4.One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Let unit place digit be x.
And the ten’s place digit be y= 3x

Original number = x + 3x × 10 = x + 30x = 31x

Condition I: 10x + 3x = 13x [interchanging the digits]

Condition II: New number + original number = 88

13x + 31x = 88

\(\Rightarrow\) 44x = 88

\(\Rightarrow\) x = 88 ÷ 44 [Transposing 44 to RHS]

\(\Rightarrow\) x = 2

Thus, we have,

The original number = 31x = 31 × 2 = 62

Hence the required number = 62