CAT 2018 Question Paper Shift 1 Solved Paper

Let the time taken by Partha to cover $60\mathit{k}\mathit{m}$ be $\mathit{x}$ hours.
As per the condition, Narayan will cover $60\mathit{k}\mathit{m}$ in $\mathit{x}-4$ hours.
Therefore, Speed of Partha $=60/\mathit{x}$ And
Speed of Narayan $=60/\left(\mathit{x}-4\right)$
It is also given that Partha reaches the mid-point of $\mathit{A}$ and $\mathit{B}$ two hours before Narayan reaches B. Hence,
$\Rightarrow \frac{30}{\frac{60}{\mathit{x}}}+2=\frac{60}{\frac{60}{\left(\mathit{x}-4\right)}}$
$\frac{\mathit{x}}{2}+2=\mathit{x}-4$
$\frac{\mathit{x}+4}{2}=\mathit{x}-4$
$\mathit{x}+4=2\mathit{x}-8$
$\mathit{x}=12$
OR Partha will take 12 hours to cross $60\mathit{k}\mathit{m}.$
$\Rightarrow$ Speed of Partha $=60∕12=5\mathrm{Kmph}$