Can anyone solve this math word problem?

Question

A column of soldiers 25 miles long marches 25 miles in the course of 1 day. One day, the soldier at the end of the line was asked to deliver a message to the soldier at the front. If the soldier at the end of the line started his delivery during the beginning of the day and returned to his position at the end of the day, what is the total distance the soldier traveled? (both the column and soldier are moving at constant speeds, but not at the same speed. They are both moving on a straight line)

Answer 1

50 isn't right.  Ab (below) is right.

Answer 2

Its not 50 miles per day because the soldier would just reach the front of the column by the end of the day.  Notice he has to travel like 60 miles per day to reach the front of the line within .7 days because he only travels 35 miles per day faster than the pack, then of course the return trip is shorter i.e 60 mpd + 25mpd.  The third equation says we have to do this in one day.

Set up 3 equations for x, y , z.

X= days to reach the front

Y = days to reach the back

Z = speed

x(z-25)=25

y(z+25)=25

x + y = 1

Solve for X, Y, Z

x= .618 days

y= .382

z= 65.45 miles

Answer 3

I have 3 answers..

maybe the soldier will not able to deliver the message because as he move forward the first soldier will also move forward....

...........................or............

if you said in front of  the last soldier,it will be very easy to deliver......

...........................or............

the last soldier will just say it to the soldier in front of him and that soldier to whom the last soldier said will tell to the soldier in front of him and that soldier will tell to the soldier in front and so on and so forth....until the message reaches the first soldier.........that's all

Answer 4

50.If there are 25 miles of soldiers and they walked for 25 miles

25+25=50.That soldier walked 50 miles.All you had to do is add

25+25.It's hard when they say the question in a harder way to understand.

Answer 5

Let  x be the messenger's speed

The solgers' speed is 25ml/day

Then it took him to reach the front one  25/(x-25) and to get back 25/(x+25)

25/(x-25) + 25/(x+25)=1 or x^2 - 50x-25^2=0

Solving x=25(1+sqrt2)ml/day

He covered S=x*1day= 25(1+sqrt2)