Three runners start running simultaneously from the same point on a 50

Question

Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

(A) 1,000
(B) 1,250
(C) 2,500
(D) 5,000
(E) 10,000

(A) 1,000
(B) 1,250
(C) 2,500
(D) 5,000
(E) 10,000

GMATinsight · Accepted Answer

Let, Speed of A = 4.4
Speed of B = 4.8
Speed of C = 5

Relative speed of A and B = 4.8-4.4 = 0.4

at their relative speed the distance between the two should increase by 500 meters for them to meet for teh first time

So time of A and B to meet on circular track = 500/.4 = 1250 second

Relative speed of C and B = 5 - 4.8 = 0.2

at their relative speed the distance between the two should increase by 500 meters for them to meet for teh first time

So time of C and B to meet on circular track = 500/.2 = 2500 seconds

i.e. All three to meet on the track times needed = LCM of 1250 and 2500 = 2500 second

Answer: Option C

abhishek31 · Answer

Hi, can you please explain why have you devided the incremental speed by 500? i get the concept of using incremental speed, but to equate with 500 i cant get

effatara · Answer

Thanks, Bunuel, for always coming up with something that exercises the little grey cells!

The times taken for the three runners(say R1, R2 and R3) to complete one revolution of the track are 500/5, 500/4.8 and 500/4.4 meters per second respectively. We can take a hint from the answer choices and deduce that, since it takes 100 seconds R1 to complete one rev and all the options are divisible by 100, the runners complete a whole number of revs in the time it takes for all three to meet at one point.

Let T seconds be the time it takes for all three to meet. Then, the number of revs completed by the runners in T seconds:
R1: 5T/500 = 25T/2500
R2: 4.8T/500 = 24T/2500
R3: 4.4T/500 = 22T/2500

All the above values must be whole numbers and since 2500 is the minimum value for T that satisfies this condition, (C) is the answer. In 2500 seconds R1, R2 and R3 complete 25, 24 and 22 revs respectively and meet at the point where they started.

Ans: C

Fdambro294 · Answer

Call the runners A (speed of 4.4 mps), B (speed of 4.8 mps), and C (speed of 5.0 mps)

Since they all start at the same point at the same instance, in order for B to “catch up” to A again such that they meet, B will have to run 1 complete lap around A and come up behind her to meet her.

All the while, A will be running in the same direction negating any progress B makes at trying to complete that extra +1
Lap (500 m) around.

For every 1 second that passes, B will move 4.8 meters. However, in that same 1 second, A will also move 4.6 m in the same direction, hindering the lead that B gets. The net effect is that every 1 second that passes, B is able to get ahead 0.2 meters. Since the total distance that B must travel AHEAD of A is 1 complete extra lap of 500 meters, it will take:

Time = T-a = 500 m / (4.8 - 4.4) mps = 1,250 seconds before B can run a lap around A and “catch up” to A from behind such that they meet again

Then, when they meet again, the process will begin again. It will take B another +1,250 seconds before she runs around A and catches up to A from behind again.

Every 1,250 seconds B will meet A like this on the circular track.

Using the same logic, it will take C the following time before C can run +1 extra lap around B and catch up to B from behind:

Time = T-b = 500 m/ (5.0 - 4.8) mps = 2,500 seconds

They will 1st simultaneously meet on the track when B laps around and catches up to A and C laps around and catches up to B.

This will happen at the L.C.M. of the time intervals at that it takes to complete a “lap around” —— L.C.M. (1,250 sec and 2,500 sec) = 2,500 sec

All the 3 will simultaneously meet after 2,500 seconds.

Answer choice -C-

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ScottTargetTestPrep · Answer

Solution:

Let’s call the runners A, B, and C and check each answer choice. 
If it takes 1,000 seconds for them to be at the same place again, then A, B, and C run 4,400, 4,800, and 5,000 meters, respectively. We see that since 5,000 is a multiple of 500, we see that C is at the starting point of the track. However, since neither 4,000 nor 4,800 is a multiple of 500, neither A nor B is at the starting point.

If it takes 1,250 seconds for them to be at the same place again, then A, B, and C run 5,500, 6,000, and 6,250 meters, respectively. We see that since both 5,500 and 6,000 are multiples of 500, we see that A and B are at the starting point of the track. However, since 6,250 is not a multiple of 500, C is not at the starting point.

If it takes 2,500 seconds for them to be at the same place again, then A, B, and C run 11,000, 12,000, and 12,500 meters, respectively. We see that since all three numbers are multiples of 500, we see that all three runners at the starting point of the track. Therefore, 2,500 is the correct answer.

Alternate Solution:

Let the three runners be all together again after t seconds. Then, the runners will run 4.4t, 4.8t, and 5t meters. The three runners are at the same point despite running different distances, so the pairwise difference between the distances they run must be multiples of 500. In other words, 5t - 4.8t = 0.2t is a multiple of 500, 4.8t - 4.4t = 0.4t is a multiple of 500, and 5t - 4.4t = 0.6t is a multiple of 500.

Let 0.2t = 500k, 0.4t = 500s and 0.6t = 500p. Thus, 500k/0.2 = 2500k, 500s/0.4 = 1250s and 500p/0.6 = 2500p/3 are all integers. The smallest value of p that makes 2500p/3 an integer is p = 3; thus t = (2500 * 3)/3 = 2500. Notice that this implies k = 1 and s = 2.

Answer: C

ScottTargetTestPrep · Answer

Solution: 
 
Let’s call the three runners A, B, and C. The easiest strategy is to check the answer choices.
If it takes 1,000 seconds for them to be at the same place again, then A, B, and C run 4,400, 4,800, and 5,000 meters, respectively. We see that since 5,000 is a multiple of 500, we see that C is at the starting point of the track. However, since neither 4,000 nor 4,800 is a multiple of 500, neither A nor B is at the starting point.

If it takes 1,250 seconds for them to be at the same place again, then A, B, and C run 5,500, 6,000, and 6,250 meters, respectively. We see that since both 5,500 and 6,000 are multiples of 500, we see that A and B are at the starting point of the track. However, since 6,250 is not a multiple of 500, C is not at the starting point.

If it takes 2,500 seconds for them to be at the same place again, then A, B and C run 11,000, 12,000, and 12,500 meters, respectively. We see that since all three numbers are multiples of 500, we see that all three runners at the starting point of the track. Therefore, 2,500 is the correct answer.

Answer: C

proxy__guy123 · Answer

Hi can u please explain why did you work with A and B; B and C;
Why not A and C?

bumpbot · Answer

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Three runners start running simultaneously from the same point on a 50

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