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(C) Let the speed of the car be $x\, km/hr$ and the speed of the train be $1.5x\, km/hr$.The distance between point $A$ and point $B$ is $75\, km$.The

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$120$

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(C) Let the speed of the car be $x\, km/hr$ and the speed of the train be $1.5x\, km/hr$.The distance between point $A$ and point $B$ is $75\, km$.The time taken by the car to travel from $A$ to $B$ is $T_c = \frac{75}{x}\, 	ext{hours}$.The time taken by the train to travel from $A$ to $B$ (excluding stops) is $T_t = \frac{75}{1.5x}\, 	ext{hours}$.Since the train stops for $12.5\, 	ext{minutes}$ (which is $\frac{12.5}{60} = \frac{5}{24}\, 	ext{hours}$),the total time taken by the train is $T_t + \frac{5}{24}$.Given that both reach at the same time,we have: $\frac{75}{x} = \frac{75}{1.5x} + \frac{5}{24}$.Subtracting $\frac{75}{1.5x}$ from both sides: $\frac{75}{x} - \frac{75}{1.5x} = \frac{5}{24}$.Factoring out $\frac{75}{x}$: $\frac{75}{x} (1 - \frac{1}{1.5}) = \frac{5}{24} \Rightarrow \frac{75}{x} (1 - \frac{2}{3}) = \frac{5}{24}$.$\frac{75}{x} (\frac{1}{3}) = \frac{5}{24} \Rightarrow \frac{25}{x} = \frac{5}{24}$.Solving for $x$: $x = \frac{25 	imes 24}{5} = 5 	imes 24 = 120\, km/hr$.

$A$ train can travel $50\%$ faster than a car. Both start from point $A$ at the same time and reach point $B$,$75\, km$ away from $A$,at the same time. On the way,however,the train lost about $12.5\, \text{minutes}$ while stopping at the stations. The speed of the car is (in $km/hr$):

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