Question

A person travels a distance AB of 15.625 m in 5 seconds. In the last four seconds, every second he covers a distance equal to one-fourth of the total distance covered upto the end of the previous second. Find the distance covered in the 1st second. (in m)​

Answer 1

Answer:

Statement of the given problem,

A person travels a distance AB of 15.625 m in 5 seconds. In the last four seconds, every second he covers a distance equal to one-fourth of the total distance covered up to the end of the previous second. What is distance covered in the 1st second?

Let D denotes the distance (in m) covered in the 1st second.

According to given problem,

Distance covered in the

1st second = D

2nd second = D/4

3rd second = (D + D/4)/4 = 5*D/16

4th second = (5*D/4 + 5*D/16)/4 = 25*D/64

5th second = (5*D/4 + 5*D/16 + 25*D/64)/4 = 125*D/256

Hence from above data we get following relation,

D + D/4 + 5*D/16 + 25*D/64 + 125*D/256 = 15.625

or 125*D/64 + 125*D/256 = 15.625

or 625*D/256 = 15.625

or D = 15.625*256/625 = 6.4 (m) [Ans]

Answer 2

Answer:

A person travels a distance AB of 15.625 m in 5 seconds. In the last four seconds, every second he covers a distance equal to one-fourth of the total distance covered upto the end of the previous second. Find the distance covered in the 1st second. (in m)