Question

the numerator of a fraction is 6 less than the denominator .if 3 is added to the numerator the fraction becomes 2/3 .wha,t is the original fraction equal to ?
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Answer 1

$\sf\small\underline\purple{Let:-}$

$\sf{\implies The\: numerator\:_{(fraction)}=N}$

$\sf{\implies The\: denominator\:_{(fraction)}=D}$

$\sf{\implies The\: original\:_{(fraction)}=\dfrac{N}{D}}$

$\sf\small\underline\purple{To\: Find:-}$

$\sf{\implies The\: original\:_{(fraction)}=?}$

$\sf\small\underline\purple{Solution:-}$

• To calculate the original fraction ,at first we have to set up equation as per the given clue in the question then calculate the value of x and y after that by putting the value of X and y to find the original fraction:-]

$\sf\small\underline\green{Given\:in\:case\:(i):}$

• The numerator of a fraction is 6 less than the denominator:-]

$\tt{\implies Numerator=Denominator-6}$

$\tt{\implies N=D-6-------(i)}$

$\sf\small\underline\green{Given\:in\:case\:(ii):}$

• If 3 is added to the numerator the fraction becomes 2/3:-]

$\tt{\implies \dfrac{N+3}{D}=\dfrac{2}{3}}$

$\tt{\implies 3(N+3)=2*D}$

$\tt{\implies 3N+9=2D}$

$\tt{\implies 3N-2D=-9-----(ii)}$

• Here putting the value of N=D-6 in eq (ii)

$\tt{\implies 3(D-6)-2D=-9}$

$\tt{\implies 3D-18-2D=-9}$

$\tt{\implies 3D-2D=-9+18}$

$\tt{\implies D=9}$

• Now putting the value of D=9 in eq (i)

$\tt{\implies N=D-6}$

$\tt{\implies N=9-6}$

$\tt{\implies N=3}$

$\sf\large{Hence,}$

$\sf{\implies The\: original\:_{(fraction)}=\dfrac{3}{9}}$

$\sf{\implies The\: original\:_{(fraction)}=\dfrac{1}{3}}$

Answer 2

☄GIVEN:-

• Numerator is 6 less than the denominator.
• when 3 is added to Numerator fraction is equals to $\frac{2}{3}$

☄FIND:-

• What will be the original fraction = ?

☄SOLUTION:-

Let the numerator be x

and the denominator be y.

Now,

$\mathbb{ \pink{ACCORDING \: TO \: QUESTION}}$

$\tt x = y - 6.......(i)$

$\tt\implies \frac{x}{y} = \frac{(y - 6)}{y}$

Now,

$\tt\implies \frac{(y - 6) + 3}{y} = \frac{2}{3} .......(ii)$

now, in eq(ii)

$\tt\implies \frac{y - 6+ 3}{y} = \frac{2}{3} \\ \tt \implies \frac{y - 3}{y} = \frac{2}{3}$

$\tt Now, \: cross \: Multiply \: it$

$\tt\implies 3(y - 3) = 2(y) \\ \tt \implies3y - 9 = 2y$

collect like terms

$\tt \implies3y - 2y = 9$

$\tt \implies y = 9$

put this value of y in eq. (i)

$\tt x = y - 6$

$\tt \implies x = 9 - 6 = 3$

$\tt \implies x = 3$

$\tt \therefore fraction = \frac{x}{y} = \frac{ \cancel3}{ \cancel9} = \frac{1}{ 3}$

$\tt Hence, \: the \: fraction \: was \boxed{ \frac{1}{3} }$