Question

The digit at ten's place of a two-digit number is three
times the digit at one's place. If the sum of this
number and the number formed by reversing its
digits is 88, find the number.​

Answer 1

Let the number at ten's place be X and one's place be y

According to the question,

x= 3y

10x+y+10y+x= 88

11x+11y=88

11(3y)+11y = 88

33y+11y= 88

44y=88

y= 2

x=3*2=6

Therefore, the number is 10*x+y= 10*6+2= 62

OR

using 1 variable,

Let the numbers at the one's place be x

Ten's place be 3x

(10*3x+x) + (10*x+3x) = 88

31x+ 13x= 88

44x= 88

x = 2

Therefore, number is 10*3x+ x= 10*3*2 + 2= 60+2= 62

Hope it helps:)