Test Index

SSC CHSL Model Paper 3

© examsnet.com

Question : 38 of 100

Marks: +1, -0

If

\frac{3}{x+y}+\frac{2}{x-y}=2

\frac{9}{x+y}-\frac{4}{x-y}=1,

then what is the value of

\frac{x}{y}

?

$\frac{3}{2}$
5
$\frac{2}{3}$
$\frac{1}{5}$

Solution:

Given,

\frac{3}{x+y}+\frac{2}{x-y}=2

...(i)

and

\frac{9}{x+y}-\frac{4}{x-y}=1

....(ii)

Let

\frac{1}{x+y}=a

and

\frac{1}{x-y}=b

∴

3a + 2b = 2

....(iii)

9a - 4b = 1

... (iv)

On multiplyin g equation ( iii) by 2 and addition of equation (iv) and new one, then we get

6a + 4b = 4

9a - 4b = 1

\text{----------}

15a = 5

⇒

a = \frac{5}{15} = \frac{1}{3}

∴

\frac{1}{x+y}=\frac{1}{3}

⇒

x + y = 3

...(v)

On putting the value of (a) in equation (iii), we get

3 \times \frac{1}{3} + 2b = 2

⇒

2b = 2 - 1 = 1

⇒

b = \frac{1}{2} \Rightarrow \frac{1}{x-y} = \frac{1}{2}

⇒

x - y = 2

....(vi)

∴

x + y = 3

\;\;\;x - y = 2

\text{----------}

2x = 5

⇒

x = \frac{5}{2}

From equation (v),

y = 3 - \frac{5}{2} = \frac{1}{2}

\therefore \frac{x}{y} = \frac{\frac{5}{2}}{\frac{1}{2}} = 5

© examsnet.com

Go to Question:

Prev Question

Next Question