Question

in a fraction's numerator is increased by 1 and the denominator is increased by 2 then the fraction becomes 2/3.but when the numerator is increased by 5 and the denominator is increased by 1then the fraction becomes 5/4.what is the value of the original fraction.
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Answer 1

Step-by-step explanation:

Fraction = 3/7

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Answer 2

Given,

The fraction obtained by increasing the numerator by 1 and denominator by 2 = $\frac{2}{3}$

The fraction obtained by increasing the numerator by 5 and denominator by 1 = $\frac{5}{4}$

To Find,

The original fraction.

Solution,

Let the original fraction be $\frac{x}{y}$.

As per the statements given in the question,

$\frac{x+1}{y+2} = \frac{2}{3}$ and $\frac{x+5}{y+1} = \frac{5}{4}$.

$\frac{x+1}{y+2} = \frac{2}{3}$

On cross multiplication,

⇒$3(x+1) = 2(y+2)$

⇒$3x+3 = 2y+4$

⇒$3x-2y = 4-3$

⇒$3x-2y = 1$→ equation 1

$\frac{x+5}{y+1} = \frac{5}{4}$

On cross multiplication,

⇒$4(x+5) = 5(y+1)$

⇒$4x+20 = 5y+5$

⇒$5y-4x = 20-5$

⇒$5y-4x = 15$→ equation 2

Multiply equation 1 with 4.

⇒$4(3x-2y = 1)$

⇒$12x-8y = 4$ → equation 3

Multiply equation 2 with 3.

⇒$3(5y-4x = 15)$

⇒$15y-12x = 45$ → equation 4

Add equation 3 and equation 4,

⇒$(12x-8y)+(15y-12x) = 45+4$

⇒$15y-8y = 49$

⇒$7y = 49$

⇒$y=\frac{49}{7}$

⇒$y = 7$

Substitute y=7 in equation 1.

⇒$3x-2y = 1$

⇒$3x-2(7) = 1$

⇒$3x-14 = 1$

⇒$3x = 15$

⇒$x = \frac{15}{3}$

⇒$x = 5$

Thus $x =5$ and $y = 7$.

Henceforth, the fraction, $\frac{x}{y}$ = $\frac{5}{7}$