Question

In a certain positive fraction, the sum of the numerator and the denominator is 8. If 2 is added to 4 both the numerator and the denominator, the fraction increases by 4/35. Find the fraction.​

Answer 1

Step-by-step explanation:

Let's be the denominator be x

Then the numerator is 8 - x

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \: The \: required \: fraction \: = \: \frac{8 \: - \: x}{x}}}}$

When 2 is added to both the numerator and denominator, the fraction becomes

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \: \frac{10 \: - \: x}{2 \: + \: x}}}}$

From the question,

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \: \frac{10 \: - \: x}{2 \: + \: x} \: \: - \: \: \frac{8 \: - \: x}{x} \: \: = \: \: \frac{4}{35}}}}$

Let's do further calculations,

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \: \frac{x(10 \: - \: x) \: - \: (8 \: - \: x)(2 \: + \: x)}{x(2 \: + \: x)} \: \: = \: \: \frac{4}{35}}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \frac{10x \: - \: x² \: - \: 16 \: - \: 6x \: + \: x²}{x(2 \: + \: x)} \: \: = \: \: \frac{4}{35}}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \frac{4x \: - \: 16}{x(2 \: + \: x)} \: \: = \: \: \frac{4}{35}}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 35(4x \: - \: 16) \: \: = \: \: 4x(2 \: + \: x)}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 35(x \: - \: 4) \: \: = \: \: x(2 \: + \: x)}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 35x \: - \: 140 \: = \: 2x \: + \: x²}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: x³ \: - \: 33x \: + \: 140 \: \: = \: \: 0}}}$

Factorisation,

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: x³ \: - \: 28x \: - \: 5x \: + \: 140 \: \: = \: \: 0}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: x(x \: - \: 28) \: - \: 5(x \: - \: 28) \: \: = \: \: 0}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: (x \: - \: 28) (x \: - \: 5) \: \: = \: \: 0}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: (x \: - \: 28) \: \: = \: \: 0 \: \: or \: \: (x \: - \: 5) \: \: = \: \: 0}}}$

${\bold{\rm{⟼ \: \: \: \: \: \: \: \: \: \: x \: = \: 28 \: \: OR \: \: x \: = \: 5}}}$

But x is not equal to 28 because 8 - 28/ 28 is not a positive fraction

x = 5 and then 8 - x = 8 - 5 = 3

${\large{\boxed{\mathfrak{\pink{So \: the \: fraction \: is \: \frac{3}{5}}}}}}$

Answer 2

Let the numerator be x and denominator be y.

The fraction is x/y.

Given that their sum = 8.

x + y = 8 ------ (1)

y = 8 - x --------- (2)

Given that if 2 is added to both numerator and denominator, the fraction is increased by 4/35.

$\frac{x+2}{y+2} = \frac{x}{y} + \frac{4}{35}$

$\frac{x+2}{8 - x+2} = \frac{x}{8 - x} + \frac{4}{35}$

$\small{35(x + 2)(8-x) = 35x(10-x)+4(8-x)(10-x)}$

$\small{-35x^2 +210x + 560 = -31x^2 +278x + 320}$

$\small{-35x^2 -68x +240=-31x^2}$

$-4x^2 -68x + 240 = 0$

$x^2 + 17x -60=0$

$x^2 - 3x +20x - 60 = 0$

$x(x - 3) + 20(x - 3) = 0$

x = 3 (or) x = -20.

Since x cannot be -ve, so x = 3.

Then y = 8 - x

= 8 - 3

= 5.

$\bold{Therefore\: the\: fraction = \frac{3}{5}}$