The “Racing magic” takes 120 seconds to circle the racing track once. The “Charging bull” makes 40 rounds of the track in an hour. If they left the starting point together, how many minutes will it take for them to meet at the starting point for the second time?
(a) 3
(b) 6
(c) 9
(d) 12
(e) 15
OA:D
What's the fastest way to solve this? Thanks!
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francopiccolo
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The Racing Magic has a rate of 1track/120seconds = 1track/2min
The Charging bull has a rate of 40track/60min = 2track/3min = 1track/(3/2)min.
We have to find the lcm of 2 and 3/2. 3/2 times the number of tracks results into an integer every 2 tracks. 3 minutes to circle twice the track, 6 minutes to circle it four times.
The lcm of 2 and 3/2 is therefore 6 minutes, so every 6 minutes they find each other, and they find for the second time at minute 12.
The Charging bull has a rate of 40track/60min = 2track/3min = 1track/(3/2)min.
We have to find the lcm of 2 and 3/2. 3/2 times the number of tracks results into an integer every 2 tracks. 3 minutes to circle twice the track, 6 minutes to circle it four times.
The lcm of 2 and 3/2 is therefore 6 minutes, so every 6 minutes they find each other, and they find for the second time at minute 12.
