CAT Exam 2021 Slot 1 Solved Paper

Speed of the fastest train $=\frac{160}{12}=\frac{40}{3}\mathrm{m}∕\mathrm{s}$
Speed of the slower train $=\frac{40}{3}-\left(6\times \frac{5}{18}\right)=\frac{35}{5}\mathrm{m}∕\mathrm{s}$
Sum of speeds (when the train travels towards each other) $=\frac{40}{3}+\frac{35}{3}=25\mathrm{m}∕\mathrm{s}$
let the slower train be $\mathit{x}$ meters long: then: $\frac{160+\mathit{x}}{25}=14$
on solving $\mathit{x}=190\mathrm{m}$