Question

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

Answer 1

Topic :-

Linear Equations

Given :-

The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of new fraction and original fraction is 29/20.

To Find :-

Original Fraction

Solution :-

Let denominator of fraction be 'x'.

Then numerator will be ' x - 3 ' as it is 3 less than denominator.

So, original fraction is

$\dfrac{x-3}{x}$

Add 2 to both numerator and denominator,

New fraction

$\dfrac{x-3+2}{x+2}$

$\dfrac{x-1}{x+2}$

Add both fractions,

$\dfrac{x-3}{x}+\dfrac{x-1}{x+2}=\dfrac{29}{20}$

$\dfrac{(x-3)(x+2)+x(x-1)}{x(x+2)}=\dfrac{29}{20}$

$\dfrac{x^2+2x-3x-6+x^2-x}{x(x+2)}=\dfrac{29}{20}$

$\dfrac{2x^2-2x-6}{x^2+2x}=\dfrac{29}{20}$

Now, cross multiply,

20( 2x² - 2x - 6 ) = 29( x² + 2x )

40x² - 40x - 120 = 29x² + 58x

40x² - 29x² - 40x - 58x - 120 = 0

11x² - 98x - 120 = 0

11x² - 110x + 12x - 120 = 0

11x( x - 10 ) + 12( x - 10 ) = 0

( 11x + 12 )( x - 10 ) = 0

11x + 12 = 0

x = -12/11 or

x - 10 = 0

x = 10

So, x = 10 or -12/11

Original Fraction

$\dfrac{x-3}{x}$

Put x = 10,

$\dfrac{10-3}{10}$

$\dfrac{7}{10}$

Put x = -12/11,

$\dfrac{\dfrac{-12}{11}-3}{\dfrac{-12}{11}}$

$\dfrac{\dfrac{-45}{11}}{\dfrac{-12}{11}}$

Answer :-

So, original fractions are

$\dfrac{7}{10}$ and

$\dfrac{\dfrac{-45}{11}}{\dfrac{-12}{11}}$

Answer 2

$\boxed{\boxed{\frak{ \bigstar \: \underline {\cal{Q}\frak{uestion }}}}} - \begin{cases}\sf{ \underline{\:\;\;The \: numerator \: of \: a \: fraction \: is \: 3 \: less \: than \: its \: }}\\\sf{ \underline{\;\;\;denominator. \: if \: 2 \: is \: added \: to \: both \: the \: numerator \: and }}\\\sf{ \underline{\:\;\;the \: denominator, \: then \: the \: sum \: of \: new \: fraction }} \\ \: \sf{ \underline{\:\;\;orignal \: fraction \: is \: \frac{29}{20} . \: find \: the \: original \: fraction.}}\end{cases}$

$\boxed{\boxed{\frak{☯ \: \underline {Given }}}} - \begin{cases}\bf{ \underline{\:\;\;The \: numerator \: of \: a \: fraction \: is \: 3 \: less \: than \: its \: }}\\\bf{ \underline{\;\;\;denominator. \: if \: 2 \: is \: added \: to \: both \: the \: numerator \: and }}\\\bf{ \underline{\:\;\;the \: denominator, \: then \: the \: sum \: of \: new \: fraction }} \\ \: \bf{ \underline{\:\;\;orignal \: fraction \: is \: \frac{29}{20} . \: }}\end{cases}$

$\boxed{\boxed{\frak{☯ \: \underline {To \: Find }}}} - \begin{cases}\boldsymbol{ \underline{\:\;\;find \: \: the \: \: original \: \: number.}}\end{cases}$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀

⠀⠀⠀⠀⠀⠀

$\bigstar\; \huge{\boldsymbol{{ \underbrace{ \underline { \: \: \: \: \: Solution \: \: \: \: \: \: }}}}}$

Let,

• $\large { \bf{ \underline{☯ \:Original\: fraction \: = \: \sf \dfrac{x - 3}{x} }}}$

$\underline{\bf{\dag} \:\mathfrak{ \bf{A} \frak {ccording \:to\:the\:condition, }} }$

• $\large { \bf{ \underline{☯\: New\: fraction \: = \: \sf \dfrac{x - 3 + 2}{x + 2} \: \: \: and \: \: \: \dfrac{x - 1}{x + 2} }}}$

Then,

⠀⠀⠀⠀

$\sf \: \dfrac{x - 3}{x} \: + \: \dfrac{x - 1}{x + 2} = \dfrac{29}{20}$

➙ 20 [ (x - 3) (x + 2) + x (x - 1)] = 29 (x² + 2x)

➙ 20 [ x² - x - 6 + x² - x] = 29x² + 58x

➙ 20 [ 2x² - 2x - 6 ] = 29x² + 58x

➙ 40x² - 40x - 120 - 29x² - 58x = 0

➙ 11x² - 98x - 120 = 0

➙ 11x² - 110x + 12x - 120 = 0

➙ 11x (x - 10) + 12 (x - 10) = 0

➙ (x - 10) + (11x + 12) = 0

➙ x = 10

• $\large { \bf{ \underline{☯ \:Original\: fraction \: = \: \sf \dfrac{x - 3}{x} }}}$

$\large { \bf{:\implies \: \sf \dfrac{10 -3}{10} }}$

$\large { \bf{:\implies \: \sf \dfrac{7}{10} }}$

$\sf\therefore \:x \: = \: \frac{ - 12}{11} \: \boldsymbol{\: \underline {is \: rejected.}}$

${\underline{\sf{\therefore \: the \; original \; no. \: \; is\; \bf{ \dfrac{7}{10} }.}}}$

★ Hope it helps u ★