Two bodies A and B start moving at the same time from P to Q, with speeds 20m/s and 30 m/s respectively. Distance between P and Q is 100m. Initially both A & B are at P. They move between P & Q only. For example, once B reaches Q, it turns around and start moving towards P, and so on. Find the distance traveled by A till their first meeting.
A)70m
B)80m
C)90m
D)95m
E)100m
Can we solve this question based on relative speed concept?
B
Two bodies A and B start moving at the same time from P...
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Last edited by resolehtmai on Tue Oct 01, 2013 1:20 am, edited 2 times in total.

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Are these wordings from official source. This question is too ambiguous. Distance travelled by A till their first meet? they will never meet as the speed of B is more than speed of A and they are starting from the same point. If they were starting from opposite point then it makes sense. Or if the question clarifies that the distance travelled by A till B finished the race to point Q.resolehtmai wrote:Two bodies A and B start moving at the same time from P to Q, with speeds 20m/s and 30 m/s respectively. Distance between P and Q is 100m. Initially both A & B are at P. Find the distance traveled by A till their first meeting.
A)70m
B)80m
C)90m
D)95m
E)100m
B
Please clarify.

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I don't know how this can be fitted into the relative speed calculation method, but the following is my approach to the problem.resolehtmai wrote:Two bodies A and B start moving at the same time from P to Q, with speeds 20m/s and 30 m/s respectively. Distance between P and Q is 100m. Initially both A & B are at P. They move between P & Q only. For example, once B reaches Q, it turns around and start moving towards P, and so on. Find the distance traveled by A till their first meeting.
A)70m
B)80m
C)90m
D)95m
E)100m
Can we solve this question based on relative speed concept?
B
By the main statement, the speed of A and B is in the ratio of 2:3, so something like this happens:
A B
20 m 30 m after 1 second
40 m 60 m after 2 seconds
60 m 90 m after 3 seconds
80 m 120m after 4 seconds this means B reached the point "Q" and turned back.
So A had travelled 80 m and B had travelled 120 m when they first met.
Hope i was of some help.
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The distance between P and Q = 100 meters.resolehtmai wrote:Two bodies A and B start moving at the same time from P to Q, with speeds 20m/s and 30 m/s respectively. Distance between P and Q is 100m. Initially both A & B are at P. They move between P & Q only. For example, once B reaches Q, it turns around and start moving towards P, and so on. Find the distance traveled by A till their first meeting.
A)70m
B)80m
C)90m
D)95m
E)100m
Can we solve this question based on relative speed concept?
B
For A and B to MEET, the SUM of their distances must be a MULTIPLE OF 100 meters.
For example, if A were to travel halfway to Q (50 meters), and B were to travel to Q and halfway back (for a total of 150 meters), A and B would MEET after traveling a combined distance of 50+150 = 200 meters  a MULTIPLE OF 100.
B's rate/A's rate = 30/20.
B's rate = (3/2)(A's rate).
Implication:
Every second, B travels 3/2 of the distance traveled by A.
We can plug in the answers, which represent the distance traveled by A:
70, 80, 90, 95, 100
Since B travels 3/2 of the distance traveled by A, the options for B's distance are 3/2 of the answer choices:
105, 120...
We can stop here.
The sum of the distances in red is a multiple of 100:
80+120 = 200.
The correct answer is B.
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Thanks for the inputs.
Can you guys also tell me what's wrong with the following approach:
Relative speed of A and B = 3020 = 10 m/s
relative distance they need to travel for the first meeting = 200m.
so relative time taken= 200/10= 20 sec.
now in 20 seconds, distance traveled by A = speed of A* 20 sec
= 20*20=40 m
but this is different from the correct answer.
where did i go wrong?
Can you guys also tell me what's wrong with the following approach:
Relative speed of A and B = 3020 = 10 m/s
relative distance they need to travel for the first meeting = 200m.
so relative time taken= 200/10= 20 sec.
now in 20 seconds, distance traveled by A = speed of A* 20 sec
= 20*20=40 m
but this is different from the correct answer.
where did i go wrong?
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When elements COMPETE, we SUBTRACT their rates.resolehtmai wrote:Thanks for the inputs.
Can you guys also tell me what's wrong with the following approach:
Relative speed of A and B = 3020 = 10 m/s
relative distance they need to travel for the first meeting = 200m.
so relative time taken= 200/10= 20 sec.
now in 20 seconds, distance traveled by A = speed of A* 20 sec
= 20*20=40 m
but this is different from the correct answer.
where did i go wrong?
But A and B are NOT competing.
Quite the opposite: they are WORKING TOGETHER to travel a COMBINED DISTANCE that is a multiple of 100.
Since they are working together, we ADD their rates:
20+30 = 50 meters per second.
Since the least possible distance that must be traveled for A and B to meet = 200 meters, the time required = 200/50 = 4 seconds.
In 4 seconds, the distance traveled by A = r*t = 20*4 = 80 meters.
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Wow. This is the approach i had been looking for.GMATGuruNY wrote: When elements COMPETE, we SUBTRACT their rates.
But A and B are NOT competing.
Quite the opposite: they are WORKING TOGETHER to travel a COMBINED DISTANCE that is a multiple of 100.
Since they are working together, we ADD their rates:
20+30 = 50 meters per second.
Since the least possible distance that must be traveled for A and B to meet = 200 meters, the time required = 200/50 = 4 seconds.
In 4 seconds, the distance traveled by A = r*t = 20*4 = 80 meters.
Struck like a bullet in my head.
perfect explanation Mitch.
I had been banging my head over this thing for hours.

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There is an error in your approach too. You have calculated the total time 20seconds by taking relative speed 10m/s....this speed is good when they are going in the same direction from P to Q. but when Q has already reached and turned around and now walking back to P their relative speed will add up and it will become 30+20=50.resolehtmai wrote:Thanks for the inputs.
Can you guys also tell me what's wrong with the following approach:
Relative speed of A and B = 3020 = 10 m/s
relative distance they need to travel for the first meeting = 200m.
so relative time taken= 200/10= 20 sec.
now in 20 seconds, distance traveled by A = speed of A* 20 sec
= 20*20=40 m
but this is different from the correct answer.
where did i go wrong?
So to answer your question here is the error. they travelled 200m but all those 200 meters their relative speed was not the same.

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ok. i understand now that why my approach wasn't giving me the correct answer.rakeshd347 wrote: There is an error in your approach too. You have calculated the total time 20seconds by taking relative speed 10m/s....this speed is good when they are going in the same direction from P to Q. but when Q has already reached and turned around and now walking back to P their relative speed will add up and it will become 30+20=50.
But as per Mitch, there can be one relative speed considered for all those 200 m,rakeshd347 wrote: So to answer your question here is the error. they travelled 200m but all those 200 meters their relative speed was not the same.
Do you still think we can't consider a single relative speed for those 200m? Coz the above approach by Mitch got us the correct answer.When elements COMPETE, we SUBTRACT their rates.
But A and B are NOT competing.
Quite the opposite: they are WORKING TOGETHER to travel a COMBINED DISTANCE that is a multiple of 100.
Since they are working together, we ADD their rates:
20+30 = 50 meters per second.
Since the least possible distance that must be traveled for A and B to meet = 200 meters, the time required = 200/50 = 4 seconds.
In 4 seconds, the distance traveled by A = r*t = 20*4 = 80 meters.
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Another Approach to solve this question:
Let x be delta distance from Q
Since, travel time is same
Time of A = Time of B
(100x)/20 = (100+x)/30
300  3x = 200 + 2x
x = 20
A's Distance: 100  20 = 80
Answer [spoiler]{B}[/spoiler]
resolehtmai, the method you applied on which Rakesh corrected you and the method which Mitch suggested is bit different. Mitch has nicely converted this question similar to Work, Rate & Time Question, where A & B do some work.
Workdone by A + Workdone by B = Total workdone
20(t) + 30(t) = 200 [time is same]
50(t) = 200 = 4 Seconds
A's distance = 20*4 = 80M
Let x be delta distance from Q
Since, travel time is same
Time of A = Time of B
(100x)/20 = (100+x)/30
300  3x = 200 + 2x
x = 20
A's Distance: 100  20 = 80
Answer [spoiler]{B}[/spoiler]
resolehtmai, the method you applied on which Rakesh corrected you and the method which Mitch suggested is bit different. Mitch has nicely converted this question similar to Work, Rate & Time Question, where A & B do some work.
Workdone by A + Workdone by B = Total workdone
20(t) + 30(t) = 200 [time is same]
50(t) = 200 = 4 Seconds
A's distance = 20*4 = 80M
R A H U L

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You are welcome!!!.. We all gave you so many different approaches... now you can select the one you found easy to understand and apply in future.resolehtmai wrote:i hoped to get just 1 method clear.
instead i got so many approaches.
thanks members.
R A H U L