Question

The sum of two digits number and the number obtained by reversing it's digit is 88 find the number if it's units place digit it's greater than tens place digit by 2​

Answer 1

Answer:

$\huge\underline\frak{Answer★}$

Let the digit at tens place be x and the digit unis place be y.

Then the original number will be 10x+y.

The digit at tens place is three times the digit at the units place.

i.e. x=3y ....(1)

The sum of this number and the number formed by reversing its digits is 88.

(10x+y)+(10y+x)=88

⇒11x+11y=88

⇒x+y=8 ....(2)

Substitute the value of equation (1) in equation (2).

Therefore, 3y+y=8

⇒4y=8

⇒y=2

Therefore, x=3y=3×2=6

Therefore, the original number is 62.