Question

the number obtained by reversing a two digit number is 18 more than original number if the sum of its digits place is 8 .find the number​

Answer 1

Step-by-step explanation:

$\bf\underline{\underline{\red{Given:-}}}$

• The number obtained by reversing the digits of a two digit number is 18 more than the original number.

• The sum of the digits of the number is 8.

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• The original number.

$\bf\underline{\underline{\green{Solution:-}}}$

Let the tens digit of the number be x

And ones digit be y

The original number = 10x + y

The number obtained by reversing the digits = 10y + x

Case 1:-

The sum of digits is 8

→ x + y = 8......(i)

Case 2:-

The number obtained by reversing the digits of a two digit number is 18 more than the original number.

→ Reversed number - 18 = Original number

→ 10y + x - 18 = 10x + y

→ 10y + x - 10x - y = 18

→ 9y - 9x = 18

Dividing the whole equation by 2

→ y - x = 2......(ii)

Adding equation (i) and (ii)

→ x + y + (y - x) = 8 + 2

→ x + y + y - x = 10

→ 2y = 10

→ $\rm y = \dfrac{10}{2}$

→ y = 5

Substituting y = 5 in equation (i)

→ x + y = 8

→ x + 5 = 8

→ x = 8 - 5

→ x = 3

The original number

= 10x + y

= 10(3) + 5

= 30 + 5

= 35

$\underline{\boxed{\purple{\rm \therefore The \: original \: number = 35}}}$