Interest

Name: _____________________

Date: _____________________

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

Question 1:

Asmita invest an amount of Rs. 9534 @ 4 p.c.p.a to obtain a total amount of Rs. 11442 on simple interest after a certain period. For how many years did she invest the amount to obtain the total sum?

A. 2 years
B. 4 years
C. 5 years
D. 10 years
Answer: _________
Question 2:

What sum will amount to Rs. 7000 in 5 years at $${ ext{3}}frac{1}{3}$$ % simple interest ?

A. Rs. 6300
B. Rs. 6500
C. Rs. 6000
D. Rs. 5000
Answer: _________
Question 3:

Mohan lent some amount of money at 9% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was Rs. 760, what amount was lent in each case ?

A. Rs. 1700
B. Rs. 1800
C. Rs. 1900
D. Rs. 2000
Answer: _________
Question 4:

If the simple interest on a sum of money for 15 months at $${ ext{7}}frac{1}{2}$$ % per annum exceeds the simple interest on the same sum for 8 months at $${ ext{12}}frac{1}{2}$$ % per annum by Rs. 32.50, then the sum of money ( In Rs.) is ?

A. Rs. 312
B. Rs. 312.50
C. Rs. 3120
D. Rs. 3120.50
Answer: _________
Question 5:

Deepak invested an amount of Rs. 21250 for 6 years. At what rate of simple interest will be obtain the total amount of Rs. 26350 at the end of 6 years?

A. 5 p.c.p.a
B. 6 p.c.p.a
C. 8 p.c.p.a
D. 12 p.c.p.a
Answer: _________
Question 6:

At what rate of simple interest per annum can an amount of Rs. 1553.40 be obtained on the principal amount of Rs. 8630 after 3 years?

A. 4 p.c.p.a
B. 5 p.c.p.a
C. 7 p.c.p.a
D. 8 p.c.p.a
Answer: _________
Question 7:

If simple interest on Rs. 600 for 4 years and on Rs. 600 for 2 years combined together is Rs.180, find the rate of interest.

A. 4%
B. 5%
C. 5.5%
D. 6.25%
Answer: _________
Question 8:

A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double itself at the same rate of simple interest ?

A. 8 years
B. 10 years
C. 12 years
D. 14 years
Answer: _________
Question 9:

A person borrows some money for 5 years and ratio of loan amount : total interest amount is 5 : 2. Then find the ratio of loan amount : interest rate is equal to = ?

A. 2 : 25
B. 2 : 1
C. 5 : 2
D. 25 : 2
Answer: _________
Question 10:

The effective annual rate of interest, corresponding to a nominal rate of 6% per annum payable half yearly is = ?

A. 6.06%
B. 6.07%
C. 6.08%
D. 6.09%
Answer: _________
Question 11:

The simple interest on a certain sum is one-eighth of the sum when the number of years is equal to half of the rate percentage per annum. Find the simple interest (in Rs.) on Rs. 15,000 at the same rate of simple interest for 8 years.

A. 5,800
B. 5,000
C. 6,000
D. 5,250
Answer: _________
Question 12:

A person borrows Rs. 1,00,000 from a bank at 10% per annum simple interest and clears the debt in five years. If the instalment paid at the end of the first, second, third and fourth years to clear the debt are Rs. 10,000, Rs. 20,000, Rs. 30,000 and Rs. 40,000, respectively, what amount should be paid at the end of the fifth year to clear the debt?

A. Rs. 38,250
B. Rs. 39,490
C. Rs. 40,450
D. Rs. 36,450
Answer: _________
Question 13:

A mobile phone is available for Rs. 25,000 or Rs. 5,200 down payment, followed by 4 equal monthly instalments. If the rate of interest is 25% p.a. simple interest, calculate the amount of each instalment.

A. Rs. 5,020
B. Rs. 4,950
C. Rs. 5,362
D. Rs. 5,200
Answer: _________
Question 14:

At which rate of simple interest does an amount become double in 12 years?

A. $$7frac{4}{5}\% $$
B. $$8\% $$
C. $$8frac{1}{3}\% $$
D. $$7frac{1}{2}\% $$
Answer: _________
Question 15:

On simple interest, a certain sum becomes Rs. 59,200 in 6 years and Rs. 72,000 in 10 years. If the rate of interest had been 2% more, then in how many years would the sum have become Rs. 76,000?

A. 10
B. 9
C. 8
D. 7
Answer: _________
Question 16:

A sum is deposited in a bank which gives simple interest. The sum becomes 1.25 times in 3 years. If there is a requirement of Rs. 7,60,000 after seven years, how much amount (in Rs.) should one deposit to fulfil the requirement?

A. 5,20,000
B. 5,70,000
C. 4,80,000
D. 6,00,000
Answer: _________
Question 17:

A certain sum is lent at 4% per annum for 3 years 8% per annum for next 4 years and 12% per annum beyond 7 years. If for a period of 11 years, the simple interest obtained is Rs. 27,600, then the sum is (in Rs.):

A. 25,000
B. 32,000
C. 27,000
D. 30,000
Answer: _________
Question 18:

If the amount obtained by A by investing Rs. 9,100 for three years at a rate of 10% p.a. on simple interest is equal to the amount obtained by B by investing a certain sum of money for five year at a rate of 8% p.a. on simple interest, then 90% of the sum invested by B (in Rs.) is:

A. 7,800
B. 8,450
C. 8,540
D. 7,605
Answer: _________
Question 19:

A person deposited Rs. 15,600 in a fixed deposit at 10% per annum
simple interest. After every second year he adds his interest earned to
the principal. The interest at the end of 4 years is:

A. Rs. 6,655
B. Rs. 6,864
C. Rs. 3,975
D. Rs. 3,744
Answer: _________
Question 20:

A person borrowed 1,200 at 8% per annual and Rs. 1,800 at 10% per annum, as simple interest for the same period. He had to pay Rs. 1380 in all as interest. Find the time? (in year)

A. 4 years
B. 5 years
C. 6 years
D. 10 years
Answer: _________
Question 21:

A sum was put at simple interest at a certain rate for 3 years. Had it been put at 1% higher rate, it would have fetched Rs. 5100 more. The sum is

A. Rs. 1, 20, 000
B. Rs. 1, 25, 000
C. Rs. 1, 50, 000
D. Rs. 1, 70, 000
Answer: _________
Question 22:

What equal installment of annual payment will discharge a debt which is due as Rs. 848 at the end of 4 years at 4% per annum simple interest ?

A. Rs. 200
B. Rs. 212
C. Rs. 225
D. Rs. 250
E. Rs. 212
F. Rs. 200
G. Rs. 250
H. Rs. 225
Answer: _________
Question 23:

A man buys a TV priced at Rs. 16000. He pays Rs. 4000 at once and the rest after 15 months on which he is charges a simple interest at the rate of 12% per year. The total amount he pays for TV is = ?

A. Rs. 18200
B. Rs. 17200
C. Rs. 17800
D. Rs. 16800
Answer: _________
Question 24:

If the ratio of principal and the simple interest of 5 years is 10 : 3, then the rate of interest is = ?

A. 6%
B. 8%
C. 3%
D. 5%
Answer: _________
Question 25:

Mr. Dutta desired to deposit his retirement benefit of Rs. 3 lacs partly to a post office and partly to a bank at 10% and 6% simple interests respectively. If his monthly income was Rs. 2000, then the difference of his deposits in the post office and in the bank was = ?

A. Rs. 100000
B. Rs. 40000
C. Rs. 50000
D. Rs. Nil
Answer: _________
Question 26:

A sum of Rs. 10 is lent to be returned in 11 monthly instalments of Rs. 1 each, interest being simple. The rate of interest is:

A. $$9frac{1}{{11}}$$ %
B. 10%
C. 11%
D. $$21frac{9}{{11}}$$ %
Answer: _________
Question 27:

A computer is available for Rs. 39000 cash or Rs. 17000 as cash down payment followed by five monthly instalments of Rs. 4800 each. What is the rate of interest under the instalment plan?

A. 35.71 % p.a.
B. 36.71 % p.a.
C. 37.71 % p.a.
D. 38.71 % p.a.
Answer: _________
Question 28:

If the rate increases by 2%, the simple interest received on a sum of money increases by Rs. 108. If the time period is increased by 2 years, the simple interest on the same sum increases by Rs. 180. The sum is:

A. Rs. 1800
B. Rs. 3600
C. Rs. 5400
D. Data inadequate
Answer: _________
Question 29:

A boy aged 12 years is left with Rs. 100000 which is under a trust. The trustees invest the money at 6% per annum and pay the minor boy a sum of Rs. 2500, for his pocket money at the end of each year. The expenses of trust come out to be Rs. 500 per annum. Find the amount that will be handed over to the minor boy after he attains the age of 18 years ?

A. Rs. 125000
B. Rs. 118000
C. Rs. 150000
D. Rs. 120000
Answer: _________
Question 30:

The simple interest on Rs. 36000 for the period from 5 th January to 31 st May, 2013 at 9.5% per annum is = ?

A. Rs. 1368
B. Rs. 1338
C. Rs. 1425
D. Rs. 1400
Answer: _________
Question 31:

In a certain time, a sum becomes 4 times of itself on simple interest at the rate of 10% per annum. What is the rate of interest if the same sum becomes 7 times of itself in the same duration?

A. 15%
B. 10%
C. 5%
D. 20%
Answer: _________
Question 32:

Rs. 2,500, when invested for 8 years at a given rate of simple interest per year, amounted to Rs. 3,725 on maturity. What was the rate of simple interest that was paid per annum?

A. 6%
B. 6.125%
C. 6.25%
D. 5.875%
Answer: _________
Question 33:

In how many least number of complete years a sum of money become more than four times of itself at the rate of 50% per annum on simple interest?

A. 9 years
B. 7 years
C. 6 years
D. 5 years
Answer: _________
Question 34:

A sum of Rs. 8,400 amounts to Rs. 11,046 at 8.75% p.a. simple interest in a certain time. What will be the simple interest (in Rs.) on a sum of Rs. 10,800 at the same rate for the same time?

A. 3,204
B. 3,024
C. 3,402
D. 3,420
Answer: _________
Question 35:

The simple interest on Rs. 7,300 at 15% p.a. from 28 th April to 4 th November, approximately is

A. Rs. 570.00
B. Rs. 547.50
C. Rs. 1,095.00
D. Rs. 750.00
Answer: _________
Question 36:

In how many years will the simple interest on a sum of money be equal to the principle at rate of $$12frac{2}{4}\% $$ xa0per annum?

A. 7 years
B. 5 years
C. 8 years
D. 6 years
Answer: _________
Question 37:

What is the present value of Rs. 10,000 received in 2 years, if the interest rate is 12% per year discounted semi-annually?

A. Rs. 7,020.94
B. Rs. 7,920.90
C. Rs. 7,920.94
D. Rs. 7,900.94
Answer: _________
Question 38:

An amount of Rs. 12,000 was borrowed at some rate of simple interest. After four months, Rs. 6,000 more were added to it and the rate of interest on the total principal was doubled from the previous rate. At the end of the year, Rs. 2,800 was paid as interest, calculate the rate of interest initially charged?

A. 14%
B. 16%
C. 12%
D. 10%
Answer: _________
Question 39:

Damani purchased an item costing Rs. 7,500 and paid Rs. 3,500 as a down payment for the same. If the simple interest charged for the remaining amount is 9% per annum and Damani cleared all dues after 4 months of the purchase, how much did Damani pay after 4 months as interest?

A. Rs. 120
B. Rs. 100
C. Rs. 132
D. Rs. 125
Answer: _________
Question 40:

A sum of money invested at simple interest becomes 17/10 of itself in 2 years and 6 months. What is the rate of interest per annum?

A. 22%
B. 16%
C. 28%
D. 34%
Answer: _________
Question 41:

A person invests Rs. 12000 as fixed deposit at a bank at the rate of 10% per annum simple interest. But due to some pressing needs he has to withdraw the entire money after three years, for which the bank allowed him a lower rate of interest. If he gets Rs. 3320 less than what he would have got at the end of 5 years, the rate of interest allowed by the bank is = ?

A. $${ ext{7}}frac{5}{9}$$ %
B. $${ ext{7}}frac{4}{9}$$ %
C. $${ ext{7}}frac{8}{9}$$ %
D. $${ ext{8}}frac{7}{9}$$ %
Answer: _________
Question 42:

A certain scheme of investment in simple interest declares that it triples the investment in 8 years. If you want to quadruple the money through that scheme for how many years you have to invest for = ?

A. 11 years 6 months
B. 10 years 8 months
C. 10 years
D. 12 years
Answer: _________
Question 43:

A person deposits Rs. 500 in 4 years and Rs. 600 for 3 years at the same rate of simple interest in a bank. Altogether he received Rs. 190 as interest. The rate of simple interest per annum was = ?

A. 4%
B. 5%
C. 2%
D. 3%
Answer: _________
Question 44:

If the simple interest for 6 years be equal to 30% of the principal, it will be equal to the principal after

A. 10 years
B. 20 years
C. 22 years
D. 30 years
Answer: _________
Question 45:

Simple interest on a certain sum at a certain annual rate of interest is $$frac{1}{9}$$ of the sum. If the numbers representing rate percent and time in years be equal, then the rate of interest is -

A. $$3frac{1}{3}$$ %
B. 5%
C. $$6frac{2}{3}$$ %
D. 10%
Answer: _________
Question 46:

If x, y, z are three sum of money such that y is the simple interest on x and z is the simple interest on y for the same time and at the same rate of interest, then what is the relation between x, y, z we have = ?

A. x 2 = xy
B. xyz = 1
C. x 2 = yz
D. y 2 = zx
Answer: _________
Question 47:

Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, the money borrowed by him (in Rupees) was?

A. Rs. 12000
B. Rs. 6000
C. Rs. 8000
D. Rs. 10000
Answer: _________
Question 48:

A sum of Rs. 800 amounts to Rs. 920 in 3 years at the simple interest rate. If the rate is increased by 3% p.a. What will be the sum amount to in the same period ?

A. Rs. 992
B. Rs. 962
C. Rs. 942
D. Rs. 982
Answer: _________
Question 49:

Simple interest on a certain amount is $$frac{9}{{16}}$$ of the principal. If the numbers representing the rate of interest in percent and time in years be equal, then time, for which the principal is lent out, is -

A. $$5frac{1}{2}$$ years
B. $$6frac{1}{2}$$ years
C. 7 years
D. $$7frac{1}{2}$$ years
Answer: _________
Question 50:

A lends Rs. 2500 to B and a certain sum to C at the same time at 7% p.a. simple interest. If after 4 years, A altogether receives Rs. 1120 as interest from B and C, then the sum lent to C is :

A. Rs. 700
B. Rs. 1500
C. Rs. 4000
D. Rs. 6500
Answer: _________
Question 51:

Find the simple interest on Rs. 5200 for 2 years at 6% per annum.

A. Rs. 450
B. Rs. 524
C. Rs. 600
D. Rs. 624
Answer: _________
Question 52:

Rs. 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.

A. Rs. 1380
B. Rs. 1290
C. Rs. 1470
D. Rs.1200
Answer: _________
Question 53:

Interest obtained on a sum of Rs. 5000 for 3 years is Rs. 1500. Find the rate percent.

A. 8%
B. 9%
C. 10%
D. 11%
Answer: _________
Question 54:

Rs. 2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years.

A. Rs. 2300
B. Rs. 2315.25
C. Rs. 2310
D. Rs. 2320
Answer: _________
Question 55:

Find the difference between the simple interest and the compound interest at 5% per annum for 2 years on principal of Rs. 2000.

A. 5
B. 10.5
C. 4.5
D. 5.5
Answer: _________
Question 56:

Find the rate of interest if the amount after 2 years of simple interest on a capital of Rs. 1200 is Rs. 1440.

A. 8%
B. 9%
C. 10%
D. 11%
Answer: _________
Question 57:

What is the difference between the simple interest on a principal of Rs. 500 being calculated at 5% per annum for 3 years and 4% per annum for 4 years?

A. Rs. 5
B. Rs. 10
C. Rs. 20
D. Rs. 40
Answer: _________
Question 58:

What is the simple interest on a sum of Rs. 700 if the rate of interest for the first 3 years is 8% per annum and for the last 2 years is 7.5% per annum?

A. Rs. 269.5
B. Rs. 283
C. Rs. 273
D. Rs. 280
Answer: _________
Question 59:

Find the compound interest on Rs. 1000 at the rate of 20% per annum for 18 month when interest is compounded half yearly.

A. Rs. 331
B. Rs. 1331
C. Rs. 320
D. Rs. 325
Answer: _________
Question 60:

Find the principal if the interest compounded at the rate of 10% per annum for two years is Rs. 420.

A. Rs. 1000
B. Rs. 2200
C. Rs. 2000
D. Rs. 1100
Answer: _________
Question 61:

Tushar borrowed a sum of Rs. 12000 at 15% per annum from a money - lender on 13 th January, 1987 and return the amount on 8 th June, 1987 to clear his debt. Then the amount paid by Tushar to the money - lender to clear his debt was = ?

A. Rs. 15000
B. Rs. 11394
C. Rs. 12720
D. Rs. 13650
Answer: _________
Question 62:

Vishwas borrowed a total amount of Rs. 30000, part of it on simple interest rate of 12 p.c.p.a. and remaining on simple interest rate of 10 p.c.p.a. If at the end of 2 year she paid in all Rs. 36480 to settle the loan amount, what was the amount borrowed at 12 p.c.p.a ?

A. Rs. 12000
B. Rs. 16000
C. Rs. 17500
D. Rs. 18000
Answer: _________
Question 63:

A sum of Rs. 18750 is left by a will by a father to be divided between the two sons, 12 and 14 years of age, so that when they attain maturity at 18, the amount (principal + interest) received by each at 5 percent simple interest will be the same. Find the sum alloted at present to each son.

A. Rs. 9500, Rs. 9250
B. Rs. 8000, Rs. 1750
C. Rs. 9000, Rs. 9750
D. None of these
Answer: _________
Question 64:

I had Rs. 10000 with me. Out of this money I lent some money to A for 2 years @ 15% simple interest. I lent the remaining money to B for an equal number of years @18% simple interest. After 2 years, I found that A had given me Rs. 360 more as interest as compared to B. The amount of money which I had lent to B must have been.

A. Rs. 2000
B. Rs. 3000
C. Rs. 4000
D. Rs. 5000
Answer: _________
Question 65:

A certain sum of money amount to Rs 2200 at 5% interest Rs 2320 at 8% interest in the same period of time. The period of time is = ?

A. 2 years
B. 4 years
C. 5 years
D. 3 years
Answer: _________
Question 66:

For 2 years, a sum was put at SI at a certain rate. If the rate was 3% higher, it would have fetched Rs. 300 more. What will be the sum ?

A. Rs. 5000
B. Rs. 4800
C. Rs. 2500
D. Rs. 4484
Answer: _________
Question 67:

A money lender claims to lend money at the rate of 10% per annum simple interest. However, he takes the interest in advance when he lends a sum for one year. At what interest rate does he lend the money actually ?

A. 10%
B. $$10frac{1}{9}$$ %
C. 11%
D. $$11frac{1}{9}$$ %
Answer: _________
Question 68:

A sum of Rs. 1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is:

A. 5 : 8
B. 8 : 5
C. 16 : 15
D. 31 : 6
Answer: _________
Question 69:

A sum of Rs. 1440 is lent out in three parts in such away that the interests on first part at 2% for 3 years, second part at 3% for 4 years and third part at 4% for 5 years are equal. Then the difference between the largest and the smallest sum is -

A. Rs. 200
B. Rs. 400
C. Rs. 460
D. Rs. 560
Answer: _________
Question 70:

A borrows a sum of Rs. 90,000 for 4 years at 5% simple interest. He lends it to B at 7% for 4 years at simple interest. What is his gain (in Rs.)?

A. 8,000
B. 9,000
C. 7,200
D. 7,500
Answer: _________
Question 71:

A person borrowed 2,000 at 5% annual simple interest repayable in 3 equal annual installments. What will be the annual installment?

A. $${ ext{Rs}}{ ext{.}},730frac{{10}}{{63}}$$
B. $${ ext{Rs}}{ ext{.}},840frac{9}{{61}}$$
C. $${ ext{Rs}}{ ext{.}},640frac{{11}}{{63}}$$
D. $${ ext{Rs}}{ ext{.}},250frac{{10}}{{63}}$$
Answer: _________
Question 72:

A milkman borrowed a sum of Rs. 2,500 from two money lenders for one loan on simple interest at the rate of 5% and 7%, respectively, for a period of 2 years. If the total interest paid by him was Rs. 276, then how much did he borrow at the rate of 7% p.a.?

A. Rs. 700
B. Rs. 650
C. Rs. 675
D. Rs. 550
Answer: _________
Question 73:

A sum of Rs. 12,800 is invested separately at 15% per annum and the remaining at 12% per annum simple interest. If the total interest at the end of 3 years is Rs. 5,085, then how much money was invested at 15% per annum?

A. Rs. 5,200
B. Rs. 7,500
C. Rs. 5,800
D. Rs. 5,300
Answer: _________
Question 74:

A Sum of Rs. 10,500 amounts to Rs. 13,825 in $$3frac{4}{5}$$xa0years at a certain rate percent per annum simple interest. What will be the simple interest on the same sum for 5 years at double the earlier rate?

A. Rs. 8,670
B. Rs. 8,750
C. Rs. 8,470
D. Rs. 8,560
Answer: _________
Question 75:

A man has Rs. 10,000. He lent a part of it at 15% simple interest and the remaining at 10% simple interest. The total Interest he received after 5 years amount to Rs. 6,500. The difference between the parts of
the amounts he lent is:

A. Rs. 2,000
B. Rs. 2,500
C. Rs. 1,500
D. Rs. 1,750
Answer: _________
Question 76:

Simple interest on a certain sum one-fourth of the sum and the interest rate percentage per annum is 4 times the numbers of years. If the rate of interest increases by 2%, then what will be the simple interest (in Rs.) on Rs. 5,000 for 3 years?

A. 300
B. 1,500
C. 2,000
D. 1,800
Answer: _________
Question 77:

In a certain duration of time, a sum become 2 times itself at the rate of 5% per annum simple interest. What will be the rate of interest if the same sum becomes 5 times itself in the same duration?

A. 20%
B. 16%
C. 10%
D. 18%
Answer: _________
Question 78:

A car with a price of Rs. 6,50,000 is bought by making some down payment. On the balance, a simple interest of 10% is charged in lump sum and the money is to be paid in 20 equal annual instalments of Rs. 25,000. How much is the down payment?

A. Rs. 1,55,945
B. Rs. 1,95,455
C. Rs. 1,94,555
D. Rs. 1,45,955
Answer: _________
Question 79:

Brajesh had Rs. 8,600 which he invested in two parts. Simple interest received on the first part at 15% p.a. in 4 years is equal to the simple interest received on the second part at 20% p.a. in 3 years. Find the difference in the two parts.

A. Rs. 0
B. Rs. 210
C. Rs. 190
D. Rs. 290
Answer: _________
Question 80:

What will be the simple interest earned on an amount of Rs. 16,800 in 9 months at the rate of $$6frac{1}{4}$$ % p.a. ?

A. Rs. 787.50
B. Rs. 812.50
C. Rs. 860
D. Rs. 887.50
Answer: _________
Question 81:

The simple interest on a sum of money is $$frac{4}{9}$$ of the principal and the number of years is equal to the rate percent per annum. The rate per annum is = ?

A. 5%
B. $$6frac{2}{3}$$ %
C. 6%
D. $$7frac{1}{5}$$ %
Answer: _________
Question 82:

At what rate percent per annum will the simple interest on a sum of money be $$frac{2}{5}$$ of the principal amount in 10 years ?

A. 4%
B. 6%
C. $$5frac{2}{3}$$ %
D. $$6frac{2}{3}$$ %
Answer: _________
Question 83:

A sum of Rs. 1750 is divided into two parts such that the interests on the first part at 8% simple interest per annum and that on the other part at 6% simple interest per annum are equal. The interest on each part ( in Rupees) is ?

A. Rs. 60
B. Rs. 65
C. Rs. 70
D. Rs. 40
Answer: _________
Question 84:

A person borrows Rs. 5000 for 2 year at 4% p.a. simple interest. He immediately lends it to another person at $$6frac{1}{4}$$ % p.a. for 2 years. Find his gain in the transaction per year.

A. Rs. 112.50
B. Rs. 125
C. Rs. 150
D. Rs. 167.50
Answer: _________
Question 85:

Ramakant invested amounts in two different schemes A and B for five years in the ratio of 5 : 4 respectively. Scheme A offers 8% simple interest and bonus equal to 20% of the amount of interest earned in 5 years on maturity. Scheme B offers 9% simple interest. If the amount invested in scheme A was Rs. 20000, what was the total amount received on maturity from both the schemes?

A. Rs. 50800
B. Rs. 51200
C. Rs. 52800
D. Rs. 58200
Answer: _________
Question 86:

A sum of Rs. 1550 was lent partly at 5% and partly at 8% simple interest. The total interest received after 3 years is Rs. 300. The ratio of money lent at 5% to that at 8% is = ?

A. 5 : 8
B. 8 : 5
C. 31 : 6
D. 16 : 15
Answer: _________
Question 87:

What sum of money will amount to Rs. 520 in 5 years and to Rs. 568 in 7 years at simple interest?

A. Rs. 400
B. Rs. 120
C. Rs. 510
D. Rs. 220
Answer: _________
Question 88:

A money lender finds that due to fall in the annual rate of interest from 8% to $${ ext{7}}frac{3}{4}{ ext{%,}}$$xa0 his yearly income diminishes by Rs. 61.50. His capital is = ?

A. Rs. 22400
B. Rs. 23800
C. Rs. 24600
D. Rs. 26000
Answer: _________
Question 89:

How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A. 3.5 years
B. 4 years
C. 4.5 years
D. 5 years
Answer: _________
Question 90:

A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?

A. 9 years
B. 8 years
C. 27 years
D. 12 years
Answer: _________
Question 91:

Divide Rs. 6000 into two parts so that simple interest on the first part for 2 years at 6% p.a. may be equal to the simple interest on the second part for 3 years at 8% p.a.

A. Rs. 4000, Rs. 2000
B. Rs. 5000, Rs. 1000
C. Rs. 3000, Rs. 3000
D. None of these
Answer: _________
Question 92:

A sum of money becomes $$frac{7}{4}$$ of itself in 6 years at a certain rate of simple interest. Find the rate of interest.

A. 12%
B. $$12frac{1}{2}$$ %
C. 8%
D. 14%
Answer: _________
Question 93:

If a sum of Rs. 13040 is to be repaid in two equal installments at $$3frac{3}{4}$$ % per annum, what is the amount of each installment?

A. 7045
B. 8000
C. 65067
D. 6889
Answer: _________
Question 94:

What is the amount of equal installment, if a sum of Rs. 1428 due 2 years hence has to be completely repaid in 2 equal annual installments starting next year?.

A. 700
B. 800
C. 650
D. cannot be determined
Answer: _________
Question 95:

A milkman sells cow milk at the rate of Rs. 55 litre including a profit 12 per cent. He also sells buffalo milk at the rate of Rs. 36 per litre including a profit of 20%. How much profit will he earn in five days if he sells 8 litres of cow milk and 10 litres of buffalo milk per day?

A. Rs. 632
B. Rs. 624
C. Rs. 646
D. Rs. 642
Answer: _________
Question 96:

The population of vultures in a particular locality is decreases by certain rate of interest (compounded annually). If the current population of vultures be 29160 and the ratio of decrease in population for second year and 3rd year be 10 : 9. What was the population of vultures 3 years ago?

A. 30000
B. 35000
C. 40000
D. 50000
Answer: _________
Question 97:

A man had 1000 hens at the beginning of year 2001 and the number of hens each year increases by 10% by giving birth. At the end of each year we double the no. of hens by purchasing the same no. of hens as there is the no. of hens with us at the time. What is the no. of hens at the beginning of 2004?

A. 10600
B. 10648
C. 8848
D. 8226
Answer: _________
Question 98:

The difference between simple and compound interest for the fourth year is Rs. 7280 at 20% p.a. What is the principal sum?

A. 10000
B. 50000
C. 70000
D. 40000
Answer: _________
Question 99:

Rahul borrowed a sum of Rs. 1150 from Amit at the simple interest rate of 6 p.c.p.a. for 3 Years. He then added some more money to the borrowed sum and lent it to Sachin for the same time at 9 p.c.p.a simple interest. If Rahul gains Rs. 274.95 by way of interest on borrowed sum as well as his own amount from the whole transaction, then what is the sum lent by him to Sachin ?

A. Rs. 1200
B. Rs. 1285
C. Rs. 1690
D. Rs. 1785
Answer: _________
Question 100:

The amount Rs. 2100 become Rs. 2352 in 2 years at simple interest. If the interest rate is decreased by 1% , what is the new interest ?

A. Rs. 210
B. Rs. 220
C. Rs. 242
D. Rs. 252
Answer: _________
Question 101:

Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amounts in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is Rs. 1350 then the sum deposited in each case is = ?

A. Rs. 3000
B. Rs. 4000
C. Rs. 6500
D. Rs. 5000
Answer: _________
Question 102:

A some of money lent out at simple interest amount to Rs. 720 after 2 years and Rs. 1020 after a further period of 5 years. Find the principal ?

A. Rs. 6000
B. Rs. 600
C. Rs. 1740
D. Rs. 120
Answer: _________
Question 103:

A person invested some account at the rate of 12% simple interest and a certain amount at rate of 10% simple interest. He received yearly interest of Rs. 130. But if he had interchanged the amounts invested,he would have received Rs. 4 more as interest. How much did he invest at 12% simple interest ?

A. Rs. 400
B. Rs. 500
C. Rs. 700
D. Rs. 800
Answer: _________
Question 104:

Two equal sums of money are lent at the same time at 8% and 7% per annum simple interest. The former is recovered 6 months earlier than the latter and the amount in each case is Rs. 2560. The sum and the time for which the sums of money are lent out are.

A. Rs. 2000, 3.5 years and 4 years
B. Rs. 1500, 3.5 years and 4 years
C. Rs. 2000, 4 years and 5.5 years
D. Rs. 3000, 4 years and 4.5 years
Answer: _________
Question 105:

A sum of Rs. 7930 is divided into 3 parts and given at loan at 5% simple interest to A, B and C for 2, 3 and 4 years respectively. If the amounts of all three are equal after their respective periods of loan, then the A received a loan of = ?

A. Rs. 2800
B. Rs. 3050
C. Rs. 2750
D. Rs. 2760
Answer: _________
Question 106:

The principal which gives Rs 1 interest per day at a rate of 5% simple interest per annum is = ?

A. Rs. 5000
B. Rs. 35500
C. Rs. 7300
D. Rs. 3650
Answer: _________
Question 107:

Arvind deposited a sum of money with a bank on 1 st january, 2012 at 8% simple interest per annum. He received an amount 3144 on 7 th August, 2012. The money he deposited with the bank was = ?

A. Rs. 3080
B. Rs. 2500
C. Rs. 3000
D. Rs. 3100
Answer: _________
Question 108:

A man invested Rs. 5000 at some rate of simple interest and Rs. 4000 at 1 percent higher rate of interest. If the interest in both the cases after 4 years is same, the rate of interest in the former case is

A. 4% p.a.
B. 5% p.a.
C. $$6frac{1}{4}$$ % p.a.
D. $$8frac{1}{3}$$ % p.a.
Answer: _________
Question 109:

Simple interest on Rs. 500 for 4 years at 6.25% per annum is equal to the simple interest on Rs. 400 at 5% per annum for a certain period of time. The period of time is = ?

A. 4 years
B. 5 years
C. $$6frac{1}{4}$$ years
D. $$8frac{2}{3}$$ years
Answer: _________
Question 110:

With a given rate of simple interest, the ratio of principal and amount for a certain period of time is 4 : 5. After 3 years with the same rate of interest, the ratio of the principal and amount becomes 5 : 7. The rate of interest is = ?

A. 4%
B. 6%
C. 5%
D. 7%
Answer: _________
Question 111:

If x, y, z are three sums of money such that y is the simple interest on x, z is the simple interest on y for the same time and at the same rate of interest, then we have.

A. x 2 = yz
B. y 2 = xz
C. z 2 = xy
D. xyz = 1
Answer: _________
Question 112:

Arun borrowed a sum of money from Jayant at the rate of 8% per annum simple interest for the first four years, 10% per annum for the next 6 years and 12% per annum for the period beyond 10 years. If he pays a total of Rs. 12160 as interest only at the end of 15 years, how much money did he borrow?

A. Rs. 8000
B. Rs. 9000
C. Rs. 10000
D. Rs. 12000
Answer: _________
Question 113:

Kruti took a loan at simple interest rate of 6 p.c.p.a. in the first year and it increased by 1.5 p.c.p.a. every year. If she pays Rs. 8190 as interest at the end of 3 years, what was her loan amount ?

A. Rs. 35400
B. Rs. 36000
C. Rs. 36800
D. Cannot be determined
Answer: _________
Question 114:

A sum of money at a certain rate per annum of simple interest doubles in the 5 years and at a different rate becomes three times in 12 years. The lower rate of interest per annum is = ?

A. 15%
B. 20%
C. $${ ext{15}}frac{3}{4}$$ %
D. $$16frac{2}{3}$$ %
Answer: _________
Question 115:

If Rs. 12000 is divided into two parts such that the simple interest on the first part for 3 years at 12% per annum is equal to the simple interest on the second part for $$4frac{1}{2}$$ years at 16% per annum, the greater part is = ?

A. Rs. 8000
B. Rs. 6000
C. Rs. 7000
D. Rs. 7500
Answer: _________
Question 116:

An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes

A. 10%
B. 10.25%
C. 10.5%
D. None of these
Answer: _________
Question 117:

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is

A. Rs. 650
B. Rs. 690
C. Rs. 698
D. Rs. 700
Answer: _________
Question 118:

In what time will Rs. 3300 becomes Rs. 3399 at 6% per annum interest compounded half-yearly?

A. 6 months
B. 1 year
C. $$1frac{1}{2}$$ years
D. 3 months
Answer: _________
Question 119:

What will be the simple interest on Rs. 700 at 9% per annum for the period from February 5, 1994 to April 18, 1994?

A. Rs. 12.60
B. Rs. 11.30
C. Rs. 15
D. Rs. 13
Answer: _________
Question 120:

A sum was invested at simple interest at a certain interest for 2 years. It would have fetched Rs. 60 more had it been invested at 2% higher rate. What was the sum?

A. Rs. 1500
B. Rs. 1300
C. Rs. 2500
D. Rs. 1000
Answer: _________
Question 121:

The difference between simple and compound interest on a sum of money at 20% per annum for 3 years is Rs. 48. What is the sum?

A. Rs. 550
B. Rs. 500
C. Rs. 375
D. Rs. 400
Answer: _________
Question 122:

In what time will the simple interest on Rs. 1750 at 9% per annum be the same as that on Rs. 2500 at 10.5% per annum in 4 years?

A. 6 years and 8 months
B. 7 years and 3 months
C. 6 years
D. 7 years and 6 months
Answer: _________
Question 123:

Raju lent Rs. 400 to Ajay for 2 years and Rs. 100 to Manoj for 4 years and received together from both Rs. 60 as interest. Find the rate of interest, simple interest being calculated.

A. 5%
B. 6%
C. 8%
D. 9%
Answer: _________
Question 124:

Suresh lent out a sum of money to Rakesh for 5 years at simple interest. At the end of 5 years. Rakesh paid 9/8 of the sum to Suresh to clear out the amount. Find the rate of simple interest per annum.

A. 3% p.a.
B. 2% p.a.
C. 3.5% p.a.
D. 2.5% p.a.
Answer: _________
Question 125:

A sum at a simple interest of 8% p.a. becomes $$frac{7}{5}$$ of itself in how many years?

A. 5
B. $$2frac{1}{2}$$
C. $$3frac{1}{2}$$
D. 2
Answer: _________
Question 126:

A man takes a loan of some amount at some rate of simple interest. After three years, the loan amount is doubled and rate of interest is decreased by 2%. After 5 years, if the total interest paid on the whole is Rs. 13,600, which is equal to the same when the first amount was taken for $$11frac{1}{3}$$xa0years, then the loan taken initially is:

A. Rs. 13,600
B. Rs. 12,500
C. Rs. 10,000
D. Rs. 12,000
Answer: _________
Question 127:

A person borrows Rs. 7,000 for 3 year's at 5% p.a. simple interest. He immediately lends it to another person at $$6frac{1}{3}\% $$ xa0p.a. for 3 years. Find the gain in the transaction per year.

A. Rs. 90
B. Rs. 93.33
C. Rs. 92
D. Rs. 95.33
Answer: _________
Question 128:

What annual instalment will discharge a debit of Rs. 5,664 in 4 years at 12% simple interest?

A. Rs. 1,230
B. Rs. 1,210
C. Rs. 1,200
D. Rs. 1,220
Answer: _________
Question 129:

A sum at simple interest becomes two times in 8 years at a certain rate of interest p.a. The time in which the same sum will be 4 times at the same rate of interest at simple interest is:

A. 30 years
B. 25 years
C. 24 years
D. 20 years
Answer: _________
Question 130:

What annual installment will discharge a debt of Rs. 3,270 due in 3 years at 9% per annum simple interest?

A. Rs. 1,090
B. Rs. 1,050
C. Rs. 1,000
D. Rs. 1,075
Answer: _________
Question 131:

If the annual rate of simple interest increase from 11% to $$17frac{1}{2}\% ,$$ xa0then a person's yearly income increase by Rs. 1,071.20. The simple interest (in Rs.) on the same sum at 10% for 5 year is:

A. 16,480
B. 9,120
C. 7,250
D. 8,240
Answer: _________
Question 132:

A sum of Rs. 36,000 is divided into two parts A and B, such that the simple interest at the rate of 15% p.a. on A and B after two years and four years, respectively, is equal. The total interest (in Rs.) received from A is:

A. 7,200
B. 5,400
C. 3,600
D. 1,800
Answer: _________
Question 133:

If the simple interest on a sum of Rs. P at 5% per annum for three years is thrice the simple interest received on Rs. Q at 7% per annum for four years, then find the relation between P and Q.

A. P = 5.6Q
B. Q = 2.1P
C. P = 4.8Q
D. Q = 3.1P
Answer: _________
Question 134:

If a sum doubles in 16 years, how much will it be in 8 years ?

A. $$1frac{1}{2}$$ times
B. $$1frac{1}{3}$$ times
C. $$1frac{1}{4}$$ times
D. $$1frac{3}{4}$$ times
Answer: _________
Question 135:

Consider the following statements If a sum of money is lent at simple interest, then the I - money gets doubled in 5 years if the rate of interest is $$16frac{2}{3}$$ % II - money gets doubled in 5 years if the rate of interest is 20%. III - money becomes four times in 10 years if it gets doubled in 5 years.

A. I and III are correct
B. II alone is correct
C. III alone is correct
D. II and III are correct
Answer: _________
Question 136:

In a certain time, the ratio of a certain principal and interest obtained from it are in the ratio 10 : 3 at 10% interest per annum. The number of years for which the money was invested is = ?

A. 1 year
B. 3 year
C. 5 year
D. 7 year
Answer: _________
Question 137:

Jhon invested a sum of money at an annual simple interest rate of 10%. At the end of four years the amount invested plus interest earned was Rs. 770. The amount invested was = ?

A. Rs. 650
B. Rs. 350
C. Rs. 550
D. Rs. 500
Answer: _________
Question 138:

In what time will Rs. 1860 amount to 2641.20 at simple interest 12% per annum ?

A. 3 years
B. $$3frac{1}{2}$$ years
C. 4 years
D. $$4frac{1}{2}$$ years
Answer: _________
Question 139:

The simple interest on a sum of money at 8% per annum for 6 years is half the sum. The sum is:

A. Rs. 4800
B. Rs. 6000
C. Rs. 8000
D. Date inadequate
Answer: _________
Question 140:

In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?

A. $$1frac{1}{4}$$ years
B. $$1frac{3}{4}$$ years
C. $$2frac{1}{4}$$ years
D. $$2frac{3}{4}$$ years
Answer: _________
Question 141:

The population of a village decreases at the rate of 20% per annum. If its population 2 years ago was 10000, the present population is = ?

A. 4600
B. 6400
C. 7600
D. 6000
Answer: _________
Question 142:

Rs. 12000 is divided into two parts such that simple interest on the first part for 3 years at 12% per annum may be equal to the simple interest on the second part for $$4frac{1}{2}$$ years at 16% per annum. The ratio of the first part to the second part is = ?

A. 2 : 1
B. 1 : 2
C. 2 : 3
D. 3 : 2
Answer: _________
Question 143:

A person who pays income tax at the rate of 4 paise per rupee, find that fall of interest rate (income tax) from 4% to 3.75% diminishes his net yearly income by Rs. 48. What is his capital ?

A. Rs. 24000
B. Rs. 25000
C. Rs. 20000
D. Rs. 18000
Answer: _________
Question 144:

Veena obtained an amount of Rs. 8376 as simple interest on a certain amount at 8 p.c.p.a. after 6 years. what is the amount invested by veena?

A. Rs. 16660
B. Rs. 17180
C. Rs. 17450
D. Rs. 18110
Answer: _________
Question 145:

At which sum the simple interest at the rate of $$3frac{3}{4}$$ % per annum will be Rs. 210 in $$2frac{1}{3}$$ years?

A. Rs. 1580
B. Rs. 2400
C. Rs. 2800
D. None of these
Answer: _________
Question 146:

If the simple interest for 6 years be equal to 30% of the principal, it will be equal to the principal after = ?

A. 20 years
B. 30 years
C. 10 years
D. 22 years
Answer: _________
Question 147:

A person invests money in three different schemes for 6 years, 10 years and 12 years at 10%, 12% and 15% simple interest respectively. At the completion of each scheme, he gets the same interest. The ratio of his investment is = ?

A. 6 : 3 : 2
B. 2 : 3 : 4
C. 3 : 4 : 6
D. 3 : 4 : 2
Answer: _________
Question 148:

Rs. 1000 is invested at 5% per annum simple interest. If the interest is added to the principal after every 10 years, the amount will become Rs. 2000 after = ?

A. 15 years
B. 18 years
C. 20 years
D. $$16frac{2}{3}$$ years
Answer: _________
Question 149:

The simple interest at x% for x years will be Rs. x on a sum of:

A. Rs. x
B. Rs. $$frac{{100}}{x}$$
C. Rs. 100x
D. Rs. $$frac{{100}}{{{x^2}}}$$
Answer: _________
Question 150:

Rs. 6200 amounts to Rs. 9176 in 4 years at simple interest. If the interest rate is increased by 3% it would amount to how much?

A. Rs. 8432
B. Rs. 9820
C. Rs. 9920
D. Rs. 10920
Answer: _________
Question 151:

The simple interest accrued on a certain principal in 5 years at the rate of 12 p.c.p.a. is Rs.1536. what amount of simple interest would one get if one invests Rs.1000 more than the previous principal for 2 years and at the same rate p.c.p.a. ?

A. Rs. 614.40
B. Rs. 845.40
C. Rs. 1536
D. Rs. 2136
Answer: _________
Question 152:

The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs. 840. At what rate of interest the same account of interest can be received on the same sum after 5 years?

A. 6%
B. 8%
C. 9%
D. 10%
Answer: _________
Question 153:

Rs. 6000 becomes Rs. 7200 in 4 years. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be = ?

A. Rs. 8000
B. Rs. 8250
C. Rs. 9250
D. Rs. 9000
Answer: _________
Question 154:

A sum becomes its double in 10 years. Find the annual rate of simple interest.

A. 8%
B. 5%
C. 10%
D. 20%
Answer: _________
Question 155:

The interest earned on Rs. 15000 in 3 years at simple interest is Rs. 5400. Find the rate of interest per annum.

A. 11.5%
B. 12%
C. 12.5%
D. 15%
Answer: _________
Question 156:

The sum invested in scheme B is thrice the sum invested in scheme A. The investment in scheme A is made for 4 years at 8% p.a. simple interest and in scheme B for 2 years at 13% p.a. simple interest. The total interest earned from both the schemes is Rs. 1320. How much amount was invested in scheme A?

A. Rs. 1200
B. Rs. 1140
C. Rs. 960
D. Rs. 1500
Answer: _________
Question 157:

In simple interest rate per annum a certain sum amounts to Rs. 5182 in 2 years and Rs. 5832 in 3 years. The principal in rupees is = ?

A. Rs. 2882
B. Rs. 5000
C. Rs. 3882
D. Rs. 4000
Answer: _________
Question 158:

If a sum of money becomes Rs. 4000 in 2 years and Rs. 5500 in 4 years 6 months at the same rate of simple interest per annum. Then the rate of simple interest is = ?

A. $${ ext{21}}frac{3}{7}$$ %
B. $${ ext{21}}frac{2}{7}$$ %
C. $${ ext{21}}frac{1}{7}$$ %
D. $${ ext{21}}frac{5}{7}$$ %
Answer: _________
Question 159:

Rs. 260200 is divided between Ram and Shyam so that the amount that Ram receives in 3 years is the same as that Shyam receives in 6 years. If the interest is compounded annually at the rate of 4% per annum then Ram's share is = ?

A. Rs. 12500
B. Rs. 137745.02
C. Rs. 15200
D. Rs. 108200
Answer: _________
Question 160:

A person invested a total of Rs. 57,500 at 4%, 5% and 8% per annum simple interest. At the end of the year, he received equal interest in all the three cases. The amount invested at 5% was:

A. Rs. 20,000
B. Rs. 20,750
C. Rs. 25,000
D. Rs. 37,500
Answer: _________
Question 161:

A digital note-pad is available for Rs. 25,000 cash or Rs. 2,500 down payment followed by 4 equal monthly instalments. If the rate of interest charged is 24% per annum simple interest, what is the monthly instalment (in ., rounded off to the nearest tens)?

A. Rs. 5,800
B. Rs. 5,790
C. Rs. 5,890
D. Rs. 5,900
Answer: _________
Question 162:

The simple interest on a certain sum for 3 years at 14% p.a. is 4,200 less than the simple interest on the same sum for 5 years at the same rate. Find the sum.

A. Rs. 16,000
B. Rs. 10,000
C. Rs. 15,000
D. Rs. 12,000
Answer: _________
Question 163:

A sum of Rs. 5,000 divided into two parts such that the simple interest on the first part for $$4frac{1}{5}$$xa0years at $$6frac{2}{3}\% $$ xa0p.a. is double the simple interest on the second part for $$2frac{3}{4}$$xa0years at 4% p.a. The ratio of the second part to the first part is:

A. 11 : 13
B. 13 : 11
C. 14 : 11
D. 11 : 14
Answer: _________
Question 164:

What will be the simple interest on a sum of Rs. 12000 at the rate of 15% per annum of three years?

A. Rs. 5400
B. Rs. 6000
C. Rs. 7200
D. Rs. 4500
Answer: _________
Question 165:

A sum amounts to Rs. 14,395.20 at 9.25% p.a. simple interest in 5.4 years. What will be the simple interest on the same sum at 8.6% p.a. in 4.5 years?

A. Rs. 3,715.20
B. Rs. 3,627
C. Rs. 3,797.76
D. Rs. 3,672
Answer: _________
Question 166:

In how much time will the simple interest on a certain sum of money be $$frac{6}{5}$$ times of the sum of 20% per annum?

A. 7 years
B. 8 years
C. 5 years
D. 6 years
Answer: _________
Question 167:

A sum of Rs. 10 is lent by a child to his friend to be returned in 11 monthly instalment a of Rs. 1 each, the interest being simple. The rate of interest is:

A. $$11frac{9}{{11}}\% $$
B. $$21frac{9}{{11}}\% $$
C. $$10frac{2}{{11}}\% $$
D. $$9frac{1}{{11}}\% $$
Answer: _________
Question 168:

A sum of money at simple interest amounts of Rs. 6,000 in 4 years and to Rs. 6,750 in 7 years at the same rate percent p.a. of interest. The sum (in Rs.) is:

A. 5,100
B. 4,800
C. 4,000
D. 5,000
Answer: _________
Question 169:

Sum Rs. 20000 and Rs. 40000 are given on simple interest at the rate of 10% and 15% per annum respectively for three years. What will be the total simple interest?

A. Rs. 36000
B. Rs. 32000
C. Rs. 24000
D. Rs. 28000
Answer: _________
Question 170:

Find the simple interest on Rs. 2,700 for 8 months at 5 paisa per rupee per month?

A. Rs. 950
B. Rs. 720
C. Rs. 540
D. Rs. 1,080
Answer: _________
Question 171:

A sum of Rs. 50,250 is divided into two parts such that he simple interest on the first part for $$7frac{1}{2}$$xa0years at $$8frac{1}{3}\% $$ xa0p.a. $$frac{5}{2}$$ is times the simple interest on the second part for $$5frac{1}{4}$$xa0years at 8% p.a. What is the difference (in Rs.) between the two parts?

A. 12,750
B. 12,570
C. 15,270
D. 10,275
Answer: _________
Question 172:

A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of the interest be 10% compounded annually, then the value of each installment is:

A. Rs. 121
B. Rs. 127
C. Rs. 210
D. Rs. 225
Answer: _________
Question 173:

There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?

A. Rs. 7,200
B. Rs. 7,862
C. Rs. 8,000
D. Rs. 8,500
Answer: _________
Question 174:

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A. Rs. 650
B. Rs. 690
C. Rs. 698
D. Rs. 700
Answer: _________
Question 175:

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

A. Rs. 6400
B. Rs. 6500
C. Rs. 7200
D. Rs. 7500
Answer: _________
Question 176:

A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

A. Rs. 4462.50
B. Rs. 8032.50
C. Rs. 8900
D. Rs. 8925
Answer: _________
Question 177:

How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A. 3.5 years
B. 4 years
C. 4.5 years
D. 5 years
Answer: _________
Question 178:

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

A. 3.6
B. 6
C. 18
D. Cannot be determined
Answer: _________
Question 179:

A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

A. 3%
B. 4%
C. 5%
D. 6%
Answer: _________
Question 180:

An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:

A. 10%
B. 10.25%
C. 10.5%
D. None of these
Answer: _________
Question 181:

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:

A. 5%
B. 7%
C. 9%
D. 10%
Answer: _________
Question 182:

A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:

A. Rs. 2000
B. Rs. 10,000
C. Rs. 15,000
D. Rs. 20,000
Answer: _________
Question 183:

A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:

A. 5%
B. 8%
C. 12%
D. 15%
Answer: _________
Question 184:

What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?

A. 1 : 3
B. 1 : 4
C. 2 : 3
D. Data inadequate
Answer: _________
Question 185:

A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?

A. Rs. 35
B. Rs. 245
C. Rs. 350
D. Cannot be determined
Answer: _________
Question 186:

A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6$$frac{1}{4}$$ p.a for 2 years. Find his gain in the transaction per year.

A. Rs. 112.50
B. Rs. 125
C. Rs. 150
D. Rs. 167.50
Answer: _________
Question 187:

A sum of money becomes $$frac{7}{6}$$ of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?

A. $$5frac{5}{9}$$ %
B. $$6frac{5}{9}$$ %
C. 18 %
D. 25 %
Answer: _________
Question 188:

The difference between the simple interest received from two different sources on Rs. 1500 for 3 years is Rs. 13.50. The difference between their rates of interest is ?

A. 0.1%
B. 0.2%
C. 0.3%
D. 0.4%
Answer: _________
Question 189:

A sum of Rs. 1600 gives a simple interest of Rs. 252 in 2 years and 3 months. The rate of interest per annum is = ?

A. $$5frac{1}{2}$$ %
B. 8 %
C. 7 %
D. 6 %
Answer: _________
Question 190:

Ram borrows Rs. 520 from Gaurav at a simple interest of 13% per annum. What amount of money should Ram pay to Gaurav after 6 months to be absolved of the debt?.

A. Rs. 353.80
B. Rs. 453.80
C. Rs. 552.80
D. Rs. 553.80
Answer: _________
Question 191:

The simple interest on a sum of money is $$frac{1}{4}$$ of the principal and the number of years is equal to rate percent per annum. The rate percent is = ?

A. 2.5%
B. 5%
C. 7.5%
D. 10%
Answer: _________
Question 192:

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both as interest. The rate of interest per annum is = ?

A. 7%
B. 5%
C. $$7frac{1}{8}$$ %
D. 10%
Answer: _________
Question 193:

What annual installment will discharge a debt of Rs. 6450 due in 4 years at 5% simple interest = ?

A. Rs. 1500
B. Rs. 1835
C. Rs. 1935
D. Rs. 1950
Answer: _________
Question 194:

A sum of money lent out at simple interest amounts to Rs. 720 after 2 years and to Rs.1020 after a further period of 5 years. The sum is

A. Rs. 500
B. Rs. 600
C. Rs. 700
D. Rs. 710
Answer: _________
Question 195:

A sum of money becomes Rs. 20925 in 2 years and Rs. 24412.50 in 5 years. Find the rate of interest and the sum of money.

A. 6.25%, Rs. 18600
B. 6.75%, Rs. 17775
C. 7%, Rs. 18000
D. 8%, Rs. 17560
Answer: _________
Question 196:

In how many years will a sum of money double itself at 18.75% per annum simple interest?

A. 4 years 5 months
B. 5 years 4 months
C. 6 years 2 months
D. 6 year 5 months
Answer: _________
Question 197:

In how many years will the simple interest on a sum of money be equal to the principal at the rate of $$16frac{2}{3}$$ % per annum ?

A. 4 years
B. 5 years
C. 6 years
D. 8 years
Answer: _________
Question 198:

A sum of money was invested at a certain rate of simple interest for 2 years. Had it been invested at 1% higher rate, it would have fetched Rs. 24 more interest. The sum of money is ?

A. Rs. 1200
B. Rs. 1050
C. Rs. 1000
D. Rs. 9600
Answer: _________
Question 199:

A man invests half of his capital at the rate of 10% per annum, one - third at 9% and the rest at 12% per annum. The average rate of interest per annum, which he gets is = ?

A. 9%
B. 10%
C. 10.5%
D. 12%
Answer: _________
Question 200:

A sum of money at simple interest doubles in 7 years. It will become four times in:

A. 18 years
B. 21 years
C. 38 years
D. 42 years
Answer: _________
Question 201:

A certain sum doubles in 7 years at simple interest. The same sum under the same interest rate will become 4 times in how many years = ?

A. 14 years
B. 28 years
C. 21 years
D. 10 years
Answer: _________
Question 202:

On a certain sum the simple interest for $$12frac{1}{2}$$ year is $$frac{3}{4}$$ of the sum. Then the rate of interest is = ?

A. 5% per year
B. 6% per year
C. 7% per year
D. 8% per year
Answer: _________
Question 203:

If the simple interest on Rs. 1 for 1 month is 1 paisa, then the rate percent per annum will be = ?

A. 10%
B. 8%
C. 12%
D. 6%
Answer: _________
Question 204:

A person invests money in three different schemes for 6 years, 10 years and 12 years at 10 percent, 12 percent and 15 percent simple interest respectively. At the completion of each scheme, he gets the same interest. The ratio of his investment is

A. 2 : 3 : 4
B. 4 : 3 : 2
C. 3 : 4 : 6
D. 6 : 3 : 2
Answer: _________
Question 205:

Find the amount to be received after 2 years 6 months at the rate of 5% p.a. of simple interest on a sum of Rs. 3200.

A. Rs. 3800
B. Rs. 3500
C. Rs. 3600
D. Rs. 3900
Answer: _________
Question 206:

The simple interest on a certain sum of money at the rate of 5% per annum for 8 years is Rs. 840. Rate of interest for which the same amount of interest can be received on the same sum after 5 years is = ?

A. 7%
B. 8%
C. 9%
D. 10%
Answer: _________
Question 207:

If a sum of money double itself in 8 years, then the interest rate in percentage is ?

A. $$8frac{1}{2}$$ %
B. 10%
C. $$10frac{1}{2}$$ %
D. $$12frac{1}{2}$$ %
Answer: _________
Question 208:

Alipta got some amount of money from her father. In how many years will the ratio of the money and the interest obtained from it be 10 : 3 at 6% simple interest per annum ?

A. 7 years
B. 3 years
C. 5 years
D. 4 years
Answer: _________
Question 209:

A sum of Rs. 3000 yields an interest of Rs. 1080 at 12% per annum simple interest in how many years ?

A. 4 years
B. 3 years
C. 5 years
D. $$2frac{1}{2}$$ years
Answer: _________
Question 210:

A sum lent out at simple interest amounts to Rs. 6,076 in 1 year and
Rs. 7,504 in 4 years. The sum and the rate of interest p.a. are respectively:

A. Rs. 5,600 and 9%
B. Rs. 5,600 and 8.5%
C. Rs. 5,400 and 9%
D. Rs. 5,400 and 10%
Answer: _________
Question 211:

A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

A. 3.6%
B. 4.5%
C. 5%
D. 6%
Answer: _________
Question 212:

A man invested $$frac{{ ext{1}}}{{ ext{3}}}$$ of his capital at 7%

$$frac{{ ext{1}}}{{ ext{4}}}$$ at 8% and the remainder at 10%. If his annual income is Rs. 561, the capital is -

A. Rs. 5400
B. Rs. 6000
C. Rs. 6600
D. Rs. 7200
Answer: _________
Question 213:

Rahul purchased a Maruti van for Rs. 1, 96,000 and the rate of depreciation is $$14frac{2}{7}\% $$xa0 per annum. Find the value of the van after two years.

A. Rs. 1,40,000
B. Rs.1,44,000
C. Rs. 1,50,000
D. Rs. 1,60,000
Answer: _________
Question 214:

What is the rate of simple interest for the first 4 years if the sum of Rs. 360 becomes Rs. 540 in 9 years and the rate of interest for the last 5 years is 6%?

A. 4%
B. 5%
C. 3%
D. 6%
Answer: _________
Question 215:

Asif borrows Rs. 1500 from two moneylenders. He pays interest at the rate of 12% per annum for one loan and at the rate of 14% per annum for the other. The total interest he pays for the entire year is Rs. 186. How much does he borrow at the rate of 12%

A. Rs. 1200
B. Rs.1300
C. Rs. 1400
D. Rs. 300
Answer: _________
Question 216:

A sum becomes 4 times at simple interest in 10 years. What is the rate of interest?

A. 10%
B. 20%
C. 30%
D. 40%
Answer: _________
Question 217:

If a certain sum of money becomes doubles at simple interest in 12 years, what would be the rate of interest per annum?

A. $$8frac{1}{3}$$
B. 10
C. 12
D. 14
Answer: _________
Question 218:

An amount of Rs. 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and second, 11% p.a. If the total interest at the end of one year is $$9frac{3}{4}$$ %, then the amount invested in each share was -

A. Rs. 52, 500 Rs. 47, 500
B. Rs. 62, 500 Rs. 37, 500
C. Rs. 72, 500 Rs. 27, 500
D. Rs. 82, 500 Rs. 17, 500
Answer: _________

Answer Key

1: C
Solution: $$eqalign{
& { ext{P}} = { ext{Rs}}{ ext{. 9534}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}}left( {{ ext{11442}} - { ext{9534}}}
ight) cr
& ,,,,,,,,, = { ext{Rs}}{ ext{.1908}} cr
& { ext{R}} = { ext{4}}\% cr
& herefore { ext{Time}} = left( {frac{{{ ext{100}} imes { ext{1908}}}}{{{ ext{9534}} imes { ext{4}}}}}
ight){ ext{years}} cr
& = left( {frac{{{ ext{47700}}}}{{{ ext{9534}}}}}
ight){ ext{years}} approx { ext{5 years}} cr} $$
2: C
Solution: $$eqalign{
& { ext{Amount Rs 7000}} cr
& { ext{Total interest in 5 years}} cr
& { ext{ = 5}} imes frac{{10}}{3}\% = frac{{50}}{3}\% = frac{1}{6} cr} $$ Principal Amount 6 (6 + 1) ↓ × 1000 ↓ × 1000 6000 7000 Hence required principal
= Rs. 6000
3: D
Solution: Let the amount invested = Rs. P According to the questions, $$eqalign{
& Leftrightarrow frac{{{ ext{P}} imes 9 imes { ext{2}}}}{{100}} + frac{{{ ext{P}} imes { ext{10}} imes { ext{2}}}}{{100}} = 760 cr
& Leftrightarrow frac{{ {18{ ext{P + 20P}}} }}{{100}} = 760 cr
& Leftrightarrow 38{ ext{P = 76000}} cr
& Leftrightarrow { ext{P = 2000}} cr} $$ Alternate Total interest percent $$eqalign{
& { ext{ = }}left( {9 imes 2}
ight)\% + left( {10 imes 2}
ight)\% cr
& Rightarrow 38\% = 760 cr
& Rightarrow 100\% = 2000 cr} $$ Hence required principal = Rs. 2000
4: C
Solution: $$eqalign{
& {{ ext{T}}_1} = 15operatorname{months} cr
& ,,,,,, = frac{{15}}{{12}}years cr
& {R_1} = 7frac{1}{2}\% = frac{{15}}{2}\% cr
& {{ ext{T}}_2} = 8operatorname{months} cr
& ,,,,,,, = frac{8}{{12}}years cr
& {{ ext{R}}_2} = 12frac{1}{2}\% = frac{{25}}{2}\% cr
& { ext{Let the principal}} = { ext{P}} cr
& { ext{According to the question,}} cr
& Leftrightarrow frac{{{ ext{P}} imes { ext{15}} imes { ext{15}}}}{{12 imes 2 imes 100}} - frac{{{ ext{P}} imes 25 imes 8}}{{12 imes 2 imes 100}} = 32.50 cr
& Rightarrow frac{{225{ ext{P}}}}{{2400}} - frac{{200{ ext{P}}}}{{2400}} = 32.50 cr
& Rightarrow frac{{25{ ext{P}}}}{{2400}} = 32.50 cr
& Rightarrow { ext{P = Rs 3120}} cr
& { ext{Hence required principal}} cr
& { ext{ = Rs}}{ ext{. 3120}} cr} $$
5: N/A
Solution: $$eqalign{
& { ext{P}} = { ext{Rs}}{ ext{.}},{ ext{21250}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}},left( {{ ext{26350}} - { ext{21250}}}
ight) cr
& ,,,,,,,,,, = { ext{Rs}}{ ext{.}},5100 cr
& { ext{T}} = { ext{6 years}} cr
& herefore { ext{Rate}} = left( {frac{{100 imes 5100}}{{{ ext{21250}} imes { ext{6}}}}}
ight)\% cr
& ,,,,,,,,,,,,,,,,,,, = 4\% cr} $$
6: N/A
Solution: $$eqalign{
& { ext{P}} = { ext{Rs.}}{ ext{ 8630}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs.}}{ ext{ 1553}}{ ext{.40}} cr
& { ext{T}} = { ext{3 years}} cr
& herefore { ext{Rate}} cr
& = left( {frac{{100 imes 1553.40}}{{8630 imes 3}}}
ight)\% cr
& = 6\% cr} $$
7: B
Solution: $$eqalign{
& { ext{Let the rate of be R}}\% { ext{p}}{ ext{.a}}{ ext{.}} cr
& Then, cr
& {frac{{600 imes R imes 4}}{{100}}} + {frac{{600 imes R imes 2}}{{100}}} = 180 cr
& Rightarrow 2400{ ext{R}} + 1200{ ext{R}} = 18000 cr
& Rightarrow 3600{ ext{R}} = 18000 cr
& Rightarrow { ext{R}} = 5\% cr} $$
8: B
Solution: $$eqalign{
& { ext{Let principal = P}} cr
& herefore { ext{amount = 3P}} cr
& { ext{Interest = 3P}} - { ext{P}} cr
& ,,,,,,,,,,,,,,,,,,,,,{ ext{ = 2P}} cr
& { ext{According to the question,}} cr
& { ext{2P = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{20}}}}{{100}} cr
& Rightarrow { ext{R = 10% }} cr} $$ Let after t years it will become double $$eqalign{
& { ext{Hence,}} cr
& { ext{Interest = 2P}} - { ext{P = P}} cr
& Rightarrow { ext{P = }}frac{{{ ext{P}} imes 10 imes { ext{t}}}}{{100}} cr
& Rightarrow { ext{t = 10 years}} cr} $$
9: D
Solution: $$eqalign{
& { ext{Loan Amount}}:{ ext{Total Interest}} cr
& { ext{5}}:{ ext{2}} cr
& { ext{Principal}} o { ext{5}} cr
& { ext{Interest for 1 year}} Rightarrow { ext{2}} cr
& { ext{Interest for 5 year}} Rightarrow { ext{10}} cr
& { ext{10}} = frac{{5 imes { ext{R}} imes 5}}{{100}} cr
& { ext{R}}\% = 40\% cr
& = frac{{40}}{{100}} = frac{2}{5} cr
& { ext{P}}:{ ext{R}} = 5:frac{2}{5} cr
& ,,,,,,,,,,,,,, = 25:2 cr} $$
10: D
Solution: $$eqalign{
& { ext{Half yearly rate}} cr
& { ext{ = }}frac{6}{2} = 3\% cr
& { ext{Effective rate % }} cr
& { ext{ = 3}} + { ext{3}} + frac{{3 imes 3}}{{100}} cr
& = 6.09\% cr} $$
11: C
Solution: $$eqalign{
& { ext{Let sum}} = P cr
& frac{{P imes r imes t}}{{100}} = frac{1}{8}P cr
& rt = frac{{100}}{8} = frac{{25}}{2},.....,left( { ext{i}}
ight) cr
& { ext{Since }}t = frac{r}{2} cr
& { ext{From equation}}left( { ext{i}}
ight) cr
& r imes frac{r}{2} = frac{{25}}{2} cr
& r = 5\% cr
& { ext{If }}P = 1500 cr
& t = 8{ ext{ years}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{P imes r imes t}}{{100}} cr
& = frac{{15000 imes 8 imes 5}}{{100}} cr
& = { ext{Rs}}{ ext{. }}6,000 cr} $$
12: B
Solution: $$eqalign{
& { ext{Interest at the end of }}{{ ext{1}}^{{ ext{st}}}}{ ext{ year}} cr
& = frac{{1,00,000 imes 10}}{{100}} cr
& = 10,000 cr
& { ext{Paid amount}} = 10,000 cr
& { ext{Rest amount}} cr
& = 1,10,000 - 10,000 cr
& = 1,00,000 cr
& { ext{Interest for }}{{ ext{2}}^{{ ext{nd}}}}{ ext{ year}} = 10,000 cr
& { ext{Amount paid}} = 20,000 cr
& { ext{Remaining amount}} cr
& = 1,10,000 - 20,000 cr
& = 90,000 cr
& { ext{Interest for }}{{ ext{3}}^{{ ext{rd}}}}{ ext{ year}} = 9,000 cr
& { ext{Amount paid}} = 30,000 cr
& { ext{Remaining amount}} cr
& = 99,000 - 30,000 cr
& = 69,000 cr
& { ext{Interest for }}{{ ext{4}}^{{ ext{th}}}}{ ext{ year}} = 6,900 cr
& { ext{Amount paid}} = 40,000 cr
& { ext{Remaining amount}} cr
& = 75,900 - 40,000 cr
& = 35,900 cr
& { ext{Interest for }}{{ ext{5}}^{{ ext{th}}}}{ ext{ year}} cr
& = frac{{35,900 imes 10}}{{100}} cr
& = 3,590 cr
& { ext{Amount paid at the end of }}{{ ext{5}}^{{ ext{th}}}}{ ext{ year}} cr
& = 35,900 + 3,590 cr
& = 39,490 cr} $$
13: D
Solution: $$eqalign{
& { ext{Cash payment}} = 25000 cr
& { ext{Down payment}} = 5200 cr
& { ext{Remaining}} = 19800 cr
& { ext{Installment}}left( x
ight) = 4 cr} $$ [x08egin{array}{*{20}{c}}
Rightarrow &{19800}&{ - x} \
{}&{19800}&{ - 2x} \
{}&{mathop {19800}limits_{\_\_\_\_\_\_\_\_\_\_\_} }&{mathop { - 3x}limits_{\_\_\_\_\_\_\_\_\_} } \
{}&{79200}&{ - 6x}
end{array}] $$eqalign{
& Rightarrow { ext{Interest}} = 4x - 19800 cr
& Rightarrow frac{{left( {79200 - 6x}
ight) imes 25}}{{12 imes 100}} = 4x - 19800 cr
& Rightarrow 1920x - 50400 = 79200 - 60x cr
& Rightarrow 1980x = 1029600 cr
& Rightarrow x = 5200 cr} $$
14: C
Solution: $$eqalign{
& 1 = frac{{1 imes 12 imes r}}{{100}} cr
& r = 8frac{1}{3}\% cr} $$
15: B
Solution: [x08egin{gathered}
Pxrightarrow{{{ ext{6 years}}}}59,200 hfill \
Pxrightarrow{{{ ext{10 years}}}}72,000 hfill \
end{gathered} ] $$eqalign{
& { ext{Difference 4 years}} cr
& = left( {72,000 - 59,200}
ight) cr
& = 12,800 cr
& { ext{1 year}} = frac{{12,800}}{4} = 3,200 cr
& herefore { ext{6 years S}}{ ext{.I}}{ ext{.}} cr
& = 3,200 imes 6 = 19,200 cr
& P = 59,200 - 19,200 = 40,000 cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{P imes r imes t}}{{100}} cr
& 19,200 = frac{{40,000 imes r imes 6}}{{100}} cr
& r = 8\% cr
& { ext{If }}r = 8\% + 2\% = 10\% cr
& P = 40,000 cr
& t = ? cr
& A = 76,000 cr
& herefore { ext{S}}{ ext{.I}}{ ext{.}} = 36,000 cr
& 36,000 = frac{{40,000 imes 10 imes t}}{{100}} cr
& t = 9{ ext{ years}} cr} $$
16: C
Solution: Let the sum = 100 $$eqalign{
& { ext{Rate}} = frac{{25}}{3}\% cr
& Rightarrow { ext{Total amount in}} cr
& = 100 + frac{{25}}{3} imes 7 cr
& = 100 + frac{{175}}{3} cr
& = frac{{475}}{3}mu o 76000 cr
& 1mu o 4800 cr
& 100mu o 480000 cr} $$
17: D
Solution: $$eqalign{
& 27600 = frac{{Pleft( {3 imes 4 + 4 imes 8 + 4 imes 12}
ight)}}{{100}} cr
& frac{{27600 imes 100}}{{12 + 32 + 48}} = P cr
& P = frac{{27600 imes 100}}{{92}} cr
& P = { ext{Rs}}{ ext{. }}30000 cr} $$
18: D
Solution: Amount obtained by A is equal to amount obtained by B Amount = Principal + Simple Interest $$eqalign{
& 9100 + frac{{9100 imes 10 imes 3}}{{100}} = B + frac{{B imes 8 imes 5}}{{100}} cr
& 9100 + 2730 = frac{{140}}{{100}}B cr
& B = 11830 imes frac{{100}}{{140}} cr
& B = 8450 cr} $$ 90% of invested amount by B $$eqalign{
& = 8450 imes frac{{90}}{{100}} cr
& = 7605 cr} $$
19: B
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{. for 2 years}} cr
& = frac{{15600 imes 2 imes 10}}{{100}} cr
& = 3120 cr
& { ext{Principal for next 2 years}} cr
& = 15600 + 3120 cr
& = 18720 cr
& { ext{S}}{ ext{.I}}{ ext{. for next 2 years}} cr
& = frac{{18720 imes 10 imes 2}}{{100}} cr
& = 3744 cr
& { ext{Total S}}{ ext{.I}}{ ext{.}} cr
& = 3120 + 3744 cr
& = 6864 cr} $$
20: B
Solution: $$eqalign{
& frac{{1200 imes t imes 8}}{{100}} + frac{{1800 imes t imes 10}}{{100}} = 1380 cr
& 96t + 180t = 1380 cr
& 276t = 1380 cr
& t = 5 cr} $$
21: D
Solution: $$eqalign{
& { ext{Let the sum be Rs}}{ ext{. }}x{ ext{ and}} cr
& { ext{original rate be R}}\% cr
& { ext{Then,}} cr
& Rightarrow frac{{x imes left( {{ ext{R}} + 1}
ight) imes 3}}{{100}} - frac{{x imes { ext{R}} imes 3}}{{100}} = 5100 cr
& Rightarrow 3{ ext{R}}x + 3x - 3{ ext{R}}x = 510000 cr
& Rightarrow 3x = 510000 cr
& Rightarrow x = 170000. cr
& { ext{Hence,}} cr
& { ext{Sum}} = { ext{Rs}}.170000 cr} $$
22: A, F
Solution: Let the annual installment be Rs. x. Then, $$ Rightarrow left[ {x + left( {frac{{x imes 3 imes 4}}{{100}}}
ight)}
ight] + $$ xa0 xa0 $$left[ {x + left( {frac{{x imes 2 imes 4}}{{100}}}
ight)}
ight] + $$ xa0 xa0 $$left[ {x + left( {frac{{x imes 1 imes 4}}{{100}}}
ight)}
ight] + $$ xa0 xa0 $$x = 848$$ $$eqalign{
& Leftrightarrow frac{{28x}}{{25}} + frac{{27x}}{{25}} + frac{{26x}}{{25}} + x = 848 cr
& Leftrightarrow 106x = 848 imes 25 cr
& Leftrightarrow 106x = 21200 cr
& Leftrightarrow x = 200 cr} $$ Short Cut Method : The annual payment that will discharge a debt of Rs. A due in t years at the rate of interest r % p.a. is. $$eqalign{
& frac{{100{ ext{A}}}}{{100t + frac{{rtleft( {t - 1}
ight)}}{2}}} cr
& herefore { ext{Annual installment}} cr
& = { ext{Rs}}{ ext{.}}left[ {frac{{100 imes 848}}{{100 imes 4 + frac{{4 imes 4 imes 3}}{2}}}}
ight] cr
& = { ext{Rs}}{ ext{.}}left( {frac{{100 imes 848}}{{424}}}
ight) cr
& = { ext{Rs}}{ ext{. }}200 cr} $$
23: C
Solution: Total price of TV = Rs. 16000 Initial payment = Rs. 4000 Remaining amount = Rs. 12000 Simple interest in 15 months for Rs. 12000 $$eqalign{
& Rightarrow { ext{S}}{ ext{.I}}{ ext{. = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} cr
& Rightarrow { ext{S}}{ ext{.I}}{ ext{. = }}frac{{12000 imes 12 imes 15}}{{100 imes 12}} cr
& Rightarrow { ext{S}}{ ext{.I}}{ ext{. = Rs}}{ ext{. 1800}} cr} $$ ⇒ With S.I. total amount to be paid for principal amount Rs. 12000 = Rs. (12000 + 1800) = Rs. 13800 = Therefore, total amount he pays for the TV is = 4000 + 13800 = Rs. 17800
24: A
Solution: $$eqalign{
& frac{{ ext{P}}}{{{ ext{S}}{ ext{.I}}{ ext{.}}}} = frac{{10}}{3} cr
& { ext{Let Principal = 10}} cr
& { ext{S}}{ ext{.I}}{ ext{. for 5 years = 3}} cr
& { ext{S}}{ ext{.I}}{ ext{. for 1 year = 0}}{ ext{.6}} cr
& { ext{Rate = }}frac{{{ ext{S}}{ ext{.I}}{ ext{.}}}}{{{ ext{Principal}}}} imes 100 cr
& { ext{Rate = }}frac{{0.6}}{{10}} imes 100 cr
& ,,,,,,,,,,,, = 6\% cr} $$
25: D
Solution: 10% of Rs. 3 Lacs = 30000 6% of Rs. 3 Lacs = 18000 1 month interest income = 2000 ∴ 1 year interest income = 2000 × 12 = 24000 Profit of Bank = 24000 - 18000 = 6000 Profit of Post Office = 30000 - 24000 = 6000 ∴ Ratio of profit = 6000 : 6000 = 1 : 1 So, amount deposited = Rs. 150000 each And difference = 0
26: D
Solution: ⇒ Rs. 10 + S.I. on Rs. 10 for 11 months = Rs. 11 + S.I. on Rs. 1 for (1 + 2 + 3 + 4 + ........... + 10) months ⇒ Rs. 10 + S.I. on Rs. 1 for 110 months = Rs. 11 + S.I. on Rs. 1 for 55 months S.I. on Rs. 1 for 55 months = Rs. 1 $$eqalign{
& herefore { ext{Rate}} = left( {frac{{100 imes 12}}{{1 imes 55}}}
ight)\% cr
& ,,,,,,,,,,,,,,,,,,, = 21frac{9}{{11}}\% cr} $$
27: D
Solution: Total cost of the computer = Rs. 39000 Down payment = Rs. 17000 Balance = Rs. (39000 - 17000) = Rs. 22000. Let the rate of interest be R% p.a. Amount of Rs. 22000 for 5 months $$eqalign{
& = { ext{Rs}}{ ext{.}}left( {22000 + 22000 imes frac{5}{{12}} imes frac{{ ext{R}}}{{100}}}
ight) cr
& = { ext{Rs}}{ ext{.}}left( {22000 + frac{{275{ ext{R}}}}{3}}
ight) cr} $$ The customer pays the shopkeeper Rs. 4800 after 1 month, Rs. 4800 after 2 months, ...... and Rs. 4800 after 5 months. Thus, the shopkeeper keeps Rs. 4800 for 4 months, Rs. 4800 for 3 months, Rs. 4800 for 2 months, Rs. 4800 for 1 months and Rs. 4800 at the end. ∴ sum of the amounts of these installments = (Rs. 4800 + S.I. on Rs 4800 for 4 months) + (Rs. 4800 + S.I. on Rs. 4800 for 3 months) + ...... + (Rs. 4800 + S.I. on Rs. 4800 for 1 month) + Rs. 4800 = Rs. (4800 × 5) + S.I. on Rs. 4800 for (4 + 3 + 2 + 1) months = Rs. 24000 + S.I. on Rs. 4800 for 10 months $$ = { ext{Rs}}{ ext{.}}left( {24000 + { ext{4800}} imes { ext{R}} imes frac{{10}}{{12}} imes frac{1}{{100}}}
ight) = $$ xa0 xa0 xa0 xa0 xa0$${ ext{Rs}}{ ext{.}}left( {24000 + 40{ ext{R}}}
ight)$$ $$eqalign{
& herefore 22000 + frac{{275{ ext{R}}}}{3} = 24000 + 40{ ext{R}} cr
& Rightarrow frac{{155}}{3} = 2000 cr
& Rightarrow { ext{R}} = frac{{2000 imes 3}}{{155}} cr
& ,,,,,,,,,,,,,, = 38.71\% ,{ ext{p}}{ ext{.a}}{ ext{.}} cr} $$
28: D
Solution: $$eqalign{
& { ext{Let the sum be Rs}}{ ext{. }}x cr
& { ext{Rate be R}}\% { ext{ p}}{ ext{.a}}{ ext{.}} cr
& { ext{Time be T years}}{ ext{.}} cr
& { ext{Then,}} cr
& left[ {frac{{x imes left( {{ ext{R}} imes 2}
ight) imes { ext{T}}}}{{100}}}
ight] - left( {frac{{x imes { ext{R}} imes { ext{T}}}}{{100}}}
ight) = 108 cr
& Leftrightarrow 2x{ ext{T}} = 10800,........(i) cr
& And, cr
& left[ {frac{{x imes { ext{R}} imes left( {{ ext{T}} + 2}
ight)}}{{100}}}
ight] - left( {frac{{x imes { ext{R}} imes { ext{T}}}}{{100}}}
ight) = 108 cr
& Leftrightarrow 2x{ ext{R}} = 18000,.......(ii) cr} $$ Clearly, from (i) and (ii), we cannot the find the value of x. So, the data is inadequate.
29: B
Solution: $$eqalign{
& { ext{Sum of the 12 years age }} cr
& { ext{ = Rs}}{ ext{. 100000}} cr
& { ext{Sum of the 18 years age }} cr
& = { ext{P}} + frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} cr
& = { ext{100000}} + frac{{100000 imes 6 imes 6}}{{100}} cr
& = { ext{100000}} + { ext{36000}} cr
& = { ext{136000}} cr} $$ Total expenses = 2500 + 500 = 3000 per year Total expenses ( 6 years ) = 3000 × 6 = Rs. 18000 Amount obtained = 136000 - 18000 = 118000
30: A
Solution: Number of days = 26 + 28 + 31 + 30 + 30 + 31 = 146 days $$eqalign{
& Rightarrow { ext{SI = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} cr
& Rightarrow { ext{SI = }}frac{{36000 imes 9.5 imes 146}}{{100}} cr
& Rightarrow { ext{SI = Rs}}{ ext{. 1368}} cr} $$
31: D
32: B
Solution: $$eqalign{
& { ext{Simple interest}} = 3725 - 2500 = 1225 cr
& { ext{SI}} = frac{{P imes r imes t}}{{100}} cr
& 1225 = frac{{2500 imes r imes 8}}{{100}} cr
& r = frac{{49}}{8} cr
& r = 6.125\% cr} $$
33: C
Solution: $$eqalign{
& { ext{SI}} = frac{{{ ext{PRT}}}}{{100}} cr
& 3x = frac{{x imes 50 imes { ext{T}}}}{{1.0}} cr
& { ext{T}} = frac{{3 imes 100}}{{50}} cr
& { ext{T}} = 6 cr} $$ In T = 6 yrs it will become 4 times For making it more than 4 time check what the time least time after 6 years in option.
34: C
Solution: $$eqalign{
& { ext{A}} = { ext{P}} + { ext{S}}{ ext{.I}}{ ext{.}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} cr
& 11046 = 8400 + frac{{8400 imes 8.75 imes { ext{T}}}}{{100}} cr
& 2646 = frac{{84 imes 875 imes { ext{T}}}}{{100}} cr
& 2646 = 21 imes 35 imes { ext{T}} cr
& { ext{T}} = 3.6{ ext{ years}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{10800 imes 8.75 imes 3.6}}{{100}} cr
& = frac{{108 imes 875 imes 36}}{{1000}} cr
& = 3402 cr} $$
35: A
Solution: $$eqalign{
& P = 7300 cr
& R = 15\% { ext{ per annum}} cr
& = frac{{15}}{{365}}{ ext{ per day}} cr
& t = { ext{2}}{{ ext{8}}^{{ ext{th}}}}{ ext{ April to }}{{ ext{4}}^{{ ext{th}}}}{ ext{ November}} cr
& t = 190{ ext{ days}} cr
& { ext{SI}} = frac{{PRT}}{{100}} cr
& = frac{{7300}}{{100}} imes frac{{15}}{{365}} imes 190 cr
& = 570{ ext{ Answer}} cr} $$
36: C
Solution: $$eqalign{
& 1 = 1 imes frac{{50}}{{400}} imes t cr
& t = 8{ ext{ years}} cr} $$
37: C
Solution: $$eqalign{
& { ext{Amount received in 2 years}} = { ext{Rs}}{ ext{. }}10,000 cr
& { ext{Interest rate}} = 12\% { ext{ per year}} cr
& { ext{Time}} = 2{ ext{ years}} cr
& {x08f{Formula ,used:}} cr
& { ext{Amount}} = { ext{Principal}}{left[ {1 + left( {frac{R}{{100}}}
ight)}
ight]^n} cr
& {x08f{Calculation:}} cr
& A = 10000 cr
& R = 6\% { ext{ }}left( {{ ext{semi - annually}}}
ight) cr
& n = 4\% { ext{ }}left( {{ ext{semi - annually}}}
ight) cr
& { ext{Thus,}} cr
& Rightarrow 10000 = P{left[ {1 + left( {frac{6}{{100}}}
ight)}
ight]^4} cr
& Rightarrow 10000 = P{left[ {frac{{106}}{{100}}}
ight]^4} cr
& Rightarrow 10000 = P imes {left( {1.06}
ight)^4} cr
& Rightarrow P = frac{{10000}}{{{{1.06}^4}}} cr
& Rightarrow P = 7920.9426 cr
& Rightarrow P = { ext{Rs}}{ ext{. }}7920.94 cr
& { ext{Hence, the correct answer is Rs}}{ ext{. 7,920}}{ ext{.94}} cr} $$
38: D
Solution: $$eqalign{
& { ext{Let, Rate}} = 12x\% { ext{ per year}} cr
& frac{{12000 imes 4x}}{{100}} + frac{{18000 imes 16x}}{{100}} = 2800 cr
& 480x + 2880x = 2800 cr
& 3360x = 2800 cr
& x = frac{{10}}{{12}} cr
& { ext{Rate}} = 12x cr
& = frac{{10}}{{12}} imes 12 cr
& = 10\% cr} $$
39: A
Solution: $$eqalign{
& { ext{She paid}} = 3500 cr
& { ext{Left amount}} = 7500 - 3500 = 4000 cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{4000 imes 9 imes 4}}{{100 imes 12}} = 120 cr} $$
40: C
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{P imes r imes t}}{{100}} cr
& 7 = frac{{10 imes r imes 5}}{{2 imes 100}} cr
& r = 28\% cr} $$
41: B
Solution: Principal = Rs. 12000 Rate % = 10% Interest paid by the person in 5 years $$eqalign{
& = frac{{12000 imes 10 imes 5}}{{100}} cr
& = { ext{Rs}}{ ext{. 6000}} cr} $$ Interest received by the person after 3 years $$eqalign{
& = { ext{Rs}}{ ext{. }}left( {6000 - 3320}
ight) cr
& = { ext{Rs}}{ ext{. 2680}} cr
& { ext{By using formula,}} cr
& { ext{Rate}}\% cr
& { ext{ = }}frac{{2680}}{{12000}} imes frac{{100}}{3} cr
& = frac{{67}}{9} cr
& = 7frac{4}{9}\% cr
& { ext{Hence required rate}}\% cr
& { ext{ = 7}}frac{4}{9}\% cr} $$
42: D
Solution: $$eqalign{
& { ext{P}} + frac{{P imes { ext{r}} imes { ext{t}}}}{{100}} = 3{ ext{P}} cr
& Rightarrow 1 + frac{{rt}}{{100}} = 3 cr
& Rightarrow frac{{{ ext{rt}}}}{{100}} = 2 cr
& Rightarrow { ext{r}} = frac{{2 imes 100}}{8} = 25\% cr
& { ext{so, }},left( {1 + frac{{{ ext{rt}}}}{{100}}}
ight) = 4 cr
& Rightarrow frac{{{ ext{rt}}}}{{100}} = 3 cr
& Rightarrow { ext{t}} = frac{{3 imes 100}}{{25}} cr
& Rightarrow { ext{t}} = 12,{ ext{years}} cr} $$
43: B
Solution: Let rate of interest = R% According to the question, $$eqalign{
& frac{{500 imes 4 imes { ext{R}}}}{{100}} + frac{{600 imes 3 imes { ext{R}}}}{{100}} = 190 cr
& Rightarrow 20{ ext{R + 18R = 190}} cr
& Rightarrow 38{ ext{R = 190}} cr
& Rightarrow { ext{R = 5% }} cr} $$ Hence required rate % = 5% Alternate Note : In such type of questions to save your valuable time follow the given below method. Let rate of interest = 1% $$eqalign{
& { ext{Case (I): Interest (}}{{ ext{I}}_1}{ ext{)}} cr
& { ext{ = }}frac{{500 imes 4 imes 1}}{{100}} cr
& = 20 cr
& { ext{Case (II): Interest (}}{{ ext{I}}_2}{ ext{)}} cr
& { ext{ = }}frac{{{ ext{600}} imes 3 imes 1}}{{100}} cr
& = 18 cr} $$ According to the question, Interest Rate % 38 1 ↓×5 ↓×5 190 5% Hence required rate % = 5%
44: B
Solution: $$eqalign{
& { ext{Let sum}} = { ext{Rs}}{ ext{. }}x cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = 30\% ,{ ext{of}},{ ext{Rs}}{ ext{.}},x cr
& ,,,,,,,,, = { ext{Rs}}{ ext{.}}frac{{3x}}{{10}} cr
& { ext{Time}} = 6,{ ext{years}}{ ext{.}} cr
& herefore { ext{Rate}} = left( {frac{{100 imes 3x}}{{10 imes x imes 6}}}
ight)\% cr
& ,,,,,,,,,,,,,,,,, = 5\% cr
& { ext{Now, sum}} = { ext{Rs}}{ ext{. }}x cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{. }}x cr
& { ext{Rate}} = 5\% cr
& herefore { ext{Time}} = left( {frac{{100 imes x}}{{x imes 5}}}
ight){ ext{years}} cr
& ,,,,,,,,,,,,,,,,,, = 20,{ ext{years}} cr} $$
45: A
Solution: $$eqalign{
& { ext{Let sum}} = x cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{x}{9}. cr
& { ext{Let rate}} = { ext{R}}\% ,{ ext{and}} cr
& { ext{time}} = { ext{R}},{ ext{years}}{ ext{.}} cr
& herefore left( {frac{{x imes { ext{R}} imes { ext{R}}}}{{100}}}
ight) = frac{x}{9} cr
& Rightarrow {{ ext{R}}^2} = frac{{100}}{9} cr
& Rightarrow { ext{R}} = frac{{10}}{3} = 3frac{1}{3} cr
& { ext{Hence, rate}} = 3frac{1}{3}\% cr} $$
46: D
Solution: Let the time is 't' years and the rate of interest is R% According to the question, $$eqalign{
& {x08f{Case(I):}} cr
& y = frac{{x imes { ext{R}} imes { ext{t}}}}{{100}}.....(i) cr
& {x08f{Case(II):}} cr
& z = frac{{y imes { ext{R}} imes { ext{t}}}}{{100}}.....(ii) cr} $$ By dividing equation (i) by equation (ii) $$eqalign{
& frac{y}{z} = frac{{x imes { ext{R}} imes { ext{t}}}}{{y imes { ext{R}} imes { ext{t}}}} cr
& Rightarrow {y^2} = zx cr} $$
47: C
Solution: Let the money borrowed by Nitin = Rs. P According to the question, $$eqalign{
& frac{{{ ext{P}} imes 6 imes 3}}{{100}} + frac{{{ ext{P}} imes 9 imes 5}}{{100}} + frac{{{ ext{P}} imes 13 imes 3}}{{100}} = { ext{Rs}}{ ext{. }}8160 cr
& Rightarrow frac{{{ ext{18P}}}}{{100}} + frac{{{ ext{45P}}}}{{100}} + frac{{{ ext{39P}}}}{{100}} = { ext{Rs}}{ ext{. 1860}} cr
& Rightarrow frac{{{ ext{102P}}}}{{100}} = { ext{Rs}}{ ext{. 8160}} cr
& Rightarrow { ext{P = Rs}}{ ext{.}}frac{{8160 imes 100}}{{102}} cr
& ,,,,,,,,,,,,{ ext{ = Rs}}{ ext{. 8000}} cr} $$ Alternate Note : In such type of questions to save your valuable time follow the given below method. Let principal = Rs. 100 Total Interest $$frac{{100 imes 6 imes 3}}{{100}} + frac{{100 imes 9 imes 5}}{{100}} + $$ xa0 xa0xa0 $$frac{{100 imes 13 imes 3}}{{100}}$$ $$eqalign{
& = 18 + 45 + 39 cr
& = 102{ ext{ units}} cr
& { ext{According to the question,}} cr
& { ext{102 units = Rs}}{ ext{. 8160}} cr
& { ext{1 unit = Rs}}{ ext{. }}frac{{8160}}{{102}} cr
& ,,,,,,,,,,,,,,,,{ ext{ = Rs}}{ ext{. 80}} cr
& { ext{100 units = Rs}}{ ext{. 8000}} cr
& { ext{Hence sum = Rs}}{ ext{. 8000}} cr} $$ Alternate Total rate of interest in 11 years $$eqalign{
& left( {6 imes 3}
ight)\% + left( {5 imes 9}
ight)\% + left( {3 imes 13}
ight)\% { ext{ }} cr
& Rightarrow 102\% = 8160 cr
& Rightarrow 100\% = 8000 cr
& herefore { ext{Sum = Rs}}{ ext{. 8000}} cr} $$
48: A
Solution: $$eqalign{
& { ext{Increased interest in 3 years}} cr
& { ext{ = 3}} imes { ext{3}} cr
& { ext{ = 9% }} cr
& { ext{Hence increased amount}} cr
& { ext{ = }}frac{{800 imes 9}}{{100}} cr
& = { ext{Rs}}{ ext{. }}72 cr
& { ext{Total amount}} cr
& { ext{ = }}left( {920 + 72}
ight) cr
& = { ext{Rs}}{ ext{. }}992 cr} $$
49: D
Solution: $$eqalign{
& { ext{Let}},{ ext{sum}} = x. cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{ ext{9}}}{{{ ext{16}}}}x cr
& { ext{Let rate}} = { ext{R}}\% , ext{and} cr
& ext{time} = R, ext{years} cr
& herefore left( {frac{{x imes { ext{R}} imes { ext{R}}}}{{100}}}
ight) = frac{{9x}}{{16}} cr
& Rightarrow {{ ext{R}}^2} = frac{{900}}{{16}} cr
& { ext{R}} = frac{{30}}{4} = 7frac{1}{2} cr
& { ext{Hence,}} cr
& { ext{time}} = 7frac{1}{2} ext{years} cr} $$
50: B
Solution: Let the sum lent to C be Rs. x. Then, $$eqalign{
& left( {frac{{2500 imes 7 imes 4}}{{100}}}
ight) + left( {frac{{x imes 7 imes 4}}{{100}}}
ight) = 1120 cr
& Rightarrow frac{{7x}}{{25}} = left( {1120 - 700}
ight) cr
& Rightarrow x = left( {frac{{420 imes 25}}{7}}
ight) cr
& ,,,,,,,,,,,, = 1500. cr} $$
51: D
Solution: $$eqalign{
& { ext{P = Principal}},{ ext{Amount}} cr
& { ext{T = }},{ ext{Time}},{ ext{period}} cr
& { ext{R = Rate}},{ ext{of}},{ ext{Interest}} cr
& I = frac{{P imes T imes R}}{{100}} cr
& I = frac{{5200 imes 2 imes 6}}{{100}} cr
& I = 624. cr} $$
52: A
Solution: $$eqalign{
& A = P + I cr
& A = 1200 + {frac{{PTR}}{{100}}} cr
& A = {1200 + {frac{{1200 imes 5 imes 3}}{{100}}} } cr
& { ext{Amount}},,A = Rs.,1380 cr} $$
53: C
Solution: $$eqalign{
& { ext{Let rate is }}R\% cr
& { ext{We have}}, cr
& I = frac{{PTR}}{{100}} cr
& { ext{Here}},1500 = frac{{5000 imes 3 imes R}}{{100}} cr
& { ext{Thus}},R = 10\% cr} $$
54: B
Solution: $$eqalign{
& { ext{We}},{ ext{can}},{ ext{use}},{ ext{formula}},{ ext{of}},{ ext{compound}},{ ext{interest}} cr
& A = P imes {left[ {1 + left( {frac{r}{{100}}}
ight)}
ight]^n} cr
& A = 2100 imes {left[ {1 + left( {frac{5}{{100}}}
ight)}
ight]^2} cr
& A = 2100 imes {left[ {frac{{105}}{{100}}}
ight]^2} cr
& A = frac{{ {2100 imes 11025} }}{{10000}} cr
& { ext{Hence,}},{ ext{Amount}},A = Rs.,2315.25 cr} $$
55: A
Solution: $$eqalign{
& { ext{The}},{ ext{difference}},{ ext{between}},{ ext{compound}},{ ext{interest}},{ ext{and}} cr
& { ext{simple}},{ ext{interest}},{ ext{over}},{ ext{two}},{ ext{years}},{ ext{is}},{ ext{given}},{ ext{by}} cr
& frac{{{{Pr }^2}}}{{{{100}^2}}},or,P{left( {frac{r}{{100}}}
ight)^2} cr
& { ext{Here,}},{ ext{Principal}},left( P
ight) = Rs.,2000 cr
& { ext{Rate}},left( r
ight) = 5\% cr
& { ext{Now}},{ ext{difference}}, cr
& D = frac{{ {2000 imes 5 imes 5} }}{{ {100 imes 100} }} cr
& D = Rs.,5 cr} $$
56: C
Solution: $$eqalign{
& { ext{Amount}},,A = Rs.,1440 cr
& { ext{Principal}},,P = Rs.,1200 cr
& { ext{Interest}},,I = Rs.,left( {1440 - 1200}
ight) = 240 cr
& R = frac{{ {240 imes 100} }}{{ {1200 imes 2} }} = 10\% cr
& cr
& { ext{Alternatively}}, cr
& { ext{We}},{ ext{can}},{ ext{go}},{ ext{through}},{ ext{a}},{ ext{thought}},{ ext{process}},i.e. cr
& 1200, == 20\% uparrow left( {240,{ ext{in}},2,{ ext{years}}}
ight) ⇒ 1400 cr
& { ext{That}},{ ext{means}},10\% ,{ ext{rise}},{ ext{in}},{ ext{each}},{ ext{year}} cr} $$
57: A
Solution: $$eqalign{
& {I_1} = frac{{P{T_1}{R_1}}}{{100}} cr
& {I_1} = frac{{ {500 imes 3 imes 5} }}{{100}} cr
& ,,,,,,, = Rs.{kern 1pt} 75 cr
& {I_2} = frac{{P{T_2}{R_2}}}{{100}} cr
& {I_2} = frac{{ {500 imes 4 imes 4} }}{{100}} cr
& ,,,,,,,, = Rs.{kern 1pt} 80 cr
& { ext{Difference}} = 80 - 75 cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, = { ext{Rs}}{ ext{.}},5 cr} $$ Alternatively The interest is calculated simply and then it will have a rise of 15% in 1 st case and 16% in 2 nd case. Difference = 1% on 500 = Rs. 5 Otherway, 500 == 15% $$ uparrow $$ ⇒ 575 (1 st case) 500 == 16% $$ uparrow $$ ⇒ 580 (2 nd case) We can see clear difference of Rs. 5
58: C
Solution: $$eqalign{
& {1^{{ ext{st}}}}{kern 1pt} { ext{case}}: cr
& {I_1} = frac{{700 imes 3 imes 8}}{{100}} = { ext{Rs}}{ ext{. }}168 cr
& {2^{{ ext{nd}}}}{kern 1pt} { ext{case}}: cr
& {I_2} = frac{{700 imes 2 imes 7.5}}{{100}} = { ext{Rs}}{ ext{. }}105 cr} $$ Then total interest for five years $$ = left( {{I_1} + {I_2}}
ight) = { ext{Rs}}{ ext{. }}273$$ Alternatively, As interest is calculated as simple interest So, we can add up rates for all given 5 years and calculate it easily i.e. For the five years rate = (8 × 3 + 7.5 × 2) = 39% Now, 700 = 39% $$ uparrow $$ ⇒ 973 Interest = Rs. 273 The thought can go this way we internally calculated $$eqalign{
& {kern 1pt} 10\% ,{ ext{of}},700 = {frac{{700}}{{10}}} = 70 cr
& { ext{Then}}, cr
& 39\% ,{ ext{of}},700 cr
& = left( {40\% - 1\% }
ight),{ ext{of}},700 cr
& = left( {280 - 7}
ight) cr
& = 273 cr} $$
59: A
Solution: $$eqalign{
& { ext{Given,}},{ ext{Principal}},,P = Rs.,1000 cr
& { ext{Compoind}},{ ext{rate}},,R = 20\% ,{ ext{per}},{ ext{annum}} cr
& = frac{{20}}{2} = 10\% ,{ ext{half - yearly}} cr
& { ext{Time}} = 18,{ ext{month}} = 3,{ ext{half - years}} cr
& { ext{Amount}}, cr
& A = left{ {P imes {{left[ {1 + left( {frac{R}{{100}}}
ight)}
ight]}^n}}
ight} cr
& = left{ {1000 imes {{left[ {1 + left( {frac{{10}}{{100}}}
ight)}
ight]}^3}}
ight} cr
& = { {frac{{1000 imes 11 imes 11 imes 11}}{{10 imes 10 imes 10}}} } cr
& A = Rs.,1331 cr
& { ext{Hence,}},{ ext{compound}},{ ext{interest}} = Rs.,331 cr} $$
60: C
Solution: $$eqalign{
& { ext{Given}}, cr
& { ext{Compound}},{ ext{rate}},,R = 10\% ,{ ext{per}},{ ext{annum}} cr
& { ext{Time}} = 2,{ ext{years}} cr
& CI - Rs.,420 cr
& { ext{Let}},P,{ ext{be}},{ ext{the}},{ ext{required}},{ ext{principal}} cr
& A = left( {P + CI}
ight) cr
& { ext{Amount}},A = left{ {P imes {{left[ {1 + left( {frac{R}{{100}}}
ight)}
ight]}^n}}
ight} cr
& left( {P + CI}
ight) = left{ {P imes {{left[ {1 + frac{{10}}{{100}}}
ight]}^2}}
ight} cr
& left( {P + 420}
ight) = P imes {left[ {frac{{11}}{{10}}}
ight]^2} cr
& P - 1.21P = - 420 cr
& 0.21P = 420 cr
& { ext{Hence}},P = frac{{420}}{{0.21}} = Rs.,2000 cr} $$
61: C
Solution: Time = 18 + 28 + 31 + 30 + 31 + 8 = 146 days $$eqalign{
& { ext{Simple Interest}} cr
& { ext{ = }}frac{{12000 imes 15 imes 146}}{{365 imes 100}} cr
& { ext{ = Rs}}{ ext{. 720}} cr
& herefore { ext{Amount will be }} cr
& { ext{ = Rs}}{ ext{.}}left( {12000 + 720}
ight){ ext{ }} cr
& { ext{ = Rs}}{ ext{. 12720}} cr} $$
62: A
Solution: Let the sum borrowed at 12% p.a. be Rs. x and that borrowed at 10% p.a. be Rs. (30000 - x) S.I. at the end of 2 years = Rs. (36480 - 30000) = Rs. 6480 $$ herefore left( {frac{{x imes 12 imes 2}}{{100}}}
ight) + $$ xa0xa0 $$left[ {frac{{left( {30000 - x}
ight) imes 10 imes 2}}{{100}}}
ight]$$ xa0 xa0 $$ = 6480$$ $$eqalign{
& Leftrightarrow 24x + 600000 - 20x = 648000 cr
& Leftrightarrow 4x = 48000 cr
& Leftrightarrow x = 12000 cr} $$
63: C
Solution: Let the two sums be Rs. x and Rs. (18750 - x). Then, $$ = x + frac{{x imes 5 imes 6}}{{100}} = left( {{ ext{18750}} - x}
ight) + $$ xa0 xa0 xa0 $$frac{{left( {{ ext{18750}} - x}
ight) imes 5 imes 4}}{{100}}$$ $$eqalign{
& Leftrightarrow x + frac{{30x}}{{100}} = left( {{ ext{18750}} - x}
ight) + 3750 - frac{{20x}}{{100}} cr
& Leftrightarrow 2x + frac{x}{2} = 22500 cr
& Leftrightarrow frac{{5x}}{2} = 22500 cr
& Leftrightarrow x = left( {frac{{22500 imes 2}}{5}}
ight) cr
& Leftrightarrow x = 9000 cr} $$ So the other sum will be = ( 18750 - 9000) = 9750 Hence, The two sums are Rs. 9000, Rs. 9750
64: C
Solution: Let the sum lent to A be Rs. x. and that lent to B be Rs. (10000 - x) Then, $$eqalign{
& Rightarrow frac{{x imes 15 imes 2}}{{100}} - frac{{left( {10000 - x}
ight) imes 18 imes 12}}{{100}} = 360 cr
& Rightarrow 30x - 360000 + 36x = 36000 cr
& Rightarrow 66x = 396000 cr
& Rightarrow x = 6000 cr} $$ Hence, Sum lent to B = Rs. (10000 - 6000) = Rs. 4000
65: A
Solution: $$eqalign{
& { ext{P}} + { ext{SI = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} + { ext{P}} cr
& Rightarrow 2200{ ext{ = }}frac{{{ ext{P}} imes 5 imes { ext{T}}}}{{100}} + { ext{P}} cr} $$ ⇒ 2200 × 100 = 5PT + 100P ......... (i) $$ Rightarrow 2320 = frac{{{ ext{P}} imes 8 imes { ext{T}}}}{{100}} + { ext{P}}$$ ⇒ 2320 × 100 = 8PT + 100P ⇒ 2320 × 100 = 3PT + 5PT + 100P ............(ii) Value of equation (i) put equation (ii) ⇒ 2320 × 100 = 3PT + 2200 × 100 ⇒ 3PT = 120 × 100 ⇒ PT = 4000 Value of PT in equation (i) ⇒ 2200 × 100 = 5 × 4000 + 100P ⇒ 220000 - 20000 = 100P $$eqalign{
& Rightarrow { ext{P = }}frac{{200000}}{{100}} cr
& Rightarrow { ext{P = Rs 2000}} cr
& { ext{Using this formula}} cr
& left( {x08ecause { ext{SI = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}}}
ight) cr
& herefore { ext{200}} = frac{{2000 imes 5 imes { ext{T}}}}{{100}} cr
& Rightarrow { ext{T}} = frac{{200}}{{100}}{ ext{ = 2 years}} cr
& cr
& { ext{ }}{x08f{Alternate:}} cr
& left( {8 - 5}
ight)\% = 2320 - 2200 cr
& Rightarrow 3\% = 120 cr
& Rightarrow 1\% = 40 cr
& Rightarrow 5\% = 200 cr
& { ext{Principal = 2200 - 200}} cr
& ,,,,,,,,,,,,,,,,,,,,,,{ ext{ = Rs}}{ ext{. 2000}} cr
& { ext{SI = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} cr
& Rightarrow { ext{200}} = frac{{2000 imes 5 imes { ext{T}}}}{{100}} cr
& Rightarrow { ext{T}} = frac{{200}}{{100}}{ ext{ = 2 years}} cr} $$
66: A
Solution: $$eqalign{
& { ext{Let principal is P}} cr
& { ext{then,}} cr
& { ext{300 = }}frac{{{ ext{P}} imes 3 imes 2}}{{100}} cr
& { ext{P = 5000}} cr} $$
67: D
Solution: $$eqalign{
& { ext{Amount}} o { ext{10}} cr
& ,,,,, downarrow cr
& ,,,,,90 o frac{{10}}{{90}} imes 100 cr
& ,,,,,,,,,, = 11frac{1}{9}\% cr} $$
68: C
Solution: Let the sum lent at 5% be Rs. x and that lent 8% be Rs. (1550 - x). Then, $$eqalign{
& left( {frac{{x imes 5 imes 3}}{{100}}}
ight) + left[ {frac{{left( {1550 - x}
ight) imes 8 imes 3}}{{100}}}
ight] = 300 cr
& Leftrightarrow 15x - 24x + left( {1550 imes 24}
ight) = 30000 cr
& Leftrightarrow 9x = 7200 cr
& Leftrightarrow x = 800. cr
& herefore { ext{Required ratio}} = 800:750 cr
& = 16:15 cr} $$
69: D
Solution: Let the parts be Rs. x, Rs. y and Rs. [1440 - (x + y)]. Then, $$eqalign{
& = frac{{x imes 2 imes 3}}{{100}} = frac{{y imes 3 imes 4}}{{100}} cr
& = frac{{left[ {1440 - left( {x + y}
ight)}
ight] imes 4 imes 5}}{{100}} cr
& herefore 6x = 12y,or,x = 2y. cr
& So,frac{{x imes 2 imes 3}}{{100}} = frac{{left[ {1440 - left( {x + y}
ight)}
ight] imes 4 imes 5}}{{100}} cr
& Leftrightarrow 12y = left( {1440 - 3y}
ight) imes 20 cr
& Rightarrow 72y = 28800 cr
& Rightarrow y = 400 cr
& { ext{First part}} = x = 2y = { ext{Rs}}{ ext{. }}800, cr
& { ext{Second part}} = { ext{Rs}}{ ext{. }}400 cr
& { ext{Third part}} cr
& = { ext{Rs}}.left[ {1440 - left( {800 + 400}
ight)}
ight] cr
& = { ext{Rs}}{ ext{. }}240. cr
& herefore { ext{Required difference}} cr
& { ext{ = Rs}}{ ext{.}}left( {800 - 240}
ight) cr
& = { ext{Rs}}.560 cr} $$
70: C
Solution: $$eqalign{
& { ext{A's,}},{ ext{Principal}} = { ext{Rs}}{ ext{. }}90,000 cr
& { ext{Rate}} = 5\% cr
& { ext{Time}} = 4{ ext{ years}} cr
& herefore { ext{S}}{ ext{.I}} = frac{{90,000 imes 5 imes 4}}{{100}} = { ext{Rs}}{ ext{. }}18,000 cr
& { ext{B's,}},{ ext{Principal}} = { ext{Rs}}{ ext{. }}90,000 cr
& { ext{Rate}} = 7\% cr
& { ext{Time}} = 4{ ext{ years}} cr
& herefore { ext{S}}{ ext{.I}} = frac{{90,000 imes 7 imes 4}}{{100}} = { ext{Rs}}{ ext{. }}25,200 cr
& herefore { ext{Profit of A}} cr
& = left( {25,200 - 18,000}
ight) = { ext{Rs}}{ ext{. }}7,200 cr} $$
71: A
Solution: $$eqalign{
& { ext{Amount after 3 years}} cr
& = 2000 + frac{{2000 imes 5 imes 3}}{{100}} cr
& = 2000 + 300 cr
& = 2300 cr
& { ext{EMI}} = frac{{A imes 100}}{{100 imes t + rtleft( {frac{{t - 1}}{2}}
ight)}} cr
& = frac{{2300 imes 100}}{{100 imes 3 + 5 imes 3left( {frac{{3 - 1}}{2}}
ight)}} cr
& = frac{{2300 imes 100}}{{315}} cr
& = frac{{2300 imes 20}}{{63}} cr
& = frac{{46000}}{{63}} cr
& = 730frac{{10}}{{63}} cr} $$
72: B
Solution: $$eqalign{
& { ext{Let, loan taken at rate of }}7\% = x cr
& { ext{Then, }}frac{{left( {2500 - x}
ight) imes 5}}{{100}} + frac{{x imes 7}}{{100}} = frac{{276}}{2} cr
& 12500 - 5x + 7x = 13800 cr
& 2x = 13800 - 12500 cr
& 2x = 1300 cr
& x = 650,{ ext{Answer}} cr} $$
73: D
Solution: $$eqalign{
& { ext{Sum}} = 12,800 cr
& {{ ext{I}}^{{ ext{st}}}}{ ext{ part}} = x cr
& { ext{I}}{{ ext{I}}^{{ ext{nd}}}}{ ext{ part}} = left( {12,800 - x}
ight) cr
& x imes frac{{15}}{{100}} + left( {12,800 - x}
ight)frac{{12}}{{100}} = frac{{5,085}}{3} cr
& frac{{3x}}{{100}} + 128 imes 12 = 1,695 cr
& frac{{3x}}{{100}} = 1,695 - 1,536 cr
& frac{{3x}}{{100}} = 159 cr
& { ext{then }}x = 5,300 cr} $$
74: B
Solution: 3 4 5 yr → S .I 133 1 yr → S .I 35 Ifxa0ratexa0isxa0double 1 yr → S .I 70 5 yr → S .I 350 420 → Unit 10500 1 → Unit 25 350 → Unit 350 × 25 = 8750
75: A
Solution: $$eqalign{
& {x08f{Given,}} cr
& { ext{Principal}} = { ext{Rs}}{ ext{. }}10,000 cr
& { ext{Some part of money lent at }}15\% cr
& { ext{Some part of it lent at }}10\% cr
& { ext{Total interest earned}} = { ext{Rs}}{ ext{. }}6,500 cr
& { ext{Time}} = 5{ ext{ years}} cr
& {x08f{Formula}},{x08f{used:}} cr
& { ext{Interest}} = frac{{{ ext{Principal}} imes { ext{Rate}} imes { ext{Time}}}}{{100}} cr
& {x08f{Calculation:}} cr
& { ext{Let the money lent at }}15\% { ext{ be }}x cr
& { ext{Then money lent at }}10\% { ext{ be }}left( {10,000 - x}
ight) cr
& 6,500 = frac{{x imes 15 imes 5}}{{100}} + frac{{left( {10,000 - x}
ight) imes 10 imes 5}}{{100}} cr
& Rightarrow 6,500 = frac{{3x}}{4} + frac{{10,000 - x}}{2} cr
& Rightarrow 6,500 = frac{{3x + 20,000 - 2x}}{4} cr
& Rightarrow 26,000 = 20,000 + x cr
& Rightarrow x = { ext{Rs}}{ ext{. }}6,000 cr
& { ext{Then,}} cr
& left( {10,000 - x}
ight) = left( {10,000 - 6,000}
ight) = { ext{Rs}}{ ext{. }}4000 cr
& { ext{Difference between the two sums lent}} cr
& = { ext{Rs}}{ ext{. }}6,000 - { ext{Rs}}{ ext{. }}4,000 cr
& = { ext{Rs}}{ ext{. }}2,000 cr
& herefore { ext{The difference between the sums lent is Rs}}{ ext{. }}2,000 cr} $$
76: D
Solution: [frac{1}{4}x08egin{array}{*{20}{c}}
{,,,, leftarrow { ext{S}}{ ext{.I}}{ ext{.}}} \
{ leftarrow P}
end{array}] $$eqalign{
& { ext{Let }}P = 4x{ ext{ and S}}{ ext{.I}}{ ext{.}} = x cr
& { ext{Time}} = t{ ext{ years}} cr
& { ext{Rate}} = 4t\% cr
& herefore { ext{S}}{ ext{.I}}{ ext{.}} = frac{{P imes R imes T}}{{100}} cr
& x = frac{{4x imes 4t imes t}}{{100}} cr
& {t^2} = frac{{100}}{{16}} cr
& {t^2} = frac{{25}}{4} cr
& herefore t = frac{5}{2} = 2.5{ ext{ years}} cr
& herefore R = 4 imes 2.5{ ext{ years}} = 10\% cr
& { ext{New rate}} = 10\% + 2\% cr
& P = { ext{Rs}}{ ext{. }}5000 cr
& T = 3{ ext{ years}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{P imes R imes T}}{{100}} cr
& = frac{{5000 imes 12 imes 3}}{{100}} cr
& = { ext{Rs}}{ ext{. }}1800 cr} $$
77: A
78: B
Solution: Cash payment = 650000 Each installment = 25000 Total installment = 20 Sum = 25000 × 20 = 500000 Rate = 10% = $$frac{1}{{10}}$$ ⇒ Down payment = 650000 - 454545.45 = 195454.55 ≈ 195455
79: A
Solution: Difference between both part $$eqalign{
& = left( {4 imes 15\% - 3 imes 20\% }
ight){ ext{ of }}8600 cr
& = 0\% { ext{ of }}8600 cr
& = 0{ ext{ Answer}} cr} $$
80: A
Solution: $$eqalign{
& { ext{P}} = { ext{Rs}}{ ext{. }}16800 cr
& { ext{R}} = 6frac{1}{4}\% = frac{{25}}{4}\% cr
& { ext{T}} = 9,{ ext{mths}} = frac{3}{4}{ ext{yr}}{ ext{.}} cr
& herefore { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}}left( {16800 imes frac{{25}}{4} imes frac{3}{4} imes frac{1}{{100}}}
ight) cr
& = { ext{Rs}}{ ext{.}},787.50 cr} $$
81: B
Solution: $$eqalign{
& { ext{Let the principal}} cr
& { ext{ = 9 units }} cr
& { ext{Hence simple interest}} cr
& { ext{ = }}frac{4}{9} imes { ext{9 = 4 units}} cr
& { ext{Let,}} cr
& { ext{Rate of interest = R% }} cr
& { ext{R = T (given)}} cr
& { ext{By using formula}} cr
& x08oxed{{ ext{SI = }}frac{{{ ext{P}} imes { ext{T}} imes { ext{R}}}}{{100}}} cr
& 4 = frac{{9 imes { ext{R}} imes { ext{R}}}}{{100}} cr
& {{ ext{R}}^2} = frac{{400}}{9} cr
& { ext{R = }}frac{{20}}{3} = 6frac{2}{3}\% cr} $$
82: A
Solution: $$eqalign{
& { ext{Let principal = 5 units}} cr
& { ext{Hence interest}} cr
& { ext{ = 5}} imes frac{2}{5} cr
& { ext{ = 2 units}} cr
& { ext{Time = 10 years}} cr
& { ext{By using formula, }} cr
& { ext{Rate% = }}frac{2}{5} imes frac{{100}}{{10}} cr
& ,,,,,,,,,,,,,,,,,,,,,{ ext{ = 4% }} cr} $$
83: A
Solution: $$eqalign{
& { ext{Principal = Rs}}{ ext{. 1750}} cr
& { ext{Let the first part = }}x cr
& { ext{Hence second part}} cr
& { ext{ = }}left( {1750 - x}
ight) cr
& { ext{According to the question,}} cr
& Rightarrow x imes frac{8}{{100}} imes 1 = left( {1750 - x}
ight) imes frac{6}{{100}} imes 1 cr
& Rightarrow 4x = 5250 - 3x cr
& Rightarrow 7x = 5250 cr
& Rightarrow x = 750 cr
& { ext{First part = Rs}}{ ext{. 750}} cr
& herefore { ext{ Second part}} cr
& { ext{ = Rs}}{ ext{. }}left( {1750 - 750}
ight) cr
& { ext{ = Rs}}{ ext{. 1000}} cr
& { ext{Required interest}} cr
& { ext{ = 750}} imes frac{8}{{100}} cr
& { ext{ = Rs}}{ ext{. 60}} cr} $$ Alternate Note : In such type of questions to save your valuable time follow the given below method. Let, Principal = 100 units in both cases 1 st part 2 nd part Total Principal → 100 ×3 : 100 ×4 +→ 700 units Interest → 8 ×3 : 6 ×4 Note : Interest is same in both cases According to the question, $$eqalign{
& { ext{700 units = Rs}}{ ext{. 1750}} cr
& { ext{1 unit = Rs}}{ ext{. }}frac{{1750}}{{700}} cr
& { ext{24 units = Rs}}{ ext{. }}frac{{1750}}{{700}} imes 24 cr
& ,,,,,,,,,,,,,,,,,,,, = { ext{Rs}}{ ext{. 60}} cr
& { ext{Hence, required interest}} cr
& { ext{ = Rs}}{ ext{. 60}} cr} $$
84: A
Solution: $$eqalign{
& { ext{Gain in 2 years}} cr
& = {left( {5000 imes frac{{25}}{4} imes frac{2}{{100}}}
ight) - left( {frac{{5000 imes 4 imes 2}}{{100}}}
ight)} cr
& = { ext{Rs}}{ ext{.}}left( {625 - 400}
ight) cr
& = { ext{Rs}}{ ext{. }}225 cr
& herefore { ext{Gain 1 year}} = { ext{Rs}}{ ext{.}}left( {frac{{225}}{2}}
ight) cr
& = { ext{Rs}}{ ext{. }}112.50 cr} $$
85: C
Solution: Let the amounts invested in schemes A and B be Rs. 5x and 4x respectively. then, 5x = 20000 ⇒ x = 4000 ∴ Amount invested in scheme B = Rs. 16000 Total interest received on maturity $$ = { ext{Rs}}{ ext{.}}left[ {{ ext{120}}\% { ext{ of}}left( {frac{{{ ext{20000}} imes { ext{8}} imes { ext{5}}}}{{{ ext{100}}}}}
ight) + left( {frac{{{ ext{16000}} imes { ext{9}} imes { ext{5}}}}{{{ ext{100}}}}}
ight)}
ight]$$ $$eqalign{
& = { ext{Rs}}{ ext{.}},left( {{ ext{120}}\% ,{ ext{of}},{ ext{8000}} + { ext{7200}}}
ight) cr
& = { ext{Rs}}{ ext{.}},left( {{ ext{9600}} + { ext{7200}}}
ight) cr
& = { ext{Rs}}{ ext{.}},{ ext{16800}} cr
& herefore { ext{Total amount}} cr
& = { ext{Rs}}{ ext{.}},left( {{ ext{20000}} + { ext{16000}} + { ext{16800}}}
ight) cr
& = { ext{Rs}}{ ext{.}},{ ext{52800}} cr} $$
86: D
Solution: Let the sum at 8% be Rs. x Let the sum at 5% be 1550 - x According to the given condition, $$eqalign{
& frac{{{ ext{x}} imes 8 imes 3}}{{100}} + frac{{left( {1550 - { ext{x}}}
ight) imes 5 imes 3}}{{100}} = 300 cr
& Rightarrow frac{{24{ ext{x}} + left( {1550 - { ext{x}}}
ight) imes 15}}{{100}} = 300 cr
& Rightarrow frac{{24{ ext{x}} + 23250 - 15{ ext{x}}}}{{100}} = 300 cr
& Rightarrow 24{ ext{x}} - 15{ ext{x}} = 30000 - 23250 cr
& Rightarrow 24{ ext{x}} - 15{ ext{x}} = 6750 cr
& Rightarrow 9{ ext{x}} = 6750 cr
& Rightarrow { ext{x}} = frac{{6750}}{9} = 750 cr
& frac{{{ ext{Money lent at 5}}\% }}{{{ ext{Money lent at 8}}\% }} cr
& = frac{{1550 - 750}}{{750}} cr
& = frac{{800}}{{750}} cr
& = frac{{16}}{{15}} cr
& = 16:15 cr} $$
87: A
Solution: (Principal + Interest) in 5 years → 520 (Principal + Interest) in 7 years → 568 $$eqalign{
& herefore { ext{Interest in 2 years = 48}} cr
& herefore { ext{I = }}frac{{{ ext{48}}}}{2}{ ext{ }} cr
& ,,,,,,,,,{ ext{ = Rs}}{ ext{. 24 in one year}} cr
& herefore { ext{P = 520 - 24}} imes { ext{5}} cr
& ,,,,,,,,,,{ ext{ = Rs}}{ ext{. 400}} cr} $$
88: C
Solution: Let initial capital = 100 units According to the question, $$eqalign{
& frac{1}{4}{ ext{unit = Rs}}{ ext{. 61}}{ ext{.50}} cr
& { ext{1 unit = Rs}}{ ext{. 61}}{ ext{.50}} imes { ext{4}} cr
& ,,,,,,,,,,,,,,,,,,{ ext{ = Rs}}{ ext{. 246}} cr
& { ext{100 units = Rs}}{ ext{. 24600}} cr} $$ Hence, Required capital = Rs. 24600 Alternate Difference in percentage $$eqalign{
& { ext{ = 8% }} - frac{{31}}{4}\% cr
& Leftrightarrow frac{1}{4}\% = 61.50 cr
& Leftrightarrow 100\% = 24600 cr} $$
89: B
Solution: $$eqalign{
& { ext{P}} = { ext{Rs}}{ ext{.}},{ ext{450}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{. 81}} cr
& { ext{R}} = { ext{4}}{ ext{.5}}\% cr
& { ext{Time}} cr
& = left( {frac{{100 imes 81}}{{450 imes 4.5}}}
ight){ ext{years}} cr
& = 4,{ ext{years}}{ ext{}} cr} $$
90: D
Solution: $$eqalign{
& { ext{Let}}, cr
& { ext{Principal}} = Rs.,100 cr
& { ext{Amount}} = Rs.,200 cr
& { ext{Rate}} = r\% cr
& { ext{Time}} = 4,{ ext{years}} cr
& { ext{Now}}, cr
& A = P imes {left[ {1 + left( {frac{r}{{100}}}
ight)}
ight]^n} cr
& 200 = 100 imes {left[ {1 + left( {frac{r}{{100}}}
ight)}
ight]^4} cr
& 2 = {left[ {1 + left( {frac{r}{{100}}}
ight)}
ight]^4} - - - - - - left( i
ight) cr
& { ext{If}},{ ext{sum}},{ ext{become}},{ ext{8}},{ ext{times}},{ ext{in}},{ ext{the}},{ ext{time}},n,{ ext{years}} cr
& { ext{then,}} cr
& 8 = {left( {1 + left( {frac{r}{{100}}}
ight)}
ight)^n} cr
& {2^3} = {left( {1 + left( {frac{r}{{100}}}
ight)}
ight)^n} - - - - - - left( {ii}
ight) cr
& { ext{Using}},{ ext{eqn}},left( i
ight)inleft( {ii}
ight),,{ ext{we}},{ ext{get}} cr
& {left( {{{left[ {1 + left( {frac{r}{{100}}}
ight)}
ight]}^4}}
ight)^3} = {left( {1 + left( {frac{r}{{100}}}
ight)}
ight)^n} cr
& {left[ {1 + left( {frac{r}{{100}}}
ight)}
ight]^{12}} = {left( {1 + left( {frac{r}{{100}}}
ight)}
ight)^n} cr
& { ext{Thus}},,n = 12,{ ext{years}}. cr} $$
91: A
Solution: $$eqalign{
& { ext{Let}},{1^{st}},{ ext{part}},{ ext{is}},x,{ ext{and}},{2^{nd}},{ ext{part}},{ ext{is}},left( {6000 - x}
ight) cr
& { ext{According}},{ ext{to}},{ ext{question}}, cr
& {frac{{x imes 2 imes 6}}{{100}}} = frac{{ {left( {6000 - x}
ight) imes 3 imes 8} }}{{100}} cr
& 12x = 144000 - 24x cr
& Or,,36x = 144000 cr
& Or,,x = frac{{144000}}{{36}} = Rs.,4000 cr
& {1^{st}},{ ext{part}} = Rs.,4000 cr
& {2^{nd}},{ ext{part}} = Rs.,2000 cr} $$
92: B
Solution: $$eqalign{
& { ext{Let}},{ ext{sum}},is,{ ext{Rs}}.,100, cr
& { ext{Then}},{ ext{it}},{ ext{become}},frac{7}{4},{ ext{times}} cr
& i.e.,{ ext{Rs}}.,frac{{700}}{4},{ ext{in}},6,{ ext{years}}. cr
& { ext{Interest}} = {frac{{700}}{4}} - 100 = Rs.,frac{{300}}{4} cr
& { ext{Hence}}, cr
& { ext{Rate}} = frac{{{ ext{Total}},{ ext{interest}}}}{{{ ext{Given}},{ ext{time}}}} cr
& ,,,,,,,,,,,,,, = frac{{300}}{4} imes 6 cr
& ,,,,,,,,,,,,,, = 12 {frac{1}{2}} \% cr} $$
93: D
Solution: $$eqalign{
& { ext{Let}},{ ext{each}},{ ext{installment}},{ ext{be}},P cr
& { ext{Hence}}, cr
& {frac{x}{{ {left( {frac{{100}}{{100 + r}}}
ight) + {{left( {frac{{100}}{{100 + r}}}
ight)}^2}} }}} cr
& Or,,frac{x}{{ {1 + {frac{{15}}{{400}}} } }} + frac{x}{{ {1 + {{left( {frac{{15}}{{400}}}
ight)}^2}} }} = Rs.,13040 cr
& { ext{On}},{ ext{solving,}},{ ext{it}},{ ext{gives}},x = Rs.,6889 cr} $$
94: D
Solution: As short - cut of installment for 2 installments is given by Installment = $$ {frac{{ ext{P}}}{{ {left( {frac{{100}}{{100}} + { ext{R}}}
ight) + {{left( {frac{{100}}{{100}} + { ext{R}}}
ight)}^2}} }}} $$ There is the need of rate (R) which is unavailable in the question. so, we cannot determine the answer
95: N/A
Solution: $$eqalign{
& { ext{Total}},{ ext{cow}},{ ext{milksold}},{ ext{in}},{ ext{five}},{ ext{days}} cr
& = 5 imes 8 = 40,{ ext{litres}} cr
& { ext{Total}},{ ext{buffalo}},{ ext{milk}},{ ext{sold}},{ ext{in}},{ ext{five}},{ ext{days}} cr
& = 5 imes 10 = 50,{ ext{litres}} cr
& { ext{Hence}}, cr
& { ext{SP}},{ ext{of}},{ ext{40}},{ ext{litres}},{ ext{cow}},{ ext{milk}} cr
& = 40 imes 55 = Rs.,2200 cr
& { ext{Hence}}, cr
& { ext{Profit}},{ ext{on}},{ ext{cow}},{ ext{milk}} cr
& = frac{{ {2200 imes 12} }}{{112}} = Rs.,235.71 cr
& { ext{Profit}},{ ext{on}},{ ext{buffalo}},{ ext{milk}} cr
& = frac{{ {50 imes 36 imes 20} }}{{120}} = Rs.,300 cr
& { ext{Thus}}, cr
& { ext{Total}},{ ext{Profit}} cr
& = 235.71 + 300 = Rs.,535.71 cr} $$
96: C
Solution: $$eqalign{
& frac{{{ ext{Decrease in second year}}}}{{{ ext{Decrease in third year}}}} cr
& = frac{{100}}{{left( {100 - r}
ight)}} cr
& = frac{{10}}{9} o r = 10\% cr} $$ Let the population of vultures 3 years ago be P, then $$eqalign{
& P imes {left[ {1 - left( {frac{{10}}{{100}}}
ight)}
ight]^3} = 29160 cr
& o P = 40000 cr} $$
97: B
Solution: 1000 == 10%↑ ⇒ 1100 == Double ⇒ 2200 == 10%↑ ⇒ 2420 == Double ⇒ 4840 == 10%↑ ⇒ 5324 == Double ⇒ 10648
98: B
Solution: Difference between Compound interest and Simple interest for the fourth year is Rs. 7280 $$eqalign{
& P{left( {frac{6}{5}}
ight)^3}left( {frac{1}{5}}
ight) - frac{P}{5} = 7280 cr
& frac{P}{5}left[ {{{left( {frac{6}{5}}
ight)}^3} - 1}
ight] = 7280 cr
& frac{P}{5}left[ {frac{{216}}{{125}} - 1}
ight] = 7280 cr
& frac{P}{5} imes frac{{91}}{{125}} = 7280 cr
& herefore P = 50000 cr} $$
99: D
Solution: Let the money added by Rahul be Rs. x Then, $$ Rightarrow frac{{left( {1150 + x}
ight) imes 9 imes 3}}{{100}} - $$ xa0 xa0 $$frac{{1150 imes 6 imes 3}}{{100}} = $$ xa0xa0 $$274.95$$ ⇒ 1150 × 27 + 27x - 1150 × 18 = 27495 ⇒ 27x + 1150 × (27 - 18) = 27495 ⇒ 27x = 27495 - 10350 ⇒ 27x = 17145 ⇒ x = 635 So, sum lent by Rahul to Sachin = Rs. ( 1150 + 635 ) = Rs. 1785
100: A
Solution: $$eqalign{
& { ext{According to the question,}} cr
& { ext{Principal}} = { ext{Rs}}{ ext{. }}2100 cr
& { ext{Amount}} = { ext{Rs}}{ ext{. }}2352 cr
& { ext{SI}} = { ext{A}} - { ext{P}} cr
& ,,,,,,, = 2352 - 2100 cr
& ,,,,,,, = { ext{Rs}}{ ext{. }}252 cr
& { ext{Time = 2 years,}} cr
& { ext{Let rate = R% }} cr
& { ext{R = }}frac{{252}}{{2100}} imes frac{{100}}{2}{ ext{ = 6% }} cr
& { ext{New rate of interest}} cr
& { ext{ = (6}} - { ext{1)}} cr
& { ext{ = 5% }} cr
& { ext{New interest}} cr
& { ext{ = }}frac{{2100 imes 5 imes 2}}{{100}} cr
& { ext{ = Rs}}{ ext{. 210}} cr
& { ext{Hence required interest}} cr
& { ext{ = Rs}}{ ext{. 210}} cr} $$
101: D
Solution: Difference between their rates he gained from both boys $$eqalign{
& Rightarrow (15 imes 5)\% - (12 imes 4)\% cr
& Rightarrow 75\% - 48\% cr
& Rightarrow 27\% = 1350{ ext{ }}({ ext{given)}} cr
& Rightarrow 100\% = { ext{Rs}}{ ext{. 5000}} cr} $$
102: B
Solution: According to the question, Principal + SI for 2 year = Rs. 720 ......(i) Principal + SI for 7 year = Rs. 1020 ......(ii) Subtracting equation (i) from (ii) ⇒ SI for 5 years = (1020 - 720) = Rs. 300 ⇒ SI for 1 years = Rs. 60 ⇒ SI for 2 years = 60 × 2 = Rs. 120 ⇒ Principal amount = (Amount after 2 years - 2 years SI) = (720 - 120) ⇒ Principal amount = Rs. 600
103: B
Solution: Let the amount invested at 12% be Rs. x and that invested at 10% be Rs. y $$eqalign{
& ext{Then,} cr
& o 12\% ,{ ext{of }}x + 10\% ,{ ext{of }}y = 130 cr
& Rightarrow 12x + 10y = 13000 cr
& Rightarrow 6x + 5y = 6500......{ ext{(i)}} cr
& { ext{And,}} cr
& o 10\% ,{ ext{of }}x + 12\% ,{ ext{of }}y = 134 cr
& Rightarrow 10x + 12y = 13400 cr
& Rightarrow 5x + 6y = 6700......{ ext{(ii)}} cr
& { ext{Adding (i) and (ii), we get:}} cr
& 11left( {x + y}
ight) = 13200 cr
& Rightarrow x + y = 1200.......({ ext{iii}}) cr
& { ext{Subtracting (i) from (ii),}} cr
& { ext{we get: }} - x + y = 200.......({ ext{iv}}) cr
& { ext{Adding (iii) and (iv), }} cr
& { ext{we get}}:2y = 1400,or,y = 700 cr
& { ext{Hence,}} cr
& { ext{Amount invested at 12%}} cr
& = left( {1200 - 700}
ight) cr
& = { ext{Rs}}{ ext{. 500}} cr} $$
104: A
Solution: $$eqalign{
& { ext{Let each sum}} = { ext{Rs}}{ ext{. }}x. cr
& { ext{Let the first sum be invested for}} cr
& left( {T - frac{1}{2}}
ight){ ext{years and}} cr
& { ext{the second sum for }}T{ ext{ years}}{ ext{.}} cr
& { ext{Then,}} cr
& x + frac{{x imes 8 imes left( {T - frac{1}{2}}
ight)}}{{100}} = 2560 cr
& Rightarrow 100x + 8xT - 4x = 256000 cr
& Rightarrow 96x + 8xT = 256000....(i) cr
& { ext{And,}} cr
& x + frac{{x imes 7 imes T}}{{100}} = 2560 cr
& Rightarrow 100x + 7xT = 256000....(ii) cr
& { ext{From(i) and (ii), we get:}} cr
& 96x + 8xT = 100x + 7xT cr
& Rightarrow 4x = xT cr
& Rightarrow T = 4 cr
& { ext{Putting }}T = { ext{4 in (i),we get:}} cr
& 96x + 32x = 256000 cr
& Rightarrow 128x = 256000 cr
& Rightarrow x = 2000 cr
& { ext{Hence,}} cr
& { ext{each sum}} = { ext{Rs}}{ ext{. 2000}} cr
& { ext{time periods}} = cr
& { ext{4 years and }}3frac{1}{2}{ ext{years}} cr} $$
105: D
Solution: According to the question, $${ ext{A}} + left( {frac{{{ ext{A}} imes { ext{5}} imes { ext{2}}}}{{{ ext{100}}}}}
ight) = $$ xa0 xa0 $${ ext{B}} + left( {frac{{{ ext{B}} imes { ext{5}} imes { ext{3}}}}{{{ ext{100}}}}}
ight) = $$ xa0 xa0 $${ ext{C}} + left( {frac{{{ ext{C}} imes { ext{5}} imes { ext{4}}}}{{{ ext{100}}}}}
ight)$$ 110A xa0 = xa0 115B xa0 = xa0 120C 22A xa0 = xa0 xa0 23B xa0 = xa0 24X Ratio of amount ( by using L.C.M. of 22, 23 and 24) $$eqalign{
& { ext{276 : 264 : 253}} cr
& { ext{A's loan = }}frac{{276}}{{793}} imes { ext{7930}} cr
& ,,,,,,,,,,,,,,,,,,{ ext{ = Rs}}{ ext{. 2760}} cr} $$
106: C
Solution: $$eqalign{
& { ext{According to the question,}} cr
& { ext{Interest = Rs}}{ ext{. 1 per day}} cr
& herefore { ext{Interest in one year}} cr
& { ext{ = 1}} imes { ext{365 = Rs}}{ ext{. 365}} cr
& herefore { ext{S}}{ ext{.I}}{ ext{. = }}frac{{{ ext{P}} imes { ext{T}} imes { ext{R}}}}{{100}} cr
& Rightarrow 365 = frac{{{ ext{P}} imes 5 imes 1}}{{100}} cr
& Rightarrow { ext{P}} = frac{{365 imes 100}}{5} cr
& Rightarrow { ext{P}} = { ext{Rs}}{ ext{. 7300}} cr} $$
107: C
Solution: $$eqalign{
& { ext{According to the question,}} cr
& { ext{Amount = Rs}}{ ext{. 3144}} cr
& { ext{Rate = 8}}\% cr
& { ext{Let, Principal = Rs}}{ ext{. }}x cr
& herefore { ext{Time = }} cr
& frac{{30 + 29 + 31 + 30 + 31 + 30 + 31 + 7}}{{366}} cr
& = frac{{219}}{{366}} cr
& herefore { ext{SI = }}frac{{{ ext{P}} imes { ext{R}} imes { ext{T}}}}{{100}} cr
& Rightarrow 3144 - x = frac{{x imes 8 imes 219}}{{100 imes 366}} cr
& = { ext{Rs}}{ ext{. 3000}} cr
& cr
& {x08f{Alternate}} cr
& Rightarrow { ext{P}} + frac{{{ ext{P}} imes 8 imes frac{{219}}{{365}}}}{{100}} = 3144 cr
& Rightarrow { ext{P}} + frac{{{ ext{P}} imes 8 imes frac{3}{5}}}{{100}} = 3144 cr
& Rightarrow 100{ ext{P}} + frac{{24{ ext{P}}}}{5} = 314400 cr
& Rightarrow frac{{524{ ext{P}}}}{5} = 314400 cr
& Rightarrow { ext{P = }}frac{{314400 imes 5}}{{524}} cr
& Rightarrow { ext{P}} = { ext{600}} imes { ext{5}} cr
& Rightarrow { ext{P = Rs}}{ ext{.}} , 3000 cr} $$
108: A
Solution: Let the rates of interest in the former and latter cases be R% and (R + 1) % p.a. Then, $$eqalign{
& 5000 imes { ext{R}} imes 4 = 4000 imes left( {{ ext{R}} + 1}
ight) imes 4 cr
& Rightarrow frac{{{ ext{R}} + 1}}{{ ext{R}}} = frac{{5000 imes 4}}{{4000 imes 4}} cr
& Rightarrow 1 + frac{1}{{ ext{R}}} = 1 + frac{1}{4} cr
& Rightarrow { ext{R}} = 4 cr
& { ext{Hence,}} cr
& { ext{Required rate}} = 4\% ,{ ext{p}}{ ext{.a}}{ ext{.}} cr} $$
109: C
Solution: $$eqalign{
& { ext{Let the required time = t years }} cr
& { ext{According to the question,}} cr
& Leftrightarrow frac{{500 imes 4 imes 6.25}}{{100}} = frac{{400 imes 5 imes { ext{t}}}}{{100}} cr
& Leftrightarrow 5 imes 4 imes 625 = 400 imes 5 imes { ext{t}} cr
& Leftrightarrow { ext{t = }}frac{{625}}{{100}} = frac{{25}}{4} cr
& Leftrightarrow { ext{t}} = 6frac{1}{4}{ ext{years}} cr} $$
110: C
Solution: $$eqalign{
& frac{{{ ext{Principal}}}}{{{ ext{Amount}}}} = frac{{4 imes 5}}{{5 imes 5}} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,, = frac{{20}}{{25}} cr
& { ext{After three year}} cr
& frac{{ ext{P}}}{{ ext{A}}} = frac{{5 imes 4}}{{7 imes 4}} cr
& ,,,,,,,, = frac{{20}}{{28}} cr
& { ext{In three year S}}{ ext{.I}}{ ext{.}} cr
& = 28x - 25x cr
& = 3x cr
& ext{So, the required interest will be} cr
& 3x = frac{{20x imes { ext{R}} imes 3}}{{100}} cr
& { ext{R}} = 5\% cr} $$
111: B
Solution: Let time be T years and rate be R% p.a. $$eqalign{
& { ext{Then, }}y{ ext{ is the S}}{ ext{.I}}{ ext{. on x}} cr
& Rightarrow frac{{x{ ext{RT}}}}{{100}} = y......(i) cr
& { ext{And, }}z{ ext{ is the S}}{ ext{.I}}{ ext{. on y}} cr
& Rightarrow frac{{y{ ext{RT}}}}{{100}} = z cr
& Rightarrow y = frac{{100z}}{{RT}}......(ii) cr
& { ext{From (i) and (ii) we have:}} cr
& frac{{x{ ext{RT}}}}{{100}} = frac{{100z}}{{{ ext{RT}}}} cr
& Rightarrow frac{{x{{ ext{R}}^2}{{ ext{T}}^2}}}{{{{left( {100}
ight)}^2}}} = z cr
& Rightarrow frac{{{y^2}}}{x} = z cr
& Rightarrow {y^2}= xz cr }$$
112: A
Solution: $$eqalign{
& { ext{Let the sum be Rs}}{ ext{. }}x cr
& { ext{Then,}} cr} $$ $$ {frac{{x imes 8 imes 4}}{{100}}} + {frac{{x imes 10 imes 6}}{{100}}} ,+ $$ xa0 xa0 $$ {frac{{x imes 12 imes 5}}{{100}}} $$ xa0 $$ = 12160$$ $$eqalign{
& Rightarrow 32x + 60x + 60x = 1216000 cr
& Rightarrow 152x = 1216000 cr
& Rightarrow x = 8000 cr} $$
113: N/A
Solution: Let the loan amount be Rs. x $$eqalign{
& { ext{Then,}} cr
& Rightarrow frac{{6x}}{{100}} + frac{{7.5x}}{{100}} + frac{{9x}}{{100}} = 8190 cr
& Rightarrow 22.5x = 819000 cr
& Rightarrow x = 36400 cr} $$
114: D
Solution: Rate of interest $$ = frac{{100(x - 1)}}{t}\% $$ Where x is the no. of times the sum becomes of itself. Here the sum is getting 3 times. Therefore, x = 3 Where t is the time taken by sum to become x times of itself. Here, t = 12 years By the short trick approach, we get $$eqalign{
& = frac{{100(3 - 1)}}{{12}} cr
& = frac{{200}}{{12}} cr
& = frac{{50}}{3} = 16frac{2}{3}\% cr} $$ Alternative Method : Let the principal be P and in the 2nd scenario, SI = 2P $$eqalign{
& { ext{Rate}} = frac{{{ ext{SI}} imes 100}}{{{ ext{Principal}} imes { ext{Time}}}} cr
& = frac{{2{ ext{P}} imes 100}}{{{ ext{P}} imes 12}} cr
& = frac{{50}}{3} cr
& = 16frac{2}{3}\% cr} $$
115: A
Solution: $$eqalign{
& { ext{Let the first part = Rs}}{ ext{. }}x cr
& herefore { ext{Hence second part}} cr
& { ext{ = Rs}}{ ext{. }}left( {12000 - x}
ight) cr
& { ext{According to the question,}} cr
& Rightarrow frac{{x imes 12 imes 3}}{{100}} = frac{{left( {12000 - x}
ight) imes 9 imes 16}}{{2 imes 100}} cr
& Rightarrow 36x = 72left( {12000 - x}
ight) cr
& Rightarrow x = 24000 - 2x cr
& Rightarrow 3x = 24000 cr
& Rightarrow x = { ext{Rs}}{ ext{. 8000}} cr
& {{ ext{1}}^{{ ext{st}}}}{ ext{ part = Rs}}{ ext{. 8000}} cr
& {{ ext{2}}^{{ ext{nd}}}}{ ext{ part}} cr
& { ext{ = Rs}}{ ext{. }}left( {12000 - 8000}
ight) cr
& = { ext{Rs}}{ ext{. 4000 }} cr
& { ext{Hence maximum part}} cr
& { ext{ = Rs}}{ ext{. 8000}} cr} $$ Alternate Note : In such type of questions to save your valuable time follow the given below method. Let two parts P 1 and P 2 respectively According to the question, $$eqalign{
& Rightarrow {{ ext{P}}_1} imes frac{{36}}{{100}} imes 1 = {{ ext{P}}_2} imes frac{9}{2} imes frac{{16}}{{100}} imes 1 cr
& Rightarrow {{ ext{P}}_1} imes 4 = 8{{ ext{P}}_2} cr
& Rightarrow frac{{{{ ext{P}}_1}}}{{{{ ext{P}}_2}}} = frac{2}{1} cr
& Rightarrow {{ ext{P}}_1}{ ext{:}}{{ ext{P}}_2} = 2:1 cr
& { ext{Hence greater part}} cr
& { ext{ = }}frac{{12000}}{{left( {2 + 1}
ight)}} imes 2 cr
& = { ext{Rs}}{ ext{. }}8000 cr} $$
116: B
Solution: $$eqalign{
& { ext{Let the sum be Rs}}{ ext{.100}}{ ext{}} cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.for first 6 months}} cr
& = { ext{Rs}}{ ext{.}}left( {frac{{100 imes 10 imes 1}}{{100 imes 2}}}
ight) cr
& = { ext{Rs}}{ ext{. }}5 cr
& { ext{S}}{ ext{.I}}{ ext{.for last 6 months}} cr
& = { ext{Rs}}{ ext{.}}left( {frac{{105 imes 10 imes 1}}{{100 imes 2}}}
ight) cr
& = { ext{Rs}}{ ext{. }}5.25 cr
& So, cr
& { ext{Amount at the end of 1year}} cr
& = { ext{Rs}}{ ext{.}}left( {100 + 5 + 5.25}
ight) cr
& = { ext{Rs}}{ ext{.}},110.25 cr
& herefore { ext{Effective rate}} cr
& = left( {110.25 - { ext{100}}}
ight) cr
& = 10.25\% cr} $$
117: C
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{. for 1 year}} cr
& = { ext{Rs}}.left( {854 - 815}
ight) cr
& = { ext{Rs}}.39 cr
& { ext{S}}{ ext{.I}}{ ext{. for 3 years}} cr
& = { ext{Rs}}{ ext{.}}left( {39 imes 3}
ight) cr
& = { ext{Rs}}{ ext{. }}117 cr
& herefore ext{Principal} cr
& = { ext{Rs}}{ ext{.}}left( {854 - 117}
ight) cr
& = { ext{Rs}}{ ext{. }}698 cr} $$
118: A
Solution: $$eqalign{
& P = Rs.,3300 cr
& A = Rs.,3399 cr
& R = 6\% ,{ ext{per}},{ ext{annum}} cr
& { ext{Let}},{ ext{the}},{ ext{time}},{ ext{be}},{ ext{n}},{ ext{years}}{ ext{.}} cr
& { ext{Compound}},{ ext{interest}},{ ext{is}},{ ext{taken}},{ ext{half - yearly}}. cr
& A = P imes {left[ {1 + left( {frac{R}{2} imes 100}
ight)}
ight]^{2n}} cr
& 3399 = 3300{left( {1 + frac{3}{{100}}}
ight)^{2n}} cr
& {left( {1.03}
ight)^{2n}} = frac{{3399}}{{3300}} cr
& {left( {1.03}
ight)^{2n}} = {left( {1.03}
ight)^1} cr
& Thus,,2n = 1,year cr
& n = frac{1}{2}{ ext{year}} = 6,{ ext{months}} cr} $$
119: A
Solution: $$eqalign{
& { ext{Here,}},{ ext{time}},{ ext{interval}},{ ext{is}},{ ext{given}},{ ext{as}}, cr
& { ext{February}},5,,1994,{ ext{to}},{ ext{April}},18,,1994 cr
& = 73,{ ext{days}} = frac{{73}}{{365}} = 0.2,{ ext{years}}. cr
& { ext{Now}},{ ext{interest}} = frac{{PTR}}{{100}} cr
& = frac{{ {700 imes 9 imes 0.2} }}{{100}} cr
& = Rs.,12.60 cr} $$
120: A
Solution: Let the rate be R at which Principal P is invested for 2 years. According to question, {Interest at Rate (R + 2)}% - (interest at rate R%) = Rs. 60 $$frac{{left( {P imes 2 imes left( {R + 2}
ight)}
ight)}}{{100}} - $$ xa0 xa0 $$frac{{left( {P imes 2 imes R}
ight)}}{{100}}$$ xa0 $$ = 60$$ $$eqalign{
& frac{{ {2PR + 4P - 2PR} }}{{100}} = 60 cr
& 4P = 60 imes 100 cr
& { ext{Or}},P = frac{{60 imes 100}}{4} cr
& { ext{Hence}},P = { ext{Rs}}{ ext{.}},1500 cr} $$
121: C
Solution: $$eqalign{
& { ext{Let}},{ ext{sum}},{ ext{is}},P. cr
& { ext{The}},{ ext{difference}},{ ext{between}},{ ext{compound}},{ ext{interest}},{ ext{and}} cr
& ,{ ext{simple}},{ ext{interest}},{ ext{over}},{ ext{three}},{ ext{years}},{ ext{is}},{ ext{given}},{ ext{by}} cr
& = Pleft( {frac{r}{{100}}}
ight)2 imes left{ {left( {frac{r}{{100}}}
ight) + 3}
ight} cr
& 48 = P imes left( {frac{{20}}{{100}}}
ight)2 imes left{ {left( {frac{{20}}{{100}}}
ight) + 3}
ight} cr
& 48 = P imes frac{4}{{100}} imes frac{{16}}{5} cr
& 48 = P imes frac{{64}}{{500}} cr
& { ext{Or}},,64P = 48 imes 500 cr
& { ext{Hence}},,P = Rs.,375 cr} $$
122: A
Solution: $$eqalign{
& { ext{Let}},{ ext{time}},{ ext{is}},T,{ ext{years}}. cr
& { ext{According}},{ ext{to}},{ ext{questions}}, cr
& frac{{1750 imes 9 imes T}}{{100}} = frac{{left( {2500 imes 10.5 imes 4}
ight)}}{{100}} cr
& Or,,T = frac{{ {2500 imes 10.5 imes 4} }}{{1750 imes 9}} cr
& Or,,T = 6.66 = 6,{ ext{years}},{ ext{and}},8,{ ext{months}} cr} $$
123: A
Solution: $$eqalign{
& { ext{Let}},{ ext{rate}},{ ext{is}},R\% cr
& { ext{According}},{ ext{to}},{ ext{the}},{ ext{question}}, cr
& left[ {frac{{400 imes 2 imes R}}{{100}}}
ight] + left[ {frac{{100 imes 4 imes R}}{{100}}}
ight] = 60 cr
& 8R + 4R = 60 cr
& { ext{Hence}},,R = 5\% cr} $$
124: D
Solution: $$eqalign{
& 9/8 cr
& { ext{Principle}} = 8{ ext{ unit}} cr
& { ext{Amount}} = 9{ ext{ unit}} cr
& { ext{Interest}} = 9.8{ ext{ unit}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{8 imes 5 imes R}}{{100}} cr
& 1 = frac{{40 imes R}}{{100}} cr
& R = frac{{10}}{4} = 2.5\% { ext{ p}}{ ext{.a}}{ ext{.}} cr} $$
125: A
Solution: $$eqalign{
& 2P = frac{{5P imes 8 imes T}}{{10}} cr
& T = 5{ ext{ years}} cr} $$
126: C
Solution: $$eqalign{
& frac{{x imes r imes 3}}{{100}} + frac{{2x imes left( {r - 2}
ight)}}{{100}} imes 5 = frac{{x imes r}}{{100}} imes frac{{34}}{3} cr
& frac{{2x imes left( {r - 2}
ight) imes 5}}{{100}} = frac{{x imes r}}{{100}} imes frac{{34}}{3} - frac{{x imes r imes 3}}{{100}} cr
& frac{{2x imes left( {r - 2}
ight) imes 5}}{{100}} = frac{{left( {34r - 9r}
ight) imes x}}{{300}} cr
& 6left( {r - 2}
ight) imes 5 = 25r cr
& frac{{36x}}{{100}} + frac{{2x imes 10}}{{100}} imes 5 = 13600 cr
& frac{{136x}}{{100}} = 13600 cr
& x = { ext{Rs}}{ ext{. }}10000 cr} $$
127: B
Solution: $$eqalign{
& {x08f{Given:}} cr
& {P_1} = 7000,,{T_1} = 3{ ext{ years, }}{R_1} = 5\% cr
& {P_1} = 7000,,{T_2} = 3{ ext{ years, }}{R_2} = 6frac{1}{3}\% cr
& {x08f{Formula}},{x08f{used:}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{{ ext{Principal amout}} imes { ext{Rate of interest}} imes { ext{Time}}}}{{100}} cr
& {x08f{Calculation:}} cr
& { ext{S}}{ ext{.I}}{{ ext{.}}_1} = frac{{{P_1} imes {R_1} imes {T_1}}}{{100}} cr
& = frac{{7000 imes 5 imes 3}}{{100}} cr
& = 1050 cr
& { ext{S}}{ ext{.I}}{{ ext{.}}_1}{ ext{ for one year}} = frac{{1050}}{3} = 350 cr
& { ext{S}}{ ext{.I}}{{ ext{.}}_2} = frac{{{P_1} imes {R_2} imes {T_2}}}{{100}} cr
& = frac{{7000 imes frac{{19}}{3} imes 3}}{{100}} cr
& = 1330 cr
& { ext{S}}{ ext{.I}}{{ ext{.}}_2}{ ext{ for one year}} = frac{{1330}}{3} = 443.33 cr
& { ext{Gain for one year}} = 443.33 - 350 = 93.33 cr
& herefore { ext{He gain in the transaction per year 93}}{ ext{.33}} cr} $$
128: C
Solution: $$eqalign{
& { ext{Annual instalment}} = frac{{{ ext{due debt}} imes 100}}{{100 imes t + rtleft( {frac{{t - 1}}{2}}
ight)}} cr
& = frac{{5664 imes 100}}{{100 imes 4 + 4 imes 12 imes frac{3}{2}}} cr
& = frac{{5664 imes 100}}{{472}} cr
& = { ext{Rs}}{ ext{. }}1200 cr} $$
129: C
Solution: Formula used: A = $${ ext{P}}frac{{left( {1 + { ext{RT}}}
ight)}}{{100}}$$ A → Amount P → Principal R → Rate T → Time Calculation: A sum at simple interest becomes two times in 8 years ⇒ 2P = $$frac{{{ ext{P}}left( {1 + { ext{R}} imes { ext{8}}}
ight)}}{{100}}$$ ⇒ R = $$frac{{100}}{8}$$ ⇒ R = 12.5 ∴ The rate of interest is 12.5% The time in which the same sum will be 4 times ⇒ 4P = $$frac{{{ ext{P}}left( {1 + 12.5 imes { ext{T}}}
ight)}}{{100}}$$ ⇒ 300 = 12.5T ⇒ T = 24 ∴ The sum will be 4 times in 24 years
130: C
Solution: $$eqalign{
& R = 9\% cr
& P = 3270 cr
& t = 3 cr
& { ext{EMI}} = frac{{P imes 100}}{{100t + left[ {left( {t - 1}
ight) + left( {t - 2}
ight),...}
ight] imes R}} cr
& = frac{{3270 imes 100}}{{100 imes 3 + left[ {2 + 1}
ight] imes 9}} cr
& = frac{{3270 imes 100}}{{327}} cr
& = 1000 cr} $$
131: D
Solution: $$eqalign{
& 6frac{1}{2}\% o 1071.20 cr
& 1\% o 164.80 cr
& 100\% o 16480 cr
& { ext{SI}} = frac{{16480 imes 5 imes 10}}{{100}} = 8240 cr} $$
132: A
Solution: [x08egin{array}{*{20}{c}}
A&B \
x&{36000 - x}
end{array}] $$eqalign{
& { ext{As per question}} cr
& x imes 30 = left( {36000 - x}
ight) imes 60 cr
& x = 72000 - 2x cr
& x = 24000 cr
& { ext{So interest from }}A cr
& = 24000 imes frac{{30}}{{100}} cr
& = 7200 cr} $$
133: A
Solution: $$eqalign{
& left( {frac{{P imes 5 imes 3}}{{100}}}
ight) = 3left( {frac{{Q imes 7 imes 4}}{{100}}}
ight) cr
& P = Q imes frac{{28}}{5} cr
& P = 5.6Q cr} $$
134: A
Solution: Let Sum = Rs. x. Then, S.I. = Rs. x, Time = 16 years $$eqalign{
& herefore { ext{Rate}} = left( {frac{{100 imes x}}{{x imes 16}}}
ight)\% = {frac{25}{4}}\% = {6frac{1}{4}}\% cr
& { ext{Now, sum}} = { ext{Rs}}{ ext{. }}x, cr
& { ext{Time}} = 8{kern 1pt} { ext{years}} cr
& { ext{Rate}} = 6frac{1}{4}\% cr
& herefore { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}}left( {frac{{x imes 25 imes 8}}{{100 imes 4}}}
ight) cr
& ,,,,,,,,,,,,,,,, = { ext{Rs}}{ ext{. }}frac{x}{2} cr
& { ext{So,}} cr
& { ext{Amount}} = { ext{Rs}}{ ext{.}}left( {x + frac{x}{2}}
ight) cr
& ,,,,,,,,,,,,,,,,,,,,,, = { ext{Rs}}{ ext{. }}frac{{3x}}{2} cr
& ,,,,,,,,,,,,,,,,,,,,,, = 1frac{1}{2}{ ext{ times}} cr} $$
135: B
Solution: $$eqalign{
& { ext{Let sum be x}}{ ext{.}} cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = x cr
& { ext{I - Time}} cr
& = frac{{100 imes x}}{{x imes frac{{50}}{3}}} cr
& = 6,{ ext{years(false)}} cr
& { ext{II}} - { ext{Time}} cr
& = frac{{100 imes x}}{{x imes 20}} cr
& = 5,{ ext{years(True)}} cr
& { ext{III}} - { ext{Suppose sum}} = x. cr
& { ext{Then, S}}{ ext{.I}}{ ext{. }} = x cr
& { ext{Time }} = { ext{5 }}{ ext{years}}{ ext{.}} cr
& { ext{Rate}} = left( {frac{{100 imes x}}{{x imes 5}}}
ight)\% cr
& ,,,,,,,,,,,, = 20\% . cr
& { ext{Now, sum}} = x,,{ ext{S}}{ ext{.I}}{ ext{.}} = 3x,{ ext{and}},{ ext{Rate}} = 20\% cr & herefore { ext{Time}} = left( {frac{{100 imes 3x}}{{x imes 20}}}
ight){ ext{years}} cr
& ,,,,,,,,,,,,,,,,,, = 15,{ ext{years}}( ext{false}) cr
& { ext{So, B alone is correct}}{ ext{.}} cr} $$
136: B
Solution: According to the question, Principal Interest 10 3 $$eqalign{
& { ext{Rate }}\% { ext{ = 10}}\% cr
& { ext{Time = }}frac{3}{{10}} imes frac{{100}}{{10}} cr
& ,,,,,,,,,,,,, = 3,{ ext{years}} cr} $$
137: C
Solution: Let the amount invested = Rs. P According to the question, $$eqalign{
& { ext{P}} + frac{{{ ext{P}} imes 10 imes 4}}{{100}} = 770 cr
& Rightarrow { ext{P}} + frac{{4{ ext{P}}}}{{10}} = 770 cr
& Rightarrow frac{{14{ ext{P}}}}{{10}} = 770 cr
& Rightarrow { ext{P}} = frac{{770 imes 10}}{{14}} cr
& Rightarrow { ext{P}} = { ext{Rs 550}} cr} $$ Hence, required invested amount = Rs. 550 Alternate $$eqalign{
& { ext{10}}\% { ext{ = }}frac{{1 o { ext{Interest}}}}{{10 o { ext{Principal}}}} cr
& { ext{Interest in 4 years}} cr
& { ext{ = 1}} imes { ext{4}} cr
& = { ext{4}} cr
& { ext{Amount = }} cr
& = left( {{ ext{Interest + Principal}}}
ight) cr
& = 4 + 10 cr
& = 14 cr
& { ext{According to the question,}} cr
& { ext{14 units = 770}} cr
& { ext{1 unit = }}frac{{770}}{{14}} cr
& { ext{10 units = }}frac{{770}}{{14}} imes { ext{10}} cr
& ,,,,,,,,,,,,,,,,,,{ ext{ = Rs}}{ ext{. 550 }} cr
& { ext{The amount invested}} cr
& { ext{ = Rs}}{ ext{. 550}} cr} $$
138: B
Solution: $$eqalign{
& { ext{Rate }}\% = { ext{12}}\% cr
& { ext{Principal = Rs}}{ ext{. 1860}} cr
& { ext{Amount = Rs}}{ ext{. 2641}}{ ext{.20}} cr
& { ext{Interest}} cr
& { ext{ = Rs}}{ ext{. }}left( {2641.20 - 1860}
ight) cr
& = { ext{Rs}}{ ext{. 781}}{ ext{.20}} cr
& { ext{By using formula,}} cr
& { ext{Required time }} cr
& = frac{{781.20 imes 100}}{{1860 imes 12}} cr
& = 3frac{1}{2}{ ext{ years}} cr} $$
139: D
Solution: $$eqalign{
& { ext{Let sum}} = x{ ext{.}} cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{x}{2} cr
& herefore frac{x}{2} = frac{{x imes 8 imes 6}}{{100}} cr} $$ Clearly, data is inadequate.
140: A
Solution: $$eqalign{
& { ext{Let sum}} = x. cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = 0.125x = frac{1}{8}x cr
& { ext{R}} = 10\% cr
& herefore ext{Rate} cr
& = left( {frac{{100 imes x}}{{x imes 8 imes 10}}}
ight){ ext{years}} cr
& = frac{5}{4}{ ext{years}} cr
& = { ext{1}}frac{1}{4}{ ext{years}} cr} $$
141: B
Solution: $$eqalign{
& { ext{Present Population}} cr
& = { ext{P}}{left( {frac{{1 - { ext{R}}}}{{100}}}
ight)^n} cr
& = 10000{left( {frac{{1 - 20}}{{100}}}
ight)^2} cr
& = 10000{left( {frac{{100 - 20}}{{100}}}
ight)^2} cr
& = 10000{left( {frac{{80}}{{100}}}
ight)^2} cr
& = 10000{left( {frac{4}{5}}
ight)^2} cr
& = 10000 imes frac{{16}}{{25}} cr
& = 400 imes 16 cr
& = 6400 cr} $$
142: A
Solution: Let two parts are P 1 and P 2 respectively According to the question $$eqalign{
& frac{{{{ ext{P}}_1} imes 3 imes 12}}{{100}} = frac{{{{ ext{P}}_2} imes 9 imes 16}}{{2 imes 100}} cr
& 36{{ ext{P}}_1} = 72{{ ext{P}}_2} cr
& frac{{{{ ext{P}}_1}}}{{{{ ext{P}}_2}}} = frac{{72}}{{36}} = frac{2}{1} cr
& {{ ext{P}}_1}{ ext{:}}{{ ext{P}}_2} = 2:1 cr
& { ext{Hence,}} cr
& { ext{Required ratio = 2 : 1}} cr} $$
143: C
Solution: Capital after paying income tax $$eqalign{
& Rightarrow { ext{4}}\% - { ext{3}}{ ext{.75}}\% = { ext{48}} cr
& Rightarrow { ext{0}}{ ext{.25}}\% { ext{ = 48}} cr
& { ext{100}}\% { ext{ = }}frac{{48}}{{0.25}} imes 100 cr
& ,,,,,,,,,,,,,,, = 19200 cr} $$ ⇒ Capital without paying income tax ⇒ 19200 = Capital × 96% Net capital = 20000
144: C
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}},{ ext{8376}} cr
& { ext{R}} = 8\% cr
& { ext{T}} = { ext{6}}{kern 1pt} { ext{years}} cr
& herefore { ext{Sum}} = { ext{Rs}}{ ext{.}}left( {frac{{100 imes 8376}}{{8 imes 6}}}
ight) cr
& ,,,,,,,,,,,,,,,,,, = { ext{Rs}}{ ext{. }}17450 cr} $$
145: B
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{. 210}} cr
& { ext{R}} = 3frac{3}{4}\% = frac{{15}}{4}\% cr
& { ext{T}} = { ext{2}}frac{{ ext{1}}}{{ ext{3}}}{ ext{years}} = frac{7}{3}{ ext{years}} cr
& herefore { ext{Sum}} = { ext{Rs}}{ ext{.}}left( {frac{{100 imes 210}}{{frac{{15}}{4} imes frac{7}{3}}}}
ight) cr
& = { ext{Rs}}{ ext{.}}left( {frac{{100 imes 210 imes 4 imes 3}}{{15 imes 7}}}
ight) cr
& = { ext{Rs}}{ ext{. }}2400 cr} $$
146: A
Solution: $$eqalign{
& { ext{Let principal = 10P}} cr
& { ext{Interest = 10P}} imes frac{{30}}{{100}} cr
& ,,,,,,,,,,,,,,,,,,,{ ext{ = 3P}} cr
& { ext{According to the question,}} cr
& { ext{Case (I)}} cr
& Rightarrow 3{ ext{P = }}frac{{{ ext{10P}} imes { ext{R}} imes { ext{6}}}}{{100}} cr
& Rightarrow { ext{R = 5}}\% cr
& { ext{Case (II)}} cr
& { ext{Interest = Principal = 10P}} cr
& Rightarrow { ext{10P = }}frac{{{ ext{10P}} imes { ext{5}} imes { ext{t}}}}{{100}} cr
& Rightarrow { ext{t = 20 years}} cr} $$
147: A
Solution: Let the principal in each case = 100 units
According to the question, 1 st part 2 nd part 3 rd part Principal → 100 x6 100 x3 100 x2 Rate % → 10 12 15 Time → 6 10 12 Interest → 60 x6 120 x3 180 x2 Interest Interest is same in each, so equal the interest. Hence required ratio = 600 : 300 : 200 of sum = 6 : 3 : 2
148: D
Solution: $$eqalign{
& { ext{Principal = Rs}}{ ext{. 1000 }} cr
& { ext{Rate = 5}}\% cr
& { ext{Interest for first 10 years}} cr
& = frac{{1000 imes 5 imes 10}}{{100}} cr
& = { ext{Rs}}{ ext{. 500}} cr
& { ext{After 10 years principal}} cr
& = { ext{(1000}} + { ext{500)}} cr
& { ext{ = Rs}}{ ext{. 1500}} cr
& { ext{Remaining interest}} cr
& { ext{ = Rs}}{ ext{. (2000}} - { ext{1500)}} cr
& { ext{ = Rs}}{ ext{. 500}} cr
& { ext{Required time }} cr
& { ext{ = }}frac{{500}}{{1500}} imes frac{{100}}{5} cr
& = frac{{100}}{15} cr
& = frac{{20}}{3} cr
& = 6frac{2}{3}{ ext{ years}} cr
& { ext{Total time}} cr
& = left( {10 + 6frac{2}{3}}
ight){ ext{years}} cr
& { ext{ = 16}}frac{2}{3}{ ext{ years}} cr} $$
149: B
Solution: $$eqalign{
& { ext{Sum}} = {frac{{100 imes S.I.}}{{R imes T}}} cr
& = { ext{Rs}}{ ext{.}},, {frac{{100 imes x}}{{x imes x}}} cr
& = { ext{Rs}}{ ext{.}},, {frac{{100}}{x}} cr} $$
150: C
Solution: $$eqalign{
& { ext{P}} = { ext{Rs}}{ ext{. 6200}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} cr
& = { ext{Rs}}{ ext{.}}left( {{ ext{9176}} - { ext{6200}}}
ight) cr
& = { ext{Rs}}{ ext{. }}2976 cr
& { ext{T}} = 4,{ ext{years}} cr
& herefore { ext{Rate}} cr
& = left( {frac{{100 imes 2976}}{{6200 imes 4}}}
ight)\% cr
& = 12\% cr
& { ext{New }}{ ext{rate}} cr
& = left( {12 + 3}
ight)\% cr
& = 15\% cr
& { ext{New}},{ ext{S}}{ ext{.I}}{ ext{.}} cr
& = { ext{Rs}}{ ext{.}}left( {frac{{{ ext{6200}} imes 15 imes 4}}{{100}}}
ight) cr
& = ext{Rs.} ,3720 cr
& { ext{New}},{ ext{amount}} cr
& = { ext{Rs}}{ ext{.}}left( {{ ext{6200}} + 3720}
ight) cr
& = { ext{Rs}}{ ext{.}},9920 cr} $$
151: N/A
Solution: $$eqalign{
& { ext{Sum}} = { ext{Rs}}{ ext{.}}left( {frac{{100 imes 1536}}{{12 imes 5}}}
ight) cr
& = { ext{Rs}}{ ext{. }}2560 cr
& Now, cr
& P = { ext{Rs}}{ ext{.}}left( {2560 + 1000}
ight) cr
& ,,,,,, = { ext{Rs}}{ ext{. }}3560 cr
& { ext{T}} = { ext{2years}} cr
& { ext{R}} = { ext{12}}\% cr
& herefore S.I. cr
& = { ext{Rs}}{ ext{.}}left( {frac{{3560 imes 12 imes 2}}{{100}}}
ight) cr
& = { ext{Rs}}{ ext{.}},854.40 cr} $$
152: B
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{. 840}} cr
& { ext{R}} = { ext{5}}\% cr
& { ext{T}} = 8,{ ext{years}} cr
& { ext{Principal}} cr
& = { ext{Rs}}{ ext{. }}left( {frac{{100 imes 840}}{{5 imes 8}}}
ight) cr
& = { ext{Rs}}{ ext{.}},2100 cr
& { ext{Now,}} cr
& { ext{P}} = { ext{Rs}}{ ext{.}},2100 cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}},{ ext{840}} cr
& { ext{T}} = { ext{5 years}}{ ext{}} cr
& herefore { ext{Rate}} cr
& = left( {frac{{100 imes 840}}{{2100 imes 5}}}
ight)\% cr
& = { ext{8}}\% cr} $$
153: B
Solution: $$eqalign{
& { ext{Principal}},,,,,,,{ ext{Amount}} cr
& underbrace {,,,,,{ ext{6000}},,,,,,,,,,,,,,{ ext{7200}},,,,,}_{ + 1200} cr
& { ext{By using formula,}} cr
& { ext{Rate }}\% cr
& = frac{{1200}}{{6000}} imes frac{{100}}{4} cr
& { ext{ = 5}}\% cr
& { ext{New rate}}\% cr
& = { ext{5}} imes frac{3}{2} = { ext{7}}{ ext{.5}}\% cr
& { ext{Interest after 5 years}} cr
& = frac{{6000 imes 7.5 imes 5}}{{100}} cr
& { ext{ = Rs}}{ ext{. 2250}} cr
& { ext{Hence,}} cr
& { ext{amount }} cr
& { ext{ = Rs}}{ ext{. }}left( {6000 + 2250}
ight) cr
& { ext{ = Rs 8250}} cr} $$
154: C
Solution: Let Sum be Rs. x and Let S.I. = Rs. x $$eqalign{
& { ext{Time}} = 10,{ ext{years}} cr
& herefore { ext{Rate}} = frac{{{ ext{S}}{ ext{.I}}{ ext{.}} imes { ext{100}}}}{{{ ext{Principal}} imes { ext{Time}}}} cr
& = frac{{x imes 100}}{{x imes 10}} cr
& = 10\% ,{ ext{per}},{ ext{annum}}{ ext{.}} cr} $$
155: B
Solution: $$eqalign{
& { ext{Rate}} = frac{{{ ext{S}}{ ext{.I}}{ ext{.}} imes { ext{100}}}}{{{ ext{Principal}} imes { ext{Time}}}} cr
& ,,,,,,,,,,,,, = frac{{5400 imes 100}}{{15000 imes 3}} cr
& = 12\% ,{ ext{Per annum}}{ ext{.}} cr} $$
156: A
Solution: Let the amount invested in scheme A be Rs. x and that in B be Rs. 3x. Then, $$eqalign{
& = frac{{x imes 4 imes 8}}{{100}} + frac{{3x imes 2 imes 13}}{{100}} = 1320 cr
& or,,frac{{32x}}{{100}} + frac{{78x}}{{100}} = 1320 cr
& or,,frac{{110x}}{{110}} = 1320 cr
& herefore x = frac{{1320 imes 100}}{{110}} cr
& = { ext{Rs}}{ ext{. 1200}}. cr} $$
157: C
Solution: S.I for 1 year = 5832 – 5182 = Rs. 650 S.I for 2 years = Rs. 1300 P = 5182 – 1300 = Rs. 3882
158: A
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{. of 2}}{ ext{.5 years}} cr
& { ext{ = 5500}} - { ext{4000}} cr
& { ext{ = 1500}} cr
& { ext{S}}{ ext{.I}}{ ext{. of 2 years}} cr
& { ext{ = }}frac{{1500}}{{2.5}} imes { ext{2}} cr
& { ext{ = 1200}} cr
& { ext{So, Principal}} cr
& { ext{ = 4000}} - { ext{SI of 2 years}} cr
& Rightarrow { ext{Principal = 4000}} - 1200 cr
& Rightarrow { ext{Principal = 2800}} cr
& { ext{r}}\% { ext{ = }}frac{{1200}}{{2800 imes 2}} imes { ext{100}} cr
& ,,,,,,,,{ ext{ = 21}}frac{3}{7}\% cr} $$
159: B
Solution: Let Ram get Rs. x and Shyam get Rs. (260200 - x) Then Amount get by Ram after 3 years $${ ext{ = }}x imes {left( {1 + frac{4}{{100}}}
ight)^3}$$ and Amount get by Shyam after 6 years $$ = left( {260200 - x}
ight) imes {left( {1 + frac{4}{{100}}}
ight)^6}$$ But both get equal amount $$ herefore x imes {left( {1 + frac{4}{{100}}}
ight)^3} = left( {260200 - x}
ight) imes $$ xa0 xa0 xa0 $${left( {1 + frac{4}{{100}}}
ight)^6}$$ $$eqalign{
& Rightarrow frac{x}{{2600200 - x}} = frac{{17576}}{{15625}} cr
& Rightarrow 15625x = 4573275200 - 17576x cr
& Rightarrow 33201x = 4573275200 cr
& Rightarrow x = 137745.022 cr} $$ So, Ram will get Rs. 137745.022
160: A
Solution: $$eqalign{
& frac{1}{4}:frac{1}{5}:frac{1}{8} cr
& 10:8:5 cr
& { ext{Amount invested at }}5\% cr
& = frac{8}{{23}} imes 57,500 cr
& = 20,000 cr} $$
161: D
Solution: $$eqalign{
& { ext{Cash amount}} = 25000 cr
& { ext{Down payment}} = 2500 cr
& { ext{Remaining amount}} = 22500 cr
& { ext{Rate}} = 24\% ,{ ext{p}}{ ext{.a}} cr
& { ext{12 month}} o 24\% cr
& o frac{{24}}{{12}} imes 100 = frac{1}{{50}} cr
& 4{ ext{ month}} o 8\% cr
& Rightarrow { ext{Total amount}} = 22500 imes frac{{108}}{{100}} cr
& = 24300 cr} $$ [x08egin{array}{*{20}{c}}
{50}& - &{50} \
{50}& - &{51} \
{50}& - &{52} \
{50}& - &{mathop {53}limits_{\_\_\_\_\_\_} } \
{}&{}&{206}
end{array}] $$eqalign{
& { ext{EMI}} = frac{{24300 imes 50}}{{206}} cr
& = 5898.05 cr
& = 5900 cr} $$
162: C
Solution: $$eqalign{
& { ext{3 years total rate}} = 14 imes 3 = 42\% cr
& { ext{5 years total rate}} = 14 imes 5 = 70\% cr
& { ext{Gap}} = 28\% o 4200 cr
& { ext{Sum}} = frac{{4200}}{{28}} imes 100 = { ext{Rs}}{ ext{. }}15,000 cr} $$
163: C
Solution: $$eqalign{
& { ext{S}}{ ext{.}}{{ ext{I}}_1} = 2 imes { ext{S}}{ ext{.}}{{ ext{I}}_2} cr
& frac{{x imes 20 imes 21}}{{100 imes 3 imes 5}} = 2 imes frac{{y imes 4 imes 11}}{{100 imes 4}} cr
& 28x = 22y cr
& x:y = 22:28 cr
& { ext{Ratio of second part of first part}} cr
& y:x = 28:22 = 14:11 cr} $$
164: A
Solution: $${ ext{S}}{ ext{.I}}{ ext{.}} = frac{{12000 imes 15 imes 3}}{{100}} = { ext{Rs}}{ ext{. }}5400$$
165: A
Solution: $$eqalign{
& frac{{P imes 5.4 imes 9.25}}{{100}} = left( {14395.20 - P}
ight) cr
& P = 9600 cr
& frac{{9600 imes 8.6 imes 4.5}}{{100}} = { ext{SI}} cr
& { ext{SI}} = 3715.20 cr} $$
166: D
Solution: $$eqalign{
& 6 = 5 imes frac{{20}}{{100}} imes t cr
& t = 6{ ext{ years}} cr} $$
167: B
Solution: Let the rate of interest be R% per annum = Amount to be paid. If paid at the end of 11 months $$ Rightarrow 10 + frac{{10 imes R imes frac{{11}}{{12}}}}{{100}} = 10 + frac{{11R}}{{120}}$$ Total effective payments = (Rs. 1 + Interest on Rs. 1 for 10 months) + (Rs. 1 + Interest on Rs. 1 for 9 months) + . . . . . + (Rs. 1 + Interest on Rs. 1 for 1 month) + Rs. 1 $$eqalign{
& = left( {1 + frac{{1 imes R imes frac{{10}}{{12}}}}{{100}}}
ight) + left( {1 + frac{{1 imes R imes frac{9}{{12}}}}{{100}}}
ight) + ..... + left( {1 + frac{{1 imes R imes frac{1}{{12}}}}{{100}}}
ight) + 1 cr
& = left( {1 + frac{{10R}}{{1200}}}
ight) + left( {1 + frac{{9R}}{{1200}}}
ight) + ..... + left( {1 + frac{R}{{1200}}}
ight) + 1 cr
& = 11 + frac{{Rleft( {frac{{10 imes 11}}{2}}
ight)}}{{1200}} cr
& = 11 + frac{{11R}}{{240}} cr
& { ext{Now we have}} cr
& 10 + frac{{11R}}{{120}} = 11 + frac{{11R}}{{240}} cr
& frac{{11R}}{{240}} = 1 cr
& R = frac{{240}}{{11}} cr
& x08oxed{R = 21frac{9}{{11}}\% } cr} $$
168: D
Solution: Simple interest for 3 years = 6750 - 6000 = 750 Simple interest for 1 year = 750 ÷ 3 = 250 Simple interest for 4 years = 4 × 250 = 1000 Sum = 6000 - 1000 = 5000
169: C
Solution: Simple interest on 20000 for 3 years at 10% rate $${ ext{SI}} = frac{{20000 imes 3 imes 10}}{{100}} = 6000$$ Simple interest on 40000 for 3 years at 15% rate $${ ext{SI}} = frac{{40000 imes 3 imes 15}}{{100}} = 18000$$ Total interest = 6000 + 18000 = 24000
170: D
Solution: $$eqalign{
& { ext{Rate}} = frac{{5{ ext{ paisa}}}}{{1{ ext{ rupee}}}} = 5\% cr
& { ext{Time}} = 8{ ext{ months}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{2700 imes 5 imes 8}}{{100}} = 1080 cr} $$
171: A
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{{ ext{.}}_1} = frac{5}{2} imes { ext{S}}{ ext{.I}}{{ ext{.}}_2} cr
& frac{{x imes 25 imes 15}}{{100 imes 3 imes 2}} = frac{5}{2} imes frac{{y imes 8 imes 21}}{{100 imes 4}} cr
& 25 imes 5 imes x = 5 imes 2 imes 21 imes y cr
& x:y = 210:125 = 42:25 cr
& = 42 - 25 = 17{ ext{ units}} cr
& 67{ ext{ units}} = 50250 cr
& 1{ ext{ unit}} = 750 cr
& 17{ ext{ units}} = 750 imes 17 = 12750 cr
& { ext{Difference between both parts}} = 12750 cr} $$
172: A
Solution: Let X = equal installment at the end of one year( rate% annually) .
Now 1st year, P =210, Interest = $$frac{{{ ext{PTR}}}}{{100}}$$xa0 = 210 * 0.1 = 21. Let X is to be paid as an equal installment. Now, at the beginning of 2nd year, P = 210 + 21 - X, Interest at the end of 2nd year, = (231 - X) * 0.1 = 23.1 - 0.1X. Hence,total installment, 2X = 210 + 21 + 23.1 - 0.1X, X = $$frac{{{ ext{254}}{ ext{.1}}}}{{2.1}}$$xa0 = 121.
173: C
Solution: $$A = P {left( {1 - frac{R}{{100}}}
ight)^n}$$ Where A = Value of goods after n years P = Initial Price R = Rate of depriciation $$eqalign{
& herefore P = frac{{5832}}{{{{left( {1 - frac{{10}}{{100}}}
ight)}^3}}} cr
& Rightarrow P = frac{{5832}}{{{{left( {1 - frac{1}{{10}}}
ight)}^3}}} cr
& Rightarrow P = frac{{5832}}{{{{left( {frac{9}{{10}}}
ight)}^3}}} cr
& Rightarrow P = 5832 imes frac{{10}}{9} imes frac{{10}}{9} imes frac{{10}}{9} cr
& Rightarrow P = 8000 cr} $$
174: C
Solution: S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Therefore Principal = Rs. (815 - 117) = Rs. 698.
175: A
Solution: Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x ) Then, $$ {frac{{x imes 14 imes 2}}{{100}}} + $$ xa0 $$ {frac{{left( {13900 - x}
ight) imes 11 imes 2}}{{100}}} $$ xa0 xa0 $$ = 3508$$ ⇒ 28x - 22x = 350800 - (13900 x 22) ⇒ 6x = 45000 ⇒ x = 7500 So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400
176: D
Solution: $$eqalign{
& { ext{Principal}} cr
& = Rs.,left( {frac{{100 imes 4016.25}}{{9 imes 5}}}
ight) cr
& = Rs.,left( {frac{{401625}}{{45}}}
ight) cr
& = Rs.,8925 cr} $$
177: B
Solution: $$eqalign{
& { ext{Time}} = left( {frac{{100 imes 81}}{{450 imes 4.5}}}
ight){ ext{years}} cr
& ,,,,,,,,,,,,,, = 4,{ ext{years}} cr} $$
178: B
Solution: $$eqalign{
& { ext{Let}},{ ext{rate}}, = R\% ,{ ext{and}} cr
& { ext{Time}},{ ext{ = }},{ ext{R}},{ ext{years}}{ ext{}} cr
& { ext{Then,}}, {frac{{1200 imes R imes R}}{{100}}} = 432 cr
& Rightarrow 12{R^2} = 432 cr
& Rightarrow {R^2} = 36 cr
& Rightarrow R = 6 cr} $$
179: D
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{.}} = Rs.,left( {15500 - 12500}
ight) cr
& ,,,,,,,,,, = Rs.,3000 cr
& { ext{Rate}} = left( {frac{{100 imes 3000}}{{12500 imes 4}}}
ight)\% cr
& ,,,,,,,,,,,,,, = 6\% cr} $$
180: B
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{sum}},{ ext{be}},{ ext{Rs}}{ ext{.}},{ ext{100}}{ ext{.}},{ ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}},{ ext{for}},{ ext{first}},{ ext{6}},{ ext{months}} cr
& = Rs.,left( {frac{{100 imes 10 imes 1}}{{100 imes 2}}}
ight) cr
& = Rs.,5 cr
& { ext{S}}{ ext{.I}}{ ext{.}},{ ext{for}},{ ext{last}},{ ext{6}},{ ext{months}} cr
& = Rs.,left( {frac{{105 imes 10 imes 1}}{{100 imes 2}}}
ight) cr
& = Rs.,5.25 cr} $$ So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25 ∴ Effective rate = (110.25 - 100) = 10.25%
181: D
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{rate}},{ ext{be}},R\% ,{ ext{p}}{ ext{.a}}{ ext{.}},{ ext{Then,}} cr
& {frac{{5000 imes R imes 2}}{{100}}} + {frac{{3000 imes R imes 4}}{{100}}} cr
& = 2200. cr
& Rightarrow 100R + 120R = 2200 cr
& Rightarrow R = {frac{{2200}}{{220}}} = 10 cr
& herefore { ext{Rate}} = 10\% cr} $$
182: C
Solution: $$eqalign{
& { ext{Principal}} = Rs.,left( {frac{{100 imes 5400}}{{12 imes 3}}}
ight) cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,, = Rs.,15000 cr} $$
183: C
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{. }}{kern 1pt} { ext{for 3 years}} cr
& = { ext{Rs}}{ ext{.}}left( {12005 - 9800}
ight) cr
& = { ext{Rs}}{ ext{. }}2205 cr
& { ext{S}}{ ext{.I}}{ ext{. for 5 years}} cr
& = { ext{Rs}}{ ext{.}},left( {frac{{2205}}{3} imes 5}
ight) cr
& = { ext{Rs}}{ ext{.}},3675 cr
& herefore { ext{Principal}} cr
& = { ext{Rs}}{ ext{.}},(9800 - 3675) cr
& = { ext{Rs}}{ ext{.}},6125 cr
& { ext{Hence,}},{ ext{rate}} = left( {frac{{100 imes 3675}}{{6125 imes 5}}}
ight)\% cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, = 12\% cr} $$
184: C
Solution: Let the principal be P and rate of interest be R%. Therefore Required ratio $$eqalign{
& = frac{{ {frac{{P imes R imes 6}}{{100}}} }}{{ {frac{{P imes R imes 9}}{{100}}} }} cr
& = frac{{6PR}}{{6PR}} cr
& = frac{6}{9} cr
& = 2:3 cr} $$
185: D
Solution: We need to know the S.I., principal and time to find the rate. Since the principal is not given, so data is inadequate.
186: A
Solution: $${ ext{Gain in 2 years}}$$ $$ = { ext{Rs}}{ ext{.}},left[ {left( {5000 imes frac{{25}}{4} imes frac{2}{{100}}}
ight) - left( {frac{{5000 imes 4 imes 2}}{{100}}}
ight)}
ight]$$ $$eqalign{
& = { ext{Rs}}{ ext{.}},left( {625 - 400}
ight) cr
& = { ext{Rs}}.225 cr
& herefore { ext{Gain in 1 year}} cr
& = { ext{Rs}}{ ext{.}},left( {frac{{225}}{2}}
ight) cr
& = { ext{Rs}}{ ext{.}},112.50 cr} $$
187: A
Solution: $$eqalign{
& { ext{Let principal = 6P}} cr
& { ext{Hence Amount}} cr
& { ext{ = 6P}} imes frac{7}{6} = 7{ ext{P}} cr
& herefore { ext{SI = 7P}} - { ext{6P = P}} cr
& { ext{Time = 3 years}} cr
& x08oxed{{ ext{SI = }}frac{{{ ext{P}} imes { ext{T}} imes { ext{R}}}}{{100}}} cr
& Rightarrow { ext{P = }}frac{{6{ ext{P}} imes { ext{R}} imes { ext{3}}}}{{100}} cr
& Rightarrow { ext{R = }}frac{{100}}{{18}} cr
& ,,,,,,,,,,,, = frac{{50}}{9} cr
& ,,,,,,,,,,,, = 5frac{5}{9}\% cr} $$ Alternate: Note: In such type of questions to save your valuable time try to think like that. [x08egin{gathered}
x08egin{array}{*{20}{c}}
{{ ext{Amount}}}&{{ ext{Principal}}}
end{array} hfill \
{ ext{ }}underbrace {7,,,,,,,,,,,,,,,,,,,,,,,,,,,,{ ext{6}}}_{ + 1} hfill \
end{gathered} ] $$eqalign{
& { ext{Required rate % }} cr
& { ext{ = }}frac{1}{6} imes frac{{100}}{3} cr
& = 5frac{5}{9}\% cr} $$
188: C
Solution: Let the rate of interest for two different sources is r 1 and r 2 respectively. $$eqalign{
& frac{{1500 imes {{ ext{r}}_1} imes 3}}{{100}} - frac{{1500 imes {{ ext{r}}_2} imes 3}}{{100}} cr
& = 13.50 cr
& 4500{{ ext{r}}_1} - 4500{{ ext{r}}_2} = 1350 cr
& left( {{{ ext{r}}_1} - {{ ext{r}}_2}}
ight) = frac{{1350}}{{4500}} = 0.3\% cr} $$ Hence required difference in rates = 0.3%
189: C
Solution: $$eqalign{
& { ext{Time = 2 years 3 months}} cr
& { ext{ = 2 + }}frac{3}{{12}}{ ext{ = }}frac{9}{2}{ ext{ years}} cr
& { ext{We know }}x08oxed{{ ext{SI = }}frac{{{ ext{P}} imes { ext{T}} imes { ext{R}}}}{{100}}} cr
& { ext{P = Rs 1600,}} cr
& { ext{T = }}frac{9}{4}{ ext{years,}} cr
& { ext{SI = Rs 252}} cr
& { ext{Put values in the formula ,}} cr
& Rightarrow 252 = frac{{1600 imes { ext{R}} imes 9}}{{100}} cr
& Rightarrow 252 = 36{ ext{R}} cr
& Rightarrow { ext{R = }}frac{{252}}{{36}} = 7\% cr} $$
190: D
Solution: $$eqalign{
& { ext{P}} = { ext{Rs}}.,520, cr
& { ext{R}} = 13\% cr
& { ext{T}} = frac{1}{2}yr. cr
& herefore { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}}left( {frac{{520 imes 13}}{{100 imes 2}}}
ight) cr
& ,,,,,,,,,,,,,, = { ext{Rs}}{ ext{. }}33.80 cr
& { ext{Hence, amount after 6 months}} cr
& = { ext{Rs}}{ ext{. }}left( {520 + 33.80}
ight) cr
& = { ext{Rs}}{ ext{. }}553.80 cr} $$
191: B
Solution: $$eqalign{
& { ext{Principal}},,,,,{ ext{Interest}} cr
& underbrace {,,,,,,,4{ ext{P}},,,,,,,,,,,,,,,,,,,,{ ext{P}},,,,,,,,,,}_{} cr
& { ext{Time = Rate }}\% { ext{ (given)}} cr
& { ext{Now by using formula , }} cr
& { ext{P = }}frac{{4{ ext{P}} imes { ext{R}} imes { ext{R}}}}{{100}} cr
& Rightarrow {{ ext{R}}^2} = frac{{100}}{4} cr
& Rightarrow { ext{R = }}frac{{10}}{2} cr
& Rightarrow { ext{R = 5}}\% cr} $$
192: D
Solution: $$eqalign{
& { ext{Let the rate }}\% { ext{ = R}} cr
& { ext{According to the question,}} cr
& frac{{5000 imes 2 imes { ext{R}}}}{{100}} + frac{{3000 imes 4 imes { ext{R}}}}{{100}} = 2200 cr
& Rightarrow 100{ ext{R}} + 120{ ext{R}} = 2200 cr
& Rightarrow 220{ ext{R}} = 2200 cr
& Rightarrow { ext{R}} = 10\% cr
& { ext{Hence required rate}}\% cr
& = 10\% cr} $$
193: A
Solution: $$eqalign{
& { ext{By using formula,}} cr
& { ext{Installment}} cr
& { ext{ = }}frac{{6450 imes 100}}{{4 imes 100 + left( {3 + 2 + 1}
ight) imes 5}} cr
& { ext{ = }}frac{{6450 imes 100}}{{4 imes 100 + left( 6
ight) imes 5}} cr
& = frac{{6450 imes 100}}{{4 imes 100 + 30}} cr
& = frac{{6450 imes 100}}{{430}} cr
& = { ext{Rs}}{ ext{. 1500}} cr
& { ext{Hence value of installment}} cr
& { ext{ = Rs}}{ ext{. 1500}} cr} $$
194: B
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{. for 5 years}} cr
& = { ext{Rs}}{ ext{.}}left( {1020 - 720}
ight) cr
& = { ext{Rs}}{ ext{. 300}} cr
& { ext{S}}{ ext{.I}}{ ext{. for 2 years}} cr
& = { ext{Rs}}{ ext{.}}left( {frac{{300}}{5} imes 2}
ight) cr
& = { ext{Rs}}{ ext{. }}120 cr
& herefore { ext{Principal}} cr
& = { ext{Rs}}{ ext{.}}left( {{ ext{720}} - 120}
ight) cr
& = { ext{Rs}}{ ext{. }}600 cr} $$
195: A
Solution: $$eqalign{
& { ext{S}}{ ext{.I}}{ ext{. }}{ ext{for 3 years}} cr
& = { ext{Rs}}{ ext{.}}left( {24412.50 - 20925}
ight) cr
& = { ext{Rs}}{ ext{. }}3487.50 cr
& { ext{S}}{ ext{.I}}{ ext{. }}{ ext{for 2 years}} cr
& = { ext{Rs}}{ ext{.}}left( {frac{{3487.50}}{3} imes 2}
ight) cr
& = { ext{Rs}}{ ext{. }}2325 cr
& herefore ext{Principal} cr
& = { ext{Rs}}{ ext{.}}left( {20925 - 2325}
ight) cr
& = { ext{Rs}}{ ext{. }}18600 cr
& { ext{Hence,}} cr
& { ext{rate}} = left( {frac{{100 - 2325}}{{18600 imes 2}}}
ight)\% cr
& = 6.25\% cr} $$
196: B
Solution: $$eqalign{
& { ext{Let sum = Rs}}{ ext{. }}x{ ext{}} cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{. = Rs}}{ ext{. }}x{ ext{}} cr
& herefore ext{Time} = left( {frac{{100 imes { ext{S}}{ ext{.I}}{ ext{.}}}}{{{ ext{P}} imes { ext{R}}}}}
ight) cr
& = left( {frac{{100 imes x}}{{x imes 18.75}}}
ight){ ext{years}} cr
& = frac{{26}}{3}{ ext{years}} cr
& = 5frac{1}{3}{ ext{years}} cr
& = { ext{5 years 4 months}}{ ext{}} cr} $$
197: C
Solution: $$eqalign{
& { ext{16}}frac{2}{3} = frac{{1 o { ext{ Interest}}}}{{6 o { ext{ Principal }}}} cr
& { ext{Let principal = 6}} cr
& { ext{Interest = 6}} cr
& { ext{Time = t years}} cr
& { ext{By using formula }} cr
& { ext{6}} = frac{{6 imes 50 imes { ext{t}}}}{{3 imes 100}} cr
& Rightarrow { ext{t}} = 6,{ ext{years}} cr} $$ Alternate Note : In such type of questions to save your valuable time think like the given way. $$eqalign{
& { ext{Rate}}\% cr
& { ext{ = 16}}frac{2}{3}\% = frac{{1 o { ext{ Interest}}}}{{6 o { ext{ Principal }}}} cr
& { ext{Represent for 1 years}} cr
& { ext{According to the question,}} cr
& { ext{Principal = Interest}} cr
& { ext{6 = 1}} imes { ext{6}} cr
& { ext{Hence,}} cr
& { ext{Time = 1}} imes { ext{6}} cr
& ,,,,,,,,,,,,{ ext{ = 6 years}} cr} $$ Note : If interest will be six times then time will also be six times.
198: A
Solution: More interest paid in 2 years $$eqalign{
& { ext{ = 2}} imes { ext{1}} = { ext{2}}\% cr
& { ext{According to the question, }} cr
& { ext{2}}\% { ext{ of sum = Rs}}{ ext{. 24}} cr
& { ext{1}}\% { ext{ of sum = Rs}}{ ext{.}}frac{{24}}{2} cr
& { ext{Total sum}} cr
& { ext{ = Rs}}{ ext{. }}frac{{24}}{2} imes 100 cr
& = { ext{Rs}}{ ext{. }}1200 cr} $$
199: B
Solution: Let the total amount = Rs. 6 Total average rate of interest $$eqalign{
& { ext{ = }}frac{{left( {3 imes 10\% }
ight) + left( {2 imes 9\% }
ight) + left( {1 imes 12\% }
ight)}}{6} cr
& = frac{{left( {30 + 18 + 12}
ight)}}{6} \% cr
& = 10\% cr} $$
200: B
Solution: $$eqalign{
& { ext{Let sum}} = { ext{Rs}}{ ext{. }}x cr
& { ext{Then,}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = { ext{Rs}}{ ext{.}},x cr
& herefore ext{Rate},\% cr
& = left( {frac{{100 imes x}}{{x imes 7}}}
ight)\% cr
& = frac{{100}}{7}\% cr
& { ext{Now, sum}} = { ext{Rs}}{ ext{. }}x cr
& { ext{S}}{ ext{.I}}. = { ext{Rs}}{ ext{. }}3x cr
& ext{Rate} = frac{{100}}{7}\% cr
& herefore { ext{Total Time}} cr
& = left( {frac{{100 imes 3x}}{{x imes frac{{100}}{7}}}}
ight){ ext{years}} cr
& = 21,{ ext{years}} cr} $$
201: C
Solution: $$eqalign{
& frac{{left( {n - 1}
ight)}}{{{t_1}}} = frac{{left( {m - 1}
ight)}}{{{t_2}}} cr
& frac{1}{7} = frac{3}{{{t_2}}} cr
& {t_2} = 21{ ext{ years}} cr} $$
202: B
Solution: $$eqalign{
& { ext{Let sum = }}x cr
& { ext{Interest = }}frac{3}{4}x{ ext{ }} cr
& { ext{Interest = }}frac{{{ ext{PTR }}}}{{100}} cr
& Rightarrow frac{3}{4}x = frac{{x imes { ext{R}} imes 12.5}}{{100}} cr
& { ext{R = 6}}\% cr} $$
203: C
Solution: $$eqalign{
& { ext{t}} = { ext{1 month = }}frac{1}{{12}}{ ext{year}} cr
& { ext{SI = 1 paisa = Rs}}{ ext{. }}frac{1}{{100}} cr
& { ext{r}}\% = frac{{{ ext{SI}} imes { ext{100}}}}{{{ ext{P}} imes { ext{T}}}} = frac{{1 imes 100 imes 12}}{{100 imes 1 imes 1}} cr
& { ext{r}}\% = 12\% cr} $$
204: D
Solution: Let the three amounts be Rs. x, Rs. y and Rs. z, Then, $$eqalign{
& frac{{x imes 10 imes 6}}{{100}} = frac{{y imes 12 imes 10}}{{100}} = frac{{z imes 15 imes 12}}{{100}} cr
& Rightarrow 60x = 120y = 180z cr
& Rightarrow x = 2y = 3z = k(say) cr
& Rightarrow x = k,y = frac{k}{2},z = frac{k}{3} cr
& Rightarrow x:y:z = k:frac{k}{2}:frac{k}{3} cr
& Rightarrow x:y:z = 1:frac{1}{2}:frac{1}{3} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,, = 6:3:2 cr} $$
205: C
Solution: $$eqalign{
& { ext{Rate of interest}} = { ext{5}}\% { ext{ p}}{ ext{.a}}{ ext{.}} cr
& { ext{sum}} = { ext{Rs}}{ ext{. 3200}} cr
& { ext{Time}} = { ext{2 years 6 months}} cr
& ,,,,,,,,,,,,,,, = 2frac{1}{2}{ ext{years}} cr
& ,,,,,,,,,,,,,,, = frac{5}{2}{ ext{years}} cr
& { ext{S}}{ ext{.I}}{ ext{.}} = frac{{{ ext{Prinicipal}} imes { ext{Time}} imes { ext{Rate}}}}{{100}} cr
& ,,,,,,,,,, = frac{{3200 imes 5 imes 5}}{{2 imes 100}} cr
& ,,,,,,,,,, = { ext{Rs}}{ ext{. }}400 cr
& herefore { ext{Amount}} = { ext{Rs}}{ ext{. sum}} + { ext{S}}{ ext{.I}}. cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,, = left( {3200 + 400}
ight) cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,, = { ext{Rs}}{ ext{. 3600}}{ ext{}} cr} $$
206: B
Solution: $$eqalign{
& {x08f{Case - I:}} cr
& { ext{SI}}\% = { ext{R}}\% imes { ext{t}} cr
& { ext{SI}}\% = { ext{5}}\% imes { ext{8}} cr
& = { ext{40}}\% cr
& {x08f{Case - II:}} cr
& { ext{SI}}\% = { ext{5}} imes { ext{r}}\% { ext{ }} cr
& { ext{According to the question}} cr
& Rightarrow { ext{40}}\% = { ext{5}} imes { ext{r}}\% cr
& Rightarrow { ext{r}}\% = { ext{8}}\% cr} $$
207: D
Solution: Money is double itself means interest is equal to money. $$eqalign{
& { ext{P = }}frac{{{ ext{P}} imes { ext{8}} imes { ext{r}}}}{{100}} cr
& Rightarrow { ext{r = }}frac{{100}}{8} cr
& Rightarrow { ext{r = 12}}frac{1}{2}\% cr} $$
208: C
Solution: $$eqalign{
& { ext{Let principal and SI is}} cr
& { ext{ = }}10x,{ ext{ }}3x{ ext{ }} cr
& { ext{and time is = }}t cr
& Rightarrow 3x = frac{{10x imes 6 imes t}}{{100}} cr
& Rightarrow t = 5,{ ext{years}} cr} $$
209: B
Solution: $$eqalign{
& { ext{By using formula }} cr
& Leftrightarrow frac{{{ ext{3000}} imes 12 imes { ext{T}}}}{{100}} = 1080 cr
& Leftrightarrow { ext{T = }}frac{{108}}{{36}}{ ext{ = 3 years}} cr} $$
210: B
Solution: SI in three years = 7504 - 6076 = 1428 SI in one year $$ = frac{{1428}}{3} = 476$$ Now we can go through option

from option B $$frac{{5600 imes 8.5}}{{100}} = 476$$
211: N/A
Solution: Let the original rate be R%. Then, new rate = (2R)%. Note: Here, original rate is for 1 year(s)

the new rate is for only 4 months i.e. $$frac{1}{3}$$ year(s). $$eqalign{
& herefore {frac{{725 imes R imes 1}}{{100}}} + {frac{{362.50 imes 2R imes 1}}{{100 imes 3}}} cr
& = 33.50 cr
& Rightarrow left( {2175 + 725}
ight)R = 33.50 imes 100 imes 3 cr
& Rightarrow left( {2175 + 725}
ight)R = 10050 cr
& Rightarrow left( {2900}
ight)R = 10050 cr
& Rightarrow R = frac{{10050}}{{2900}} = 3.46 cr
& herefore ext{Original rate} = 3.46\% cr} $$
212: C
Solution: Let total capital be Rs. x Then, $$ Rightarrow left( {frac{x}{3} imes frac{7}{{100}} imes 1}
ight) + left( {frac{x}{4} imes frac{8}{{100}} imes 1}
ight) + $$ xa0 xa0 xa0 $$left( {frac{{5x}}{{12}} imes frac{{10}}{{100}} imes 1}
ight)$$ xa0xa0 $$ = 561$$ $$eqalign{
& Rightarrow frac{{7x}}{{300}} + frac{x}{{50}} + frac{x}{{24}} = 561 cr
& Rightarrow 51x = left( {561 imes 600}
ight) cr
& Rightarrow x = left( {frac{{561 imes 600}}{{51}}}
ight) cr
& ,,,,,,,,,,,,,, = 6600 cr} $$
213: B
Solution: $$eqalign{
& { ext{Value}},{ ext{of}},{ ext{maruti}},{ ext{Van}},, cr
& {V_0} = Rs.,196000 cr
& { ext{Rate}},{ ext{of}},{ ext{depreciation}},, cr
& r = 14 {frac{2}{7}} \% = frac{{100}}{7}\%
cr
& { ext{Time}},,t = 2,{ ext{years}} cr
& { ext{Let}},{V_1},{ ext{is}},{ ext{the}},{ ext{value}},{ ext{after}},{ ext{depreciation}}. cr
& {V_1} = {V_0} imes {left[ {1 - left( {frac{r}{{100}}}
ight)}
ight]^t} cr
& {V_1} = 196000 imes {left[ {1 - left( {frac{{left( {frac{{100}}{7}}
ight)}}{{100}}}
ight)}
ight]^2} cr
& {V_1} = 196000 imes {left( {1 - {frac{1}{7}}}
ight)^2} cr
& {V_1} = 196000 imes {left( {frac{7-1}{7}}
ight)^2} cr
& {V_1} = 196000 imes {left( {frac{6}{7}}
ight)^2} cr
& {V_1} = frac{{left( {196000 imes 36}
ight)}}{{49}} cr
& {V_1} = Rs.,144000 cr} $$
214: B
Solution: $$eqalign{
& { ext{Interest}},{ ext{for}},{ ext{the}},{ ext{last}},{ ext{5}},{ ext{years}} cr
& = frac{{PTR}}{{100}} cr
& = frac{{360 imes 5 imes 6}}{{100}} = Rs.,108 cr
& { ext{Interest}},{ ext{for}},{ ext{year}} = 540 - 360 = 180 cr
& { ext{So,}},{ ext{interest}},{ ext{for}},{ ext{first}},{ ext{four}},{ ext{years}} cr
& = 180 - 108 = Rs.,72
cr
& { ext{Now,}},{ ext{rate}},{ ext{for}},{ ext{first}},{ ext{four}},{ ext{years}} cr
& = frac{{ {72 imes 100} }}{{360 imes 4}} cr
& = 5\% cr} $$
215: A
Solution: Let Asif lent Rs. X at 14% per year. Hence, Money lent at 12% = (1500 - x)
Given, total interest = Rs. 186 $$ {frac{{left( {x imes 14 imes 1}
ight)}}{{100}}} , + $$ xa0 xa0$$ {frac{{left[ {left( {1500 - x}
ight) imes 12 imes 1}
ight]}}{{100}}} $$ xa0 xa0xa0 = 186 $$eqalign{
& frac{{14x}}{{100}} + frac{{ {18000 - 12x} }}{{100}} = 186 cr
& 14x + 18000 - 12x = 186 imes 100 cr
& 2x = 18600 - 18000 cr
& x = frac{{600}}{2} = { ext{Rs}}{ ext{. }}300 cr
& { ext{Hence, money lent}}{kern 1pt} { ext{at }}12\% cr
& = 1500 - 300 cr
& = { ext{Rs}}{ ext{.}},1200 cr} $$
216: C
Solution: $$eqalign{
& {1^{{ ext{st}}}},{ ext{Method}}: cr
& { ext{Let rate is }}R\% cr
& { ext{Now}}, cr
& P = 100, cr
& A = 400, cr
& I = 400 - 100 = 300, cr
& { ext{Time}},,T = 10,{ ext{years}} cr
& I = frac{{PTR}}{{100}} cr
& { ext{Or}},R = frac{{ {100 imes I} }}{{PT}} cr
& { ext{Or}},R = frac{{ {100 imes 300} }}{{ {100 imes 10} }} cr
& { ext{Hence}},{kern 1pt} R = 30\% cr} $$ 2 nd Method : Here, the sum become 4 times that means 100, become 400. Rate of such question is given by $$R = frac{{{ ext{interest}}}}{{{ ext{time}}}} = frac{{300}}{{10}} = 30\% $$ 3 rd Method : Here, 300% of rise in the sum so $$eqalign{
& 100 - - - 300\% uparrow - - - {kern 1pt} 400
cr
& R = {frac{{{ ext{total}},{ ext{percentage rise}}}}{{{ ext{given time}}}}} cr
& ,,,,,,,, = frac{{300\% }}{{10}} cr
& ,,,,,,,, = 30\% cr} $$
217: A
Solution: Let,
Principal, P = Rs. 100
Amount, A = Rs. 200
Time = 12 years
Interest = Rs. 100
Rate of interest $$eqalign{
& = frac{{{ ext{Total Interest}}}}{{{ ext{Given Time}}}} cr
& = frac{{100}}{{12}} cr
& = 8frac{1}{3}\% cr} $$
218: B
Solution: Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then, $$eqalign{
& = left( {frac{{x imes 9 imes 1}}{{100}}}
ight) + left[ {frac{{left( {100000 - x}
ight) imes 11 imes 1}}{{100}}}
ight] cr
& = left( {100000 imes frac{{39}}{4} imes frac{1}{{100}}}
ight) cr
& Leftrightarrow frac{{9x + 1100000 - 11x}}{{100}} = frac{{39000}}{4} = 9750 cr
& Leftrightarrow 2x = left( {1100000 - 975000}
ight) = 125000 cr
& Leftrightarrow x = 62500 cr
& herefore { ext{Sum invested at 9}}\% cr
& = { ext{Rs}}{ ext{. }}62500 cr
& { ext{Sum invested at 11% }} cr
& { ext{ = Rs}}{ ext{.}}left( {100000 - 62500}
ight) cr
& = { ext{Rs}}{ ext{. }}37500 cr} $$