English, asked by ishubhiisinghh, 9 days ago

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 ?

Answers

Answered by mddilshad11ab
87

\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}}

Answered by Anonymous
32

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} }

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