How fast is the village clock. 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.
 A: 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 $C$.
A: $2s+s=\dfrac{d+\dfrac{5d}{4}}{T}$ 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 $C$.
A: 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.
