Question

The denominator of a fraction exceeds Its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get `3/2`. Find the original fraction.

Answer 1

Let the numerator & denominator be ‘n’ & ‘d’

Given that denominator exceeds numerator by 8

∴ d = n + 8  ...(1)

If numerator increased by 17 & denominator decreased by 1,

it becomes (n + 17) & (d – 1), fraction is `3/2`

i.e `("n" + 17)/("d" - 1) = 3/2` by cross multiplying, we get

`("n" + 17)/("d" - 1) = 3/2`

2(n + 17) = 3(d – 1)

2n + 2 × 17 = 3d – 3

∴ 34 + 3 = 3d – 2n

∴ 3d – 2n = 37  ...(2)

Substituting equation (1) in (2), we get,

3 × (n + 8) – 2n = 37

3n + 3 × 8 – 2n = 37

∴ n = 37 – 24 = 13

d = n + 8 = 13 + 8 = 21

The fraction is `"n"/"d" = 13/21`