Question

5. The digit at unit's place of a two digit number is 1 more than the digit at ten's
place. If the number is 3 more than 5 times of the sum of the digits, find the
number.​

Answer 1

Answer :-

Number is 78.

Explanation :-

Let the digit at ten's place be x

Digit at unit's place = 1 more than the digit at unit's place = (x + 1)

Number formed = 10(Digit at ten's place) + Digit at unit's place = 10(x) + (x + 1) = 10x + x + 1 = 11x + 1

Given

Number formed = 3 more than 5 times the sum of digits

⇒ 11x + 1 = 5{x + (x + 1) + 3

⇒ 11x + 1 = 5(x + x + 1) + 3

⇒ 11x + 1 = 5(2x + 1) + 3

⇒ 11x + 1 = 10x + 5 + 3

⇒ 11x + 1 = 10x + 8

⇒ 11x - 10x = 8 - 1

⇒ x = 7

i.e Digit at ten's place = x = 7

Digit at unit's place = (x + 1) = (7 + 1) = 8

Number = 10(x) + (x + 1) = 10(7) + 8 = 70 + 8 = 78

∴ the number is 78.