Problems On Numbers

Name: _____________________

Date: _____________________

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

Question 1:

The sum of two numbers is 40 and their product is 375. What will be the sum of their reciprocals ?

A. $$frac{1}{40}$$
B. $$frac{8}{75}$$
C. $$frac{75}{4}$$
D. $$frac{75}{8}$$
Answer: _________
Question 2:

A two-digit number is 7 times the sum of its two digits. The number that is formed by reversing its digits is 18 less than the original number. What is the number ?

A. 42
B. 52
C. 62
D. 72
Answer: _________
Question 3:

If the difference between the reciprocal of a positive proper fraction and the fraction itself be $$frac{9}{20}$$, then the fraction is :

A. $$frac{3}{5}$$
B. $$frac{4}{5}$$
C. $$frac{5}{4}$$
D. $$frac{3}{10}$$
Answer: _________
Question 4:

The sum of two numbers is 75 and their difference is 25. The product of the two numbers is :

A. 1350
B. 1250
C. 125
D. 1000
Answer: _________
Question 5:

Three-forth of a number is 60 more than its one-third. The number is :

A. 84
B. 108
C. 144
D. None of these
Answer: _________
Question 6:

The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is :

A. $$frac{1}{289}$$
B. $$frac{289}{290}$$
C. $$frac{290}{289}$$
D. $$289$$
Answer: _________
Question 7:

A number whose fifth part increase by 4 is equal to its fourth part diminished by 10, is :

A. 240
B. 260
C. 270
D. 280
Answer: _________
Question 8:

If $$2frac{1}{2}$$ is added tp a number and the sum multiplied by $$4frac{1}{2}$$ and 3 is added to the product and the sum is divided by $$1frac{1}{5}$$, the quotient becomes 25. What is the number ?

A. $$2frac{1}{2}$$
B. $$3frac{1}{2}$$
C. $$4frac{1}{2}$$
D. $$5frac{1}{2}$$
Answer: _________
Question 9:

The product of two numbers is 120 and the sum of their square is 289. The sum of the numbers is :

A. 20
B. 23
C. 169
D. None of these
Answer: _________
Question 10:

If the digit in the unit's place of a two-digit number is halved and the digit in the ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true ?

A. Sum of the digits is a two-digit number
B. Digit in the unit's place is half of the digit in the ten's place
C. Digit in the unit's place and the ten's place are equal
D. Digit in the unit's place is twice the digit in the ten's place
Answer: _________
Question 11:

If one-third of one-fourth of a number is 15, then three-tenth of that number is:

A. 35
B. 36
C. 45
D. 54
Answer: _________
Question 12:

Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

A. 9
B. 11
C. 13
D. 15
Answer: _________
Question 13:

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

A. 3
B. 4
C. 9
D. Cannot be determined
Answer: _________
Question 14:

The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?

A. 4
B. 8
C. 16
D. None of these
Answer: _________
Question 15:

A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

A. 18
B. 24
C. 42
D. 81
Answer: _________
Question 16:

The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

A. 69
B. 78
C. 96
D. Cannot be determined
Answer: _________
Question 17:

The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. None of these
Answer: _________
Question 18:

A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:

A. 3
B. 5
C. 9
D. 11
Answer: _________
Question 19:

In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:

A. 24
B. 26
C. 42
D. 46
Answer: _________
Question 20:

Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

A. 3
B. 10
C. 17
D. 20
Answer: _________
Question 21:

If a number is multiplied by two-third of itself the value so obtained is 864. What is the number ?

A. 34
B. 36
C. 38
D. 44
Answer: _________
Question 22:

The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is :

A. 380
B. 395
C. 400
D. 425
Answer: _________
Question 23:

If the product of three consecutive integers is 120, then the sum of the integers is :

A. 9
B. 12
C. 14
D. 15
Answer: _________
Question 24:

A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is :

A. 18
B. 24
C. 42
D. 81
Answer: _________
Question 25:

The sum of the squares of three numbers is 138, while the sum of their product taken two at a time is 131. Their sum is :

A. 20
B. 30
C. 40
D. None of these
Answer: _________
Question 26:

The sum of the squares of two positive integers is 100 and the difference of their squares is 28. The sum of the numbers is ?

A. 12
B. 13
C. 14
D. 15
Answer: _________
Question 27:

A number when multiplied by 13 is increased by 180. The number is :

A. 5
B. 12
C. 15
D. 45
Answer: _________
Question 28:

A positive number when decreased by 4 is equal to 21 times the reciprocal of the number. The number is ?

A. 3
B. 5
C. 7
D. 9
Answer: _________
Question 29:

The difference between two numbers is 1365. When the large number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is :

A. 240
B. 270
C. 295
D. 360
Answer: _________
Question 30:

A number of two digits has 3 for its unit's digit, and the sum of digits is $$frac{1}{7}$$ of the itself. The number is :

A. 43
B. 63
C. 53
D. 73
Answer: _________
Question 31:

The sum of the numerator and denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from the denominator, it becomes $$frac{2}{3}$$. The fraction is :

A. $$frac{5}{6}$$
B. $$frac{6}{5}$$
C. $$frac{3}{8}$$
D. $$frac{8}{3}$$
Answer: _________
Question 32:

In a Mathematics examination the number scored by 5 candidates are 5 successive odd integers. If their total marks are 185, the highest score is :

A. 39
B. 43
C. 41
D. 47
Answer: _________
Question 33:

Three numbers are in in the ratio of 3 : 4 : 6 and their product is 1944. The largest of these numbers is :

A. 6
B. 12
C. 18
D. None of these
Answer: _________
Question 34:

The sum of the squares of two numbers is 3341 and the difference of their squares is 891. The numbers are :

A. 25, 36
B. 25, 46
C. 35, 46
D. None of these
Answer: _________
Question 35:

In a two-digit number, if it is known that its unit's digits exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is :

A. 24
B. 26
C. 42
D. 46
Answer: _________
Question 36:

The difference between the numerator and the denominator of a fraction is 5. If 5 is added to its denominator, the fraction is decreased by $$1frac{1}{4}$$. Find the value of the fraction.

A. $$frac{1}{6}$$
B. $$2frac{1}{4}$$
C. $$3frac{1}{4}$$
D. $$6$$
Answer: _________
Question 37:

A number whose fifth part increased by 4 is equal to its fourth part diminished by 10, is :

A. 240
B. 260
C. 270
D. 280
Answer: _________
Question 38:

The ratio between a two-digit number and the sum of the digits of that number is 4 : 1. If the digit in the unit's place is 3 more than the digit in the ten's place, then the number is ?

A. 24
B. 36
C. 63
D. 96
Answer: _________
Question 39:

The difference between two positive integers is 3. If the sum of their squares is 369, then the sum of the numbers is :

A. 25
B. 27
C. 33
D. 81
Answer: _________
Question 40:

A number consists of two digits. If the digits interchange place and the new number is added to the original number, then the resulting number will be divisible by :

A. 3
B. 5
C. 9
D. 11
Answer: _________
Question 41:

A number consists of two digits such that the digit in the ten's place is less by 2 than the digit in the unit's place. Three times the number added to $$frac{6}{7}$$ times the number obtained by reversing the digits equals 108. The sum of the digits in the number is :

A. 6
B. 7
C. 8
D. 9
Answer: _________
Question 42:

If (73) 2 is subtracted from the square of a number, the answer so obtained is 5075. What is the number ?

A. 96
B. 98
C. 102
D. 106
Answer: _________
Question 43:

The sum of three consecutive odd numbers is 20 more than the first of these numbers. What is the middle number ?

A. 7
B. 9
C. 11
D. Data inadequate
Answer: _________
Question 44:

The product of two numbers is 192 and the sum of these two numbers is 28. What is the smaller these two numbers ?

A. 12
B. 14
C. 16
D. 18
Answer: _________
Question 45:

If the square of a two-digit number is reduced by the square of the number formed by reversing the digits of the number, the final result is :

A. Divisible by 11
B. Divisible by 9
C. Necessarily irrational
D. Both (A) and (B)
Answer: _________
Question 46:

Three times the first of three consecutive odd integers is 3 more than twice the third. The integer is :

A. 9
B. 11
C. 13
D. 15
Answer: _________
Question 47:

In a three-digit number, the digit in the unit's place is 75% of the digit in the ten's place. The digit in the ten's place is greater than the digit in the hundred's place by 1. If the sum of the digits in the ten's place and the hundred's place is 15. What is the number ?

A. 687
B. 786
C. 795
D. Cannot be determined
Answer: _________
Question 48:

The sum of two numbers is 37 and the difference of their squares is 185, then the difference between the two numbers is :

A. 10
B. 4
C. 5
D. 3
Answer: _________
Question 49:

The difference between $$frac{3}{5}$$th of $$frac{2}{3}$$rd of a number and $$frac{2}{5}$$th of $$frac{1}{4}$$th of the same number is 288. What is the number ?

A. 850
B. 895
C. 955
D. 960
Answer: _________
Question 50:

The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be :

A. $$frac{12}{35}$$
B. $$frac{1}{35}$$
C. $$frac{35}{8}$$
D. $$frac{7}{32}$$
Answer: _________
Question 51:

The difference between two numbers is 16. If one-third of the smaller number is greater than one-seventh of the larger number by 4, then the two numbers are :

A. 9 and 25
B. 12 and 28
C. 33 and 49
D. 56 and 72
Answer: _________
Question 52:

If a number of two digits is k times the sum of its digits, then the number formed by interchanging the digits is the sum of the digits multiplied by :

A. k - 1
B. 11 - k
C. 9 + k
D. 10 - k
Answer: _________
Question 53:

The product of two fractions is $$frac{14}{15}$$ and their quotient is $$frac{35}{24}$$. The greater fraction is :

A. $$frac{4}{5}$$
B. $$frac{7}{6}$$
C. $$frac{7}{4}$$
D. $$frac{7}{3}$$
Answer: _________
Question 54:

A man bought some eggs of which 10% are rotten. He gives 80% of the remainder to his neighbours. Now he is left out with 36 eggs. How many eggs he bought ?

A. 40
B. 100
C. 200
D. 72
Answer: _________
Question 55:

A number is double and 9 is added. If the resultant is trebled, it becomes 75. What is that number ?

A. 3.5
B. 6
C. 8
D. None of these
Answer: _________
Question 56:

The sum of a positive number and its reciprocal is thrice the difference of the number and its reciprocal. The number is :

A. $$sqrt 2 $$
B. $$frac{1}{{sqrt 2 }}$$
C. $$sqrt 3 $$
D. $$frac{1}{{sqrt 3 }}$$
Answer: _________
Question 57:

If the numerator of a fraction is increased by $$frac{1}{4}$$ and the denominator is decreased byy $$frac{1}{3}$$, the new fraction obtained is $$frac{33}{64}$$. What was the original fraction ?

A. $$frac{3}{7}$$
B. $$frac{5}{7}$$
C. $$frac{7}{9}$$
D. Cannot be determined
Answer: _________
Question 58:

If doubling a number and adding 20 to the result gives the same answer as multiplying the number by 8 and taking away 4 from the product, the number is :

A. 2
B. 3
C. 4
D. 6
Answer: _________
Question 59:

Out of six consecutive natural numbers if the sum of first three is 27, what is the sum of the other three ?

A. 24
B. 25
C. 35
D. 36
Answer: _________
Question 60:

The difference between a two-digit number and the number obtained by interchanging the two digits is 63. Which is the smaller of the two numbers ?

A. 29
B. 70
C. 92
D. Cannot be determined
Answer: _________
Question 61:

If the numerator of a fraction is increased by 200% and the denominator is increased by 300%, the resultant fraction is $$frac{15}{26}$$. What was the original fraction ?

A. $$frac{8}{11}$$
B. $$frac{10}{11}$$
C. $$frac{9}{13}$$
D. $$frac{10}{13}$$
Answer: _________
Question 62:

If 50 is subtracted from two-third of number, the result is equal to sum of 40 and one-fourth of that number. What is the number ?

A. 174
B. 216
C. 246
D. 336
Answer: _________
Question 63:

The sum of two numbers is 25 and their difference is 13. Find their product :

A. 104
B. 114
C. 315
D. 325
Answer: _________
Question 64:

The sum of seven consecutive numbers is 175. What is the difference between twice the largest number and thrice the smallest number ?

A. 7
B. 8
C. 10
D. 12
Answer: _________
Question 65:

The number obtained by interchanging the two digits of a two-digit number is lesser than the original number by 54. If the sum of the two digit of the number is 12, then what is the original number ?

A. 28
B. 39
C. 82
D. Cannot be determined
Answer: _________
Question 66:

A fraction is such that if the double of the numerator and the triple pf the denominator is changed by +10 percent and -30 percent respectively, then we get 11 percent of $$frac{16}{21}$$. Find the fraction :

A. $$frac{2}{25}$$
B. $$frac{3}{25}$$
C. $$frac{4}{25}$$
D. None of these
Answer: _________
Question 67:

A student was asked to divide the half of a certain number by 6 and the other half by 4 and then to add the two quantities so obtained. Instead of doing so the student divided the number by 5 and the result fell short by 4. The given number was ?

A. 240
B. 288
C. 384
D. 480
Answer: _________
Question 68:

If the sum of numbers is 33 and their difference is 15, the smaller number is ?

A. 9
B. 12
C. 15
D. 18
Answer: _________
Question 69:

In a two-digit positive number, the digit in the unit's place is equal to the square of the digit in ten's place, and the difference between the number and the number obtained by interchanging the digits is 54. What is 40% of the original number ?

A. 15.6
B. 24
C. 37.2
D. 39
Answer: _________
Question 70:

243 has been divided into three parts such that half of the first part, one-third of the second part and one-fourth of the third part are equal. The largest part is :

A. 74
B. 86
C. 92
D. 108
Answer: _________
Question 71:

If the sum of a number and its square is 182, what is the number ?

A. 15
B. 26
C. 28
D. 91
Answer: _________
Question 72:

The sum of two numbers is 40 and their difference is 4. The ratio of the numbers is :

A. 11 : 9
B. 11 : 18
C. 21 : 19
D. 22 : 19
Answer: _________
Question 73:

The sum of three consecutive odd numbers and three consecutive even numbers together is 231. Also, the smallest odd number is 11 less than the smallest even number. What is the sum of the largest odd number and the largest even number ?

A. 74
B. 82
C. 83
D. Cannot be determined
Answer: _________
Question 74:

A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is :

A. 145
B. 253
C. 370
D. 352
Answer: _________
Question 75:

The sum of four numbers is 64. If you add 3 to the first number, 3 is subtracted from the second number, the third is multiplied by 3 and the fourth is divided by 3, then all the results are equal. What is the difference between the largest and the smallest of the original numbers ?

A. 21
B. 27
C. 32
D. Cannot be determined
Answer: _________
Question 76:

The sum of five consecutive odd numbers is 575. What is the sum of the next set of five consecutive odd numbers ?

A. 595
B. 615
C. 635
D. Cannot be determined
Answer: _________
Question 77:

If the numerator of a fraction is increased by 2 and the denominator is increased by 3, the fraction becomes $$frac{7}{9}$$ and if both the numerator as well as he denominator are decreased by 1, the fraction becomes $$frac{4}{5}$$. What is the original fraction ?

A. $$frac{5}{6}$$
B. $$frac{9}{11}$$
C. $$frac{13}{16}$$
D. $$frac{17}{21}$$
Answer: _________
Question 78:

The difference of two numbers is 20% of the larger number. If the smaller number is 12, the larger one is :

A. 15
B. 16
C. 18
D. 20
Answer: _________
Question 79:

Two numbers are such that the ratio between them is 4 : 7. If each is increased by 4, the ratio becomes 3 : 5. The lager number is ?

A. 36
B. 48
C. 56
D. 64
Answer: _________
Question 80:

the difference between two numbers is 3 and the difference between their squares is 63. Which is the larger number ?

A. 9
B. 12
C. 15
D. Cannot be determined
Answer: _________
Question 81:

twenty times a positive integer is less than its square by 96. What is the integer ?

A. 20
B. 24
C. 30
D. Cannot be determined
Answer: _________
Question 82:

There are two numbers such that the sum of twice the first number and thrice the second number is 100 and the sum of thrice the first number and twice the second number is 120. Which is the larger number ?

A. 12
B. 14
C. 32
D. 35
Answer: _________
Question 83:

If one-seventh of a number exceeds its eleventh part by 100, then the number is :

A. 770
B. 110
C. 1825
D. 1925
Answer: _________
Question 84:

The sum of three numbers is 264. If the first number be twice the second and third number be be one-third of the first, then the second number is :

A. 48
B. 54
C. 72
D. 84
Answer: _________
Question 85:

A, B, C, D and E are five consecutive odd numbers. The sum of A and C is 146. What is the value of E ?

A. 71
B. 75
C. 79
D. 81
Answer: _________
Question 86:

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of the number ?

A. 3
B. 4
C. 9
D. Cannot be determined
Answer: _________
Question 87:

The digit in the unit's place of a numbers is equal to the digit in the ten's place of half of that number and the digit in the ten's place of that number is less than the digit in unit's place of half of the number by 1. If the sum of the digits of the number is 7, then what is the number ?

A. 34
B. 52
C. 162
D. Data inadequate
Answer: _________
Question 88:

Of the three number, the sum of the first two is 73, the sum of the second and the third is 77 and the sum of the third and thrice the first is 104. The third number is ?

A. 25
B. 39
C. 48
D. Cannot be determined
Answer: _________
Question 89:

The difference between a number and its three-fifths is 50, What is the number ?

A. 75
B. 100
C. 125
D. None of these
Answer: _________
Question 90:

Two-third of a positive number and $$frac{25}{216}$$ of its reciprocal are equal, The number is :

A. $$frac{5}{12}$$
B. $$frac{12}{5}$$
C. $$frac{25}{144}$$
D. $$frac{144}{25}$$
Answer: _________
Question 91:

The difference between two integers is 5. Their product is 500. Find the numbers.

A. 15, 20
B. 20, 25
C. 30, 25
D. 21, 26
Answer: _________
Question 92:

What is the sum of two consecutive even numbers, the difference of whose squares is 84 ?

A. 34
B. 38
C. 42
D. 46
Answer: _________
Question 93:

A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is ?

A. 23
B. 27
C. 32
D. 72
Answer: _________
Question 94:

If the product of two numbers is 5 and one of the number is $$frac{3}{2}$$, then sum of two numbers is :

A. $$4frac{1}{3}$$
B. $$4frac{2}{3}$$
C. $$4frac{5}{6}$$
D. $$5frac{1}{6}$$
Answer: _________
Question 95:

If a number is added to two-fifths of itself, the value so obtained is 455. What is the number ?

A. 325
B. 350
C. 400
D. 420
Answer: _________
Question 96:

Find the whole number which when increased by 20 is equal to 69 times the reciprocal of the number.

A. 2.5
B. 3
C. 5
D. 7
Answer: _________
Question 97:

The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:

A. 380
B. 395
C. 400
D. 425
Answer: _________
Question 98:

The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:

A. 20
B. 23
C. 169
D. None of these
Answer: _________
Question 99:

A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

A. 145
B. 253
C. 370
D. 352
Answer: _________
Question 100:

The sum of two number is 25 and their difference is 13. Find their product.

A. 104
B. 114
C. 315
D. 325
Answer: _________
Question 101:

What is the sum of two consecutive even numbers, the difference of whose squares is 84?

A. 34
B. 38
C. 42
D. 46
Answer: _________
Question 102:

The sum of twice a number and three times of 42 is 238. What is the sum of thrice the number and two times of 42 ?

A. 245
B. 250
C. 252
D. 264
Answer: _________
Question 103:

By how much is $$frac{3}{4}$$th of 568 lesser than $$frac{7}{8}$$th of 1008 ?

A. 444
B. 448
C. 452
D. 456
Answer: _________
Question 104:

Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is :

A. 4
B. 5
C. 6
D. 10
Answer: _________
Question 105:

What is the greater of the two numbers whose product is 1092 and the sum of the two numbers exceeds their difference by 42 ?

A. 44
B. 48
C. 52
D. 54
Answer: _________
Question 106:

The product of three consecutive even numbers when divided by 8 is 720. The product of their square roots is :

A. $$12sqrt {10} $$
B. $$24sqrt {10} $$
C. 120
D. None of these
Answer: _________

Answer Key

1: B
Solution: Let the numbers be x and y Then, x + y = 40 and xy = 375 $$eqalign{
& herefore frac{1}{x} + frac{1}{y} = frac{{x + y}}{{xy}} cr
& ,,,,,,,,,,,,,,,,,,,,,,, = frac{{40}}{{275}} cr
& ,,,,,,,,,,,,,,,,,,,,,,, = frac{8}{{75}} cr} $$
2: A
Solution: Let the ten's digit be x and the unit's digit be y Then, number = 10x + y $$eqalign{
& herefore 10x + y = 7left( {x + y}
ight) cr
& Leftrightarrow 3x = 6y cr
& Leftrightarrow x = 2y cr} $$ Number formed by reversing the digits = 10y + x $$eqalign{
& herefore left( {10x + y}
ight) - left( {10y + x}
ight) = 18 cr
& Leftrightarrow 9x - 9y = 18 cr
& Leftrightarrow x - y = 2 cr
& Leftrightarrow 2y - y = 2 cr
& Leftrightarrow y = 2 cr
& { ext{So, }}x = 2y = 4 cr} $$ Hence, ∴ Required number = 10x + y = 40 + 2 = 42
3: B
Solution: Let the fraction be $$frac{a}{1}$$ Then, $$eqalign{
& Leftrightarrow frac{1}{a} - a = frac{9}{{20}} cr
& Leftrightarrow frac{{1 - {a^2}}}{a} = frac{9}{{20}} cr
& Leftrightarrow 20 - 20{a^2} = 9a cr
& Leftrightarrow 20{a^2} + 9a - 20 = 0 cr
& Leftrightarrow 20{a^2} + 25a - 16a - 20 = 0 cr
& Leftrightarrow 5aleft( {4a + 5}
ight) - 4left( {4a + 5}
ight) = 0 cr
& Leftrightarrow left( {4a + 5}
ight)left( {5a - 4}
ight) = 0 cr
& Leftrightarrow a = frac{4}{5},,,,,,,,left[ {x08ecause a
e - frac{5}{4}}
ight] cr} $$
4: B
Solution: Let the numbers be a and b According to the question, $$eqalign{
& a + b = 75 cr
& a - b = 25 cr
& x08ecause {left( {a + b}
ight)^2} - {left( {a - b}
ight)^2} = 4ab cr
& Rightarrow {75^2} - {25^2} = 4ab cr
& Rightarrow 4ab = left( {75 + 25}
ight)left( {75 - 25}
ight) cr
& left[ {x08ecause {a^2} - {b^2} = left( {a + b}
ight)left( {a - b}
ight)}
ight] cr
& Rightarrow 4ab = 100 imes 50 cr
& Rightarrow ab = frac{{100 imes 50}}{4} cr
& Rightarrow ab = 1250 cr} $$
5: C
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow frac{3}{4}x - frac{1}{3}x = 60 cr
& Leftrightarrow frac{{5x}}{{12}} = 60 cr
& Leftrightarrow x = left( {frac{{60 imes 12}}{5}}
ight) cr
& Leftrightarrow x = 144 cr} $$
6: C
Solution: Let the numbers be a and b Then, ab = 17 ⇒ a = 1 and b = 17 So, $$eqalign{
& = frac{1}{{{a^2}}} + frac{1}{{{b^2}}} cr
& = frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} cr
& = frac{{{1^2} + {{left( {17}
ight)}^2}}}{{{{left( {1 imes 17}
ight)}^2}}} cr
& = frac{{290}}{{289}} cr} $$
7: D
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow frac{x}{5} + 4 = frac{x}{4} - 10 cr
& Leftrightarrow frac{x}{4} - frac{x}{5} = 14 cr
& Leftrightarrow frac{x}{{20}} = 14 cr
& Leftrightarrow x = 14 imes 20 cr
& Leftrightarrow x = 280 cr} $$
8: B
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow frac{{4frac{1}{2}left( {x + 2frac{1}{2}}
ight) + 3}}{{1frac{1}{5}}} = 25 cr
& Leftrightarrow frac{{frac{9}{2}left( {x + frac{5}{2}}
ight) + 3}}{{frac{6}{5}}} = 25 cr
& Leftrightarrow frac{{9x}}{2} + frac{{45}}{4} + 3 = 25 imes frac{6}{5} cr
& Leftrightarrow frac{{9x}}{2} + frac{{45}}{4} + 3 = 30 cr
& Leftrightarrow frac{{9x}}{2} = 30 - frac{{57}}{4} cr
& Leftrightarrow frac{{9x}}{2} = frac{{63}}{4} cr
& Leftrightarrow x = left( {frac{{63}}{4} imes frac{2}{9}}
ight) cr
& Leftrightarrow x = frac{7}{2} cr
& Leftrightarrow x = 3frac{1}{2} cr} $$
9: B
Solution: Let the numbers be x and y Then, xy = 120 and x 2 + y 2 = 289 $$eqalign{
& herefore {left( {x + y}
ight)^2} cr
& = {x^2} + {y^2} + 2xy cr
& = 289 + 240 cr
& = 529 cr
& herefore x + y cr
& = sqrt {529} cr
& = 23 cr} $$
10: D
Solution: Let the ten's digit be x and the unit's digit be y Then, number = 10x + y New number : $$eqalign{
& = 10 imes 2x + frac{y}{2} cr
& = 20x + frac{y}{2} cr} $$ $$eqalign{
& herefore 20x + frac{y}{2} = 10y + x cr
& Leftrightarrow 40x + y = 20y + 2x cr
& Leftrightarrow 38x = 19y cr
& Leftrightarrow y = 2x cr} $$ So, the unit's digit is twice the ten's digit.
11: D
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{number}},{ ext{be}},x cr
& { ext{Then}},,frac{1}{3},{ ext{of}},frac{1}{{4,}},{ ext{of}},x = 15cr
& Rightarrow , x = 15 imes 12 = 180 cr
& { ext{So,}},{ ext{required}},{ ext{number}} cr
& = {frac{3}{{10}} imes 180} cr
& = 54 cr} $$
12: D
Solution: Let the three integers be x , x + 2 and x + 4 Then, 3 x = 2( x + 4) + 3 xa0xa0 ⇔ x = 11 ∴ Third integer = x + 4 = 15
13: B
Solution: Let the ten's digit be x and unit's digit be y . Then, (10 x + y ) - (10 y + x ) = 36 ⇒ 9( x - y ) = 36 ⇒ x - y = 4
14: B
Solution: Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit. Let ten's and unit's digits be 2 x and x respectively. Then, (10 × 2 x + x ) - (10 x + 2 x ) = 36 ⇒ 9 x = 36 ⇒ x = 4 ∴ Required difference = (2 x + x ) - (2 x - x ) = 2 x = 8
15: B
Solution: Let the ten's and unit digit be x and $$frac{8}{x}$$ respectively $$eqalign{
& { ext{Then,}} cr
& left( {10x + frac{8}{x}}
ight) + 18 = 10 imes frac{8}{x} + x cr
& Rightarrow 10{x^2} + 8 + 18x = 80 + {x^2} cr
& Rightarrow 9{x^2} + 18x - 72 = 0 cr
& Rightarrow {x^2} + 2x - 8 = 0 cr
& Rightarrow left( {x + 4}
ight)left( {x - 2}
ight) = 0 cr
& Rightarrow x = 2 cr} $$ ∴ first digit will be 2 and second digit will be 4. i.e digit is 24.
16: D
Solution: Let the ten's digit be x and unit's digit be y Then, x + y = 15 and x - y = 3 xa0xa0or y - x = 3 Solving x + y = 15 xa0 and x - y = 3, we get: x = 9, y = 6 Solving x + y = 15 xa0 and y - x = 3, we get: x = 6, y = 9 So, the number is either 96 or 69 Hence, the number cannot be determined.
17: A
Solution: Let the numbers be a , b and c Then, a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 (a + b + c) 2 = a 2 + b 2 + c 2 + 2( ab + bc + ca ) = 138 + 2 x 131 = 400 ⇒ ( a + b + c ) = $$sqrt {400} $$xa0 = 20
18: D
Solution: Let the ten's digit be x and unit's digit be y . Then, number = 10 x + y Number obtained by interchanging the digits = 10 y + x ∴ (10 x + y ) + (10 y + x ) = 11( x + y ), which is divisible by 11
19: A
Solution: Let the ten's digit be x Then, unit's digit = x + 2 Number = 10 x + ( x + 2) = 11 x + 2 Sum of digits = x + ( x + 2) = 2 x + 2 ∴ (11 x + 2)(2 x + 2) = 144 ⇒ 22 x 2 + 26 x - 140 = 0 ⇒ 11 x 2 + 13 x - 70 = 0 ⇒ 11 x 2 + (35 - 22)x - 70 = 0 ⇒ 11 x 2 + 35x - 22x - 70 = 0 ⇒ ( x - 2)(11 x + 35) = 0 ⇒ x = 2 Hence, required number = 11 x + 2 = 24
20: A
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{number}},{ ext{be}},x cr
& { ext{Then}},,x + 17 = frac{{60}}{x} cr
& Rightarrow {x^2} + 17x - 60 = 0 cr
& Rightarrow (x + 20)(x - 3) = 0 cr
& Rightarrow x = 3 cr} $$
21: B
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x imes frac{2}{3}x = 864 cr
& Leftrightarrow frac{2}{3}{x^2} = 864 cr
& Leftrightarrow {x^2} = left( {frac{{864 imes 3}}{2}}
ight) cr
& Leftrightarrow {x^2} = 1296 cr
& Leftrightarrow x = sqrt {1296} cr
& Leftrightarrow x = 36 cr} $$
22: C
Solution: Let the numbers be x and y Then, $$xy = 9375{ ext{ and }}frac{x}{y} = 15$$ $$eqalign{
& Leftrightarrow frac{{xy}}{{left( {frac{x}{y}}
ight)}} = frac{{9375}}{{15}} cr
& Leftrightarrow {y^2} = 625 cr
& Leftrightarrow y = 25 cr
& herefore x = 15y cr
& Rightarrow x = 15 imes 25 cr
& Rightarrow x = 375 cr} $$ ∴ Sum of the numbers : = 375 + 25 = 400
23: D
Solution: $$eqalign{
& 120: cr
& = 2 imes 2 imes 2 imes 3 imes 5 cr
& = left( {2 imes 2}
ight) imes 5 imes left( {2 imes 3}
ight) cr
& = 4 imes 5 imes 6 cr} $$ Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6 Required sum : = 4 + 5 + 6 = 15
24: B
Solution: Let the ten's and unit's digits be $$x$$ and $$frac{8}{x}$$ respectively Then, $$eqalign{
& Leftrightarrow left( {10x + frac{8}{x}}
ight) + 18 = 10 imes frac{8}{x} + x cr
& Leftrightarrow 10{x^2} + 8 + 18x = 80 + {x^2} cr
& Leftrightarrow 9{x^2} + 18x - 72 = 0 cr
& Leftrightarrow {x^2} + 2x - 8 = 0 cr
& Leftrightarrow left( {x + 4}
ight)left( {x - 2}
ight) = 0 cr
& Leftrightarrow x = 2 cr} $$ So, ten's digit = 2 and unit's digit = 4 Hence, required number = 24
25: A
Solution: Let the numbers be a, b and c. Then, a 2 + b 2 + c 2 = 138 And (ab + bc + ca) = 131 ∴ (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) = 138 + 2 × 131 = 400 ⇒ (a + b + c) = $$sqrt {400} $$ xa0= 20
26: C
Solution: Let the positive integers be a and b where a > b According to the question, $$eqalign{
& {a^2} + {b^2} = 100.....( ext{i}) cr
& {a^2} - {b^2} = 28.....( ext{ii}) cr} $$ By adding (i) and (ii), we get : $$eqalign{
& herefore {a^2} + {b^2} + {a^2} - {b^2} = 100 + 28 cr
& Rightarrow 2{a^2} = 128 cr
& Rightarrow {a^2} = frac{{128}}{2} cr
& Rightarrow a = sqrt {64} cr
& Rightarrow a = 8 cr} $$ From equation (i) $$eqalign{
& Rightarrow {8^2} + {b^2} = 100 cr
& Rightarrow {b^2} = 100 - 64 cr
& Rightarrow b = sqrt {36} cr
& Rightarrow b = 6 cr
& herefore a + b = 8 + 6 cr
& ,,,,,,,,,,,,,,,,,,, = 14 cr} $$
27: C
Solution: Let the number be x Then, $$eqalign{
& Rightarrow 13x = x + 180 cr
& Rightarrow 12x = 180 cr
& Rightarrow x = frac{{180}}{{12}} cr
& Rightarrow x = 15 cr} $$
28: C
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x - 4 = frac{{21}}{x} cr
& Leftrightarrow {x^2} - 4x - 21 = 0 cr
& Leftrightarrow left( {x - 7}
ight)left( {x + 3}
ight) = 0 cr
& Leftrightarrow x = 7 cr} $$
29: B
Solution: Let the numbers be x and (x + 1365) Then, ⇒ x + 1365 = 6x + 15 ⇒ 5x = 1350 ⇒ x = 270
30: B
Solution: Let the ten's digit be x Then, number = 10x + 3 and Sum of digits = (x + 3) So, (x + 3) = $$frac{1}{7}$$(10x + 3) ⇔ 7x + 21 = 10x + 3 ⇔ 3x = 18 ⇔ x = 6 Hence, the number is : = (10x + 3) = (10 × 6 + 3) = 63
31: C
Solution: Let the fraction be $$frac{x}{y}$$ Then, $$eqalign{
& Leftrightarrow x + y = 11.....( ext{i}) cr
& Leftrightarrow frac{{x + 1}}{{y - 2}} = frac{2}{3} cr
& Leftrightarrow 3left( {x + 1}
ight) = 2left( {y - 2}
ight) cr
& Leftrightarrow 3x - 2y = - 7.....( ext{ii}) cr} $$ Solving (i) and (ii), we get: x = 3 and y = 8 So, the fraction is $$frac{3}{8}$$
32: C
Solution: Let the five successive odd number be, x, x + 2, x + 4, x + 6, x + 8 Then, according to given information, 185 = x + x + 2 + x + 4 + x + 6 + x + 8 $$eqalign{
& Leftrightarrow 185 = 5x + 20 cr
& Leftrightarrow 5x = 185 - 20 cr
& Leftrightarrow 5x = 165 cr
& Leftrightarrow x = 33 cr} $$ Highest number = 33 + 8 = 41
33: C
Solution: Let the numbers be 3x, 4x and 6x Then, $$eqalign{
& Leftrightarrow 3x imes 4x imes 6x = 1944 cr
& Leftrightarrow 72{x^3} = 1944 cr
& Leftrightarrow {x^3} = 27 cr
& Leftrightarrow x = 3 cr} $$ ∴ Largest number = 6x = 18
34: C
Solution: Let the numbers be x and y. Then, x 2 + y 2 = 3341..... (i) And, x 2 - y 2 = 891..... (ii) Adding (i) and (ii), we get : 2x 2 = 4232 or x 2 = 2116 or x = 46 Subtracting (ii) from (i), we get : 2y 2 = 2450 or y 2 = 1225 or y = 35 So, the numbers are 35 and 46
35: A
Solution: Let the ten's digit be x Then, unit's digit = x + 2 ∴ Number : = 10x + (x + 2) = 11x + 2 Sum of digits : = x + (x + 2) = 2x + 2 $$eqalign{
& herefore left( {11x + 2}
ight)left( {2x + 2}
ight) = 144 cr
& Leftrightarrow 22{x^2} + 26x - 140 = 0 cr
& Leftrightarrow 11{x^2} + 13x - 70 = 0 cr
& Leftrightarrow left( {x - 2}
ight)left( {11x + 35}
ight) = 0 cr
& Leftrightarrow x = 2 cr} $$ Hence, required number: = 11x + 2 = 11× 2 + 2 = 24
36: B
Solution: Let the denominator be x Then, numerator = x + 5 Now, $$eqalign{
& Leftrightarrow frac{{x + 5}}{x} - frac{{x + 5}}{{x + 5}} = frac{5}{4} cr
& Leftrightarrow frac{{x + 5}}{x} = frac{5}{4} + 1 cr
& Leftrightarrow frac{{x + 5}}{x} = frac{9}{4} cr
& Leftrightarrow frac{{x + 5}}{x} = 2frac{1}{4} cr} $$ So, the fraction is $$2frac{1}{4}$$
37: D
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow left( {frac{1}{5}x + 4}
ight) = left( {frac{1}{4}x - 10}
ight) cr
& Leftrightarrow frac{x}{{20}} = 14 cr
& Leftrightarrow x = 14 imes 20 cr
& Leftrightarrow x = 280 cr} $$
38: B
Solution: Let the ten's digit be x Then, units digit = x + 3 Number = 10x + (x + 3) = 11x + 3 Sum of digits = x + (x + 3) = 2x + 3 $$eqalign{
& herefore frac{{11x + 3}}{{2x + 3}} = frac{4}{1} cr
& Leftrightarrow 11x + 3 = 8x + 12 cr
& Leftrightarrow 3x = 9 cr
& Leftrightarrow x = 3 cr} $$ Hence, Required number = 11x + 3 = 11 × 3 + 3 = 36
39: B
Solution: Let the numbers be x and (x + 3) Then, $$eqalign{
& Leftrightarrow {x^2} + {left( {x + 3}
ight)^2} = 369 cr
& Leftrightarrow {x^2} + {x^2} + 9 + 6x = 369 cr
& Leftrightarrow 2{x^2} + 6x - 360 = 0 cr
& Leftrightarrow {x^2} + 3x - 180 = 0 cr
& Leftrightarrow left( {x + 15}
ight)left( {x - 12}
ight) = 0 cr
& Leftrightarrow x = 12 cr} $$ So, the numbers are 12 and 15 ∴ Required sum = (12 + 15) = 27
40: D
Solution: Let the ten's digit be x and unit's digit be y Then, number = 10x + y Number obtained by interchanging the digits = 10y + x ∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11
41: A
Solution: Let the unit's digit be x Then, ten's digit = (x - 2) $$ herefore 3left[ {10left( {x - 2}
ight) + x}
ight] + frac{6}{7}$$ xa0 xa0 $$left[ {10x + left( {x - 2}
ight)}
ight]$$ xa0 $$ = 108$$ ⇔ 231x - 420 + 66x - 12 = 756 ⇔ 297x = 1188 ⇔ x = 4 Hence, sum of the digits : = x + (x - 2) = 2x - 2 = 6
42: C
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow {x^2} - {left( {73}
ight)^2} = 5075 cr
& Leftrightarrow {x^2} - 5329 = 5075 cr
& Leftrightarrow {x^2} = 5075 + 5329 cr
& Leftrightarrow {x^2} = 10404 cr
& Leftrightarrow x = sqrt {10404} cr
& Leftrightarrow x = 102 cr} $$
43: B
Solution: Let the numbers be x, (x + 2) and (x + 4) Then, $$eqalign{
& Leftrightarrow x + left( {x + 2}
ight) + left( {x + 4}
ight) = x + 20 cr
& Leftrightarrow 2x = 14 cr
& Leftrightarrow x = 7 cr} $$ ∴ Middle number : = x + 2 = 9
44: A
Solution: Let the numbers be x and (28 - x) Then, $$eqalign{
& Leftrightarrow xleft( {28 - x}
ight) = 192 cr
& Leftrightarrow {x^2} - 28x + 192 = 0 cr
& Leftrightarrow left( {x - 16}
ight)left( {x - 12}
ight) = 0 cr
& Leftrightarrow x = 16{ ext{ or }}x = 12 cr} $$ So, the numbers are 16 and 12
45: D
Solution: Let the two-digit number be 10x + y Then, number formed by reversing the digits = 10y + x Difference of square of the numbers : $$ = {left( {10x + y}
ight)^2} - {left( {10y + x}
ight)^2}$$ $$ = left( {100{x^2} + {y^2} + 20xy}
ight) - $$ xa0 xa0 $$left( {100{y^2} + {x^2} + 20xy}
ight)$$ $$ = 99left( {{x^2} - {y^2}}
ight)$$ xa0 xa0 which is divisible by both 9 and 11
46: D
Solution: Let the three integers be x, (x + 2) and (x + 4) Then, Three times the first of three consecutive odd integers is 3 more than twice the third. ⇔ 3x = 2(x + 4) + 3 ⇔ x = 11 ∴ Third integer : = x + 4 = 11 + 4 = 15
47: B
Solution: Let the hundred's digit = x Then, ten's digit = (x + 1) Unit's digit : $$eqalign{
& = 75\% { ext{ of }}left( {x + 1}
ight) cr
& = frac{3}{4}left( {x + 1}
ight) cr} $$ $$eqalign{
& herefore left( {x + 1}
ight) + x = 15 cr
& Leftrightarrow 2x = 14 cr
& Leftrightarrow x = 7 cr} $$ So, hundred's digit = 7 Ten's digit = 8 Unit's digit : $$eqalign{
& = frac{3}{4}left( {x + 1}
ight) cr
& = frac{3}{4}left( {7 + 1}
ight) cr
& = frac{3}{4}left( 8
ight) cr
& = 6 cr} $$ Hence, required number = 786
48: C
Solution: Let the numbers be a and b, where a > b According to the question, $$eqalign{
& a + b = 37& {a^2} - {b^2} = 185 cr
& Rightarrow left( {a + b}
ight)left( {a - b}
ight) = 185 cr
& Rightarrow 37left( {a - b}
ight) = 185 cr
& Rightarrow a - b = frac{{185}}{{37}} cr
& Rightarrow a - b = 5 cr} $$
49: D
Solution: Let the number be x Then, $$eqalign{
& frac{3}{5}{ ext{of }}frac{2}{3}{ ext{of }}x - frac{2}{5}{ ext{of }}frac{1}{4}{ ext{of }}x = 288 cr
& Leftrightarrow left( {x imes frac{3}{5} imes frac{2}{3}}
ight) - left( {x imes frac{2}{5} imes frac{1}{4}}
ight) = 288 cr
& Leftrightarrow frac{2}{5}x - frac{1}{{10}}x = 288 cr
& Leftrightarrow frac{{3x}}{{10}} = 288 cr
& Leftrightarrow x = left( {frac{{288 imes 10}}{3}}
ight) cr
& Leftrightarrow x = 960 cr} $$
50: A
Solution: Let the numbers be x and y Then, $$eqalign{
& x + y = 12,, & ,, xy = 35 cr
& herefore frac{1}{x} + frac{1}{y} = frac{{x + y}}{{xy}} = frac{{12}}{{35}} cr} $$
51: C
Solution: Let the numbers be x and (x + 16) Then, $$eqalign{
& Leftrightarrow frac{x}{3} - frac{{left( {x + 16}
ight)}}{7} = 4 cr
& Leftrightarrow 7x - 3left( {x + 16}
ight) = 84 cr
& Leftrightarrow 4x = 84 + 48 cr
& Leftrightarrow 4x = 132 cr
& Leftrightarrow x = 33 cr} $$ Hence, the numbers are 33 and 49
52: B
Solution: Let the ten's digit be x and the unit's digit be y Then, number = 10x + y $$eqalign{
& herefore 10x + y = kleft( {x + y}
ight) cr
& Rightarrow k = frac{{10x + y}}{{x + y}} cr} $$ Number formed by interchanging the digits = 10y + x Let, 10y + x = h(x + y) Then, $$eqalign{
& h = frac{{10y + x}}{{x + y}} cr
& ,,,,, = frac{{11left( {x + y}
ight) - left( {10x + y}
ight)}}{{x + y}} cr
& ,,,,, = 11 - frac{{10x + y}}{{x + y}} cr
& ,,,,, = 11 - k cr} $$
53: B
Solution: Let the two fractions be a and b Then, $$eqalign{
& ab = frac{{14}}{{15}}, & , frac{a}{b} = frac{{35}}{{24}} cr
& Leftrightarrow frac{{ab}}{{left( {frac{a}{b}}
ight)}} = left( {frac{{14}}{{15}} imes frac{{24}}{{35}}}
ight) cr
& Leftrightarrow {b^2} = frac{{16}}{{25}} cr
& Leftrightarrow b = frac{4}{5} cr
& { ext{So, }} cr
& Leftrightarrow ab = frac{{14}}{{15}} cr
& Leftrightarrow a = left( {frac{{14}}{{15}} imes frac{5}{4}}
ight) cr
& Leftrightarrow a = frac{7}{6} cr} $$ Since a > b, So, greater fraction is $$frac{7}{6}$$
54: C
Solution: Let the total number of eggs bought be a 10% of eggs are rotten ∴ Remaining eggs : $$eqalign{
& = a - 10\% { ext{ of }}a cr
& = a - frac{{10a}}{{100}} cr
& = frac{{100a - 10a}}{{100}} cr
& = frac{{90a}}{{100}} cr
& = frac{{9a}}{{10}} cr} $$ Man gives 80% of $$frac{{9a}}{{100}}$$ eggs to his neighbour $$eqalign{
& = frac{{80}}{{100}} imes frac{{9a}}{{10}} cr
& = frac{{72a}}{{100}} cr} $$ Remaining eggs : $$eqalign{
& = frac{{9a}}{{10}} - frac{{72a}}{{100}} cr
& = frac{{90a - 72a}}{{100}} cr
& = frac{{18a}}{{100}} cr
& = frac{{9a}}{{50}} cr} $$ According the question, $$eqalign{
& Rightarrow frac{{9a}}{{50}} = 36 cr
& Rightarrow 9a = 36 imes 50 cr
& Rightarrow a = frac{{36 imes 50}}{9} cr
& Rightarrow a = 200 cr} $$ Hence, the total number of eggs bought be = 200
55: C
Solution: Let the number be x Then, ⇔ 3(2x + 9) = 75 ⇔ 2x + 9 = 25 ⇔ 2x = 16 ⇔ x = 8
56: A
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x + frac{1}{x} = 3left( {x - frac{1}{x}}
ight) cr
& Leftrightarrow frac{{{x^2} + 1}}{x} = 3left( {frac{{{x^2} - 1}}{x}}
ight) cr
& Leftrightarrow {x^2} + 1 = 3{x^2} - 3 cr
& Leftrightarrow 2{x^2} = 4 cr
& Leftrightarrow {x^2} = 2 cr
& Leftrightarrow x = sqrt 2 cr} $$
57: D
Solution: Let the fraction be $$frac{x}{y}$$ Then, $$eqalign{
& Leftrightarrow frac{{x + frac{1}{4}}}{{y - frac{1}{3}}} = frac{{33}}{{64}} cr
& Leftrightarrow frac{{3left( {4x + 1}
ight)}}{{4left( {3y - 1}
ight)}} = frac{{33}}{{64}} cr
& Leftrightarrow frac{{4x + 1}}{{3y - 1}} = frac{{33}}{{64}} imes frac{4}{3} cr
& Leftrightarrow frac{{4x + 1}}{{3y - 1}} = frac{{11}}{{16}} cr
& Leftrightarrow 16left( {4x + 1}
ight) = 11left( {3y - 1}
ight) cr
& Leftrightarrow 64x + 16 = 33y - 11 cr
& Leftrightarrow 64x - 33y = - 27 cr} $$ Which cannot be solved to find $$frac{x}{y}$$ Hence, the original fraction cannot be determined from the given data.
58: C
Solution: Let the number be x Then, ⇔ 2x + 20 = 8x - 4 ⇔ 6x = 24 ⇔ x = 4
59: D
Solution: Let the six numbers be, x , (x + 1), (x + 2), (x + 3), (x + 4) and (x + 5) Then, ⇔ x + (x + 1) + (x + 2) = 27 ⇔ 3x + 3 = 27 Required sum : = (x + 3) + (x + 4) + (x + 5) = 3x + 12 = (3x + 3) + 9 = 27 + 9 = 36
60: D
Solution: Let the ten's digit be x and unit's digit be y Then, $$eqalign{
& Leftrightarrow left( {10x + y}
ight) - left( {10y + x}
ight) = 63 cr
& Leftrightarrow 9left( {x - y}
ight) = 63 cr
& Leftrightarrow x - y = 7 cr} $$
61: D
Solution: Let the fraction be $$frac{x}{y}$$ Then, $$eqalign{
& Leftrightarrow frac{{x + 200\% { ext{ of }}x}}{{y + 300\% { ext{ of }}y}} = frac{{15}}{{26}} cr
& Leftrightarrow frac{{3x}}{{4y}} = frac{{15}}{{26}} cr
& Leftrightarrow frac{x}{y} = frac{{15}}{{26}} imes frac{4}{3} cr
& Leftrightarrow frac{x}{y} = frac{{10}}{{13}} cr} $$
62: B
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow frac{2}{3}x - 50 = frac{1}{4}x + 40 cr
& Leftrightarrow frac{2}{3}x - frac{1}{4}x = 90 cr
& Leftrightarrow frac{{5x}}{{12}} = 90 cr
& Leftrightarrow x = left( {frac{{90 imes 12}}{5}}
ight) cr
& Leftrightarrow x = 216 cr} $$
63: B
Solution: Let the numbers be x and y Then, x + y = 25 and x - y = 13 $$eqalign{
& Leftrightarrow 4xy = {left( {x + y}
ight)^2} - {left( {x - y}
ight)^2} cr
& Leftrightarrow 4xy = {left( {25}
ight)^2} - {left( {13}
ight)^2} cr
& Leftrightarrow 4xy = 625 - 169 cr
& Leftrightarrow 4xy = 456 cr
& Leftrightarrow xy = 114 cr} $$
64: N/A
Solution: Let the seven numbers be x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5) and (x + 6) Then, ⇔ x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) = 175 ⇔ 7x + 21 = 175 ⇔ 7x = 154 ⇔ x = 22 Required difference : = 2(x + 6) - 3x = 12 - x = 12 - 22 = -10
65: N/A
Solution: Let ten's digit = x Then, unit's digit = (12 - x) $$ herefore left[ {10x + left( {12 - x}
ight)}
ight] - $$ xa0 xa0$$left[ {10left( {12 - x}
ight) + x}
ight]$$ xa0 $$ = 54$$ $$eqalign{
& Leftrightarrow 18x - 108 = 54 cr
& Leftrightarrow 18x = 162 cr
& Leftrightarrow x = 9 cr} $$ So, ten's digit = 9 and unit's digit = 3 Hence, original number = 93
66: A
Solution: Let the fraction be $$frac{x}{y}$$ Then, $$eqalign{
& Leftrightarrow frac{{110\% { ext{ of 2}}x}}{{70\% { ext{ of 3}}y}} = 11\% { ext{ of }}frac{{16}}{{21}} cr
& Leftrightarrow frac{{22x}}{{21y}} = frac{{11}}{{100}} imes frac{{16}}{{21}} cr
& Leftrightarrow frac{x}{y} = left( {frac{{11}}{{100}} imes frac{{16}}{{21}} imes frac{{21}}{{22}}}
ight) cr
& Leftrightarrow frac{x}{y} = frac{2}{{25}} cr} $$
67: D
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow left[ {frac{{left( {frac{x}{2}}
ight)}}{6} + frac{{left( {frac{x}{2}}
ight)}}{4}}
ight] - frac{x}{5} = 4 cr
& Leftrightarrow frac{x}{{12}} + frac{x}{8} - frac{x}{5} = 4 cr
& Leftrightarrow frac{{10x + 15x - 24x}}{{120}} = 4 cr
& Leftrightarrow x = 4 imes 120 cr
& Leftrightarrow x = 480 cr} $$
68: A
Solution: Let the numbers be x and y Then, $$x + y = 33.....(i)$$ And, $$x - y = 15.....(ii)$$ Solving (i) and (ii), we get : x = 24 , y = 9 ∴ Smaller number = 9
69: A
Solution: Let ten's digit = x Then, unit's digit = x 2 Then, number = 10x + x 2 Clearly, since x 2 > x, So, the number formed by interchanging the digits is greater than the original number. $$eqalign{
& herefore left( {10{x^2} + x}
ight) - left( {10x + {x^2}}
ight) = 54 cr
& Leftrightarrow 9{x^2} - 9x = 54 cr
& Leftrightarrow {x^2} - x = 6 cr
& Leftrightarrow {x^2} - x - 6 = 0 cr
& Leftrightarrow {x^2} - 3x + 2x - 6 = 0 cr
& Leftrightarrow left( {x - 3}
ight)left( {x + 2}
ight) = 0 cr
& Leftrightarrow x = 3 cr} $$ So. ten's digit = 3, unit's digit = 3 2 = 9 ∴ Original number = 39 Required result : = 40% of 39 = 15.6
70: D
Solution: Let the three parts be A, B and C $$eqalign{
& { ext{Let }}frac{A}{2} = frac{B}{3} = frac{C}{4} = x cr
& { ext{Then, }}A = 2x,B = 3x{ ext{ and }}C = 4x cr
& { ext{So, }}A:B:C = 2:3:4 cr
& herefore { ext{Largest part :}} cr
& = left( {243 imes frac{4}{9}}
ight) cr
& = 108 cr} $$
71: N/A
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x + {x^2} = 182 cr
& Leftrightarrow {x^2} + x - 182 = 0 cr
& Leftrightarrow left( {x + 14}
ight)left( {x - 13}
ight) = 0 cr
& Leftrightarrow x = 13 cr} $$
72: A
Solution: Let the numbers be x and y Then, $$eqalign{
& Leftrightarrow frac{{x + y}}{{x - y}} = frac{{40}}{4} cr
& Leftrightarrow frac{{x + y}}{{x - y}} = 10 cr
& Leftrightarrow left( {x + y}
ight) = 10left( {x - y}
ight) cr
& Leftrightarrow 9x = 11y cr
& Leftrightarrow frac{x}{y} = frac{{11}}{9} cr
& Leftrightarrow x:y = 11:9 cr} $$
73: N/A
Solution: Let the three odd numbers be x, (x + 2), (x + 4) and The three even numbers be (x + 11), (x + 13) and (x + 15) Then, ⇔ x + (x + 2) + (x + 4) + (x + 11) + (x + 13) + (x + 15) = 231 ⇔ 6x + 45 = 231 ⇔ 6x = 186 ⇔ x = 31 ∴ Required sum : = (x + 4) + (x + 15) = 2x + 19 = 2 × 31 + 19 = 62 + 19 = 81
74: B
Solution: Let the middle digit be x Then, 2x = 10 or x = 5 So, the number is either 253 or 352 Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit. Hence, required number = 253
75: C
Solution: Let the four numbers be , A, B, C and D Let A + 3 = B - 3 = 3C = $$frac{D}{3}$$ = x Then, A = x - 3 B = x + 3 C = $$frac{x}{3}$$ D = 3x $$eqalign{
& Leftrightarrow A + B + C + D = 64 cr
& Leftrightarrow left( {x - 3}
ight) + left( {x + 3}
ight) + frac{x}{3} + 3x = 64 cr
& Leftrightarrow 5x + frac{x}{3} = 64 cr
& Leftrightarrow 16x = 192 cr
& Leftrightarrow x = 12 cr} $$ Thus, the numbers are 9, 15, 4 and 36 ∴ Required difference : = (36 - 4) = 32
76: N/A
Solution: Let the five numbers be x, (x + 2), (x + 4), (x + 6) and (x + 8) Then, ⇔ x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 575 ⇔ 5x + 20 = 575 ⇔ 5x = 555 ⇔ x = 111 ∴ Required sum : = (x + 10) + (x + 12) + (x + 14) + (x + 16) + (x + 18) = 5x + 70 = 5 × 111 + 70 = 555 + 70 = 625
77: A
Solution: Let the fraction be $$frac{x}{y}$$ Then, $$eqalign{
& Leftrightarrow frac{{x + 2}}{{y + 3}} = frac{7}{9} cr
& Leftrightarrow 9x - 7y = 3.....(i) cr
& { ext{And,}} cr
& Leftrightarrow frac{{x - 1}}{{y - 1}} = frac{4}{5} cr
& Leftrightarrow 5x - 4y = 1.....(ii) cr} $$ Solving (i) and (ii), we get : x = 5 and y = 6 Hence, the original fraction is $$frac{5}{6}$$
78: A
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x - 12 = 20\% { ext{ of }}x cr
& Leftrightarrow x - frac{x}{5} = 12 cr
& Leftrightarrow frac{{4x}}{5} = 12 cr
& Leftrightarrow x = frac{{12 imes 5}}{4} cr
& Leftrightarrow x = 15 cr} $$
79: C
Solution: Let the numbers be 4x and 7x Then, $$eqalign{
& Leftrightarrow frac{{4x + 4}}{{7x + 4}} = frac{3}{5} cr
& Leftrightarrow 5left( {4x + 4}
ight) = 3left( {7x + 4}
ight) cr
& Leftrightarrow x = 8 cr} $$ ∴ Larger number : = 7x = 7 × 8 = 56
80: B
Solution: Let the number be x and y Then, $${x^2} - {y^2} = 63,,,,& ,,,,x - y = 3$$ On dividing, we get: x + y = 21 Solving x + y = 21 and x - y = 3, We get: x = 12 and y = 9 ∴ Larger number = 12
81: B
Solution: Let the integer be x Them, $$eqalign{
& Leftrightarrow {x^2} - 20x = 96 cr
& Leftrightarrow {x^2} - 20x - 96 = 0 cr
& Leftrightarrow left( {x + 4}
ight)left( {x - 24}
ight) = 0 cr
& Leftrightarrow x = 24 cr} $$
82: C
Solution: Let the numbers be x and y Then, 2x + 3y = 100 ..... (i) And, 3x + 2y = 120 ..... (ii) Adding (i) and (ii), we get : 5x + 5y = 220 or x + y = 44 ..... (iii) Subtracting (i) from (ii): we get : x - y = 20 ..... (iv) Adding (iii) and (iv), we get : 2x = 64 or x = 32 Putting x = 32 in (iii), we get : y = 12 Hence, larger number = 32
83: D
Solution: Let the number be x Then, $$eqalign{
& Rightarrow frac{1}{7}x - frac{1}{{11}}x = 100 cr
& Rightarrow frac{{4x}}{{77}} = 100 cr
& Rightarrow x = frac{{7700}}{4} cr
& Rightarrow x = 1925 cr} $$
84: C
Solution: Let the second number be x Then, first number = 2x, And third number = $$frac{2x}{3}$$ $$eqalign{
& herefore 2x + x + frac{{2x}}{3} = 264 cr
& Leftrightarrow frac{{11x}}{3} = 264 cr
& Leftrightarrow x = left( {frac{{264 imes 3}}{{11}}}
ight) cr
& Leftrightarrow x = 72 cr} $$
85: C
Solution: Let A = x B = x + 2 C = x + 4 D = x + 6 and E = x + 8 Then, $$eqalign{
& Rightarrow A + C = 146 cr
& Rightarrow x + left( {x + 4}
ight) = 146 cr
& Rightarrow 2x = 142 cr
& Rightarrow x = 71 cr
& herefore E = x + 8 cr
& ,,,,,,,,,,, = 71 + 8 cr
& ,,,,,,,,,,, = 79 cr} $$
86: B
Solution: Let the ten's digit be x and unit's digit be y Then, $$eqalign{
& Leftrightarrow left( {10x + y}
ight) - left( {10y + x}
ight) = 36 cr
& Leftrightarrow 9left( {x - y}
ight) = 36 cr
& Leftrightarrow x - y = 4 cr} $$
87: B
Solution: Let the ten's digit be x and unit's digit be y Then, $$eqalign{
& Leftrightarrow frac{{10x + y}}{2} = 10y + left( {x + 1}
ight) cr
& Leftrightarrow 10x + y = 20y + 2x + 2 cr
& Leftrightarrow 8x - 19y = 2.....(i) cr
& { ext{And}} Leftrightarrow x + y = 7.....(ii) cr} $$ Solving (i) and (ii), we get : x = 5, y = 2 Hence, required number = 52
88: N/A
Solution: Let the numbers be x, y and z Then, $$eqalign{
& {x08f{x + y = 73}} cr
& Leftrightarrow y = 73 - x cr
& {x08f{y + z = 77 }} cr
& Leftrightarrow z = 77 - y cr
& Leftrightarrow z = 77 - left( {73 - x}
ight) cr
& Leftrightarrow z = 4 + x cr
& {x08f{3x + z = 104}} cr
& Leftrightarrow 3x + 4 + x = 104 cr
& Leftrightarrow 4x = 100 cr
& Leftrightarrow x = 25 cr
& x08ecause y = left( {73 - 25}
ight) cr
& Leftrightarrow y = 48 cr
& x08ecause z = left( {4 + 25}
ight) cr
& Leftrightarrow z = 29 cr} $$ ∴ Third number = 29
89: C
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x - frac{3}{5}x = 50 cr
& Leftrightarrow frac{2}{5}x = 50 cr
& Leftrightarrow x = left( {frac{{50 imes 5}}{2}}
ight) cr
& Leftrightarrow x = 125 cr} $$
90: A
Solution: Let the number be x Then, $$eqalign{
& frac{2}{3}x = frac{25}{216} imes frac{1}{x} cr
& Leftrightarrow {x^2} = frac{{25}}{{216}} imes frac{3}{2} cr
& Leftrightarrow {x^2} = frac{{25}}{{144}} cr
& Leftrightarrow x = sqrt {frac{{25}}{{144}}} cr
& Leftrightarrow x = frac{5}{{12}} cr} $$
91: B
Solution: Let the integers be x and (x + 5) Then, $$eqalign{
& Leftrightarrow xleft( {x + 5}
ight) = 500 cr
& Leftrightarrow {x^2} + 5x - 500 = 0 cr
& Leftrightarrow left( {x + 25}
ight)left( {x - 20}
ight) = 0 cr
& Leftrightarrow x = 20 cr} $$ So, the numbers are 20 and 25
92: C
Solution: Let the numbers be x and x + 2 Then, (x + 2) 2 - x 2 = 84 ⇔ 4x + 4 = 84 ⇔ 4x = 80 ⇔ x = 20 ∴ Required sum = x + (x + 2) = 2x + 2 = 42
93: B
Solution: Let the ten's digit be x and unit's digit be y Then, $$eqalign{
& Rightarrow 10x + y = 3left( {x + y}
ight) cr
& Rightarrow 7x - 2y = 0.....(i) cr
& 10x + y + 45 = 10y + x cr
& Rightarrow y - x = 5.....(ii) cr} $$ Solving (i) and (ii), we get : x = 2 and y = 7 ∴ Required number = 27
94: C
Solution: Let two numbers be a and b Given ab = 5 and a = $$frac{3}{2}$$ $$eqalign{
& Rightarrow b = frac{5}{a} cr
& Rightarrow b = frac{5}{{frac{3}{2}}} cr
& Rightarrow b = frac{{5 imes 2}}{3} cr
& Rightarrow b = frac{{10}}{3} cr} $$ Required sum of : ⇒ a + b = $$frac{{3}}{2}$$ + $$frac{{10}}{3}$$ L.C.M. of 2 and 3 = 6 ⇒ a + b = $$frac{{9 + 20}}{6}$$ ⇒ a + b = $$frac{{29}}{6}$$ ⇒ a + b = $$4frac{{5}}{6}$$
95: A
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow x + frac{2}{5}x = 455 cr
& Leftrightarrow frac{7}{5}x = 455 cr
& Leftrightarrow x = left( {frac{{455 imes 5}}{7}}
ight) cr
& Leftrightarrow x = 325 cr} $$
96: B
Solution: Let the required number be x Then, $$eqalign{
& Leftrightarrow x + 20 = frac{{69}}{x} cr
& Leftrightarrow {x^2} + 20x - 69 = 0 cr
& Leftrightarrow {x^2} + 23x - 3x - 69 = 0 cr
& Leftrightarrow xleft( {x + 23}
ight) - 3left( {x + 23}
ight) = 0 cr
& Leftrightarrow left( {x + 23}
ight)left( {x - 3}
ight) = 0 cr
& Leftrightarrow x = 3,,,,,,,,,left[ {x08ecause x
e - 23}
ight] cr} $$
97: C
Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{number}},{ ext{be}},x,{ ext{and}},y cr
& { ext{Then}},,xy = 9375,{ ext{and}},frac{x}{y} = 15 cr
& frac{{xy}}{{left( {x/y}
ight)}} = frac{{9375}}{{15}} cr
& Rightarrow {y^2} = 625 cr
& Rightarrow y = 25 cr
& Rightarrow x = 15y = left( {15 imes 25}
ight) = 375 cr
& herefore { ext{Sum}},{ ext{of}},{ ext{the}},{ ext{number}} cr
& = x + y = 375 + 25 = 400 cr} $$
98: B
Solution: Let the numbers be x and y Then, x y = 120 and x 2 + y 2 = 289 ∴ ( x + y ) 2 = x 2 + y 2 + 2 x y = 289 + (2 x 120) = 529 ∴ x + y = $$sqrt {529} $$xa0 = 23
99: B
Solution: Let the middle digit be x Then, 2 x = 10 or x = 5. So, the number is either 253 or 352 Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit. Hence, required number = 253
100: B
Solution: Let the numbers be x and y Then, x + y = 25 and x - y = 13 4 x y = ( x + y ) 2 - ( x - y ) 2 = (25) 2 - (13) 2 = (625 - 169) = 456 ∴ x y = 114
101: C
Solution: Let the numbers be x and x + 2. Then, ( x + 2) 2 - x 2 = 84 ⇒ 4 x + 4 = 84 ⇒ 4 x = 80 ⇒ x = 20. ∴ The required sum = x + ( x + 2) = 2 x + 2 = 42
102: C
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow 2x + 3 imes 42 = 238 cr
& Leftrightarrow 2x + 126 = 238 cr
& Leftrightarrow 2x = 112 cr
& Leftrightarrow x = 56 cr} $$ ∴ Required sum : $$eqalign{
& = 3x + 2 imes 42 cr
& = 3 imes 56 + 2 imes 42 cr
& = 168 + 84 cr
& = 252 cr} $$
103: D
Solution: $$eqalign{
& = frac{7}{8}{ ext{ of }}1008 - frac{3}{4}{ ext{ of }}568 cr
& = left( {1008 imes frac{7}{8}}
ight) - left( {568 imes frac{3}{4}}
ight) cr
& = 882 - 426 cr
& = 456 cr} $$
104: B
Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow 3{x^2} - 4x = x + 50 cr
& Leftrightarrow 3{x^2} - 5x - 50 = 0 cr
& Leftrightarrow left( {3x + 10}
ight)left( {x - 5}
ight) = 0 cr
& Leftrightarrow x = 5 cr} $$
105: C
Solution: Let the numbers be x and y Then, $$xy = 1092.....(i)$$ And, $$eqalign{
& Leftrightarrow left( {x + y}
ight) - left( {x - y}
ight) = 42 cr
& Leftrightarrow 2y = 42 cr
& Leftrightarrow y = 21 cr} $$ Putting y = 21 in (i), we get : $$x = frac{{1092}}{{21}} = 52$$ Hence, greater number = 52
106: B
Solution: Let the numbers be x, x + 2 and x + 4 Then, $$eqalign{
& Leftrightarrow frac{{xleft( {x + 2}
ight)left( {x + 4}
ight)}}{8} = 720 cr
& Leftrightarrow xleft( {x + 2}
ight)left( {x + 4}
ight) = 5760 cr
& herefore sqrt x imes sqrt {left( {x + 2}
ight)} imes sqrt {left( {x + 4}
ight)} cr
& = sqrt {xleft( {x + 2}
ight)left( {x + 4}
ight)} cr
& = sqrt {5760} cr
& = 24sqrt {10} cr} $$