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September 5th, 2018, 01:42 PM  #11  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,961 Thanks: 991  Quote:
[A@28,B@16,C@8].........C(at 8 )........B(at 16).......................A(at 28 ) B  C = 16  8 = 8 A  B = 28  16 = 12 "When A completes the first round, the distance between B and C is 2/3 the distance between A and B." "2/3 the distance between A and B" = 2/3 * 12 = 8  
September 6th, 2018, 03:52 AM  #12 
Senior Member Joined: Aug 2014 From: India Posts: 343 Thanks: 1  
September 6th, 2018, 04:40 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 20,285 Thanks: 1968 
The problem doesn't define "round", so I've assumed that it refers to a single circuit of the track by A. This means that rounds 1 and 2 take the same amount of time, and so each runner covers twice as far in rounds 1 and 2 combined as they cover in round 1. It's known that C reaches Y at the end of round 2. As C is slower than B, that must be the first occasion on which C reaches Y, and so C reached Y/2 at the end of round 1.


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circular, problem, solve, track 
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