Question

10. A two-digit number is such that its ten's place
digit is 2 more than twice the unit's place
digit. When the digits are interchanged, the
reversed number is 5 more than thrice the
sum of the digits. Find the number.
​

Answer 1

Let ten's digit number be x and unit digit be y.

Number = 10x + y

A two-digit number is such that its ten's place digit is 2 more than twice the unit's place digit.

According to question,

→ x = 2y + 2........(1)

When the digits are interchanged, the

reversed number is 5 more than thrice the

sum of the digits. Find the number.

According to question,

→ 10y + x = 3(x + y) + 5

→ 10y + x = 3x + 3y + 5

→ 10y - 3y + x - 3y = 5

→ 7y - 2x = 5

Substitute value of x above

→ 7y - 2(2y + 2) = 5

→ 7y - 4y - 4 = 5

→ 3y = 9

→ y = 3

Substitute value of y in x

→ x = 2(3) + 2

→ x = 6 + 2

→ x = 8

Therefore,

Number = 10(8) + 3 = 80 + 3 = 83

Answer 2

Let ,

tens digit = x

ones digit = y

according to the question

x = 2 y + 2  _____equation (1)

when the digits will be interchanged reversed number will be

10 y + x   that will be equal to 3 ( x + y ) + 5

so,

10 y + x = 3 ( x + y ) + 5

10 y + x = 3 x + 3 y + 5

7 y - 2 x - 5 = 0

( putting equation (1) here )

7 y - 2 ( 2 y + 2 ) - 5 = 0

7 y - 4 y - 4 - 5 = 0

3 y - 9 = 0

3 y = 9

y = 3

putting y = 3 in equation (1)

x = 2 y + 2

x = 2 ( 3 ) + 2

x = 6 + 2

x = 8

so, the number was

10 x + y  = 10 (8) + 3 = 83

so, the number was 83.