Question

28. The numberator of a fracation is 6 lessthan the denomirator. If 3 is added to the numberator. The
fracation is equal to Find the original fracation.​

Answer 1

Correct Question -

• The numberator of a fracation is 6 lessthan the denominator. If 3 is added to the numberator, the fracation is equal to 2/3. Find the original fracation.

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Solution -

• Let the denominator of the fraction be x.

Now, according to the given condition

⇢ Numerator = x - 6

Therefore, the required fraction would be

• $\sf{Required\: fraction = \dfrac{x - 6}{x}}$

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It is given that, when 3 is added to the fraction it becomes 2/3.

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According to the question

$\tt\dashrightarrow{\dfrac{(x - 6) + 3}{x} = \dfrac{2}{3}}$

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$\tt\dashrightarrow{\dfrac{x - 6 + 3}{x} = \dfrac{2}{3} }$

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$\tt\dashrightarrow{\dfrac{(x - 3)}{x} = \dfrac{2}{3}}$

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Cross multiplying

$\tt\dashrightarrow{3(x - 3) = 2(x)}$

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$\tt\dashrightarrow{3x - 9 = 2x}$

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$\tt\dashrightarrow{3x - 2x = 9}$

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$\bf\dashrightarrow{x = 9}$

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We get,

• x = 9

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Therefore,

$\tt\longrightarrow{Required\: fraction = \dfrac{9- 6}{9}}$

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$\tt\longrightarrow{Required \: fraction = \dfrac{3}{9}}$

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$\bf\longrightarrow{Required\: fraction = \dfrac{1}{3}}$

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$\: \: \: \underline{\sf{Thus,\: the\: required\: fraction\: is\: \dfrac{1}{3}}}$