Question 1106402: A piece of land is 20 miles in circumference. Three persons, A, B, and C, start from the same place and travel the same way. A travels 3 mph, B 7 mph, and C 11 mph. In what time will they be together again?
Unsure how to solve. Non-homework.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It takes 20/3 hours for the first to go around, 20/7 for the second and 20/11 for the third.
Look at the first and second
3/20 of the circumference is the distance the first goes around in an hour.
How long until A is lapped by the second, who goes around 7/20 of the circumference in an hour?
(3x/20)+1=(7x/20) is the distances traveled in one hour, adding a lap to the A to have them both meet again.
multiply everything by 20
3x+20=7x, 4x=20, x=5
In 5 hours, the first goes 15/20 or 3/4 of the way, and the second goes 35/20 around or 1 3/4 of the way. They meet.
If the first meets the second, what about the second and the third?
(7x/20)+1=(11x/20)
7x+20=11x
4x=20
x=5 hours as well.
In 5 hours, C goes 55/20 around, and that is 2 15/20 laps or 2 3/4 the way. The three of them will meet in 5 hours, 3/4 of the way around the track.
(One can do this for the planets as well).
Another way in x hours A goes (# of laps) 3x/20 miles, B 7x/20 miles, and C 11x/20 miles
B and C have to go at least 1 lap more than A.
7x/20 -1=3x/20
4x/20=1
x=5
Try that for C.
11x/20-1=3x/30
8x=20
x=2.5. C laps A in 2.5 hours and every 2.5 hours.
Try C and B
11x/20-1=7x/20
4x/20=1
x=5 hours.
C laps B in 5 hours and A in 2.5 and 5 hours, so they are all together in 5 hours.
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