Question

the numerator of a fraction is three less than the denominator. if 4 is added to both the numerator and denominator the value of the fraction increases 1/8 denominator the fraction.
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Answer 1

Given:

The numerator of a fraction is three less than the denominator.

If 4 is added to both the numerator and denominator the value of the fraction increases 1/8

To find:

The fraction

Solution:

Let's assume,

" $\bold{\frac{x}{y}}$ " ← the original fraction

So, we have

x = y - 3 ..... (i)

When 4 is added to both numerator and denominator the fraction increases by 1/8, so the equation will be as:

$\frac{x + 4}{y + 4} = \frac{x}{y} + \frac{1}{8}$

substituting the value of x from (i)

$\implies \frac{y - 3 + 4}{y + 4} = \frac{y - 3}{y} + \frac{1}{8}$

$\implies \frac{y + 1}{y + 4} = \frac{y - 3}{y} + \frac{1}{8}$

$\implies \frac{y + 1}{y + 4} =\frac{8(y - 3) + y}{8y}$

$\implies \frac{y + 1}{y + 4} =\frac{8y - 24 + y}{8y}$

$\implies \frac{y + 1}{y + 4} =\frac{9y - 24 }{8y}$

$\implies 8y(y +1 ) = (y + 4)(9y - 24)$

$\implies 8y^2 + 8y = 9y^2 + 36y - 24y - 96$

$\implies y^2 + 4y - 96 = 0$

$\implies y^2 + 12y - 8y - 96 = 0$

$\implies y(y+12)- 8(y + 12) = 0$

$\implies (y+12)(y-8)= 0$

$\implies y = -12 \:or\:y = 8$

(1). When y = -12:

from eq. (i), we get

x = y - 3 = -12 - 3 = - 15

∴ Fraction = $\frac{-15}{-12} = \frac{5}{4}$ ← thus does not satisfy the equation (i), so this is not the original fraction

(2). When y = 8:

from eq. (i), we get

x = y - 3 = 8 - 3 = 5

∴ Fraction = $\frac{5}{8}$

Thus, the original fraction is → $\underline {\bold{\Biggg{{\frac{5}{8} }}}}$.

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