EXERCISE 7 A (OBJECTIVE TYPE QUESTIONS IN 61 65 SOLUTION )

61.       A number of two digits has 3 for its unit’s digit, and the sum of digits is 1/7 of the number itself. The number is

a)      43

b)      53

c)       63

d)      73.

62.       A two – digit number exceeds the sum of the digits of that number by 18. If the digit at the unit’s double the digit in the ten’s place, what is the number?

a)      24

b)      42

c)       48

d)      Data inadequate

63.       The sum of the digits of a two – digits of 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 none of these

64.       In a two – digit number, 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

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

a)      3

b)      5

c)       9

d)      11

ANSWERS

61. C

62. A

63. D

64. A

65. D

SOLUTION 

61.       Let the ten’s digit be x. Then, number = 10x + 3 and sum of digits = (X + 3).

             So, (X + 3 ) 1/7 (10x + 3)    7X + 21  = 10 X + 3     3X = 13 18     X = 6.

             Hence, the number is 63.

62.       Let the ten’s digit be x. Then, unit’s digit = 2X.

             Number = 10X = 2x = 12X; sum of digits = X + 2X = 3x.

             12X – 3x = 18     9X = 18       X = 2.

63.       Let the ten’s digit be X and unit’s digit be y.

            Then, x +y = 15 and x – y = 3 or y – x = 3.

            Solving x +y = 15 and x – y = 3, we get: x =, y= 6.

           Solving X + y = 15 and y – x = 3, we get: x = 6, y =9.

             So, the number is either 96 or 69. Hence, the number cannot be determined.

64.       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) = 2x + 2.

         (11 X + 2) (2x + 2) = 144   22x² + 26x – 140 =    11x² + 13x – 70 = 0.

        (X – 2) (11X + 35) = 0     X =2.

65.       Let the ten’s digit be x and unit’s digit by y. Then, number = 10X + y.

               Number obtained by interchanging the digits = 10y + x.

               10 x + y ) + (10 y + x ) = 11 ( x + y), which is divisible by 11.