Question

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

Answer 1

Let the digit at ones place be y and at tens place be x respectively.

According to the question:

☛ x = 3y ...(i)

☛ 10x + y + 10y + x = 88

➜ 11x + 11y = 88

➜ 11( x + y) = 88

➜ x + y = 8 ...(ii)

Putting x = 3y in (ii) from (i)

☛ x + y = 8

➜ 3y + y = 8 { from (i) , x = 3y }

➜ 4y = 8

➜ y = 8/4

➜ y = 2

Putting value of y in (i)

☛ x = 3y

➜ x = 3×2

➜ x = 6

Here, x = 6 ; y = 2

So Required number is 10x + y

=> number = 10(6) + 2

=> number = 60 + 2

=> number = 62

Hence, The original number is 62

Answer 2

Answer:

$= > 60 + 2 \\ \\ = > > 62$