CAT 2023 Slot 2 Solved Paper

It is given that the speed of Ravi is $40\text{ kmph}$, which is equal to $\;\frac{100}{9}\text{ m/s}$. It is also known that the speed of Ashok is $50\text{ kmph}$, which is equal to $\;\frac{125}{9}\text{ m/s}$.It is known that the distance between Ravi and Ashok is 225 meters, and the relative speed of Ravi and Ashok is $\;\frac{125}{9}+\;\frac{100}{9}=25\text{ m/s}$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 $(100+54)=154$ meters in these 9 seconds.Hence, the speed of Vijay is $\;\frac{154}{9}\text{ m/s}$, which is equal to $\;\frac{154}{9} \times \;\frac{18}{5} = \;\frac{308}{5} = 61.6\text{ kmph}$.The correct option is C