Question

A fraction becomes 9 11 , if two is added to both the numerator and denominator . If 3 is added to both the numerator and denominator it becomes 5 6 . Find the fraction(

Answer 1

Given

A fraction becomes 9 / 11 , if two is added to both the numerator and denominator.

$\implies \dfrac{x+2}{y+2}=\dfrac{9}{11}$

$\implies 11(x+2)=9(y+2)$

$\implies 11x+22=9y+18$

$\implies 11x-9y=18-22$

$\implies 11x-9y=-4--(1)$

If 3 is added to both the numerator and denominator, it becomes 5 / 6 .

$\implies \dfrac{x+3}{y+3} =\dfrac{5}{6}$

$\implies 6(x+3)=5(y+3)$

$\implies 6x+18=5y+15$

$\implies 6x-5y=15-18$

$\implies 6x-5y=-3--(2)$

Solving (1) and (2) we get,

$66x-55y-33\\\underline{66x-54y=-24}\\\underline{\underline{-y=-9}}\\\implies y = 9$

Substituting y = 9 in the second equation, we get,

$6x-45=-3\\\implies 6x=-3+45\\\implies 6x=42\\\implies x = 7$

$\boxed{\boxed{\bold{Therefore, \ the \ required \ fraction \ is \ \frac{7}{9} }}}}}}}}}}}$

                                                       

Answer 2

Let , the fraction be x/y

First condition

A fraction becomes 9/11 , if two is added to both the numerator and denominator

Thus ,

$\Rightarrow \sf \frac{x + 2}{y + 2} = \frac{9}{11} \\ \\\Rightarrow \sf 11x + 22 = 9y + 18 \\ \\\Rightarrow \sf 11x - 9y = - 4 \: --- (i) \:$

Second Condition

A fraction becomes 5/6 , if three is added to both the numerator and denominator

Thus ,

$\Rightarrow \sf \frac{x + 3}{y + 3} = \frac{5}{6} \\ \\\Rightarrow \sf 6x + 18 = 5y + 15 \\ \\\Rightarrow \sf 6x - 5y = - 3 \: --- \: (ii)$

Solving eq (i) and (ii) , we get

$\Rightarrow \sf y = 9$

Substitute the value of y = 9 in eq (i) , we get

$\Rightarrow \sf 11x - 9 \times 9 = - 4 \\ \\\Rightarrow \sf 11x - 81 = - 4 \\ \\ \Rightarrow \sf 11x = 77 \\ \\\Rightarrow \sf x = 7$

$\therefore \bold{ \underline{The \: fraction \: is \: \frac{7}{9} }}$