SSC CHSL 7 Mar 2018 Shift 3 Paper

Let the number be $100\left(\mathit{x}+\mathit{y}\right)$ such that both the parts are $100\mathit{x}$ and $100\mathit{y}$
According to ques,
⇒ $\left(\frac{30}{100}\times 100\mathit{x}\right)$ $-\left(\frac{20}{100}\times 100\mathit{y}\right)=25$
⇒ $30\mathit{x}-20\mathit{y}=25$
⇒ $6\mathit{x}-4\mathit{y}=5$ -----------(i)
Similarly, $50\mathit{y}-60\mathit{x}=33.5$
⇒ $10\mathit{y}-12\mathit{x}=6.7$ -------------(ii)
Multiplying first equation by '2' and adding both equations,
⇒ $10\mathit{y}-8\mathit{y}=6.7+10$
⇒ $2\mathit{y}=16.7$
⇒ $\mathit{y}=\frac{16.7}{2}=8.35$
Similarly, $6\mathit{x}=5+4\left(8.35\right)=38.4$
⇒ $\mathit{x}=\frac{38.4}{6}=6.4$
∴ Number = $100\left(6.4+8.35\right)$
= $100\times 14.75=1475$