CAT 2023 Slot 2 Solved Paper

It is given that the speed of Ravi is $40\mathrm{kmph}$, which is equal to $\frac{100}{9}\mathrm{m}/\mathrm{}\mathit{s}$. It is also known that the speed of Ashok is $50\mathrm{kmph}$, which is equal to $\frac{125}{9}\mathrm{m}/\mathrm{}\mathit{s}$.
It is known that the distance between Ravi and Ashok is 225 meters, and the relative speed of Ravi and Ashok is $\frac{125}{9}+\frac{100}{9}=25\mathrm{m}/\mathrm{}\mathit{s}$
Hence, they will meet each other in $\frac{225}{25}=9$ seconds. The distance traveled by Ravi in these 9 seconds is $\frac{100}{9}\times 9=100$ meters.
Since Vijay was already 54 meters behind Ravi when they were starting, Vijay must travel $\left(100+54\right)=154$ meters in these 9 seconds.
Hence, the speed of Vijay is $\frac{154}{9}\mathrm{m}/\mathrm{}\mathit{s}$, which is equal to $\frac{154}{9}\times \frac{18}{5}=\frac{308}{5}=61.6\mathrm{kmph}$.
The correct option is C