CAT 2018 Question Paper Shift 1 Solved Paper

Train T starts at 3 PM and train S starts at 4 PM.
Let the speed of train $\mathit{T}$ be $\mathit{t}$.
$\Rightarrow$ Speed of train $\mathit{S}=0.75\mathit{t}$
When the trains meet, train $\mathit{t}$ would have traveled for one more hour than train $\mathit{S}$.
Let us assume that the 2 trains meet $\mathit{x}$ hours after 3 PM. Trains S would have traveled for $\mathit{x}-1$ hours.
Distance traveled by train $\mathit{T}=\mathit{x}\mathit{t}$
Distance traveled by train $\mathit{S}=\left(\mathit{x}-1\right)*0.75\mathit{t}=0.75\mathit{x}\mathit{t}-0.75\mathit{t}$
We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities.
$\Rightarrow \left(\mathit{x}\mathit{t}\right)/\left(0.75\mathit{x}\mathit{t}-0.75\mathit{t}\right)=3/2$
$2\mathit{x}\mathit{t}=2.25\mathit{x}\mathit{t}-2.25\mathit{t}$
$0.25\mathit{x}=2.25$
$\mathit{x}=9$ hours.
Train T takes 9 hours to cover three-fifths the distance.
Therefore, to cover the entire distance, train T will take $9*\left(5/3\right)=15$ hours.
Therefore, 15 is the correct answer.train