Question

If 2 is added to both numerator and denominator of a fraction it becomes 9/11 .area of the rectangle having length and breadth equal to denominator and numerator is 320 sq units.Then find the fraction

Answer 1

Hey dear,

● Answer-
16/20

● Explaination-
Let the denominator and numerator be x and y respectively.

We have,
(x+2)/(y+2) = 9/11
11x + 22 = 9y + 18
y = (11x+4)/9

We have area of rectangle
A = xy = 320
x(11x+4)/9 = 320
11x^2 + 4x - 2880 = 0

Solving quadratic eqn,
x = 16 units
y = (11x+4)/9
y = (11×16+4)/9
y = 20 units

The fraction is 16/20 i.e.4/5.

Hope this is useful...

Answer 2

Solution:

Let the fraction is

$\frac{x}{y}$
Now on adding 2 in numerator and denominator

$\frac{x + 2}{y + 2} = \frac{9}{11} \\ \\ 11x + 22 = 9y + 18 \\ \\ 11x - 9y = - 4 \: \: eq1$
area of the rectangle having length and breadth equal to denominator and numerator is 320 sq units

$xy = 320 \\ \\ y = \frac{320}{x} \: \: ..eq2$
put the value of y from eq2 into eq1

$11x - 9( \frac{320}{x} ) = - 4 \\ \\ 11 {x}^{2} + 4x - 2880 = 0 \\ \\ x_{1,2} = \frac{ - 4 ± \sqrt{16 +126720 } }{22} \\ \\ x_{1,2} = \frac{ - 4 ± 356}{22} \\ \\ x_{1} = \frac{352}{22} \\ \\ x_{1} = 16 \\ \\ or \\ \\ x_{2}= \frac{ - 360}{22} \\ \\ x_{2}= - 16$

you can neglect - value,since length or breadth can not be negative

Now
$y = \frac{320}{16} \\ \\ y = 20$
So,the fraction is

$\frac{16}{20} \\ \\ or \\ = \frac{18}{22} = \frac{9}{11}$
Hope it helps you