3

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 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

(A) 10 min slow
(B) 5 min slow
(C) 5 min fast
(D) 20 min fast

What I have tried

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

Speed is scalar quantity.

Therefore $2s+s=\frac{d+\frac{5d}{4}}{T}$ here T is the time according to the clock at home. So $T=90-25=65$ min

Now I have $$2s+s=3s=\frac{d+\frac{5d}{4}}{65} \implies s=\frac{9d}{3\cdot 4\cdot 65}=\frac{3d}{4\cdot 65}$$

Now $s=\frac{d}{t}$ where $t=45$ min (This is not clear to me)

The correct answer according to book is (C). Kindly help me.

Singh
  • 2,108

3 Answers3

3

Let $s_1$ the length of the first road, $v_1$ the speed used and $T_1$ the time required to go to the village. Then:

$$T_1 = \frac{s_1}{v_1}.$$

Let $s_2 = \frac{5}{4}s_1$ the length of the second road, $v_2 = 2v_1$ the speed used and $T_2$ the time required to come back home. Then:

$$T_2 = \frac{s_2}{v_2} = \frac{\frac{5}{4}s_1}{2v_1} = \frac{5}{8}T_1.$$

The total trip took $90$ minutes ($14:30$ to $16:00$), and it can be written (in minutes) has follows:

$$T_1 + 25 + T_2 = 90 \Rightarrow T_1 + \frac{5}{8}T_1 = 65 \Rightarrow T_1 = 40,$$ and hence $$T_2 = \frac{5}{8}40 = 25.$$

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. Your answer is $C$.

the_candyman
  • 14,064
  • 4
  • 35
  • 62
2

$2s+s=\dfrac{d+\dfrac{5d}{4}}{T}$ does not hold.

Instead, we have $$\dfrac ds+\frac{\dfrac{5d}{4}}{2s}=T=65\tag 1$$ The point is that we have to consider his way to and from village separately. Here, $\frac ds$ is the time from home to village, and $\frac{\frac{5d}{4}}{2s}$ is the time from village to home, then add these.

From $(1)$, you'll get $\frac ds=40$, i.e. the time from home to village is $40$ minutes. Thus, the answer will be $C$.

mathlove
  • 139,939
1

I wouldn't even bother with defining symbols for distance and speed, as we have neither and are only interested in time.

Let $t$ be the time taken to drive from the starting point to the village.

Then the time taken to drive home again ($\frac{5}{4}$ times as far, but twice as fast) is $\frac{5t}{8}$. So total driving time is $\frac{13t}{8}$.

We know the total trip (including a 25 minute rest) took 90 minutes, so his total driving time is 65 minutes.

$\frac{13t}{8} = 65$

$t = \frac{65×8}{13} = \frac{65}{13}×8 = 5×8 = 40$.

So the actual time when he arrived at the village was 15:10 and the clock is 5 minutes fast.

IanF1
  • 1,499