Question
Thu February 07, 2013 By: Sureshyadav

# Three cyclist start out simultaneously from the same place in one direction around a circular track 1km in length. The rates of the cyclist form, in a certain order, an arthemetic progression with commom difference of 5km/hr. After sometime, the second one catches up with the first, having made one extra circuit; 4 minutes later the third one arrives at that point, having covered the same distance that the first did at the time he caught up with the second cyclists. 1) Find the rate of the second cyclist? 2)Find the time taken by the third cyclist to complete one circle? 3) Find the time taken by the first cyclist to complete one circle? 4) If the three cyclist start at 7 am then find the time when they all meet together at the starting point?

Mon February 11, 2013
let speed of first be x hence for second=x+5 hence for third=x-5

now let after n rounds of first , second catches it therfore second completes n+1 rounds in same time. equating time taken by both we can write

n/x=(n+1)/(x+5)

or, x=5n

now to complete n rounds third takes 4 min extra.

therfore , ((n/x)+4/60)*(x-5)=n

solving both equations we get n=4 & x=20

1) rate of 2nd cyclist=x+5=25
2)time taken by third cyclist to complete one circle=1/(x-5)=1/15 hour=4 min
3)time taken by first cyclist to complete one circle=1/(x)=1/20 hr=3 min
4) first takes 3 min,second takes 12/5min,third takes 4 min to complete one circle
now takin LCM we get t=12 min
therfore they will meet again at 7.12 am

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