One evening Wilson, Xavier, Yeoman, Zenger, and Osborn made separate camps along the banks of a river. The next morning all were to meet at Osborn's camp; then each man besides Osborn was to return to his own camp.
- Wilson and Xavier camped downstream from Osborn, while Yeoman and Zenger camped upstream from Osborn.
- Wilson, Xavier, Yeoman, and Zenger each had a motor boat; each motor boat would take its owner to Osborn's camp in one hour, if there were no current in the river.
- There was a strong current in the river.
- The next morning each of the four men went in his motor boat to Osborn's camp; Wilson made the trip in 75 minutes, Xavier made the trip in 70 minutes, Yeoman made the trip in 50 minutes, and Zenger made the trip in 45 minutes.
Which one of the four men made the round trip, to and from Osborn's camp, in the shortest time?
If we define (for any given camper):
Tu= time to travel upstream leg
Td=time to travel down stream leg
Vb= speed of the campers's boat
Vr = speed of the river current
X=distance from camper's camp to Osborn's camp
then, because distance = velocity * time
Tu=X/(Vb-Vr) and Td=X/(Vb+Vr) (the current helps or hinders each motor boat)
and from statement 2., X/Vb=60 minutes for any camper
then
1/Tu + 1/Td = (Vb-Vr)/X + (Vb+Vr)/X = 2*Vb/X. But Vb/X=1/60
so
1/Tu +1/Td = 1/30.
Since either Tu or Td is known for each camper, the other can be found
Camper Tu Td
WIlson 75 50
Xavier 70 52.5
Yoeman 75 50
Zennger 90 45
So Xavier has the smallest sum of the times.
Notice that the problem formulation gives the same answer for any speed of the boats, river and distance between camps.
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Posted by Kenny M
on 2012-11-20 21:08:33 |
Solution (with explanation)
