A train,when moving at an average speed of 40 | vedclass

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(C) Let the length of the journey be $x \, km$.The time taken at $40 \, km/hr$ is $t_1 = \frac{x}{40} \, hours$.The time taken at $35 \, km/hr$ is $t_

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(C) Let the length of the journey be $x \, km$.The time taken at $40 \, km/hr$ is $t_1 = \frac{x}{40} \, hours$.The time taken at $35 \, km/hr$ is $t_2 = \frac{x}{35} \, hours$.Given that the difference in time is $15 \, minutes$,which is $\frac{15}{60} = \frac{1}{4} \, hours$.Therefore,$\frac{x}{35} - \frac{x}{40} = \frac{1}{4}$.Taking the least common multiple of $35$ and $40$,which is $280$:$\frac{8x - 7x}{280} = \frac{1}{4}$.$\frac{x}{280} = \frac{1}{4}$.$x = \frac{280}{4} = 70 \, km$.Thus,the length of the journey is $70 \, km$.

$A$ train,when moving at an average speed of $40 \, km/hr$,reaches its destination on time. When its average speed becomes $35 \, km/hr$,it reaches its destination $15 \, minutes$ late. Find the length of the journey in $km$.

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