2323\frac{2}{3} of the original fraction. The numerator of the original fraction is" /> 2323\frac{2}{3} of the original fraction. The numerator of the original fraction is" /> 2323\frac{2}{3} of the original fraction. The numerator of the original fraction is" />

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The numerator and denominator of a fraction are in the ratio 3 : 4. If 9 is subtracted from the numerator, the resulting fraction in $\frac{2}{3}$ $\frac{2}{3}$ of the original fraction. The numerator of the original fraction is

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let the fraction be \frac{3 k}{4 k} \\ ∴ \frac{(3 k - 9)}{4 k} = (\frac{2}{3}) \times (\frac{3 k}{4 k}) = \overset{⌢}{\to} A \frac{1}{2} \\ ∴ 6 k - 18 = 4 k \\ ∴ k = 9 \\ \Rightarrow Numerator = 3 k = 27 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{3k}}{{4k}} \cr & \therefore \frac{{\left( {3k - 9} \right)}}{{4k}} = \left( {\frac{2}{3}} \right) \times \left( {\frac{{3k}}{{4k}}} \right) = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}\to {A} \frac{1}{2} \cr & \therefore 6k - 18 = 4k \cr & \therefore k = 9 \cr & \Rightarrow {\text{Numerator}} = 3k = 27 \cr}$

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of the original fraction. The numerator of the original fraction is", "text": "The numerator and denominator of a fraction are in the ratio 3 : 4. If 9 is subtracted from the numerator, the resulting fraction in

\frac{2}{3}

$\frac{2}{3}$ of the original fraction. The numerator of the original fraction is", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Let the fraction be \frac{3 k}{4 k} \\ ∴ \frac{(3 k - 9)}{4 k} = (\frac{2}{3}) \times (\frac{3 k}{4 k}) = \overset{⌢}{\to} A \frac{1}{2} \\ ∴ 6 k - 18 = 4 k \\ ∴ k = 9 \\ \Rightarrow Numerator = 3 k = 27 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{3k}}{{4k}} \cr & \therefore \frac{{\left( {3k - 9} \right)}}{{4k}} = \left( {\frac{2}{3}} \right) \times \left( {\frac{{3k}}{{4k}}} \right) = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}\to {A} \frac{1}{2} \cr & \therefore 6k - 18 = 4k \cr & \therefore k = 9 \cr & \Rightarrow {\text{Numerator}} = 3k = 27 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Let the fraction be \frac{3 k}{4 k} \\ ∴ \frac{(3 k - 9)}{4 k} = (\frac{2}{3}) \times (\frac{3 k}{4 k}) = \overset{⌢}{\to} A \frac{1}{2} \\ ∴ 6 k - 18 = 4 k \\ ∴ k = 9 \\ \Rightarrow Numerator = 3 k = 27 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{3k}}{{4k}} \cr & \therefore \frac{{\left( {3k - 9} \right)}}{{4k}} = \left( {\frac{2}{3}} \right) \times \left( {\frac{{3k}}{{4k}}} \right) = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}\to {A} \frac{1}{2} \cr & \therefore 6k - 18 = 4k \cr & \therefore k = 9 \cr & \Rightarrow {\text{Numerator}} = 3k = 27 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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