Math, asked by saimnoushad, 11 months ago

a fraction becomes 9 by 11 if 2 is added to both the numerator and the denominator if 3 is added to both the numerator and denominator it becomes 5 by 6 find the fraction​

Answers

Answered by Anonymous
29

Answer :-

Fraction is 7/9.

Explanation

Let the fraction be x/y

If 2 is added to both numerator and denominator fraction becomes 9/11

⇒ (x + 2)/(y + 2) = 9/11

⇒ 11(x + 2) = 9(y + 2)

⇒ 11x + 22 = 9y + 18

⇒ 11x - 9y = 18 - 22

⇒ 11x - 9y = - 4 ---eq(1)

If 3 is added to both numerator and denominator fraction becomes 5/6

⇒ (x + 3)/(y + 3) = 5/6

⇒ 6(x + 3) = 5(y + 3)

⇒ 6x + 18 = 5y + 15

⇒ 6x - 5y = 15 - 18

⇒ 6x - 5y = - 3

⇒ 6x = - 3 + 5y

⇒ x = (-3 + 5y)/6

Substitute x = (-3 + 5y)/6 in eq(1)

 \sf 11x - 9y =  - 4 \\  \\ \\  \sf 11 \bigg( \dfrac{ - 3 + 5y}{6} \bigg) - 9y =  - 4 \\  \\  \\  \sf  \dfrac{11( - 3 + 5y)}{6} - 9y =  - 4 \\  \\  \\  \sf  \dfrac{ - 33 + 55y}{6}  - 9y =  - 4 \\  \\  \\  \sf  \dfrac{ - 33 + 55y - 54y}{6}  =  - 4 \\  \\  \\  \sf  - 33 + y =  - 4(6) \\  \\  \\  \sf  - 33 + y =  - 24 \\  \\  \\  \sf y =  - 24 + 33 \\  \\  \\  \sf y = 9

Substitute y = 9 in x = (-3 + 5y)/6

⇒ x = (-3 + 5y)/6

⇒ x ={-3 + 5(9)}/6

⇒ x = (-3 + 45)/6

⇒ x = 42/6

⇒ x = 7

Therefore fraction = x/y = 7/9.

Answered by Anonymous
58

Solution:

Let Numerator of the fraction be x.

And denominator of the fraction be y.

\sf{So,\;fraction=\dfrac{x}{y}}

According to question,

Case-1

When 2 is added to both numerator and denominator, then

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

=> 11x + 22 = 9y + 18

=> 11x = 9y - 4           ............(1)

Case-2

When 3 is added to both numerator and denominator, then

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

=> 6x + 18 = 5y + 15

=> 6x = 5y - 3

\sf{\implies x = \dfrac{5y-3}{6}\;\;\;\;.......(2)}

Put the value of x in equation (1), we get

\sf{\implies 11\bigg(\dfrac{5y-3}{6}\bigg)=9y-4}

=> 11(5y - 3) = 54y - 24

=> 55y - 33 = 54y - 24

=> y = 9

Put the value of y in equation 2, we get

\sf{\implies x = \dfrac{5y-3}{6}}

\sf{\implies \dfrac{5\times 9-3}{6}=\dfrac{42}{6}=7}

=> x = 7

\sf{\implies So,\;required\;fraction=\dfrac{7}{9}}

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