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MathRevolution
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There are 3 particles A, B, and C moving around a circumference at a c 18 Jun 2020, 02:18
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75% (02:47) correct 25% (03:01) wrong based on 261 sessions

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There are \(3\) particles \(A, B\), and \(C\) moving around a circumference at a constant speed in the same direction. Particle \(A\) runs \(15\) laps per minute, \(B\) runs \(20\) laps per minute, and \(C\) takes \(8\) seconds for one lap. If particles \(A, B\), and \(C\) start from point \(P\) at the same time, how many times will the three particles pass point \(P\) simultaneously during the next \(10\) minutes?

A. \(12\)

B. \(18\)

C. \(25\)

D. \(29\)

E. \(33\)
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C
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There are 3 particles A, B, and C moving around a circumference at a c 18 Jun 2020, 03:00
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A = 4 sec/lap
B= 3 sec/Lap
C= 8 sec/lap as given

Now they will meet at starting point for the first time at the LCM of their speed.
which is LCM(3,4,8)=24

During the course of 10mins=600 sec this event will occur 600/24=25 times.

Ans C
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There are 3 particles A, B, and C moving around a circumference at a c 18 Jun 2020, 03:50
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or another way to think might be like this , A is 15 laps per minute , B is 20 laps per minute , and C is 7.5 laps per minute . And they run for ten minutes , which is A-150 laps , B-200 laps and C- 75 laps , and they will meet each other at the lcm , which is 25 number of times
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There are 3 particles A, B, and C moving around a circumference at a c 21 Jun 2020, 19:33
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Particles \(A, B\), and \(C\) take \(4, 3\), and \(8\) seconds, respectively.

The least common multiple of \(4, 3\), and \(8\) is \(24.\)

Then those three particles meet on at point \(P\) every \(24\) seconds.

Since \(10\) minutes is \(600\) seconds, they meet at point \(P, \frac{600 }{ 24} = 25\) times.

Therefore, C is the answer.
Answer: C
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There are 3 particles A, B, and C moving around a circumference at a c 27 Jun 2025, 23:01
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Bunuel, I don't understand why we're supposed to take the seconds/lap rate instead of the laps/minute rate.

If we take laps per minute, the LCM is 600 ... and I'm unable to understand how to arrive at the answer from that point. Can you please help?
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There are 3 particles A, B, and C moving around a circumference at a c 29 Jun 2025, 02:13
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DiveshR
There are \(3\) particles \(A, B\), and \(C\) moving around a circumference at a constant speed in the same direction. Particle \(A\) runs \(15\) laps per minute, \(B\) runs \(20\) laps per minute, and \(C\) takes \(8\) seconds for one lap. If particles \(A, B\), and \(C\) start from point \(P\) at the same time, how many times will the three particles pass point \(P\) simultaneously during the next \(10\) minutes?
A. \(12\)

B. \(18\)

C. \(25\)

D. \(29\)

E. \(33\)

Bunuel, I don't understand why we're supposed to take the seconds/lap rate instead of the laps/minute rate.

If we take laps per minute, the LCM is 600 ... and I'm unable to understand how to arrive at the answer from that point. Can you please help?
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It's important to understand the difference between seconds per lap (time taken to complete one lap) and laps per minute (number of laps completed in a fixed time).

Seconds per lap links the same distance with different times. For example, if A takes 4 seconds per lap, B takes 3 seconds, and C takes 8 seconds, we can ask: after how many seconds will all three be at point P together? Since they are at P every 4, 3, and 8 seconds respectively, the answer is the LCM of 4, 3, and 8, which is 24 seconds.

On the other hand, laps per minute relates different distances in a fixed time. If A does 15 laps per minute and B does 20, we can't directly say when they will next be at point P together, because that approach doesn't track their timing at a single location. So taking the LCM of 15 and 20 (which is 60) doesn't help in figuring out when they meet at P.

P.S. Also, how did you get that the LCM of 15, 20, and 15/2 is 600? 15/2 isn’t even an integer, so that doesn’t quite make sense.
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