Question

● A fraction become
$\frac{9}{11}$
if 2 is added to both numerator and denominator but when 3 is added to both numerator and denominator it becomes
$\frac{5}{6}$
Find the fraction.
(answer it by doing roughly to your copy and with full process!)
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Answer 1

Answer :

The original fraction is 27/33

Explanation:

Let's translate in math expressions:

1) x/y=9/11

2) x/y+3/3=56

The expression 1) is equivalent to x=9/11y

then we will substitute in expression 2) and get:

9/11y+3/y+3=5/6

that is equivalent to the following:

(9/11y+3)⋅6=5(y+3)

Let's expand and get

54/11y+18=5y+15

The let's solve the equation and get y:

54/11y−5y=15−18

54−55/11y=−3

−1/11y=−3

y=33

then we will find x

x=9/11y= 9⋅33/11=27

The fraction is 27/33

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.