Question

A fraction become 9/11 if 2 is added to both numerator and denominator . If 3 is aaded to both numerator and denom
inator it becomes 5/6 find fraction

Answer 1

Answer:

7/9

Step-by-step explanation:

let the fraction be x/y

x+2/y+2 = 9/11

=> 11x + 22 = 9y + 18

=> 11x - 9y = -4 ___(i)

x+3/y+3 = 5/6

=> 6x + 18 = 5y + 15

=> 6x - 5y = -3 ___(ii)

(i) ×6 -> 66x - 54y = -24

(ii)×11->66x - 55y = -33

(-) (+) (+)

----------------------

y = 9

6x - 5y = -3

6x - 45 = -3

6x = 42

x = 7

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.