Question

6. The sum of the digits of a two digit number is 9. The digit obtained by increasing the digit in ten's place by two is one eighth of the number. Find the number.​

Answer 1

Answer:

By using 1 variable:-

Let the digit at tens place be x

one's place be (9-x)

Orignal number=10(x)+9-x

=10x+9-x

=9x+9

According to question:-

$x + 2 = \frac{1}{8} (9x + 9)$

$x + 2 = \frac{9x}{8} + \frac{9}{8}$

Removing denominator

8x+16=9x+9

9x-8x=16-9

x=7

Ten's digit=7

One's digit=(9-x)

=9-7

=2

Orignal number=9(7)+9

=63+9

=72