What should be the length of the race, if it is run on a track of 400

Question

What should be the length of the race, if it is run on a track of 400 meters and it should end exactly on the 5th meeting of A and B, both of who run clockwise with speeds of 35 m/s and 45 m/s.

A. 7 km 
B. 8 km 
C. 9 km 
D. 10 km 
E. 12 km

CareerGeek · Accepted Answer

Both of them will first meet when B travels a round more than A (400 meters).
--> Let the distance covered by A when they first meet = d & time taken be 't'
-->  t = \frac{d}{35}  ....... (1)

Also, B covers 'd + 400' meters in the same time
-->  t = \frac{(d + 400)}{45}  ....... (2)

Using (1) & (2),
-->  \frac{(d + 400)}{45} = \frac{d}{35} 
-->  35d + 35*400 = 45d 
-->  10d = 35*400 
-->  d = 35*40 = 1400

So, distance traveled by B at first meet =  d + 400 = 1400 + 400 = 1800  meters
At 5th meeting, distance traveled by B =  5*1800 = 9000  meters =  9  km

IMO Option C

Kinshook · Answer

Asked: What should be the length of the race, if it is run on a track of 400 meters and it should end exactly on the 5th meeting of A and B, both of who run clockwise with speeds of 35 m/s and 45 m/s.

Time taken to meet each time = 400 / 10 = 40 seconds
Time taken for 5th meeting = 5*40 = 200 seconds

Length of the race = distance covered in 5th meeting = 45*200 m = 9 km

IMO C

GMATGuruNY · Answer

We can PLUG IN THE ANSWERS, which represent the distance that B must travel to meet A five times.
When the correct answer is plugged in, B will travel five more 400-meter loops than A -- in other words, 2000 more meters than A.

Since B's rate = 45 meters per second, it is VERY likely that the correct distance in meters will be a MULTIPLE OF 45.
Only C -- 9000 meters -- is a multiple of 45.

C: 9000 meters
Since B's rate = 45 meters per second, the time for B to travel 9000 meters =  \frac{distance}{rate} = \frac{9000}{45} = 200  seconds.
Since A's rate = 35 meters per second, the distance traveled by A in 200 seconds = (rate)(time) = 35*200 = 7000 meters.
Success!
B travels 2000 more meters than A.

C

pierjoejoe · Answer

silly mistake here.
meeting the fifth time in a circular track, just means that the GAP distance between the 2 casrs is 5 times the length of the track. in this case 5*400 = 2000. this is the GAP that need to be "shirnked" to zero. in this case the relative speed is 45-35 = 10. meaning that the total time the race would take is 
2000/10 = 200 seconds.

NOW the minimum distance the cars should be able to travel around the track for this to happen is the distance travelled by the FASTEST in 200 seconds. So the distance is 
45*200 = 9000 meters --> 9 km.

i made a silly mistake because without reasonning at the end i just used the speed of the slowest. but using that speed i just find 7 km, that is the distance that the slowest one has travelled.

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What should be the length of the race, if it is run on a track of 400

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