Question

The digit in tens place of two digit number is three times than that in the units place. If the digits are reversed the new number will be 36 less than the original number. Find the original number.

Answer 1

Number is 62.

Let say ‘x' is digit on one's place so 3x is digit on tens' place

So number is 10*3x+x=31x

Now if reverse the possition of digit new number is

10*x+3x=13x

As given new number is less by 36 from old number

31x-36=13x

18x=36

x=2

So number is 10*3*2+2=62.

Answer 2

Given:

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed the new number will be 36 less than the original number.

To find:

The original number.

Solution:

The original number is 62.

To answer this question, we will follow the following steps:

Let the digits of a two-digit number be x and y where x is at ten's place while y is at one's place.

So,

The original number is 10x + y.

Now,

According to the question,

$x = 3y \: \: (i)$

Also,

After reversing the digits,

The number becomes 10y + x

So, according to the question,

$10y + x = 10x + y - 36$

This can be written as

$9x - 9y = 36$

$x - y = 4 \: \: (ii)$

On putting x = 3y from (i) in (ii), we get

$3y - y = 4$

$2y = 4$

$y = 2$

Now,

x = 3y = 3(2) = 6

Original number = 10x + y = 10(6) + 2 = 62

Hence, the original number is 62.