Question

A fraction becomes 9/ 11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denomerator, it becomes 5/6. Find the fraction.

Answer 1

$\bold{\bf{\red{\bigstar}}}$ AnsweR :

$\star$ Let the numerator and denominator be x and y .

$\star$ A/c " A fraction becomes 9/ 11, if 2 is added to both the numerator and the denominator "

$\implies \bold{\frac{x+2}{y+2}=\frac{9}{11} }\\\\\implies \bold{11(x+2)=9(y+2)}\\\\\implies \bold{11x+22=9y+18}\\\\\implies \bold{9y-11x=4...(1)}$

$\star$ A/c " If 3 is added to both the numerator and the denominator, it becomes 5/6 "

$\implies\bold{ \frac{x+3}{y+3} =\frac{5}{6} }\\\\\implies \bold{6(x+3)=5(y+3)}\\\\\implies \bold{6x+18=5y+15}\\\\\implies \bold{5y-6x=3...(2)}$

$\star$ Now solving 6*(1) - 11*(2) , we get ,

$\implies \bold{6(9y-11x)-11(5y-6x)=6(4)-11(3)}\\\\\implies \bold{54y-66x-55y+66x=24-33}\\\\\implies \bold{-y=-9}\\\\\implies \bold{y=9}$

$\star$ Sub. y value in (2) , we get ,

$\implies \bold{6x=5(9)-3}\\\\\implies \bold{6x=45-3}\\\\\implies \bold{6x=42}\\\\\implies \bold{x=7}$

$\star$ Hence the required fraction is   $\bold{\frac{x}{y} =\frac{7}{9} }$

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Answer 2

Answer:

Let the numerator and denominator be x and y respectively.

Therefore, the required fraction will be x/y

According to Question now,

➳ (x + 2)/(y + 2) = 9/11

➳ 11(x + 2) = 9(y + 2)

➳ 11x + 22 = 9y + 18

➳ 22 - 18 = 9y - 11x

➳ 4 = 9y - 11x

➳ 11x = 9y - 4

➳ x = 9y - 4/11.......[Equation (i)]

Now, it is given that 3 is added to both numerator and denominator it becomes 5/6 :]

➳ (x + 3)/(y + 3) = 5/6

➳ (x + 3)6 = 5(y + 3)

➳ 6x + 18 = 5y + 15

➳ 6x - 5y + 18 - 15 = 0

➳ 6x - 5y + 3 = 0

➳ 6(9y - 4)/11 - 5y + 3 = 0

➳ 54y - 24/11 = 5y - 3

➳ 54y - 24 = 11(5y - 3)

➳ 54y - 24 = 55y - 33

➳ -24 + 33 = 55y - 54y

➳ 9 = y

Now, Putting y = 9 in equation (i) we get,

➳ x = 9y - 4/11

➳ x = 9(9) - 4/11

➳ x = 81 - 4/11

➳ x = 77/11

➳ x = 7

Therefore, the required fraction is 7/9.