Question

A number consists of two digit of which tens digit exceeds the unit by 3 the number is 7 times the sum of its digits find the number

Answer 1

Hi friend!

Let the digits be $x \: and \: y$

The tens place is greater than units place by 3
$x = y + 3$

Then, the number would be
$10x + y \\ \\ = 10(y + 3) + y \\ \\ = 10y + 30 + y \\ \\ = 11y + 30$

The number is 7 times the sum of the digits
$11y + 30 = 7(x + y)$
$11y + 30 = 7(y + 3 + y) \\ 11y + 30 = 14y + 21 \\ 30 - 21 = 14y - 11y \\ 9 = 3y \\ y = \frac{9}{3} = 3$

Then,
$x = y + 3 \\ = 3 + 3 \\ = 6$

The original number
$10x + y$
$= 10(6) + 3 \\ = 60 + 3 \\ = 63$

There, you've got your answer.
Hope you found my answer useful. Keep Smiling!