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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.
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Solution
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`
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