Question

A two digit number of which tens digit exceeds unit digit by 5. the number itself is 8 times the sum of its digits. find the number.

Answer 1

Let us assume x and y are the digits of a two-digit number

Therefore, the two-digit number = 10x + y

Given:

x = y + 5 -----------1

Also given:

10x + y = 8 (x + y)

10x + y = 8x + 8y

2x = 7y --------------2

Substitute the value of x from eqn 1 in eqn 2

2 (y + 5) = 7y

2y + 10 = 7y

5y = 10

y = 2

Therefore, x = y + 5 = 2 + 5 = 7

Therefore, the two-digit number = 10x + y = (10 * 7) + 2 = 72

Answer 2

Here to help..

Let the unit digit be x

tens digit = x + 5

number = x + 10(x + 5)

= x + 10x + 50

= 11x + 50

As per the condition,

11x + 50 = 8 (x + x + 5)

= 11x + 50 = 8x + 8x + 40

= 11x + 50 = 16x + 40

= 11x - 16x = 40 - 50

= -5x = -10

= x = -10/-5

= x = 2

unit digit = 2

tens digit = 2 + 5 = 7

Required number = 72

Hope it helps...