Question

Solve the following word problem.
A two-digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.

Let the digit in unit’s place is x

and that in the ten’s place is y

∴ the number = `square` y + x

The number obtained by interchanging the digits is `square` x + y

According to the first condition two digit number + the number obtained by interchanging the digits = 143

∴ 10y + x + `square` = 143

∴ `square` x + `square` y = 143

x + y = `square` ........(I)

From the second condition,

digit in unit’s place = digit in the ten’s place + 3

∴ x = `square` + 3

∴ x − y = 3 ........(II)

Adding equations (I) and (II)

2x = `square`

x = 8

Putting this value of x in equation (I)

x + y = 13

8 + `square` = 13

∴ y = `square`

The original number is 10 y + x

= `square` + 8

= 58

Answer 1

Let the digit in unit’s place is x

and that in the ten’s place is y

∴ the number = ₁₀y + x

The number obtained by interchanging the digits is 10x + y

According to the first condition two digit number + the number obtained by interchanging the digits = 143

∴ 10y + x + 10x + y = 143

∴ 11x + 11y = 143

x + y = 13 ........(I) [Dividing both side by 11]

From the second condition,

digit in unit’s place = digit in the ten’s place + 3

∴ x = y + 3

∴ x − y = 3 ........(II)

Adding equations (I) and (II)

   x + y = 13
+ x − y = 3
2x = 16

x = `16/2`

x = 8

Putting this value of x in equation (I)

x + y = 13

8 + y = 13

y = 13 − 8

∴ y = 5

The original number is 10y + x

= 10(5) + 8

= 50 + 8

= 58