Question

The tens digit of a two digit number is three times the units digit . when the digits are
reversed, we get 36 less than the number. Find the original number​

Answer 1

Solution :-

Let the tens digit of a two digit number be x and the units digit be y

Tens digit = 3 times the unit digit

⇒ x = 3y

Number formed i.e original number = 10x + y

Number formed when digits are reversed = 10y + x

Given :-

Number formed when digits are reversed is 36 less than the original number

⇒ 10y + x = 10x + y - 36

⇒ 36 = 10x + y - (10y + x)

⇒ 36 = 10x + y - 10y - x

⇒ 36 = 9x - 9y

⇒ 36 = 9(x - y)

⇒ 36/9 = x - y

⇒ 4 = x - y

Substituting x = 3y in the above equation

⇒ 4 = 3y - y

⇒ 4 = 2y

⇒ 4/2 = y

⇒ 2 = y

⇒ y = 2

Substituting y = 2 in x = 3y we get,

⇒ x = 3(2)

⇒ x = 6

Original number= 10x + y

= 10(6) + 2

= 60 + 2

= 62

Hence, the original number is 62.

Answer 2

Answer:

Original two digit number is 62.

Step-by-step explanation:

Let the unit place digit of the two-digit number be 'x'.

Then, the tens place digit of the two digit number is 3x.

Thus, the original two digit number is 10 × 3 x + x = 31x.

When the digit of the two digit number is reversed, then the unit place and the tens place digits are 3x and x respectively.

Two digit number obtained by reversing the digits = 10 × x + 3x = 13x

According to question,

⇒ 13x = 31x - 36

⇒ 13x - 31x = -36

⇒ -18x = -36

⇒ x = 2

Thus, original two number is 31 × 2 = 62.