SSC CHSL 15 Nov 2015 Shift 1 Paper

Let total work to be done = L.C.M. (10,12,15) = 60 units
⇒ A can complete the work in 10 days, ⇒ A's efficiency = $\frac{60}{10}=6$ units/day
Similarly, B's efficiency = $\frac{60}{12}=5$ units/day
and C's efficiency = $\frac{60}{15}=4$ units/day
Let the work is completed in t days
Thus, A worked for $\left(\mathit{t}-5\right)$ days and B worked for $\left(\mathit{t}-3\right)$ days
⇒ $6\left(\mathit{t}-5\right)+5\left(\mathit{t}-3\right)+4\left(\mathit{t}\right)=60$
⇒ $\left(6\mathit{t}-30\right)+\left(5\mathit{t}-15\right)+\left(4\mathit{t}\right)=60$
⇒ $15\mathit{t}=60+30+15=105$
⇒ $\mathit{t}=\frac{105}{15}=7$days