11Did it the same way bt chintan gave me a nice and a shorter method to do the problem..:)
A man started from his home at 14:30 hrs 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 1st route at a rate twice as fast, reaching home at 16:00 hrs. 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
Ans c)
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4 Answers
11If x is the distance travelled and y is the speed of the man then,
By problem-:
2:30+xy+ 25mins + 5x8y=4.00
13x8y=65mins
or xy=40mins
If xy= 40mins hence he reaches the village at 3:10..
So the village clock is 5mins faster..:)
Thnx chintan for helping me out,by giving a short solution better than my long and solution..:P
1Can also be done in this way:
Let the length of first route be x and time taken to reach be y
Now by question while returning length = 5x/4 and speed= 2x/y
Let the time taken to come back be z
Therefore z=5x/42x/y=5y8
Now let us assume that the village clock is x minutes slower than home clock
Therefore 5(45+x)8=20-x[14:30 to 15:15 and 15:40 to 16:00]
Solving this we get x=-5
Hence our assumption was wrong and so the village clock is 5 mins faster:)
11