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In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by 114. Find the fraction. - Mathematics

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प्रश्न

In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by `1/14`. Find the fraction.

योग
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उत्तर

Let the numerator be 'x'

The denominator be 'x + 3'

So the fraction is `x/(x+3)`

According to the question,

`(x - 1)/(x + 3 - 1) = x/(x + 3) - 1/14`

Firstly let us simplify RHS

`(x - 1)/(x + 2) = (14x - x - 3)/(14 (x + 3))`

 `(x - 1)/(x + 2) = (13x - 3)/(14x + 42)`

By cross-multiplying, we get

(x – 1)(14x + 42) = (x + 2)(13x – 3)

14x2 + 42x – 14x – 42 = 13x2 – 3x + 26x – 6

14x2 + 42x – 14x – 42 – 13x2 + 3x – 26x + 6 = 0

x2 + 5x – 36 = 0

Let us factorize,

x2 + 9x – 4x – 36 = 0

x(x + 9) – 4(x + 9) = 0

(x + 9)(x – 4) = 0

So,

(x + 9) = 0 or (x – 4) = 0

x = – 9 or x = 4

So the value of x = 4  ...[Since, – 9 is a negative number]

When substitute the value of x = 4 in the fraction `x/(x + 3)`, we get

`4/(4 + 3) = 4/7`

∴ The required fraction is `4/7`

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