An army 50 miles long marches at a constant rate. A courier standing at the rear moves forward and delivers the message to the first person and then turns back and reaches the rear of the army as the army completes 50 miles. Find the distance travelled by the courier.

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FreakyPlacements

  • Jun 20th, 2008
 

I think answer is 75 miles.

 The distance between first person and the last person, the courier man, is always 50 miles. So he covers this 50 miles delivers the packet and turns back and starts walking toward the rear. Lets say he is walking at the same constant rate as the army but in the opposite direction.
 So when he is half way through the distance, he would reach the rear of the army. So totally he coves 50+25=75 miles.

maninider

  • Nov 26th, 2009
 

The answer I think is 50*root 3 as
Time taken for the army to travel 50 miles with speed a = 50 / a
This must be equal to time taken by the man with speed m, i.e.

50/a =50/(m-a) + 50/(m+a)
This gives ratio m/a = root3
 
so distance would be speed * time = m * 50/a = 50*root 3

abhinavmishra85

  • Dec 9th, 2009
 

Let the speed of the army be x and speed of the courier be y.  The time taken by the courier to reach the first person be t1 and to return be t2

eqn:
(y-x)t1 = 50--------(i)
(y+x)t2 = 50---------(ii)
x(t1+t2) = 50---------(iii)

from (iii):t1+t2=50/x-------(iv)
from (ii):t2=50/y+x-------(v)
from (iii):t1=50/y-x-------(vi)

Putting (v) & (vi) in (iv):50/x=(50/y-x)+(50/y+x) ------>   1/x=(1/y+x)+(1/y-x)----->
Solving y=(1+2^.5)x=(1+1.414)x=2.414x

Now distance travelled=(t1+t2)*y = (50/x) * 2.414x = 50*2.414=120.7 mile
Answer -> 120.7 mile

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Aries Brune Tyson

  • Jan 17th, 2010
 

The speed of the courier should be grater than the speed of the army.

The time taken by the army to travel 50 miles, is equal to the time taken by the courier, to move front person and came back to rear of army.

Let speed=distance/time
armspeed=100miles/t1;
courierspeed=d2/t1;
by equating this 50/armspeed=d2/couriersped;
d2=50*(courierspd/armspeed);
so it based on the increased speed of the courier...
might be fall b/w 50-75

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nstarno

  • May 16th, 2010
 

Let D(army)=50 miles
Let D(courier)= 50+x (x-unknown distance)
Let rate(army)=a
Let rate(courier)=a+y (y= unknown rate)
Let time for each=t

Use distance formula for army; D=rt
since D(army)=50
50=at
t=50/a
a=50/t

Use distance formula for courier; D=rt
(50+x)=(a+y)t
(50+x)/(a+y)=t

Since t is the same for both army and courier,
time of courier = time of army
(50+x)/(a+y)=50/a
a[(50+x)/(a+y)]=50
x=50y/a
Distance of courier = 50 + 50y/a

If the rate of the courier was 3 mph and the army's rate was 1 mph,  then the distance of the courier would be 50 + 50(2) or 150 miles.

If the rate of the courier was 4 mph and the army's rate was 1 mph, then the distance of the courier would be 50 +50(3) or 200 miles.

If the rate of the courier was 3mph and the army's rate was 2 mph, then the distance of the courier would be 50 + 50(1) or 100 miles.

If the rate of the courier was 4mph and the army's rate is 3 mph, then the distance of the courier would be 50 + 50(1) or 100 miles.

I don't see any one answer being correct, but it seems like the answer will be between 50 and 200 miles depending on the rates of each.

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