RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

(b) Given,
$\frac{3}{\mathit{x}+\mathit{y}}+\frac{2}{\mathit{x}\mathit{–}\mathit{y}}=2$     ... (i)
and$\frac{9}{\mathit{x}+\mathit{y}}-\frac{4}{\mathit{x}\mathit{–}\mathit{y}}=1$     ... (ii)
Let$\frac{1}{\mathit{x}+\mathit{y}}=\mathit{a}$ and
$\frac{1}{\mathit{x}\mathit{–}\mathit{y}}=\mathit{b}$
∴ 3a + 2b = 2   ... (iii)
9a – 4b = 1   ... (iv)
On multiplying equation (iii) by 2 and addition of equation (iv) and new one, then we get
6a+ 4b= 419a – 4b= 1
_________
15a  = 5
$\Rightarrow \mathit{a}=\frac{5}{15}=\frac{1}{3}$
$\Rightarrow \frac{1}{\mathit{x}+\mathit{y}}=3$   ... (v)
On putting the value of (a) in equation (iii),we get
$3\times \frac{1}{3}+2\mathit{b}=2$
⇒ 2b = 2 – 1 = 1
$\Rightarrow \mathit{b}=\frac{1}{2}\Rightarrow \frac{1}{\mathit{x}\mathit{–}\mathit{y}}=\frac{1}{2}$
⇒ x – y = 2... (vi)
∴x +y =3
x – y =2
______
2x   =5
$\Rightarrow \mathit{x}=\frac{5}{2}$
From equation (v),
$\mathit{y}=3\mathit{–}\frac{5}{2}=\frac{1}{2}$
∴$\frac{\mathit{x}}{\mathit{y}}=\frac{\frac{\frac{5}{2}}{1}}{2}=5$