Question

The numerator of a fraction is 3 less than the denominator. If 4 is added to both the numerator and denominator, the value of fraction increases by 1/8 , find the fraction.​

Answer 1

${\purple{\underline{\underline{\large{\mathtt{Answer:}}}}}}$

Given:

• We have been given that the numerator of a fraction is 3 less than the denominator.
• When 4 is added to both the numerator and denominator, the value of fraction increases by 1/8.

To Find:

• We need to find the fraction.

Solution:

Let the denominator be x.

∴ Numerator will be (x - 3)

Now, according to the question, we have

$\longrightarrow\sf{ \dfrac{x - 1}{x + 4} - \dfrac{x - 3}{x} = \dfrac{1}{8}}$

$\longrightarrow\sf{ \dfrac{x(x + 1) - (x - 3)(x - 4)}{x(x - 4)} = \dfrac{1}{8} }$

$\longrightarrow\sf{8x(x + 1) = 8(x - 3)(x + 4) + x(x + 4)}$

$\longrightarrow\sf{ {8x}^{2} + 8x = 8( {x}^{2} + x - 12) + {x}^{2} + 4x}$

$\longrightarrow\sf{8 {x}^{2} - 8 {x}^{2} - {x}^{2} + 8x - 8x - 4x + 96 = 0}$

$\longrightarrow\sf{ - {x}^{2} - 4x + 96 = 0}$

$\longrightarrow\sf{ {x}^{2} + 4x - 96 = 0}$

Using the splitting the middle term, we have

$\longrightarrow\sf{ {x}^{2} + 12x - 4x - 96}$

$\longrightarrow\sf{x(x + 12) - 8(x + 12)}$

$\longrightarrow\sf{(x + 12)(x - 8)}$

Now, Either (x + 12) = 0 or (x - 8) = 0.

When (x + 12) = 0

$\mapsto\sf{x + 12 = 0}$

$\mapsto\sf{x = - 12}$

When (x - 8) = 0

$\mapsto\sf{x - 8 = 0}$

$\mapsto\sf{x = 8}$

But, x = -12 is rejected.

∴ Fraction = $\sf{\dfrac{8 - 3}{5} = \dfrac{5}{8}}$

Hence, the required fraction is 5/8.

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• Given that numerator of a fraction is 3 less then the denominator
• When 4 is added to both numerator and denominator the value of fraction increases by 1/8

To Find:

• We have to find the given fraction

Solution:

Let the denominator of fraction = x

Numerator of fraction = ( x - 3 )

$\boxed{\sf{\pink{Original \: Fraction = \dfrac{x-3}{x}}}}$

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$\bigstar \: \: \underline{\large\mathfrak\orange{According \: to \: the \: Question:}}$

When 4 is added to both numerator and denominator the values of fraction gets increased by 1/8

$\hookrightarrow \sf{\dfrac{(x-3)+4}{x+4}=\dfrac{x-3}{x} + \dfrac{1}{8}}$

$\hookrightarrow \sf{\dfrac{x-3+4}{x+4} - \dfrac{x-3}{x}= \dfrac{1}{8}}$

$\hookrightarrow \sf{\dfrac{x-1}{x+4} - \dfrac{x-3}{x}= \dfrac{1}{8}}$

Taking LCM

$\hookrightarrow \sf{\dfrac{x(x-1) - (x-3)(x+4)}{x(x+4)}= \dfrac{1}{8}}$

Solving the Equation

$\hookrightarrow \sf{\dfrac{(x^2 - x) - (x^2 - x - 12)}{x(x+4)}= \dfrac{1}{8}}$

$\hookrightarrow \sf{\dfrac{x^2 - x - x^2 + x + 12}{x^2 +4x}= \dfrac{1}{8}}$

$\hookrightarrow \sf{\dfrac{12}{x^2+4x}= \dfrac{1}{8}}$

Cross Multiplying the Terms

$\hookrightarrow \sf{96 = x^2 + 4x }$

$\hookrightarrow \sf{ x^2 + 4x - 96 = 0}$

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Solving the Equation using middle term splitting method

$\hookrightarrow \sf{x^2 + 12x - 8x - 96 = 0}$

$\hookrightarrow \sf{x(x+ 12)- 8(x + 12) = 0}$

$\hookrightarrow \sf{(x-8)(x+12) = 0}$

Either

$\implies \sf{ x - 8 = 0}$

$\implies \sf{ x = 8}$

Or

$\implies \sf{ x + 12 = 0}$

$\implies \sf{ x = (-12)}$

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When x = 8

$\implies \sf{Fraction = \dfrac{8-3}{8}}$

$\implies \sf{\dfrac{5}{8} }$

When x = - 12

$\implies \sf{ Fraction = \dfrac{-12-3}{-12}}$

$\implies \sf{ \dfrac{-15}{-12}}$

$\implies \sf{ \dfrac{5}{4}}$

But in case of x = -12 numerator is bigger than denominator

Hence x = - 12 gets rejected

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$\underline{\huge\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}} }$

$\boxed{\sf{\red{Original \: Fraction = \dfrac{5}{8}}}}$

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