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One complete lap around a particular circular track is 400 meters. Jun and Quan each start running at the starting line and run around the track; Jun runs clockwise at 3 meters per second, and Quan runs counterclockwise at 5 meters per second. When they meet for the sixth time after starting, they stop and both walk back together along the track to the starting line. What is the shortest distance they could walk back on the track together?

Guest Nov 23, 2017
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 #1
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Let the time until they meet =T
3T + 5T =400
8T = 400
T = 400/8 =50 seconds when both will meet for the first time.
50x3 =150 m run by Jun(clockwise)
50x5 =250 m run by Quan(counterclockwise)
150 x 6 =900/400 =2.25 laps run by Jun after their 6th meet
250 x 6 =1,500/400=3.75 laps run by Quan after their 6th meet
2.25 - 2 x 400 =100m from the starting line when Jun meets Quan for the 6th time(clockwise)
3.73 - 3 x 400 =300m from the starting line when Quan meets Jun(counterclockwise) =400 -300 =100 m to the starting line.
So, the shortest distance from the starting line after their 6th meeting will be 100 meters.

Guest Nov 23, 2017
 #2
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They meet at 50 second intervals

 

To see this....Jun runs   50 * 3  = 150m and Quan runs  50 * 5  = 250m....so the total distance covered by both is 150 + 250  = 400m  = 1 lap

 

So...if they meet for the 6th time.....this must be 300 seconds after they started

 

So.....Jun has run  300* 3  = 900 m and is 100m from the starting line

And Quan has run  300 * 5 =  1500 m   and is 100m from the starting line

 

So....they will walk back 100 m together to the starting line

 

 

cool cool cool

CPhill  Nov 23, 2017

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