Distance problem

The dist is the same
the speed is in the ratio 25 / 4

time = Dist / Speed
Time for onward journey = D/25
Time for return journey = D/4
Ratio of time taken = 4/25
so if total time is 29 it takes 25 to return and 4 on onward journey
Actual tiime = 5 hr 48 nim = 5 4/5 = 29 /5
ie 5 hrs on return journey
Speed = 5*4 = 20

I did it using backward substitution.

I first picked C. (Just started somewhere in the middle)

V = d/t

So, 4km/hr = 25/t = t > 6. Which means just walking back would have taken him longer than the whole journey.

Next I picked B.

Way back = 5 hours
on the way there = 4/5 hour (25km/hr = 20/t --> t = 20/24 = 4/5)

It took him 5 hours + 48 mins --> 48/60 = 4/5

Answer B.

Correct

d/25 + d/4 = 5hr 48min = 29/5 hr
d(29/100) = 29/5 hr
d = 100/5 = 20 KM

saurabhmahajan wrote:A man travelled from village to post-office at the rate of 25 km/hr and walked back at the rate of 4 km/hr.If the whole journey took him 5 hr 48 min.Find the distance between post office and village.

a) 15
b) 20
c) 25
d) 30
e) 35


Plz help me,also is this question a 700+ question?

Thanks and regards,
Saurabh.

total time taken = 5 + 4/5 hrs = 29/5 hours.

Now, as the distance is same during the two journeys, the time taken during the two them would be in the inverse proportion to the speeds. i.e would be in the ratio 4/25 .

now, if we consider from Village -> post-office , time taken = 4/29 * (Total time) = 4/29 * 29/5 = 4/5 hrs

therefore the distance = 4/5 * 25 = 20 KMs.