CAT 2019 Shift 2 Paper

Speed of John $=6\mathit{k}\mathit{m}\mathit{p}\mathit{h}=6\times \frac{5}{18}=\frac{5}{3}\mathit{m}∕\mathit{s}$
Speed of Mary $=7.5\mathit{k}\mathit{m}\mathit{p}\mathit{h}=7.5\times \frac{5}{18}=\frac{25}{12}\mathit{m}/\mathit{s}$
Let the track length of $\mathit{A}$ and $\mathit{B}$ be $\mathit{x}$ and y respectively.
Given, $\mathit{x}+\mathit{y}=325$....(1)
Time taken by John to cover one round of $\mathit{A}=\frac{\mathit{x}}{5/3}\mathit{s}\mathit{e}\mathit{c}$
Therefore, time taken to cover 9 rounds $=9\times \frac{\mathit{x}}{5/3}=\frac{27}{5}\mathit{x}$ sec
Time taken by Mary to cover one round of $\mathit{B}=\frac{\mathit{y}}{25/12}\mathit{s}\mathit{e}\mathit{c}$
Therefore, time taken to cover 5 rounds $=5\times \frac{\mathit{y}}{25/12}=\frac{12}{5}\mathit{y}\mathit{s}\mathit{e}\mathit{c}$
As per the condition:
$\frac{27}{5}\mathit{x}=\frac{12}{5}\mathit{y}$
$\Rightarrow \frac{\mathit{x}}{\mathit{y}}=\frac{12}{27}=\frac{4}{9}$
Putting in equation (1) we get $\mathit{x}=100$ and $\mathit{y}=225$.
Time taken by Mary to cover one round of $\mathit{A}=\frac{100}{25/12}=48\mathit{s}\mathit{e}\mathit{c}$