Question

The sum of the two-digit number and the number obtained by interchanging the digits is 132. The digit in the ten’s place is 2 more than the digit in the unit’s place. Complete the activity to find the original number.

Activity: Let the digit in the unit’s place be y and the digit in the ten’s place be x.

∴ The number = 10x + y

∴ The number obtained by interchanging the digits = `square`

∴ The sum of the number and the number obtained by interchanging the digits = 132

∴ 10x + y + 10y + x = `square`

∴ x + y = `square`      .....(i)

By second condition,

Digit in the ten’s place = digit in the unit’s place + 2

∴ x – y = 2     ......(ii)

Solving equations (i) and (ii)

∴ x = `square`, y = `square`

Ans: The original number = `square`

Answer 1

Let the digit in the unit’s place be y and the digit in the ten’s place be x.

∴ The number = 10x + y

∴ The number obtained by interchanging the digits = 10y + x

The sum of the number and the number obtained by interchanging the digits = 132

∴ 10x + y + 10y + x = 132

∴ 11x + 11y = 132

∴ x + y = `132/11`    ......[Dividng both side by 11]

∴ x + y = 12      .....(i)

By second condition,

Digit in the ten’s place = digit in the unit’s place + 2

∴ x = y + 2

∴ x – y = 2     ......(ii)

Adding equations (i) and (ii), we get

   x + y = 12
+ x – y = 2     
 2x       = 14

∴ x = `14/2` = 7

Substituting x = 7 in equation (i), we get

7 + y = 12

∴ y = 12 – 7 = 5

Solving equations (i) and (ii),

∴ x = 7, y = 5

The original number = 10x + y

= 10(7) + 5

= 70 + 5

= 75