Question

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

Answer 1

let tens place of the original number=x
and unit place=y
also let x=3y
therefore the original number=10x+y=10(3y)+y=31y
on interchanging the digits, we get the new number=10y+x=10y+3y=13y
also, it is given that 31y+13y=88
i.e. 44y=88
i.e. y=2
then x=3y=6
therefore the original digit=62

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

Ones digit = 3p

Tens digit = p

Initial Number :-

= 10p + 3p

= 13p

Now,

Interchange the number :-

= 10 × 3p + p

= 30p + p

= 31p

Situation,

⇒ (Initial + New)Number = 88

⇒ 13p + 31p = 88

⇒ 44p = 88

⇒ p = 88/44

⇒ p = 2

Initial Number

= 13p

= 13 × 2

= 26

Now,

⇒p = 2

Also,

= 3p

= 3 × 2

= 6

Therefore,

Two digit number is 62 or 26