ x - $$frac{25}{100}$$x = 225 ⇒ $$frac{75x}{100}$$ = 225 ⇒ x = $$frac{225 × 100}{75}$$ ⇒ x = 300 Required increase : = (390 - 300) = 90 ∴ Increase % : $$eqalign{
& = left( {frac{{90}}{{300}} imes 100}
ight)\% cr
& = 30\% cr} $$
442: B
Solution: Let the maximum marks be = x Then, 5% of x = 296 - 259 ⇒ $$frac{5x}{100}$$ = 37 ⇒ x = $$frac{37 × 100}{5}$$ ⇒ x = 740
443: D
Solution: Let 45% of 300 + $$sqrt {x} $$ = 56% of 750 - 10% of 250 Then, $$left( {frac{{45}}{{100}} imes 300}
ight) + sqrt x = left( {frac{{56}}{{100}} imes 750}
ight)$$ xa0 xa0 xa0 $$ - left( {frac{{10}}{{100}} imes 250}
ight)$$ $$eqalign{
& Rightarrow 135 + sqrt x = 420 - 25 cr
& Rightarrow sqrt x = 260 cr
& Rightarrow x = {left( {260}
ight)^2} cr
& Rightarrow x = 67600 cr} $$
444: D
Solution: Percentage increase : $$eqalign{
& = left( {frac{{21}}{{560}} imes 100}
ight)\% cr
& = frac{{15}}{4}\% cr
& = 3.75\% cr} $$
445: C
Solution: Let the number be x Then, $$eqalign{
& Rightarrow frac{{x + 75\% { ext{ of }}x + 25\% { ext{ of }}x}}{3} = 240 cr
& Rightarrow x + frac{{75}}{{100}}x + frac{{25}}{{100}}x = 240 imes 3 cr
& Rightarrow x + frac{3}{4}x + frac{1}{4}x = 720 cr
& Rightarrow 2x = 720 cr
& Rightarrow x = 360 cr} $$
446: B
Solution: Number of runs made by running = 110 - (3 × 4 + 8 × 6) = 50 ∴ Required percentage : $$eqalign{
& = left( {frac{{50}}{{110}} imes 100}
ight)\% cr
& = 45frac{5}{{11}}\% cr} $$
447: C
Solution: Required percentage $$eqalign{
& = left( {frac{{0.01}}{{0.1}} imes 100}
ight)\% cr
& = left( {frac{1}{{10}} imes 100}
ight)\% cr
& = 10\% cr} $$
448: A
Solution: Let the total number of votes be x Then, votes polled = 90% of x Valid votes = 90% of (90% of x) ∴ 54% of [90% of (90% of x)] - 46% of [90% of (90% of x)] = 1620 ⇔ 8% of [90% of ((90% of x)] = 1620 ⇔ $$frac{8}{100}$$ × $$frac{90}{100}$$ × $$frac{90}{100}$$ × x = 1620 ⇔ x = $$left( {frac{{1620 imes 100 imes 100 imes 100}}{{8 imes 90 imes 90}}}
ight)$$ ⇔ x = 25000
449: B
Solution: Let x - 6% of x = xz Then, ⇔ 94% of x = xz ⇔ $$frac{94x}{100}$$ × $$frac{1}{x}$$ = z ⇔ z = 0.94
450: A
Solution: Let the total number of children be x Then, ⇔ x × (20% of x) = 605 ⇔ $$frac{1}{5}$$x 2 = 605 ⇔ x 2 = 3025 ⇔ x = 55 ∴ Number of sweets received by cash child = 20% of 55 = 11
451: B
Solution: Asha's monthly income : = 60% of 78000 $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {frac{{60}}{{100}} imes 78000}
ight) cr
& { ext{ = Rs}}{ ext{. 46800}} cr} $$ Let Maya's monthly income be Rs. x Then, 120% of x = 46800 $$eqalign{
& Rightarrow { ext{x}} = left( {frac{{46800 imes 100}}{{120}}}
ight) cr
& Rightarrow { ext{x}} = 39000 cr} $$
452: B
Solution: Let the distance be x km and the original time time taken be y hours. Then, Original speed = $$frac{x}{y}$$ km/hr New speed : $$eqalign{
& = left( {frac{x}{{80\% { ext{ of }}y}}}
ight){ ext{ km/hr}} cr
& = left( {frac{5}{4}.frac{x}{y}}
ight){ ext{ km/hr}} cr} $$ Increase in speed : $$eqalign{
& { ext{ = }}left( {frac{{5x}}{{4y}} - frac{x}{y}}
ight) cr
& { ext{ = }}frac{x}{{4y}} cr} $$ ∴ Increase % $$eqalign{
& { ext{ = }}left( {frac{x}{{4y}} imes frac{y}{x} imes 100}
ight)\% cr
& = 25\% cr} $$
453: D
Solution: Let us assume there are 100 students in the institute.
Then, number of boys = 60 And, number of girls = 40 Further, 15% of boys get fee waiver = 9 boys 7.5% of girls get fee waiver = 3 girls Total = 12 students who gets fee waiver But, here given 90 students are getting fee waiver. So we compare 12 = 90 So, 1 = $$frac{{90}}{{12}}$$ = 7.5 Now number of students who are not getting fee waiver = 51 boys and 37 girls 50% concession = 25.5 boys and 18.5 girls (i.e. total 44) Hence, required students = 44 × 7.5 = 330
454: A
Solution: We solve it through options. Choosing options for trial depends on mental thought. 100 == 12% up ==> 112 == 12% up ==> 125.44 == 12% Up ==> 140.49 So, answer is 12%
455: C
Solution: Let the marks obtained in five subjects be 6x, 7x, 8x, 9x and 10x. Total marks obtained = 40x Max. Marks of the five subjects = $$frac{{40{ ext{x}}}}{{0.6}}$$ [40x is 60% of total marks] Max. Marks in each subject = $$frac{{40{ ext{x}}}}{{0.6 imes 5}}$$ xa0 = 13.33x Hence, % of each subject = $$frac{{6{ ext{x}} imes 100}}{{13.33}}$$ xa0 = 45.01% Or, $$frac{{7{ ext{x}} imes 100}}{{13.33}}$$ xa0 = 52.51 In same way other percentage are 60.01%, 67.52%, 75.01%. Hence, number of subjects in which he gets more than 50% marks = 4
456: N/A
Solution: Let length, breadth and height of the room be 3, 2, 1 unit respectively. Area of walls = 2(l + b) × h = 2(3 + 2) × 1 = 10 sq. unit. Now, length, breadth and height of room will become 6, 1 and $$frac{1}{2}$$ respectively. Now, area of walls = $$2left( {6 + 1}
ight) imes frac{1}{2}$$ xa0 = 7 sq. unit. % decrease in the area of walls = $$left( {10 - 7}
ight) imes frac{{100}}{{10}}$$ xa0 = 30%
457: A
Solution: Let the number of bacteria in the 1 st generation be x, then number of bacteria in 2 nd , 3 rd , 4 th . . . . . Generation would be $$8left( {frac{{ ext{x}}}{2}}
ight),,8left( {frac{{4{ ext{x}}}}{2}}
ight),,8left( {frac{{16{ ext{x}}}}{2}}
ight)$$ xa0 xa0 . . . . And so on. As x, 4x, 16x, 64x . . . . . it is in GP with common ratio 4 Hence, 7th term of GP, x(4) 6 = 4096 or, x = 1 or 1 million.
458: C
Solution: Increase in the price of sugar = (8 + 2) = 10% Hence, price of sugar on, Jan. 1, 1996 $$eqalign{
& = frac{{20 imes 110 imes 110}}{{100 imes 100}} cr
& = { ext{Rs}}{ ext{.}},24.20 cr} $$
459: B
Solution: Passed in none = 5% Passed in all = 5% Passed in four = 20% of 90% = 18% Passed in one = 25% of 90% = 22.5% Passed in two = 24.5% Passed in three = (100 - 5 - 5 - 22.5 - 24.5 - 18) = 25% But given 300 students passed in three Hence, 25% = 300 So, 100% = 1200 1200 students must have appeared
460: B
Solution: Time lost over two weeks = 25% a week time(given that $$frac{1}{2}$$ % clock loses in first week and in the second week it gains $$frac{1}{4}$$ % on true time) A week = 168 hours Hence, clock lost = 0.42 hours = 25.2 minutes or 25 minute 12 seconds Thus, correct time = 11 : 34 : 48
461: B
Solution: $$eqalign{
& { ext{Shrinking of cloth}}, cr
& = {frac{{ {36 - 33} }}{{36}}} imes 100 cr
& = frac{{100}}{{12}}\% cr
& { ext{Second time the strip shrinks,}} cr
& = frac{{ {48 imes 100} }}{{1200}} cr
& = 4, ext{inches} cr
& { ext{hence,}},{ ext{the}},{ ext{cloth}},{ ext{remains}} cr
& = 48 - 4 cr
& = 44 cr} $$
462: C
Solution: $$eqalign{
& { ext{According}},{ ext{to}},{ ext{question,}} cr
& {frac{{XY}}{{100}}} + {frac{{YX}}{{100}}} cr
& = frac{{2XY}}{{100}} cr
& = 2\% ,of,XY cr} $$
463: C
Solution: Required % = $$frac{2}{50}$$ × 100 = 4%
464: B
Solution: Let the number be x ⇒ x × $$frac{15}{100}$$ × $$frac{45}{100}$$ = 105.3 ⇒ x = 1560 ⇒ Required answer = $$frac{24}{100}$$ × 1560 = 374.4
465: D
Solution: 60% of A = $$frac{3}{4}$$B $$frac{3}{5}$$A = $$frac{3}{4}$$B $$frac{A}{B}$$ = $$frac{5}{4}$$ ⇒ A : B = 5 : 4
466: A
Solution: Percentage of failed students = (100 - 93)% = 7% According to the question, 7% → 259 1% → 37 100% → 3700 Total students = 3700
467: C
Solution: 75% = $$frac{3}{4}$$ Let the number = 4x According to the question, 4x × $$frac{3}{4}$$ + 75 = 4x x = 75 Number = 75 × 4 = 300 Required answer : = 300 × $$frac{40}{100}$$ = 120
468: D
Solution: $$frac{15}{100}$$ (A + B) = $$frac{25}{100}$$ (A - B) ⇒ 15A + 15B = 25A - 25B ⇒ 10A = 40B ⇒ A = 4B Required % : = $$frac{A}{B}$$ × 100 = $$frac{4B}{B}$$ × 100 = 400%
469: D
Solution: Note : In percentage always assume data. Which make your Calculation easier. 60% $$frac{3}{5}$$ Let the number = 5x Accounting to the question, ⇒ 5x × $$frac{3}{5}$$ - 60 = 60 ⇒ x = $$frac{120}{3}$$ ⇒ x = 40 Hence, required number : = 5x = 5 × 40 = 200
470: B
Solution: Marks obtained by D = 320 Marks obtained by C = 320 × $$frac{125}{100}$$ = 400 Marks obtained by B = 400 × $$frac{(100 - 10)}{100}$$ xa0 = 360 Marks obtained by A = 360 × $$frac{125}{100}$$ = 450 Hence, required marks obtained by A = 450
471: A
Solution: Let initial expenditure = 100 units Required increment = $$frac{10}{90}$$ = $$frac{1}{9}$$ 1 unit = 10 apples Original consumption = 9 units = 9 × 10 = 90 apples New consumption = 10 units = 10 × 10 = 100 apples New price = $$frac{54}{100}$$ × 12 = Rs. 6.48 per dozen
472: D
Solution: $$eqalign{
& nleft( A
ight) = left( {frac{{75}}{{100}} imes 600}
ight) = 450 cr
& nleft( B
ight) = left( {frac{{45}}{{100}} imes 600}
ight) = 270 cr
& herefore nleft( {A cap B}
ight) cr} $$ $$ = nleft( A
ight) + nleft( B
ight) - nleft( {A cup B}
ight)$$ xa0 xa0 $$nleft( {A cup B}
ight)$$ xa0 $$ -, 600$$ $$eqalign{
& = left( {450 + 270 - 600}
ight) cr
& = 120 cr} $$
473: B
Solution: Let original income = Rs. 100 Then, saving = Rs. 20 and expenditure = Rs. 80 New income : $$=$$ Rs. $$116frac{2}{3}$$ $$=$$ Rs. $$frac{350}{3}$$ New expenditure : $$=$$ $$137frac{1}{2}$$ % of Rs. 80 $$=$$ Rs. $$left( {frac{{275}}{2} imes frac{1}{{100}} imes 80}
ight)$$ $$=$$ Rs. 110 New saving : $$eqalign{
& = { ext{Rs}}{ ext{.}}left( {frac{{350}}{3} - 110}
ight) cr
& = { ext{Rs}}{ ext{.}}frac{{20}}{3} cr} $$ ∴ Required percentage $$eqalign{
& = left( {frac{{20}}{3} imes frac{3}{{350}} imes 100}
ight)\% cr
& = frac{{40}}{7}\% cr
& = 5frac{5}{7}\% cr} $$
474: A
Solution: Let the total number of candidates be x Number of capable candidates who got admitted to college : = 80% of 40% of x $$eqalign{
& = left( {frac{{80}}{{100}} imes frac{{40}}{{100}} imes x}
ight) cr
& = frac{{8x}}{{25}} cr} $$ Number of incapable college students : = 25% of (100 - 40)% of x $$eqalign{
& = left( {frac{{25}}{{100}} imes frac{{60}}{{100}} imes x}
ight) cr
& = frac{{3x}}{{20}} cr} $$ Total number of candidates who got admitted to college : $$=$$ $$frac{{8x}}{{25}}$$ + $$frac{{3x}}{{20}}$$ $$=$$ $$frac{{47x}}{{100}}$$ ∴ Required percentage : $$eqalign{
& = left( {frac{{8x}}{{25}} imes frac{{100}}{{47x}} imes 100}
ight)\% cr
& = left( {frac{{3200}}{{47}}}
ight)\% cr
& = 68.09\% approx 68\% cr} $$
475: B
Solution: Required percentage : $$eqalign{
& = left( {frac{{18}}{{7200}} imes 100}
ight)\% cr
& = frac{1}{4}\% cr
& = 0.25 \% cr} $$
476: B
Solution: Let x% of 450 + 46% of 285 = 257.1 $$ Rightarrow left( {frac{x}{{100}} imes 450}
ight) + left( {frac{{46}}{{100}} imes 285}
ight)$$ xa0
xa0 $$ = 257.1$$ $$eqalign{
& Rightarrow frac{{9x}}{2} = 257.1 - 131.1 cr
& Rightarrow frac{{9x}}{2} = 126 cr
& Rightarrow x = frac{{126 imes 2}}{9} cr
& Rightarrow x = 28 cr} $$
477: B
Solution: Error = (45° 27' - 45° ) = 27' Accurate measure = 45° = (45 × 60)' = 2700' ∴ Percentage error : $$eqalign{
& = left( {frac{{27}}{{2700}} imes 100}
ight)\% cr
& = 1\% cr} $$
478: B
Solution: 15% of 578 + 22.5% of 644 $$ = left( {frac{{15}}{{100}} imes 578}
ight) + $$ xa0 $$left( {frac{{225}}{{10}} imes frac{1}{{100}} imes 644}
ight)$$ $$eqalign{
& = 86.7 + 144.9 cr
& = 231.6 cr} $$
479: D
Solution: 105.27% of 1200.11 + 11.80% of 2360.85 = 21.99% of (?) + 1420.99 ⇒ 105% of 1200 + 12% of 2360 = 22% of (?) + 1421 ⇒ $$frac{{105 imes 1200}}{{100}} + frac{{12 imes 2360}}{{100}} = $$ xa0 xa0 $$frac{{22 imes (?)}}{{100}}$$ xa0 $$ + ,1421$$ $$ Rightarrowfrac{{22 imes (?)}}{{100}} = 1260 + 283.30$$ xa0 xa0 xa0 $$ - ,1421$$ $$eqalign{
& Rightarrow frac{{22 imes (?)}}{{100}} = 1543 - 1421 cr
& Rightarrow frac{{22 imes (?)}}{{100}} = 122 cr
& herefore (?) = frac{{122 imes 100}}{{22}} cr
& Rightarrow (?) = frac{{122 imes 50}}{{11}} cr
& Rightarrow (?) = 554.5,, (approximate,, 550) cr} $$
480: C
Solution: Percentage of Non-tax paying employees = (100 - 31)% = 69% 69% of total employees = 20700 Total employees = $$frac{20700}{69}$$ × 100 = 30000
481: D
Solution: According to the question, $$frac{30A}{100}$$ + $$frac{40B}{100}$$ = $$frac{80B}{100}$$ 30A = 40B ⇒ 3A = 4B ⇒ A = $$frac{4B}{3}$$ Required %
= $$frac{B}{A}$$ × 100 = $$frac{B × 3}{4B}$$ × 100 = 75%
482: B
Solution: Let us consider total number of goats = 1000x ∴ After loss of 12% of goat in flood, remain goats is = 1000x - 12% of 1000x = 1000x - 120x = 880x Now 5% goats are died from diseases, Remain goats is = 880x - 5% of 880x = 880x - 44x = 836x Total number of remain goats = 8360 ∴ 836x = 8360 ⇒ x = 10 ∴ Number of Goats = 1000 × 10 = 10000 Alternate Solution: Let the number of goats before flood be x According to the question, $$eqalign{
& { ext{x}} imes frac{{88}}{{100}} imes frac{{95}}{{100}} = 8360 cr
& Rightarrow x = frac{{8360 imes 100 imes 100}}{{88 imes 95}} cr
& Rightarrow x = 10000 cr} $$
483: D
Solution: Difference of % = 82.5% - 62.5% = 20% Now, 1% = 100 points So, 20% = 20 × 100 = 2000 points
484: A
Solution: Required percentage $$eqalign{
& = frac{{0.01}}{{0.1}} imes 100 cr
& = 10\% cr} $$
485: D
Solution: Let the number is = x According to the question, $$frac{1}{5}$$ of $$frac{1}{2}$$ of x = 20 $$frac{1}{5}$$ × $$frac{1}{2}$$ × x = 20 x = 200 ∴ 20% of 200 : = $$frac{20}{100}$$ × 200 = 40
486: B
Solution: 20% = $$frac{1}{5}$$ = $$frac{{6,, o ,,{ ext{Girls}}}}{{5,, o ,,{ ext{Boys}}}}$$ Boys : Girls 5 : 6 According to the question, (5 + 6) units = 66 11 units = 66 1 unit = 6 Hence Boys = 6 × 5 = 30 Girls = 6 × 6 = 36 The number of girls when 4 is admitted = (36 + 4) = 40 Required ratio : = 30 : 40 = 3 : 4
487: C
Solution: According to the question, P × $$frac{P}{100}$$ = 36 ⇒ P 2 = 3600 ⇒ P = 60
488: C
Solution: A = 20% of B A = $$frac{1}{5}$$ B $$frac{A}{B} = frac{1}{5}$$ ----(i) B = 25% of C B = $$frac{1}{4}$$ C $$frac{B}{C} = frac{1}{4}$$ ----(ii) Multiply (i) and (ii) We get $$frac{A}{B} imes frac{B}{C} = frac{1}{5} imes frac{1}{4}$$ $$frac{A}{C} = frac{1}{{20}}$$ Percent of C is equal to A $$ = frac{1}{{20}} imes 100 = 5\% $$
489: A
Solution: We find a percentage value of so we take or not take real value does not matters % Profit = $$11frac{1}{9}$$% = $$frac{1}{9}$$ $$frac{{1,, o ,,{ ext{Profit}}}}{{2,, o ,,{ ext{CP}}}}$$ So, SP = 10 Means in previous at the time of beginning the discount is 1 unit in 10 units MP So, discount is = $$frac{1}{10}$$ × 100 = 10%
490: C
Solution: 20 litres juice contain 10% Tomato, i.e.
20L juice = $$frac{{20 imes 10}}{{100}}$$xa0 = 2L Tomato Tomato puree contains 80% of water and 20% tomato. This 80% tomato = 2L (which is contained by 100 puree) So,
Now this 2 L Consist 80% in puree Thus, total puree will be $$frac{2}{{.8}} = 2.5{ ext{L}}$$
491: N/A
Solution: If we take different values of x then, Let x = 150, then error% $$eqalign{
& = frac{{ {frac{{ {200 - 100} }}{{100}}} }}{{100}} cr
& = 100\% cr} $$ Again if x = 100, then error% $$eqalign{
& = frac{{ {left( {150 - 50}
ight) imes 100} }}{{100}} cr
& = 200\% cr} $$ If we take different value of x, we get different answer so we can't determine it.
492: N/A
Solution: Total student = 100
Boys = 60
Girls = 40
Boys who plays hockey = 40% = 24
There is no information about boys who play badminton.
Girls who plays Badminton = 75% = 30
No girls plays hockey. Since, we do not have information that whether the rest of the boys are playing badminton or not. So, we cannot determine the total no. of student who don't play any game.
493: C
Solution: $$eqalign{
& { ext{Let the required number of pages be }}x. cr
& 30 imes 25 imes 35 = x imes 30 imes 28 cr
& x = 31.25 approx 32 cr
& \% ,{ ext{increase}},{ ext{in}},{ ext{number}},{ ext{of}},{ ext{pages}}, cr
& = {frac{2}{{30}}} imes 100 cr
& = 6.66\% cr} $$
494: C
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{fraction}},{ ext{be}},frac{{100x}}{{100y}} cr
& { ext{Now}},{ ext{according}},{ ext{to}},{ ext{the}},{ ext{question}}, cr
& {left( {frac{{100x}}{{100y}}}
ight)^2} = frac{{125{x^2}}}{{80{y^2}}} = frac{{25{x^2}}}{{16{y^2}}} cr
& frac{{25{x^2}}}{{16{y^2}}} = frac{5}{8}left( {frac{{100x}}{{100y}}}
ight) cr
& {frac{{100x}}{{100y}}} = frac{2}{5} cr
& { ext{hence}}, cr
& { ext{product of numerator and denominator}} cr
& = 2 imes 5 = 10 cr} $$
495: C
Solution: Let X be the number which is added to 80 80% of X = 0.8X Now, 80 + 0.8X = X 0.2X = 80 X = $$frac{{80}}{{0.2}} = 400$$
496: C
Solution: $$eqalign{
& { ext{Reduction}} cr
& = {frac{7}{{107}}} imes 2568 cr
& = 168 cr} $$
497: N/A
Solution: Since, Run scored = Over × Run rate If overs is reduced by 25%, run rate will go up by 33.33%. Hence, Australia could have scored any number of runs.
498: C
Solution: In 2000, the market share was 40%, 30% and 30%, means the ratio is 4 : 3 : 3 In 2001, a new product (A) enters and has 10% market share, 90% of the remaining market is shared by the previous 3. Now, divide 90% in the ratio 4 : 3 : 3 , i.e 36%, 27%, 27%. Now the ratio is 36 : 27 : 27 : 10 In 2002, another new product (B) enters and has 10% market share, now the remaining 90% market share is distributed in the ratio, 36 : 27 : 27 : 10 and hence remaining 32.4%, 24.3%, 24.3%, 9% Thus, the market share of Margo in 2002 is 32.4%
499: A
Solution: $$eqalign{
& { ext{Let the total number of applicants be x}}. cr
& { ext{Number of eligible candidates}} cr
& = { ext{ }}95\% { ext{ }}of{ ext{ }}x cr
& { ext{Eligible candidates of other categories}}, cr
& = 15\% ,of,left( {95\% { ext{ }}of{ ext{ }}x}
ight) cr
& = {frac{{15}}{{100}}} imes {frac{{95}}{{100}}} imes x cr
& = frac{{57}}{{400}}x cr
& or,left( {frac{{57}}{{400}}}
ight)x cr
& x = frac{{ {4275 imes 400} }}{{57}} cr
& ,,,,,, = 30000 cr} $$
500: D
Solution: $$eqalign{
& 72.25mu o 6936 cr
& 100mu o frac{{6936 imes 100}}{{72.25}} cr
& 100mu o 9600 cr
& Rightarrow { ext{Valid votes }}96mu o 9600 cr
& 1mu o 100 cr
& 100mu o 10000{ ext{ Answer}} cr} $$
501: C
Solution: $$eqalign{
& 20\% { ext{ of }}885 = 17271 cr
& 855 imes frac{{20}}{{100}} = 17271 cr
& 1 o 101 cr
& herefore 1000 = 101000 cr} $$
502: C
Solution: Ratio of passed student to fails students = Pass : Fail = 25x : 4x Total number 25x + 4x = 29x New ratio of passed student to fails students = Pass : Fail = 22x : 3x = 25x $$frac{{{ ext{Fail}}}}{{{ ext{Total}}}} = frac{{4x - 3}}{{29x + 1}} = frac{3}{{25}}$$ 100x - 75 = 87x + 3 13x = 78 x = 6 So, the difference is = 25x - 4x = 21x = 21 × 6 = 126
503: B
Solution: $$eqalign{
& { ext{Let the property}} = 300,{ ext{units}} cr
& { ext{Anuja have}} = 300 imes frac{2}{3} = 200{ ext{ units}} cr
& frac{{200 imes 30}}{{100}} o 125000 cr
& 60{ ext{ units}} o { ext{125000}} cr
& { ext{1 unit}} o frac{{12500}}{6} cr
& 95\% { ext{ of }}300 = 135 cr
& 135{ ext{ units}} o { ext{135}} imes frac{{12500}}{6} cr
& = 45 imes 6250 cr
& = { ext{Rs}}{ ext{. }}281250 cr} $$
504: D
Solution: Grand mother's husband's, weight = 81 kg Grand mother's weight = (81 - 26) kg = 55 kg 20% = $$frac{1}{5}$$ Ankita's weight = 55 × $$frac{4}{5}$$ = 44 kg Ankita's brother's weight = 44 + 8 = 52 kg
505: B
Solution: $$ = frac{{3111 imes 100}}{{10000}} = 31.11\% $$
506: A
Solution: A : B = 6 : 5 Let A = 600, B = 500 After 30% increased A = 780 We have to increase = 280 Increase% $$ = frac{{280}}{{500}} imes 100 = 56\% $$
507: B
Solution: Salary = Rs. 160000 Personal and family expenses (P) = 50% of 160000 = 80000 Education loan (E) = 20% of 80000 = 16000 Taxes (T) = 15% of 16000 = 2400 Saving = 160000 - 80000 - 16000 - 2400 = 61600 New salary = 130% of 160000 = Rs. 208000 Personal and family expenses (P) = 50% of 208000 = 104000 Education loan (E) = 30% of 104000 = 31200 Taxes (T) = 20% of 31200 = 6240 Saving = 208000 - 104000 - 31200 - 6240 = 66560 Sum of the two saving = 61600 + 66560 = 128160
508: D
Solution: Number of students = 45 Number of boys = $$33frac{1}{3}\% $$ xa0of 45 = 15 Number of girls = 45 - 15 = 30 Obtain marks of girls in science subject is $$66frac{2}{3}\% $$ xa0more then the obtain marks of boys. $$eqalign{
& 66frac{2}{3}\% = frac{2}{3} cr
& frac{{{ ext{Girls}}}}{{{ ext{Boys}}}} = frac{{5x}}{{3x}} cr} $$ Total marks 15 × 3x + 30 × 5x = 45 × 78 3x + 2 × 5x = 3 × 78 13x = 3 × 78 x = 3 × 6 x = 18 5x = 18 × 5 5x = 90
509: A
Solution: [x08egin{array}{l}
35\% = frac{{35}}{{100}}\
{
m{Income}} = 100\
{
m{Saving}} = 35\
{
m{Expenditure}} = 65\
x08egin{array}{*{20}{c}}
{{
m{Income}}}& = &{{
m{Expenditure}}}& + &{{
m{Saving}}}\
{100}& = &{65}& + &{35}\
{,,,,,,,,,,,,,,,,, downarrow + 20.1}&{}&{,,,,,,,,,,,,, downarrow + 20}&{}&{}\
{120.1}& = &{78}& + &x
end{array}\
x = 120.1 - 78 = 42.1\
{
m{Increase ,in ,saving}} = 42.1 - 35 = 7.1\
{
m{Increase }}\% = frac{{7.1}}{{35}} imes 100 = 20.3\%
end{array}]
510: A
Solution: [x08egin{array}{l}
75 o frac{3}{4}\
x08egin{array}{*{20}{c}}
{{
m{Income}}}&{}&{{
m{Expenditure}}}&{{
m{Saving}}}\
{400}&{}&{300}&{100}\
{,,,,,,,,,,,,,,, downarrow + 28\% }&{}&{,,,,,,,,,,,,, downarrow + 20\% }& downarrow \
{112}& - &{60}&{ = 52\% }
end{array}\
{
m{Saving}} = 52\%
end{array}]
511: C
Solution: Given: A, B, and C spend 80%, 85%, and 75% of their incomes respectively. Savings are in the ratio 8 : 9 : 20 respectively. Difference between the income of A and C is Rs. 18,000. Concept used: Income - Expenditure = Savings Calculation: A, B and C spend 80%, 85% and 75% of their incomes respectively. Out of 100%, A spend 80%, Income : Expenditure = 100 : 80 = 5 : 4, Out of 100%, B spend 85%, Income : Expenditure = 100 : 85 = 20 : 17, Out of 100%, C spend 75%, Income : Expenditure = 100 : 75 = 4 : 3, Ratio Income : Expenditure is 5 : 4, 20 : 17, 4 : 3 respectively. Savings = Income - Expenditure Savings of A : B : C = (5 - 4) : (20 - 17) : (4 - 3) Savings of A : B : C = 1 : 3 : 1 According to the question, Savings are in the ratio 8 : 9 : 20 respectively. Adjusting the savings ratio, ⇒ Savings = 1 × 8 : 3 × 3 : 1 × 20 = 8 : 9 : 20 ⇒ Income = 5 × 8 : 20 × 3 : 4 × 20 = 40 : 60 : 80 ⇒ Expenditure = 4 × 8 : 17 × 3 : 3 × 20 = 32 : 51 : 60 According to the question, Income of C - Income of A = 18,000 ⇒ 80 - 40 = 18,000 ⇒ 1 unit = $$frac{{18,000}}{{40}}$$ xa0= 450 Income of B = 60 unit ⇒ 60 × 450 = Rs. 27,000 ∴ The income of B is Rs. 27,000.
512: B
Solution: $$eqalign{
& { ext{Let initial amount}} = n cr
& frac{{n imes 160}}{{100}} imes frac{{50}}{{100}} imes frac{{160}}{{100}} imes frac{{50}}{{100}} imes frac{{160}}{{100}} imes frac{{50}}{{100}} = 15360 cr
& frac{{n imes 16}}{{10}} imes frac{1}{2} imes frac{{16}}{{10}} imes frac{1}{2} imes frac{{16}}{{10}} imes frac{1}{2} = 15360 cr
& frac{{512n}}{{10 imes 10 imes 10}} = 15360 cr
& n = 30,000 cr} $$
513: C
Solution: $$eqalign{
& = 2 imes 1500 imes frac{{88}}{{100}} imes frac{{19}}{{20}} cr
& = { ext{Rs}}{ ext{. }}2508 cr} $$
514: B
Solution: $$eqalign{
& { ext{Richa : Rita}} cr
& ,,,,,100:85 cr
& { ext{Increase}}\% = frac{{left( {100 - 85}
ight)}}{{85}} imes 100 cr
& = frac{{15}}{{85}} imes 100 cr
& = 17frac{{11}}{{17}}\% cr} $$
515: A
Solution: $$eqalign{
& 10\% = frac{1}{{10}},,A\% = frac{A}{{100}},,20\% = frac{1}{5} cr
& 45000 imes frac{{11}}{{10}} imes frac{{100 + A}}{{100}} imes frac{6}{5} = 83160 cr
& 100 + A = 140 cr
& A = 40 cr} $$
516: A
Solution: Let, weekly income = ⇒ 6μ → 40 ⇒ 24μ → 160 Answer
517: D
Solution: Total number of votes = 5x Votes polled = 4x Valid votes = 4x - 82560 Invalid votes = 82560 Defeated candidate = 42% of valid votes Winning n = 58% of valid votes 16% of valid votes (difference) = 768400 16%(4x - 82560) = 768400 4x = 4802500 + 82560 x = 1221265 $$eqalign{
& { ext{Required}}\% = frac{{{ ext{Invalid votes}}}}{{{ ext{Number of people not voted}}}} imes 100 cr
& = frac{{82560}}{{1221265}} imes 100 cr
& = 6.76\% cong 6.8\% cr} $$
518: D
Solution: Alternate solution Given: Number is first increased by 40% and then decreased by 25%, again increased by 15% and then decreased by 20% $$eqalign{
& {x08f{Calculation:}} cr
& { ext{Let the number be }}x cr
& Rightarrow x imes frac{7}{5} imes frac{3}{4} imes frac{{23}}{{20}} imes frac{4}{5} cr
& Rightarrow x imes frac{{483}}{{500}} cr
& Rightarrow 0.966x cr
& Rightarrow 0.966x{ ext{ is less than }}x cr
& Rightarrow { ext{Decrease percent in the number}} cr
& = frac{{x - 0.966x}}{{100}} cr
& = 3.4\% cr
& herefore { ext{Net decrease percent in the number is }}3.4\% cr} $$
519: A
Solution: $$eqalign{
& frac{{25\% ,{ ext{of}}left( {50\% ,{ ext{of }}30\% ,{ ext{of }}150}
ight)}}{{40\% ,{ ext{of }}2250}} cr
& = frac{{25\% imes left( {50\% imes 30\% imes 150}
ight)}}{{40\% imes 2250}} cr
& = frac{{25}}{{40}}\% cr
& = 0.625\% cr} $$
520: B
Solution: Let the number of candidates appearing from each state be X.
Then, 7% of X - 6% of X = 80 Or, 1% of X = 80 Or, X = 80 × 100 X = 8000
521: B
Solution: $$eqalign{
& { ext{Total number all three got together is}}, cr
& = {1136 + 7636 + 11628} cr
& = 20400 cr
& \% { ext{ of vote the winning candidate got is}}, cr
& = {frac{{11628}}{{20400}}} imes 100 cr
& = 57\% cr} $$
522: B
Solution: Quantity of water in 250 kg dry grapes, $$ = frac{{10}}{{100}} imes 250 = 25,{ ext{kg}}$$ Then, pulp of grapes = 225 kg We get 20 kg pulp in 100 kg fresh grapes. To get 225 kg pulp , we need fresh grapes, $${ ext{ = }}frac{{100 imes 225}}{{20}} = 1125,{ ext{kg}}$$
523: C
Solution: Let the number of bushes originally be 100 Number of bushes after one year 100 ==10% (↑) ==> 110 After second year it becomes 110 ==8%(↑) ==> 118.8 After third year, 118.8 ==8%(↓)==> 109.3 Now, according to the question 109.3 = 26730 1 = $$frac{{26730}}{{109.3}}$$ So, 100 = $$frac{{26730}}{{109.3}} imes 100$$ xa0xa0 = 25000 Thus, number of bushes originally was 25000 NOTE: You can take number of bushes originally as x then solve for the x
524: B
Solution: $$eqalign{
& frac{{ {ax} }}{{100}} = frac{{ {by} }}{{100}} cr
& or,,ax = by cr
& Rightarrow y = frac{{ {ax} }}{b} cr
& c\% ,{ ext{of}},y = frac{{ {cy} }}{{100}} cr
& or,,frac{{ {cy} }}{{100}} = frac{{ {cax} }}{{100b}} cr
& { ext{Thus}}, cr
& c\% ,{ ext{of}},y = frac{{ca}}{b}\% ,{ ext{of}},x cr} $$
525: B
Solution: Let his original salary be Rs. 100 Salary after increment = Rs. 120 Let the tax on original salary be 20% and now tax on increased salary (Rs. 20) will be 22% i.e. Rs. 4.40 Thus, increase in tax liability $$ = frac{{4.40}}{{20}} imes 100 = 22\% $$
526: C
Solution: Let the basic salaries of A and B be x and y respectively. Now, $$eqalign{
& x + 65\% ,,{ ext{of }}x = y + 80\% ,,{ ext{of }}y cr
& Rightarrow x + frac{{ {65x} }}{{100}} = y + frac{{ {80y} }}{{100}} cr
& Rightarrow frac{x}{y} = frac{{180}}{{165}} cr
& Rightarrow x:y = 12:11 cr} $$
527: D
Solution: They covered the distance in this way together in different hours
6 + 6.5 + 7 + 7.5 + 8 + 8.5 + 9 + 9.5 + 10 = 72

Means,they'll meet at the 9 th hr. So, In that time A will cover = 4 × 9 = 36km They will meet in Midway
528: C
Solution: $$eqalign{
& { ext{Let the number of total workers}} = x cr
& { ext{Number of skilled workers}} cr
& = 75\% ,of,x = frac{{75x}}{{100}} = frac{{3x}}{4} cr
& { ext{No}}{ ext{. of unskilled workers}} cr
& = 25\% ,of,x = frac{{25x}}{{100}} = frac{x}{4} cr
& { ext{No}}{ ext{. of permanent workers}}, cr
& = {frac{{80}}{{100}}} imes {frac{{3x}}{4}} + {frac{{20}}{{100}}} imes {frac{x}{4}} cr
& = {frac{{3x}}{5}} + {frac{x}{{20}}} cr
& = frac{{13x}}{{20}} cr
& { ext{No}}{ ext{.}},{ ext{of}},{ ext{temporary}},{ ext{workers,}} cr
& = x - {frac{{13x}}{{20}}} = frac{{7x}}{{20}} cr
& { ext{Now}}, cr
& frac{{7x}}{{20}} = 126 cr
& x = 360 cr} $$
529: B
Solution: No. of population who are literate = 50% of 296000 = 148000 No. of male = 166000 No. of female = 296000 - 166000 = 130000 No. of literate male = 70% of 166000 = 116200 No. of literate women = 148000 - 116200
= 31800
530: D
Solution: Required answer = 300 × $$frac{1}{4}$$ × $$frac{1}{5}$$ = 15
531: D
Solution: Required percentage : = 0.001 × 100 = 0.1%
532: D
Solution: Required percentage : = $$frac{72}{3.6 × 1000}$$ xa0× 100 = 2%
533: C
Solution: Net % effect on revenue = - 10 + 10 - $$frac{10 × 10}{100}$$ = - 1 Hence % reduction in revenue = 1%
534: C
Solution: Decrease in area $$eqalign{
& = frac{{{x^2}}}{{100}}\% cr
& = frac{{{{left( {10}
ight)}^2}}}{{100}} cr
& = 1\% cr} $$
535: D
Solution: Let the original price = 100x Article price 1st increased by 10%. i.e price of article become 100x + 100x of 10% = 110x And then the article price increased by 20% i.e. price of article become 110x + 110x of 20% = 132x According to the question, 132x = 33 ⇒ x = $$frac{33}{132}$$ ⇒ x = $$frac{1}{4}$$ Therefore Original price of Article is = 100 × $$frac{1}{4}$$ = Rs. 25
536: B
Solution: Gain in price = 7.5 - 6 = 1.5 Required percentage reduction = $$frac{1.5}{7.5}$$ × 100 = 20%
537: B
Solution: ∴ Required % = $$frac{30}{24 × 60}$$ × 100 = 2.083
538: A
Solution: 900 × 72% = 648 700 × 80% = 560 Total scored marks = 1208 = % combined marks = $$frac{1208}{(900 + 700)}$$ xa0× 100 = 75.5%
539: C
Solution: According to the question, Pass marks = (220 + 20) = 240 40% → 240 ∴ Maximum marks (100%) : = $$frac{240}{40}$$ × 100 = 600
540: D
Solution: $$eqalign{
& x\% { ext{ of 500 }} = y\% { ext{ of 300}} cr
& Rightarrow frac{x}{{100}} imes 500 = frac{y}{{100}} imes 300 cr
& Rightarrow 5x = 3y cr
& Rightarrow y = frac{5}{3}x cr
& x\% { ext{ of }}y\% { ext{ of 200}} = 60 cr
& Rightarrow frac{x}{{100}} imes frac{y}{{100}} imes 200 = 60 cr
& Rightarrow xy = 3000 cr
& Rightarrow x imes frac{5}{3}x = 3000 cr
& Rightarrow {x^2} = 3000 imes frac{3}{5} cr
& Rightarrow {x^2} = 1800 cr
& Rightarrow x = 30sqrt 2 cr} $$
541: B
Solution: Let the total population be x Then, migrant population : $$eqalign{
& = 35\% { ext{ of }}x cr
& = left( {frac{{35}}{{100}} imes x}
ight) cr
& = frac{{7x}}{{20}} cr} $$ Local population : $$eqalign{
& = left( {x - frac{{7x}}{{20}}}
ight) cr
& = frac{{13x}}{{20}} cr} $$ Number of rural migrants : $$eqalign{
& = 20\% { ext{ of }}frac{{7x}}{{20}} cr
& = left( {frac{{20}}{{100}} imes frac{{7x}}{{20}}}
ight) cr
& = frac{{7x}}{{100}} cr} $$ Number of urban migrants : $$eqalign{
& = left( {frac{{7x}}{{20}} - frac{{7x}}{{100}}}
ight) cr
& = frac{{28x}}{{100}} cr
& = frac{{7x}}{{25}} cr} $$ Female population : $$48\% { ext{ of }}frac{{13x}}{{20}} + 30\% { ext{ of }}frac{{7x}}{{100}}$$ xa0 xa0 $$ +, 40\% { ext{ of }}frac{{7x}}{{25}}$$ $$ = left( {frac{{48}}{{100}} imes frac{{13x}}{{20}}}
ight) + left( {frac{{30}}{{100}} imes frac{{7x}}{{100}}}
ight)$$ xa0 xa0xa0 $$ + left( {frac{{40}}{{100}} imes frac{{7x}}{{25}}}
ight)$$ $$eqalign{
& = frac{{39x}}{{125}} + frac{{21x}}{{1000}} + frac{{14x}}{{125}} cr
& = frac{{445x}}{{1000}} cr} $$ ∴ Required percentage : $$eqalign{
& = left( {frac{{445x}}{{1000}} imes frac{1}{x} imes 100}
ight)\% cr
& = 44.50\% cr} $$
542: D
Solution: Let original income = Rs. 100 Then savings = Rs. 25 and expenditure = Rs. 75 New income = Rs. 120 New expenditure : $$=$$ 115% of Rs. 75 $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {frac{{115}}{{100}} imes 75}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{345}}{4} cr} $$ New saving : .
$$eqalign{
& = { ext{Rs}}{ ext{. }}left( {120 - frac{{345}}{4}}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{135}}{4} cr} $$ Increase in saving : $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {frac{{135}}{4} - 25}
ight) cr
& = { ext{Rs}}{ ext{. }}frac{{35}}{4} cr} $$ ∴ Percentage increase : $$eqalign{
& = left( {frac{{35}}{4} imes frac{1}{{25}} imes 100}
ight)\% cr
& = 35\% cr} $$
543: A
Solution: Population of 6 years ago : $$eqalign{
& = left[ {frac{{1771561}}{{{{left( {1 + frac{{10}}{{100}}}
ight)}^6}}}}
ight] cr
& = 1771561 imes {left( {frac{{10}}{{11}}}
ight)^6} cr
& = frac{{1771561 imes 1000000}}{{1771561}} cr
& = 1000000 cr} $$
544: A
Solution: Let the original quantity be x kg. Vanaspati ghee in x kg $$eqalign{
& = left( {frac{{40}}{{100}}x}
ight){ ext{kg}} cr
& = left( {frac{{2x}}{5}}
ight){ ext{kg}} cr} $$ Now, $$eqalign{
& Rightarrow frac{{frac{{2x}}{5}}}{{x + 10}} = frac{{20}}{{100}} cr
& Rightarrow frac{{2x}}{{5x + 50}} = frac{1}{5} cr
& Rightarrow 5x = 50 cr
& Rightarrow x = 10 cr} $$
545: D
Solution: Number of women = 40% of 200 = 80 Number of men = 200 - 80 = 120 Number of women lectures = $$frac{1}{5}$$ of 80 = 16 Number of men lectures = 16 × 2 = 32 ∴ Required percentage : $$eqalign{
& = left( {frac{{32}}{{120}} imes 100}
ight)\% cr
& = frac{{80}}{3}\% cr
& = 26frac{2}{3}\% approx 27\% cr} $$
546: A
Solution: Let, his income be = Rs. 100 Saving = Rs. 10 Expenditure = Rs. 90 Increased income = Rs. 120 Increased expenditure = Rs. 110 Increased in expenditure : $$eqalign{
& = frac{{110 - 90}}{{90}} imes 100 cr
& = frac{{200}}{9} cr
& = 22frac{2}{9}\% cr} $$
547: C
Solution: Quantity of alcohol in 10 c.c. solution = (20% of 6 + 60% of 4) c.c. = $$left( {frac{{20}}{{100}} imes 6 + frac{{60}}{{100}} imes 4}
ight)$$ xa0 xa0c.c. = (1.2 + 2.4) c.c. = 3.6 c.c. ∴ Required strength : $$eqalign{
& = left( {frac{{3.6}}{{10}} imes 100}
ight)\% cr
& = 36\% cr} $$
548: B
Solution: Actual price : $$= Rs. (25 + 2.5)$$ $$= Rs. 27.50$$ ∴ Saving : $$eqalign{
& = left( {frac{{2.50}}{{27.50}} imes 100}
ight)\% cr
& = frac{{100}}{{11}}\% cr
& = 9frac{1}{{11}}\% approx 9\% cr} $$
549: B
Solution: Let the worth of the house be Rs. x Then, $$frac{2}{7}$$% of x = 28000 $$eqalign{
& Rightarrow left( {frac{2}{7} imes frac{1}{{100}} imes x}
ight) = 2800 cr
& Rightarrow x = left( {frac{{2800 imes 100 imes 7}}{2}}
ight) cr
& Rightarrow x = 980000 cr} $$
550: C
Solution: Let the total number of votes = 100x Winner candidate get 57% of vote i.e. = 57x and losser will get = 100x - 57x = 43x According to the question, 57x - 43x = 42000 ⇒ 14x = 42000 ⇒ x = 3000 Total votes = 100x = 100 × 3000 = 300000
551: A
Solution: Given, 60% of work whose age is 30 above 1350 males work whose age is 30 above. 75% of workers are male whose age are 30 above. 25% of female workers whose ages are 30 above. Number of female work whoes age 30 above = $$frac{{1350}}{{75}} imes 25 = 450$$ total number of work whoes age are 30 above = 1350 + 450 = 1800(60% of the work) Total number of work = $$frac{{1800}}{{60}} imes 100 = 3000$$
552: D
Solution: Percentage increase : = $$frac{51300 - 41800}{41800}$$ xa0 × 100 = $$frac{9500}{41800}$$ × 100 = $$frac{9500}{418}$$ = $$frac{500}{22}$$ = $$frac{250}{11}$$ = $$22frac{8}{11}$$%
553: A
Solution: Let the number = x ⇒ $$frac{1}{3}$$ × x = 96 ⇒ x = 288 ⇒ Required answer : = $$frac{67}{100}$$ × 288 = 192.96
554: D
Solution: % change = $$frac{{ ext{R}}}{{100 pm { ext{R}}}}$$ xa0 × 100% Required answer : = $$frac{40}{100 - 40}$$ xa0 × 100 = $$frac{40}{60}$$ × 100 = $$frac{200}{3}$$ = $$66frac{2}{3}$$%
555: C
Solution: Let the fraction = $$frac{x}{y}$$ According to the question, $$frac{x × 120}{y × 95}$$ xa0 = $$frac{5}{2}$$ $$frac{x}{y}$$ = $$frac{5 × 95}{2 × 120}$$ $$frac{x}{y}$$ = $$frac{95}{48}$$
556: C
Solution: Percentage error : = $$frac{1.55 - 1.50}{1.50}$$ xa0 × 100 = $$frac{0.05}{1.50}$$ × 100 = $$frac{5}{150}$$ × 100 = $$frac{5 × 2}{3}$$ = $$frac{10}{3}$$ = $$3frac{1}{3}$$%
557: A
Solution: Let the initial expenditure = 100 units Increase in consumption = $$frac{20}{80}$$ = $$frac{1}{4}$$ 1 unit = 5 kg Original consumption = 5 × 4 = 20 kg New consumption = 5 × 5 = 25 kg Original price = $$frac{320}{20}$$ = 16 per kg
558: B
Solution: Increase in height = 15% = $$frac{3}{20}$$ Decrease in base radius = 10% = $$frac{1}{10}$$ Initial Final Radius 10 9 height 20 23 Area 200 207 Increment in Area = 207 - 200 = 7 Required % increase in area = $$frac{7}{200}$$ × 100 = 3.5%
559: D
Solution: Let the 3 rd number is 100 According to the question, $$eqalign{
& ,{1^{{ ext{st}}}},,,,,,,,,,,,,,,,,,,{2^{{ ext{nd}}}},,,,,,,,,,,,,,,,,,{3^{{ ext{rd}}}} cr
& 20,,,,,,,:,,,,,,,,50,,,,,,,,:,,,,,,,100 cr} $$ Required % = $$frac{20}{50}$$ × 100 = 40%
560: B
Solution: Percentage of failed students = 25% ∴ Percentage of passed students = (100 - 25)% = 75% According to the question, Total students : = $$frac{450}{75}$$ × 100 = 600
561: D
Solution: Total votes = 104000 Total valid votes : = 104000 × $$frac{(100 - 2)}{100}$$ = 104000 ×$$frac{98}{100}$$ = 101920 Votes polled in favour of the candidate : = 101920 × $$frac{55}{100}$$ = 56056
562: B
Solution: Sugar in original solution = $$frac{{75 imes 30}}{{100}}$$xa0 = 22.5 gm Let m gm of sugar be mixed. $$ herefore frac{{22.5 + { ext{m}}}}{{75 + { ext{m}}}} imes 100 = 70$$ ⇒ 2250+ 100m = 75 × 70 + 70m ⇒ 2250 + 100m = 5250 + 70m ⇒ 30m = 5250 - 2250 = 3000 ⇒ m = $$frac{{3000}}{{30}}$$xa0 = 100 gm
563: D
Solution: Total number boy = 1100 and girl = 900 Total number of boy fail (50%) = 550 Total number of girl fail ( 60%) = 540 Total failed candidates = (550 + 540) = 1090 Required percentage of failed candidates : = $$frac{1090}{(1100 + 900)}$$ xa0 × 100 = $$frac{1080 × 100}{2000}$$ = 54.5%
564: A
Solution: Total income = 100x 10% deposit to orphange = 100x of 10% = 10x Remain income 90x Now 20% deposit to bank = 90x of 20% = 18x Reamin income 90x - 18x = 72x According to question 72x = 7200 ⇒ x = $$frac{7200}{72}$$ = 100 Income of Christy = 100 × 100 = 10000
565: C
Solution: Total Mark obtain by all 50 student = 50 × 70 = 3500 Total mark obtain by 1st 25 student = 25 × 60 = 1500 Total mark obtain by next 24 student = 24 × 80 = 1920 Mark obtain by last student = 3500 - (1500 + 1920) = 80 i.e 80% mark obtain by last student.
566: B
Solution: $$frac{50}{100}$$(P - Q) = $$frac{30}{100}$$(P + Q) 5P - 5Q = 3P + 3Q 2P = 8Q P = 4Q Put value of P in given equation Q = P × $$frac{x}{100}$$ Q = 4Q × $$frac{x}{100}$$ x = 25 Hence required value of x = 25
567: B
Solution: According to the question, a × $$frac{x}{100}$$ = b × $$frac{y}{100}$$ xa = yb ⇒ b = $$frac{xa}{y}$$ Put value of b in given equation, z% of b = z% of $$frac{xa}{y}$$ = $$frac{zx}{y}$$ of a
568: D
Solution: Let the maximum marks = x According to the question, Case (i) pass marks = $$frac{20x}{100}$$ + 30 Case (ii) pass marks = $$frac{32x}{100}$$ - 42 Note : Pass marks would be same in both cases. $$frac{20x}{100}$$ + 30 = $$frac{32x}{100}$$ - 42 $$frac{12x}{100}$$ = 72 x = 600 Pass marks : = 600 × $$frac{20}{100}$$ + 30 = 150 Required percentage : = $$frac{150}{600}$$ × 100 = 25%
569: A
Solution: x × $$frac{20}{100}$$ × $$frac{20}{100}$$ = 16000 x = 400000
570: B, E
Solution: Let the capacity of the tank be 100 litres. Then, Initially : A type petrol = 100 litres After first operation : A type petrol = $$left( {frac{{100}}{{2}}}
ight)$$ = 50 litres

B type petrol = 50 litres After second operation : A type petrol = $$left( {frac{{50}}{2} + 50}
ight)$$ xa0= 75 litres

B type petrol = $$left( {frac{{50}}{2}}
ight)$$ = 25 litres After third operation : A type petrol = $$left( {frac{{75}}{2}}
ight)$$ = 37.5 litres B type petrol = $$left( {frac{{25}}{2} + 50}
ight)$$ xa0= 62.5 litres ∴ Required percentage = 37.5%
571: A
Solution: $$eqalign{
& { ext{Let,}} cr
& B = 110\% { ext{ of }}b = frac{{11b}}{{10}}{ ext{ }} cr
& C = 110\% { ext{ of }}c = frac{{11c}}{{10}} cr
& D = 110\% { ext{ of }}d = frac{{11d}}{{10}} cr
& { ext{Then,}} cr
& A = B imes frac{D}{C} cr
& A = frac{{11b}}{{10}} imes frac{{11d}}{{10}} imes frac{{10}}{{11c}} cr
& A = frac{{11bd}}{{10c}} cr
& A = frac{{11}}{{10}}a cr
& A = frac{{110}}{{100}}a cr
& A = 110\% { ext{ of }}a cr
& herefore { ext{ Increase }}\% = 10\% cr} $$
572: D
Solution: $$eqalign{
& { ext{Let the present population of the town be }}P cr
& { ext{Using compound interest formula}} cr
& { ext{Then}}, cr
& P = xleft[ {1 + left( {frac{R}{{100}}}
ight)}
ight] - - - ,left( i
ight) cr
& { ext{And}},y = Pleft[ {1 + left( {frac{R}{{100}}}
ight)}
ight] cr
& = P imes frac{P}{x} - - - - ,left( {ii}
ight) cr
& {P^2} = xy
cr
& { ext{Hence}},,P = sqrt {xy} cr} $$
573: A
Solution: Let Ashu's salary = 100
Ashu's salary after rise = 125 Then Vicky's salary = 175 Vicky's salary after rise of 40% = 245 [As 10% of Vicky's salary is 17.5 then 40% = 17.5 × 4 = 70] Difference between Vicky's salary and Ashu's salary = 245 - 125 = 120 % more Vicky's salary than Ashu's = $$frac{{120 imes 100}}{{125}} = 96\% $$
574: D
Solution: Let there is 100 kg of ore. 25% ore contains 90% off Iron that means 25 kg contains
$$frac{{25 imes 90}}{{100}} = 22.5,{ ext{kg}},{ ext{iron}}$$ 22.5 kg Iron contains 100 kg of ore. Then, 1 kg of iron contains = $$frac{{25}}{{100}}{ ext{kg}},{ ext{ore}}$$ Hence, 60 kg iron contains = $$frac{{100 imes 60}}{{22.5}}$$ = 266.66 kg ore
575: A
Solution: Let Pooja's monthly salary be = Rs. x Then, 13% of x = 8554 ⇒ $$frac{13x}{100}$$ = 8554 ⇒ x = $$frac{8554 × 100}{13}$$ ⇒ x = 65800 Total percentage of salary invested = (13 + 23 + 8)% = 44% ∴ Total amount invested monthly = 44% of Rs. 65800 = Rs. $$frac{44}{100}$$ × 65800 = Rs. 28952
576: A
Solution: $$66frac{2}{3}\% = frac{2}{3}$$ Saving = 3
Expenditure = 3 + 2 = 5 Income = 5 + 3 = 8 [x08egin{array}{*{20}{c}}
{{ ext{Income}}}& = &{{ ext{Expenditure}}}& + &{{ ext{Saving}}} \
8& = &5& + &3 \
{800}& = &{500}& + &{300} \
{,,,,,,,,,,,,, downarrow + 44\% }&{}&{,,,,,,,,,,,,, downarrow + 60\% }&{}&{} \
{352}& = &{300}& + &x
end{array}] x = 352 - 300 = 52 52 units → 1040 1 unit → 20 500 units → 500 × 20 = 10,000
577: B
Solution: Let total population = x Then, number of males = $$frac{5}{9}$$x Married males : = 30% of $$frac{5}{9}$$x = $$frac{30}{100}$$ × $$frac{5}{9}$$x = $$frac{x}{6}$$ Married females = $$frac{x}{6}$$
Number of females : = x - $$frac{5}{9}$$x = $$frac{4x}{9}$$ Unmarried females : = $$frac{4x}{9}$$ - $$frac{x}{6}$$ = $$frac{5x}{18}$$ ∴ Required percentage : = $$frac{5x}{18}$$ × $$frac{1}{x}$$ × 100% = $$27frac{7}{9}$$%
578: B
Solution: Total number of students : = 1100 + 700 = 1800 Number of students passed : = (42% of 1100 + 30% of 700) = (462 + 210) = 672 Number of failures : = 1800 - 672 = 1128 ∴ Percentage failure : = $$left( {frac{{1128}}{{1800}} imes 100}
ight)\% $$ = $$62frac{2}{3}\% $$
579: B
Solution: Let the initial expenses on Sugar was Rs. 100. Now, Price of Sugar rises 25%. So, to buy same amount of Sugar, they need to expense, = (100 + 25% of 100) = Rs. 125. But, They want to keep expenses on Sugar, so they have to cut Rs. 25 in the expenses to keep it to Rs. 100. Now, % decrease in Consumption, $$frac{{25}}{{125}} imes 100 = 20\% $$ Mind Calculation Method
100 === 25%↑ ===> → 125 === X%↓ ===> 100 Here,
X = $$frac{{25}}{{125}} imes 100 = 20\% $$
580: B
Solution: 15% of families have cow and buffalo. Families only have cow = 60 - 15 = 45% Families only have buffalo = 30 - 15 = 15% Required families which do not have a cow or a buffalo = 100 - (Families only have cow + Families only have buffalo + Families have cow and buffalo) = 100 - (45 + 15 + 15) = 25% According to the question, Required number = $$frac{96}{100}$$ × 25 = 24