Question

A man started from home at 14:30 hours and drove to a village arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes he drove back by a different route of length (5/4) times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home the village clock is
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Answer 1

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Let s be the speed of the man when he is traveling from home to village, then according to question the speed while returning is 2s.

Let dd be the distance of the route when man is traveling from home to village, then according to question distance of the returning route is $\bf{\frac{5d}{4}}$

$\boxed{\sf{Speed\;is\;scalar\;quantity.}}$

Therefore,

$2s+s = \frac{d+\frac{5d}{4}}{T}$

T is the time according to the clock at home.

Then, T = 90 - 25 → 65 Minutes

$→\;{\sf{3s = \frac{d+\frac{5d}{4}}{65}}}$

$→\;{\sf{s = \frac{9d}{3×4×65}}}$

$→\;{\sf{s = \frac{3d}{4×65}}}$

Now, $\bold{s = \frac{d}{t}}$

☞ t = 45 minutes

Now, if you started from home at 14:30 and you took 40 minutes to arrive to village, then you get there at 15:10. Since the time there was 15:15, the village clock was 5 minutes fast.

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