Question

a two digit number and the number with digits interchanged add up to 143. in the given number the digit in unit's place is 3 more than the digit in ten's place. find the original number

Answer 1

Hi there!

Let the one's digit be = y and ten's digit be = x

♦ Original number = 10x + y
♦ Number when reversed = 10y + x

ATQ,

→ y - x = 3. ---(i)

→ (10x + y) + (10y + x) = 143 

10x + y + 10y + x = 143

11x + 11y = 143

Dividing whole Eqn. by 11 :-

x + y = 13 -----(ii)

Adding eqn. (i) n' (ii)

x + y + y - x = 13 + 3

2y = 16

y = 16 / 2

∴ y = 8

Substituting y = 8 in eqn. (ii)

x + y = 13

x + 8 = 13

x = 13 - 8

∴ x = 5

Hence, The required answer is :-
Original number = 10x + y = 10(5) + 8 = 50 + 8 = 58

Hope it helps! :)