A train,after travelling 50 , km,meets with a | vedclass

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(D) Let the normal speed of the train be $v \, km/hr$.
Let the distance between the point of the first accident $(C)$ and the destination $(B)$ be $d

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(D) Let the normal speed of the train be $v \, km/hr$.
Let the distance between the point of the first accident $(C)$ and the destination $(B)$ be $d \, km$.
When the accident occurs at $C$,the train travels the remaining distance $d$ at $\frac{3}{4}v$ speed. The time taken is $\frac{d}{\frac{3}{4}v} = \frac{4d}{3v}$.
The time that would have been taken at normal speed is $\frac{d}{v}$.
The delay is $\frac{4d}{3v} - \frac{d}{v} = \frac{d}{3v} = 35 \, minutes = \frac{35}{60} \, hours = \frac{7}{12} \, hours$. So,$\frac{d}{v} = \frac{7}{12} 	imes 3 = \frac{7}{4} \, hours \dots (1)$.
When the accident occurs $72 \, km$ further at $D$,the remaining distance is $(d - 72) \, km$.
The delay is $\frac{d-72}{\frac{3}{4}v} - \frac{d-72}{v} = 15 \, minutes = \frac{15}{60} \, hours = \frac{1}{4} \, hours$.
$\frac{d-72}{v} (\frac{4}{3} - 1) = \frac{1}{4} \implies \frac{d-72}{v} 	imes \frac{1}{3} = \frac{1}{4} \implies \frac{d-72}{v} = \frac{3}{4} \dots (2)$.
Subtracting $(2)$ from $(1)$: $\frac{d}{v} - \frac{d-72}{v} = \frac{7}{4} - \frac{3}{4} = 1 \implies \frac{72}{v} = 1 \implies v = 72 \, km/hr$.

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