SSC CHSL 15 Nov 2015 Shift 2 Paper

Let maximum marks in the examination = $100\mathit{x}$ and passing marks = $\mathit{y}$
Marks secured by candidate = $\frac{20}{100}\times 100\mathit{x}=20\mathit{x}$
Thus, $20\mathit{x}=\mathit{y}-30$ -----(i)
Similarly, $32\mathit{x}=\mathit{y}+42$ -----(ii)
Subtracting equation (i) from (ii), we get :
⇒ $32\mathit{x}-20\mathit{x}=42+30$
⇒ $12\mathit{x}=72$
⇒ $\mathit{x}=\frac{72}{12}=6$
Substituting it in equation (i), ⇒ $\mathit{y}=20\left(6\right)+30$ $=120+30=150$
∴ Pass% = $\frac{\mathit{y}}{100\mathit{x}}\times 100=\frac{\mathit{y}}{\mathit{x}}$
= $\frac{150}{6}=25\%$