EXERCISE 7 A (OBJECTIVE TYPE QUESTIONS IN 71 75 SOLUTION )

71.       If the number obtained on interchanging the digits o a two – digit number is 18 more than the original number and the sum of the digits is 8, then what is the original number?

a)      26

b)      35 53

c)       Cannot be determined

d)      None of these

72.       The difference between a two – digit number and the number obtained by interchanging the digits is 36. What is the difference between the sums of 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

73.       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

74.       A two – digit number becomes five – sixth of itself when its digits are reversed. The two digits differ by one. The number is:

a)      45

b)      54

c)       56

d)      65

75.       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 6/7 time 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

ANSWERS

71.B

72.B

73.B

74.B

75.A

SOLUTION

71.       Let ten’s digit = X. Then, unit’s digit = (8 –x).

               ([10 ( 8 –X) + X] – [10X + (8 – X)] = 18    18X = 54     X=3.

                So, ten’s digit = 3 and unit’s digit =3 and unit’s digit = 5. Hence, original number = 35.

72.       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 the ten’s and unit’s digits be 2x and x respectively.

                  Then, (10 x 2X + X) – (10x + 2x= 36    9x =36      X =4.

                   Required difference = (2X + X)  - (2X – X ) = 2X = 8.

73.       Let the middle digit be X. Then,2X = 10 or X = 5. So, the number is either 253 or 352.    Since the number increases of reversing the digits, so the hundred’s digit is smaller than the unit’s digit. Hence, required number = 253.

74.       Since the number reduces on reversing the digits, so ten’s digit is greater than the unit’s  digit.

                    Let the unit’s digit be x. Then, ten’s digit = (x + 1).

                   10X + (X + 1) 5/6 [10 (x + 1) + X]     66x + 6 = 55x + 50        11X = 44       X =4.

                   Hence, required number = 54.

 75.       Let the unit’s digit be X. Then, ten’s digit = (X – 2).

                        3 [10 (x – 2) +x] + 6/7 [10 x + (x – 2)] = 108

                       231 x – 420 + 66x – 12 = 756     297 X = 1188      X = 4.

                       Hence, sum of the digits =X +(X -2) = 2X – 2 =  6.