The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$ . The fraction is [[1]]

Question

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 115. The fraction is [[1]]

Accepted Answer

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by

\frac{1}{15}

. The fraction is

\frac{2}{5}

.
Given that the numerator of a fraction is 3 less than its denominator.
Let’s assume that the numerator and denominator are

n

and

d

respectively.

n = d - 3

Then,

\frac{n}{d} = \frac{d - 3}{d}

… (1)
After adding 1 to the denominator the fraction decreased by

\frac{1}{15}

⟹ \frac{d - 3}{d + 1} = \frac{1}{15}

… (2)
To find the fraction we will equate equation (1) and equation (2),

⟹ \frac{d - 3}{d} - \frac{d - 3}{d + 1} = \frac{1}{15}

⟹ \frac{(d - 3) (d + 1) - d (d - 3)}{d (d + 1)} = \frac{1}{15}

⟹ \frac{d (d + 1) - 3 (d + 1) - d^{2} + 3 d}{d^{2} + d} = \frac{1}{15}

⟹ \frac{d^{2} + d - 3 d - 3 - d^{2} + 3 d}{d^{2} + d} = \frac{1}{15}

⟹ \frac{d - 3}{d^{2} + d} = \frac{1}{15}

⟹ 15 (d - 3) = d^{2} + d

⟹ 15 d - 45 = d^{2} + d

⟹ d^{2} - 15 d + 45 = 0

⟹ d^{2} - 9 d - 5 d + 45 = 0

⟹ d (d - 9) - 5 (d - 9) = 0

⟹ (d - 5) (d - 9) = 0

⟹ d - 5 = 0 or d - 9 = 0

∴ d = 5 or d = 9

If

d = 5

then,

⟹ n = d - 3 = 5 - 3 = 2

Then the fraction is

\frac{2}{5}

.
If

d = 9

then,

⟹ n = d - 3 = 9 - 3 = 6

Then the fraction is

\frac{6}{9} = \frac{2}{3}

∴ We will neglect this fraction because it is not satisfying the given condition.
Hence, the original fraction is

\frac{2}{5}

.

Detailed Solution