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# Brainden races

## Question

At the annual Brain Denizen picnic, there were the inevitable games, among them the ever-popular three-legged race. Three teams were formed by tying one contestant's right leg to another' s left leg. Fortunately all six contestants made it to the finish line without any broken bones! For purposes of this puzzle we assume all three teams ran the 100-meter course at constant speed.  Team 2, comprising BMAD and Thalia, were able to beat Team 3, comprising rocdocmac and DejMar by 20 meters, but lost to the winner, Team 1, comprising plasmid and plainglazed, by 20 meters as well. By how many meters did Team 1 beat Team 3?

## 5 answers to this question

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Nice, Bonanova! Definitely not the obvious answer.

Using d= r*t

Team 1 beat team 2 by 20 meters out of 100 m
Team 2 beat team 3 by 20 meters
This means
R1 = r2 * 5/4
R2 = r3 * 5/4
Therefore r1 = r3 * 25/16

D3/d1 = r3/r1 = 16/25
So when d1 = 100
D3 = 16*100/25 = 64

So team 1 beat team 3 by 36 meters

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A crosses finish 20 meters a head of B ;   Now B crosses 20 meters ahead of C  ; i.e.  B at 100 meters ...C at  80 meters;    therefore C was traveling at 8/10 the speed of B.  Therefore when B was at 80 meters ...C was 8/10  x  80 meters or 64 meters.    Thus A beat C by 36 meters

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Yes ,I agree Capt Eds' answer is not quite as straight forward as could be.    After A reaches 100 m ...20 m ahead of B , the speed of A is not significant to the solution.  The key is that when B is at 100 ,C is at 80.   Therefore C is 8/10 the speed of B and it follows that when B was at 80 m...C was at 8/10  x 80  = 64 m.   A is at 100 m  when C is at 64 therefore A beats C by 36 m

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7 hours ago, Donald Cartmill said:

Capt Eds' answer is not quite as straight forward as could be.

Both answers state correctly that (a) winning distances give speed ratios and (b) combined speed ratio gives combined winning distance. What part of that can be more (or less) straightforward?

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While the combined speed ratios provide the correct answer . I still maintain the speed ratio of A to B is not crucial to the solution. i.e.  Let us suppose that A is merely to step out at the finish line to congratulate B and C the competitors in a two man race .  A steps out when  B has run 80 meters; Congratulates  B when B completes the race and notes at the time C has completed 80 meters.  Where was C when A first made his appearance ?   The answer is   4/5 x 80 = 64 meters

While the combined speed ratios provide the correct answer . I still maintain the speed ratio of A to B is not crucial to the solution. i.e.  Let us suppose that A is merely to step out at the finish line to congratulate B and C the competitors in a two man race .  A steps out when  B has run 80 meters; Congratulates  B when B completes the race and notes at the time C has completed 80 meters.  Where was C when A first made his appearance ?   The answer is   4/5 x 80 = 64 meters                                       To beat a dead horse let A step out when B is at 85 meters;  He then congratulates B as he finishes the race and notes C is at 80 meters.  He wonders where was C , when  I first stepped out ???   The answer is of course 4/5 x 85 = 68 meters

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