In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?
A) 10 B) 20 C) 30 D) 40 E) 50
Relative speed on circular path
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akpareek...
Are they traveling in the same direction or opposite direction?
If opposite direction:
Sb = 4m/s
Therefore, B will cover 40 m in 10 sec.
Remaining distance = 10040 = 60 m.
Sa = 2 m/s
Since, they are traveling in opposite direction, we can add their speeds
Sab = 4+2 = 6 m/s
Tab = 60/6 = 10 secs
Therefore, time since B started = 10 + 10 = 20 sec...which is option B in your ans choice.
If they are traveling in the same direction, it gets a bit tricky since A's speed is lower than B's speed..which would mean B will complete one circle and then catch up with A.
which is answer choice D.
Hope this helps...
Are they traveling in the same direction or opposite direction?
If opposite direction:
Sb = 4m/s
Therefore, B will cover 40 m in 10 sec.
Remaining distance = 10040 = 60 m.
Sa = 2 m/s
Since, they are traveling in opposite direction, we can add their speeds
Sab = 4+2 = 6 m/s
Tab = 60/6 = 10 secs
Therefore, time since B started = 10 + 10 = 20 sec...which is option B in your ans choice.
If they are traveling in the same direction, it gets a bit tricky since A's speed is lower than B's speed..which would mean B will complete one circle and then catch up with A.
which is answer choice D.
Hope this helps...
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 Brent@GMATPrepNow
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Are these people traveling in opposite directions?akpareek wrote:In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?
A) 10 B) 20 C) 30 D) 40 E) 50
I will assume that they are.
Here's one way to visualize this.
Rather than have them on a circular track with circumference 100 meters, picture them on a straight track that's 100 meters long (all we're doing here is straightening the track).
Runner A is at one end and Runner B is at the other end.
First: Runner B runs for 10 seconds at 4 m/s. In this time, he/she travels 40 meters.
At this point, the runners are only 60 meters apart.
Second: Runner A starts.
Important: The runners are now closing the gap between each other at a rate of 6 m/s (2 m/s + 4 m/s = 6 m/s)
So, at a rate of 6 m/s, we can see that it will take 10 seconds to close the 60 meter gap between them.
This 10 seconds represents the time AFTER runner A starts.
Since Runner B had already be running for 10 seconds, the total time is 20 seconds
Answer = B
Cheers,
Brent
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 Brent@GMATPrepNow
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RUNNING IN THE SAME DIRECTIONakpareek wrote:In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?
A) 10 B) 20 C) 30 D) 40 E) 50
If the two runners are going in the same direction, then we can see that Runner A (the slower runner) will never catch up to Runner B. Instead, the first time they meet will be when Runner B laps Runner A. When this occurs, Runner B will have run 100 meters more than Runner A.
So, let's start with a "word equation"
(Distance B travels) = (Distance A travels) + 100
Let t = number of seconds that elapse after Runner B begins running.
Distance Runner B travels = (time)(rate)
= (t)(4)
= 4t
Note: Since Runner A starts running 10 seconds after Runner B, Runner A's time spent running will equal t  10
Distance Runner A travels = (time)(rate)
= (t10)(2)
= 2t  20
So, we get...
(Distance B travels) = (Distance A travels) + 100
4t = (2t  20) + 100
4t = 2t + 80
2t = 80
t = 40
Answer = D
Cheers,
Brent

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Another way to solve this,Brent@GMATPrepNow wrote:RUNNING IN THE SAME DIRECTIONakpareek wrote:In a circular track of 100 meters A with the speed of 2m/s starts after 10 seconds of start of B with the speed of 4m/s. After how many seconds will they meet first time after the start of B ?
A) 10 B) 20 C) 30 D) 40 E) 50
If the two runners are going in the same direction, then we can see that Runner A (the slower runner) will never catch up to Runner B. Instead, the first time they meet will be when Runner B laps Runner A. When this occurs, Runner B will have run 100 meters more than Runner A.
So, let's start with a "word equation"
(Distance B travels) = (Distance A travels) + 100
Let t = number of seconds that elapse after Runner B begins running.
Distance Runner B travels = (time)(rate)
= (t)(4)
= 4t
Note: Since Runner A starts running 10 seconds after Runner B, Runner A's time spent running will equal t  10
Distance Runner A travels = (time)(rate)
= (t10)(2)
= 2t  20
So, we get...
(Distance B travels) = (Distance A travels) + 100
4t = (2t  20) + 100
4t = 2t + 80
2t = 80
t = 40
Answer = D
Cheers,
Brent
B  4m/s
A 2m/s
Thus extra speed B is traveling with = 2m/s
Now as A starts 10secs later, distance covered by B = 40m.
Now as B is traveling faster and started earlier than A, B only can catch A.
So, remaining distance between A and B= 60m
Relative speed of B =2m/s
So time taken = 30secs.
Total time taken= 10+30=40 secs.
Answer D