Question

the numerator of a fraction is 4 less than the denominator.if 1 is added to both its numerator and denominator,it becomes 1/2.find the fraction.​

Answer 1

Answer:

let the denominator of the fraction is x

then neumerator is x-4

according to question,

(x-4+1) /(x+1)=1/2

=> (x-3)/(x+1)=1/2

=> 2(x-3)=x+1

=> 2x-6=x+1

=> 2x-x=1+6

=> x=7

so,denominator is 7

numerator =7-4=3

the fraction is 3/7

Answer 2

❍ Let's Consider that, the numerator of the fraction be (x – 4) & denominator of the fraction be x respectively.

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$\underline{\bigstar\boldsymbol{According\;to\;the\;Question\;:}}$

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• If one is added to both the numerator & denominator, the fraction becomes ½.

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Therefore,

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$\begin{gathered}:\implies\sf{\bigg\{\dfrac{x-4+1}{x+1}\bigg\}=\dfrac{1}{2}}\\\\\\:\implies\sf{2(x-3)=x+1}\\\\\\:\implies\sf{2x-6=x+1}\\\\\\:\implies\sf{2x-x=1+6}\\\\\\:\implies\underline{\boxed{\frak{\pink{x=7}}}}\;\bigstar\end{gathered}$

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Hence,

• Numerator = (x – 4) = (7 – 4) = 3
• Denominator = x = 7

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$\therefore\;{\underline{\sf{Hence,\;the\;required\;fraction\;is\;{\bf{\dfrac{3}{7}}}}.}}$

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V E R I F I C A T I O N :

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• It is given that, when one is added to both the numerator and denominator of the fraction, it becomes ½. So, let's verify —

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$\twoheadrightarrow\sf{\dfrac{3+1}{7+1}=\dfrac{1}{2}}\\\\\\\twoheadrightarrow\sf{\cancel{\dfrac{4}{8}}=\dfrac{1}{2}}\\\\\\\twoheadrightarrow\underline{\boxed{\frak{\dfrac{1}{2}=\dfrac{1}{2}}}}$

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⠀⠀⠀⠀$\;\;\;\;\;\qquad\therefore\;{\underline{\textsf{\textbf{Hence,\;Verified}}.}}$⠀⠀⠀