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17 Sep 2019, 11:28
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Two people are walking around a circular track. They start at 8 A.M. from the same point but travel in opposite directions. The first person walks at a speed of two rounds per hour and the second person walks at a speed of three rounds per hour. How many times will they cross each other before the clock shows 9:30 A.M.?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
17 Sep 2019, 11:45
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First person's speed is 1/30
and second one's is 1/20.
So, (1/20 + 1/30)×X =1
5/60*x=1
X=60/5=12
That is they meet each other first time, after 12 minutes.
9:30-8=1:30 or 90 minutes
90÷12=~7
Option C
Posted from my mobile device
and second one's is 1/20.
So, (1/20 + 1/30)×X =1
5/60*x=1
X=60/5=12
That is they meet each other first time, after 12 minutes.
9:30-8=1:30 or 90 minutes
90÷12=~7
Option C
Posted from my mobile device
17 Sep 2019, 17:25
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Assume the perimeter of the track=x
Speed of first person= \(\frac{2*x}{1}\)=2x
Speed of second person= \(\frac{3*x}{1}\)=3x
Time taken by them to meet first time=\(\frac{x}{(2x+3x)}\)=\(\frac{1}{5}\) hrs= 12 minutes
Number of times they cross each other before 9:30AM= [\(\frac{1.5*60}{12}\)]=[7.5]=7
Speed of first person= \(\frac{2*x}{1}\)=2x
Speed of second person= \(\frac{3*x}{1}\)=3x
Time taken by them to meet first time=\(\frac{x}{(2x+3x)}\)=\(\frac{1}{5}\) hrs= 12 minutes
Number of times they cross each other before 9:30AM= [\(\frac{1.5*60}{12}\)]=[7.5]=7
Hovkial
Two people are walking around a circular track. They start at 8 A.M. from the same point but travel in opposite directions. The first person walks at a speed of two rounds per hour and the second person walks at a speed of three rounds per hour. How many times will they cross each other before the clock shows 9:30 A.M.?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
