Question

The numerator of a fraction is 6 less than the denominator. If3 is added to the numerator.
the fraction is equal to 23. What is the original fraction equal to?​

Answer 1

Answer:

$\frac{20}{26}$

Solution:-

Let the numerator and denominator be x and x respectively

$therefore \: fraction = \frac{x}{x}$

Now according to the question

The numerator of fraction is 6 less than its denominator

$i.e \: \frac{x - 6}{x}$

And 3 added to the numerator the fraction is equal to 23

$\frac{x - 6 + 3}{x} = 23$

$(x - 6 + 3)= 23x$

$x - 3 = 23x$

$x = 26$

The original fraction

$\frac{(x - 6)}{x}$

$\frac{26 - 6}{26} = \frac{20}{26}$

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Therefore the original fraction is equal to 20/26

Answer 2

• Let numberator be N and denominator be D.

» The numerator of a fraction is 6 less than the denominator.

=> N = D - 6 ______ (eq 1)

» If 3 is added to the numerator.

the fraction is equal to 23.

=> $\dfrac{N\:+\:3}{D}$ = 23

=> N + 3 = 23D

=> D - 6 + 3 = 23D [From (eq 1)]

=> D - 3 = 23D

=> - 3 = 22D

=> D = $\dfrac{-3}{22}$

Put value of D in (eq 1)

=> N = $\dfrac{-3}{22}$ - 6

=> N = $\dfrac{-\:3\:-\:132}{22}$

=> N = $\dfrac{-\:135}{22}$

Now..

$\dfrac{N}{D}$ = $\dfrac{ \frac{-135}{22} }{\frac{-3}{22} }$

=> $\dfrac{135}{3}$

=> 45

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The original number of fraction equal to 45.

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