Question

The digit at the tens place of a 2- digit number is three times the digit at the ones place if the sum of this number and the number formed by reversing its digits is 88 and find the number

Answer 1

let the tens place digit no. be x
and ones place digit no. be y
So, two digit no.=10*x+y
=10x+y
by reversing order.
two digit no.=10y+x
according to question
x=3y. 1
10x+y+10y+x=88
11x+11y=88
11(x+y)=88
x+y=8
from 1
3y+y=8
4y=8
y=2. 2
from 1 and 2
x=6
So,two digit no.=10*6+2
=60+2
=62