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31 Dec 2020, 12:30
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D
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Difficulty:
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55% (03:15) correct
45%
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wrong
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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?
A. m/10 and m/5 only
B. m/5 and m/4 only
C. m/20, m/10, m/5, and m/4
D. m/20, m/10, 3m/20, and m/5
E. m/20, m/10, 3m/20, m/5, and m/4
A. m/10 and m/5 only
B. m/5 and m/4 only
C. m/20, m/10, m/5, and m/4
D. m/20, m/10, 3m/20, and m/5
E. m/20, m/10, 3m/20, m/5, and m/4
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31 Dec 2020, 15:14
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Given line
0----M/5--M/4--2M/5----2M/4--3M/5----3M/4--4M/5----M
So, length between 0 and M/5= M/5 (possible)
length between M/4 and M/5= M/20 (possible)
length between 3M/4 and 3M/5= 3M/20 (possible)
length between 3M/5 and 2M/4= M/10 (possible)
But, we can't get M/4 between two consecutive markings. So, I think D.
0----M/5--M/4--2M/5----2M/4--3M/5----3M/4--4M/5----M
So, length between 0 and M/5= M/5 (possible)
length between M/4 and M/5= M/20 (possible)
length between 3M/4 and 3M/5= 3M/20 (possible)
length between 3M/5 and 2M/4= M/10 (possible)
But, we can't get M/4 between two consecutive markings. So, I think D.
Deepakjhamb
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31 Dec 2020, 19:52
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answer is E). m/20, m/10, 3m/20, m/5, and m/4
we can take m as 20 since we have 4 and 5 in denominator and by deploying number line we can find that e ) is possible
and hence solution
we can take m as 20 since we have 4 and 5 in denominator and by deploying number line we can find that e ) is possible
and hence solution
