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Question

# A fraction becomes $\frac{1}{3}$ when 2 is subtracted from the numerator and it becomes $\frac{1}{2}$ when 1 is subtracted from the denominator. Find the fraction.

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Solution

## Let the numerator of the fraction be x and the denominator be y. According to the question, $\frac{x-2}{y}=\frac{1}{3}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\frac{x}{y-1}=\frac{1}{2}...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}3\left(x-2\right)=y\phantom{\rule{0ex}{0ex}}⇒3x-6=y...\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{x}{3x-6-1}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒2\left(x\right)=3x-6-1\phantom{\rule{0ex}{0ex}}⇒2x=3x-7\phantom{\rule{0ex}{0ex}}⇒2x-3x=-7\phantom{\rule{0ex}{0ex}}⇒-x=-7\phantom{\rule{0ex}{0ex}}⇒x=7...\left(\mathrm{iv}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(\mathrm{iii}\right)\mathrm{and}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}x=7\mathrm{and}y=15\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{fraction}\mathrm{is}\frac{7}{15}.$

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