Math, asked by narasegowdavb, 2 months ago

a fraction becomes 9/11 if 2 is added to both the numerator and denominator become 5 /6​

Answers

Answered by chandbhalodia2005
2

Answer:

I hope my answer will help you.

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Answered by varadad25
6

Correct Question:

A fraction becomes \displaystyle{\sf\:\dfrac{9}{11}} if 2 is added to both the numerator and denominator. If 3 is added to both the numerator and denominator, it becomes \displaystyle{\sf\:\dfrac{5}{6}}. Find the fraction.

Answer:

\displaystyle{\underline{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{7}{9}}}}}

Step-by-step-explanation:

Let the numerator of the fraction be x.

And the denominator of the fraction be y.

\displaystyle{\sf\:The\:fraction\:=\:\dfrac{x}{y}}

From the first condition,

\displaystyle{\sf\:\dfrac{x\:+\:2}{y\:+\:2}\:=\:\dfrac{9}{11}}

\displaystyle{\implies\sf\:11\:\times\:(\:x\:+\:2\:)\:=\:9\:\times\:(\:y\:+\:2\:)}

\displaystyle{\implies\sf\:11x\:+\:22\:=\:9y\:+\:18}

\displaystyle{\implies\sf\:11x\:-\:9y\:=\:18\:-\:22}

\displaystyle{\implies\sf\:11x\:-\:9y\:=\:-\:4\:\:\:-\:-\:-\:(\:1\:)}

From the second condition,

\displaystyle{\sf\:\dfrac{x\:+\:3}{y\:+\:3}\:=\:\dfrac{5}{6}}

\displaystyle{\implies\sf\:6\:\times\:(\:x\:+\:3\:)\:=\:5\:\times\:(\:y\:+\:3\:)}

\displaystyle{\implies\sf\:6x\:+\:18\:=\:5y\:+\:15}

\displaystyle{\implies\sf\:6x\:-\:5y\:=\:15\:-\:18}

\displaystyle{\implies\sf\:6x\:-\:5y\:=\:-\:3\:\:\:-\:-\:-\:(\:2\:)}

Multiplying equation ( 1 ) by 5 and equation ( 2 ) by 9, we get,

\displaystyle{\sf\:11x\:-\:9y\:=\:-\:4\:\:\:-\:-\:-\:(\:1\:)}

\displaystyle{\implies\sf\:5\:\times\:(\:11x\:-\:9y\:)\:=\:-\:4\:\times\:5}

\displaystyle{\implies\sf\:55x\:-\:45y\:=\:-\:20\:\:\:-\:-\:(\:3\:)}

\displaystyle{\sf\:6x\:-\:5y\:=\:-\:3\:\:\:-\:-\:-\:(\:2\:)}

\displaystyle{\implies\sf\:9\:\times\:(\:6x\:-\:5y\:)\:=\:-\:3\:\times\:9}

\displaystyle{\implies\sf\:54x\:-\:45y\:=\:-\:27\:\:\:-\:-\:-\:(\:4\:)}

By subtracting equation ( 4 ) from equation ( 3 ), we get,

\displaystyle{\sf\:55x\:-\:45y\:-\:(\:54x\:-\:45y\:)\:=\:-\:20\:-\:(\:-\:27\:)}

\displaystyle{\implies\sf\:55x\:-\:\cancel{45y}\:-\:54x\:+\:\cancel{45y}\:=\:-\:20\:+\:27}

\displaystyle{\implies\sf\:55x\:-\:54x\:=\:7}

\displaystyle{\implies\boxed{\red{\sf\:x\:=\:7}}}

By substituting x = 7 in equation ( 2 ), we get,

\displaystyle{\sf\:6x\:-\:5y\:=\:-\:3\:\:\:-\:-\:-\:(\:2\:)}

\displaystyle{\implies\sf\:6\:\times\:7\:-\:5y\:=\:-\:3}

\displaystyle{\implies\sf\:42\:-\:5y\:=\:-\:3}

\displaystyle{\implies\sf\:-\:5y\:=\:-\:3\:-\:42}

\displaystyle{\implies\sf\:\cancel{-}\:5y\:=\:\cancel{-}\:45}

\displaystyle{\implies\sf\:5y\:=\:45}

\displaystyle{\implies\sf\:y\:=\:\cancel{\dfrac{45}{5}}}

\displaystyle{\implies\boxed{\red{\sf\:y\:=\:9}}}

\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{7}{9}}}}}

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