Distance Covered by Letter Delivery Person

What was the puzzle?

Let's suppose the first person of the platoon move X meters ahead by the time the last person reaches to him.

To reach at him, the last person has to cover distance of 50 + x.

Assume M be the speed of last person and P be the speed of platoon i.e. of first person.

If T1 is time needed for the last person to get to at the first person,

T1 = (50 + X)/M

T1 = X/P

(50 + X)/M = X/P

M/P = (50 + X)/X  ..........(1)

As per given data, by the time the last person gets back to it's original position (i.e. end of platoon), the platoon moves 50m ahead from it's position that was when the last person started his journey towards first person.

That is end of platoon is now at position at which the start of platoon was initially.

Since, the first person has already moved X meters ahead, he has to move only 50 - X meter to lead the platoon 50m ahead of it's original position.

And, the last person has to move only X meters to get back to  original position i.e. the end of platoon.

If T2 is the time taken by last person to get back to original position (i.e. time taken by first person to move ahead 50 - x) then,

T2 = X/M

T2 = (50 - X)/P

X/M = (50 - X)/P

M/P = X/(50 - X)  ..........(2)

Equating (1) and (2),

(50 + X)/X = X/(50 - X)

X^2 = (50 + X)(50 - X)

X^2 = 2500 - X^2

X^2 = 1250

X = 35.355 meters.

So,

the distance traveled by the last person = (50 + X) + X 
                                                          
                                                          = (50 +35.355) + 35.355 

the distance traveled by the last person = 120.71 meters.

The ratio of their speeds = M/P = (50 + X)/X = (50 + 35.355)/35.355 = 2.41

M = 2.41 P

That is the speed of last man is 2.41 times the speed of first man or the speed of platoon itself.

And that's the speed the last man needed to reach at the first person of the platoon.