SSC CHSL 6 Dec 2015 Shift 2 Paper

Let scores of A, B and C are and respectively.
According to ques, ⇒ $\mathit{a}+\mathit{b}=120$ ----(i)
$\mathit{b}+\mathit{c}=130$ -----(ii)
$\mathit{c}+\mathit{a}=140$ -----(iii)
Adding above equations, we get : $2\left(\mathit{a}+\mathit{b}+\mathit{c}\right)$ $=\left(120+130+140\right)$
⇒ $\mathit{a}+\mathit{b}+\mathit{c}=\frac{390}{2}=195$
Substituting value from equation (i) in above equation,
⇒ $120+\mathit{c}=195$
⇒ $\mathit{c}=195-120=75$