Question

The numerator of fraction is 3 less than its denominator if 2 is added to both numerator and denominator then sum of refraction and original fraction is 29upon 30 find the original fraction

Answer 1

$\large\bf\underline{Correct\: Question:-}$

The numerator of fraction is 3 less than its denominator if 2 is added to both numerator and denominator then sum of refraction and original fraction is 29/20 find the original fraction.

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$\large\bf\underline{Given:-}$

• Numerator of fraction is 3 less than it's denominator
• If 2 is added to both numerator and denominator then sum of fraction and original fraction is 29/20

$\large\bf\underline {To \: find:-}$

• original fraction

$\huge\bf\underline{Solution:-}$

$\begin{cases} \tt \: let \: the \:denominator \: be \: x \\ \tt \: then \: numerator \: =( x - 3) \end{cases}$

• Fraction = x-3/x

$\blacktriangleright \: \underbrace{ \text{According to question}}$

If 2 is added to both numerator and denominator then sum of fraction and original fraction is 29/20.

Adding 2 to both numerator and denominator then fraction = x-3+2/x +2 = x-1/x + 2

$\small \rightarrowtail \tt \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20} \\ \\ \small \rightarrowtail \tt \frac{(x + 2)(x - 3) +x (x - 1)}{x(x + 2)} = \frac{29}{20} \\ \\ \small \rightarrowtail \tt \frac{x(x - 3) + 2(x - 3) + {x}^{2} - x }{ {x}^{2} + 2x } = \frac{29}{20} \\ \\ \small \rightarrowtail \tt \frac{ {x}^{2} - 3x + 2x - 6 + {x}^{2} - x }{ {x}^{2} + 2x } = \frac{29}{20} \\ \\ \small \rightarrowtail \tt \:20( 2 {x}^{2} - 2x - 6) = 29( {x}^{2} + 2x) \\ \\ \small \rightarrowtail \tt40 {x}^{2} - 40x - 120 = 29 {x}^{2} + 58x$

$\small \rightarrowtail \tt40 {x}^{2} - 29 {x}^{2} - 120 = 58x + 40x \\ \\ \small \rightarrowtail \tt11 {x}^{2} - 120 = 98x \\ \\ \small \rightarrowtail \tt11 {x}^{2} - 98x - 120 \\ \\ \small \rightarrowtail \tt {11x}^{2} - 110x + 12x - 120 \\ \\ \small \rightarrowtail \tt \: 11x(x - 10) + 12(x - 10) \\ \\ \small \rightarrowtail \tt \:(11x + 12)(x - 10) \\ \\ \small \rightarrowtail \tt \:x = \frac{ - 12}{11} \: or \: x = 10$

So, we will take x = 10 because x is denominator and denominator is always greater then 0.

• So x = 10

Now,

• ✝️ Numerator x - 3 = 10 - 3 = 7
• Denominator x = 10

So,

• »★ Original Fraction = 7/10

Answer 2

Correct Question:

The numerator of fraction is 3 less than its denominator. If 2 is added to both numerator and denominator then sum of refraction and original fraction is 29 upon 20. Find the original fraction.

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Answer:-

$\sf{The \ original \ fraction \ is \ \frac{7}{10}}$

Given:

• The numerator of fraction is 3 less than its denominator.

• If 2 is added to both numerator and denominator then sum of refraction and original fraction is 29 upon 20.

To find:

• The fraction.

Solution:

$\sf{Let \ the \ denominator \ of \ the \ fraction \ be \ x.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{\implies{Numerator=x-3}}$

$\sf{Original \ fraction=\frac{x-3}{x}}$

$\sf{Refraction=\frac{x-1}{x+2}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{\frac{x-3}{x}+\frac{x-1}{x+2}=\frac{29}{30}}$

$\sf{\therefore{\frac{(x-3)(x+2)+(x-1)(x)}{(x)(x+2)}=\frac{29}{20}}}$

$\sf{\therefore{\frac{x^{2}+-x+6+x^{2}-x}{x^{2}+2x}=\frac{29}{20}}}$

$\sf{\therefore{20(2x^{2}-2x+6)=29(x^{2}+2x)}}$

$\sf{\therefore{40x^{2}-40x+120=29x^{2}+58x}}$

$\sf{\therefore{40x^{2}-29x^{2}-40x-58x+120=0}}$

$\sf{\therefore{11x^{2}-98x+120=0}}$

$\sf{\therefore{11x-110x+12x+120=0}}$

$\sf{\therefore{11x(x-10)+12(x-10)=0}}$

$\sf{\therefore{(x-10)(11x+12)=0}}$

$\sf{\therefore{(x=10 \ or \ \frac{-12}{11}}}$

$\sf{But \ x \ is \ denominator \ so \ x \ can't}$

$\sf{be \ smaller \ than \ zero.}$

$\sf{\therefore{x=10}}$

$\sf{Fraction=\frac{x-3}{x}=\frac{10-3}{10}}$

$\sf{\therefore{Original \ fraction=\frac{7}{10}}}$

$\sf\purple{\tt{\therefore{The \ original \ fraction \ is \ \frac{7}{10}.}}}$