GMAT Club Around The World!!! For the convenience of walkers and runne

IMO D

let the length of the track m = 100
markers placed at every fourth of its length => markers will be placed at: {0,25,50,75,100}
markers placed at every fifth of its length => markers will be placed at: {0,20,40,60,80,100}
=> consecutive markers present are present at the locations: {0,20,25,40,50,60,75,80,100}
=> distance between consecutive markers: {20,5,15,0,0,5,5,20}

A. m/10 and m/5 only -Keep
m=100
=> m/10 = 10 --- travel between consecutive markers 40 and 50 or 50 and 60- possible
and m/5=20 --- travel between consecutive markers 0 and 20 or 80 and 100- possible
m/20 and 3m/20 are also possible as mentioned in option D- discard this option

B. m/5 and m/4 only
--- m/10 is also possible - wrong option (calculated in option A)

C. m/20, m/10, m/5, and m/4
m=100
m/20 = 5 -- travel between consecutive markers 25 and 20 - possible
m/4 = 25 -- maximum possible distance between any two markers = 20 --- not possible

D. m/20, m/10, 3m/20, and m/5
m=100
m/20 - possible (verified in option C)
m/10 - possible (verified in option A)
m/5 - possible (verified in option A)
3m/20 - 15 - possible (between markers 80 and 75 or 40 and 25)
Correct

E. m/20, m/10, 3m/20, m/5, and m/4
m=100
m/4 = 25 - not possible (calculated in option C)

The fourth's markers will be at 1/4, 2/4, 3/4
The fifth's markers will be at 1/5, 2/5, 3/5, 4/5

To find the distance a runner can run between markers, find the distance between each consecutive marker.

1/5 - 0 = 1/5
1/4 - 1/5 = 1/20
2/5 - 1/4 = 3/20
2/4 - 2/5 = 2/20 = 1/10
3/5 - 2/4 = 2/20 = 1/10
3/4 - 3/5 = 3/20
4/5 - 3/4 = 1/20
1 - 4/5 = 1/5

Hence only 4 different lengths can be run.

IMO Option D

Correct answer D

For the convenience of walkers and runners, a linear exercise track m meters in length has a distance marker placed at the beginning and end of the track, as well as every fourth of its length and at every fifth of its length. If a runner were to travel the total length of a section of the track between consecutive markers, which of the following expresses the distance the runner could travel, in meters?

Assuming a total distance of 100m

The marker will be at
0m 25m , 50 m and 75m, 100m
and
0m 20 m , 40m , 60m, 80 m, 100m

difference between marker: 20m, 5 m, 10m, 10m, 15m, 20m

As given the distance is m, however, assumed for simplicity 100m

divide the difference by 100= m/5, m/20, m/10 and 3m/20

The length of linear exercise track = m meters
The track is marked at every fourth of its length and every fifth of its length
So the markers will be at distance m/4, m/2, 3m/4, m/5, 2m/5, 3m/5, 4m/5 and the end of the track
To find the distance the runner could travel between consecutive markers we have to find the difference of length in between the markers,
So the length will be m/5, m/20, 3m/20, m/10

So the answer will be D

According to the question, markers are placed at the start, the end, every fourth, and every fifth of the length. thus, at the following lengths markers are set:

m/4, 2m/4, 3m/4, m

m/5, 2m/5, 3m/5, 4m/5, m

thus, at the distance between markers m/20, 2m/20, 3m/20, 4m/20, and m/4 (the last one also because the marker is present at the start).

IMO, option E is the correct answer.

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