Question

the numerator of the given fraction is 3 lesser than its denominator. If the sum of given fraction and the fraction obtained by adding 2 to its numerator and the denominator is 29÷30 find the given fraction
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Answer 1

Correct Question :

The numerator of the given fraction is 3 lesser than its denominator. If the sum of given fraction and the fraction obtained by adding 2 to its numerator and the denominator is 29/ 20. Find the Fraction.

AnswEr :

$\bold{Let}\begin{cases}\sf{Numerator=(n - 3)} \\ \sf{Denominator=n}\\ \sf{Original \:Fraction= \dfrac{n - 3}{n} }\end{cases}$

⋆ After Adding 2 to Both Numerator and Denominator, Fraction will be :

$\bold{Now}\begin{cases}\sf{Numerator=(n - 3) + 2 =(n - 1) }\\ \sf{Denominator=(n +2)}\\ \sf{New\:Fraction= \dfrac{(n - 1)}{(n + 2)} }\end{cases}$

• According to the Question Now :

$\implies \displaystyle \tt Original \:Fraction + New \:Fraction= \frac{29}{20} \\ \\ \implies \displaystyle \tt \frac{(n - 3)}{n} + \frac{(n - 1)}{(n + 2)} = \frac{29}{20} \\ \\ \implies \displaystyle \tt \frac{(n - 3)(n + 2) + n(n - 1)}{n(n + 2)} = \frac{29}{20} \\ \\ \implies \displaystyle \tt \frac{ {n}^{2} + 2n - 3n - 6 + {n}^{2} - n}{ {n}^{2} + 2n} = \frac{29}{20} \\ \\\implies \tt20( 2{n}^{2} - 2n - 6) = 29( {n}^{2} + 2n) \\ \\ \implies \tt40 {n}^{2} - 40n - 120 = 29 {n}^{2} + 58n \\ \\ \implies \tt40 {n}^{2} - 29 {n}^{2} - 40n - 58n - 120 = 0 \\ \\ \implies \tt11 {n}^{2} - 98n - 120 = 0 \\ \\ \implies \tt11 {n}^{2} - 110n + 12n - 120 = 0 \\ \\ \implies \tt11n(n - 10) + 12(n - 10) = 0 \\ \\ \implies \tt(11n + 12) = 0 \: \: or \: \: (n - 10) = 0 \\ \\ \implies \tt \red{n = \dfrac{ - 12}{11}} \: \: or \: \: \green{n = 10}$

$\rule{300}{1}$

we will neglect Negative Value, and take Value of n as 10.

◗ NUMERATOR = (n - 3) = (10 - 3) = 7

◗ DENOMINATOR = n = 10

⠀

$\therefore \Large\boxed{ \tt Original\:Fraction = \frac{7}{10} }$

Answer 2

Answer:

Let the denominator be x

            Numerator  is x-3

The fraction is (x-3)/x

when 2 is added to numerator and denominator

we have fraction as (x-1)/(x+2)

• ATQ

   ⇒    (x-3)/x +(x-1)/(x+2)= 29/20

   ⇒    (x-3)((x+2) +x(x-1)=x(x+2)*29/20

   ⇒    (x²-6-x + x²-x)20= 29x²+58x

   ⇒    40x²-120-40x=29x²+58x

   ⇒    11x²-98x- 120=0

   ⇒    11x² - 110x +12x -120=0

   ⇒     (11x+12)(x-10)=0

Canceling the negative fraction value  of x we have x = 10

Then the fraction (x-3)/x = 7/10