Question

the denominator of a rational number is greater than the numerator by 5. if 1 is subtracted from the numerator and 3 is added ro the denominator,the new number become¼. find the original number.​

Answer 1

$\sf Given \begin{cases} & \sf{Denominator = \bf{x + Numerator}} \\ \\ & \sf{\dfrac{Numerator - 1}{Denominator + 3} = \bf{ \dfrac{1}{4}}} \end{cases}$

To find: The original number.

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☯ Let Numerator and Denominator of a rational number be x and y.

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Given that,

• The denominator of a rational number is greater than the numerator by 5.

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Therefore,

$:\implies\sf y = x + 5\qquad\qquad\bigg\lgroup\bf eq\: (1)\bigg\rgroup$

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• If 1 is subtracted from the numerator and 3 is added to the denominator, the new number become 1/4.

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$:\implies\sf \dfrac{x - 1}{y + 3} = \dfrac{1}{4}\\ \\ \\ :\implies\sf \dfrac{x - 1}{(x + 5) + 3} = \dfrac{1}{4}\\ \\ \\ :\implies\sf \dfrac{x - 1}{x + 8} = \dfrac{1}{4}\\ \\ \\ :\implies\sf (x + 8) = 4(x - 1)\\ \\ \\ :\implies\sf (x + 8) = 4x - 4\\ \\ \\ :\implies\sf 8 + 4 = 4x - x\\ \\ \\ :\implies\sf 12 = 3x\\ \\ \\ :\implies\sf x = \cancel{\dfrac{12}{3}}\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{x = 4}}}}}\;\bigstar$

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Therefore,

• $\sf Numerator\:of\:rational\:number\:,x = \bf{4}$
• $\sf Denominator\:of\:rational\:number\:,(x + 5) = 4 + 5 = \bf{9}$

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$\therefore\:{\underline{\sf{The\: original\:number\;is\: \bf{ \dfrac{4}{9}}.}}}$

Answer 2

Answer:

Given :-

• The denominator of a rational number is greater than the numerator by 5. If 1 is subtracted from the numerator and 3 is added to the denominator then the new number becomes ¼.

To Find :-

• What is the original number.

Solution :-

Let, the numerator be x

And, the denominator will be x + 5

Hence, the fraction will be $\sf\dfrac{x}{x + 5}$

According to the question,

⇒ $\sf\dfrac{x - 1}{x + 5 + 3} =\: \dfrac{1}{4}$

⇒ $\sf\dfrac{x - 1}{x + 8} =\: \dfrac{1}{4}$

By doing cross multiplication we get,

⇒ 4(x - 1) = x + 8

⇒ 4x - 4 = x + 8

⇒ 4x - x = 8 + 4

⇒ 3x = 12

⇒ x = $\sf\dfrac{\cancel{12}}{\cancel{3}}$

➠ x = 4

Hence, the required original fraction will be,

↦ $\sf\dfrac{x}{x + 5}$

↦ $\sf\dfrac{4}{4 + 5}$

➦ $\sf\dfrac{4}{9}$

$\therefore$ The original fraction is $\boxed{\bold{\large{\dfrac{4}{9}}}}$ .