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Two bodies A and B start moving at the same time from P...

by resolehtmai » Tue Oct 01, 2013 12:55 am
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
Last edited by resolehtmai on Tue Oct 01, 2013 1:20 am, edited 2 times in total.
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by rakeshd347 » Tue Oct 01, 2013 1:05 am
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
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.
Please clarify.
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by resolehtmai » Tue Oct 01, 2013 1:22 am
Modified the question for clarity. Please try now.
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by vinay1983 » Tue Oct 01, 2013 2:16 am
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
I don't know how this can be fitted into the relative speed calculation method, but the following is my approach to the problem.


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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by GMATGuruNY » Tue Oct 01, 2013 2:52 am
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
The distance between P and Q = 100 meters.
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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by resolehtmai » Tue Oct 01, 2013 3:10 am
Thanks for the inputs.
Can you guys also tell me what's wrong with the following approach:
Relative speed of A and B = 30-20 = 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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by GMATGuruNY » Tue Oct 01, 2013 3:27 am
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 = 30-20 = 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?
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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by resolehtmai » Tue Oct 01, 2013 3:46 am
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.
Wow. This is the approach i had been looking for.
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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by rakeshd347 » Tue Oct 01, 2013 5:00 am
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 = 30-20 = 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?
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.
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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by resolehtmai » Wed Oct 02, 2013 12:19 am
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.
ok. i understand now that why my approach wasn't giving me the correct answer.
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.
But as per Mitch, there can be one relative speed considered for all those 200 m,
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.
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.
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by theCodeToGMAT » Wed Oct 02, 2013 1:12 am
Another Approach to solve this question:

Let x be delta distance from Q

Since, travel time is same

Time of A = Time of B
(100-x)/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
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by resolehtmai » Wed Oct 02, 2013 4:56 am
i hoped to get just 1 method clear.
instead i got so many approaches.
thanks members. :)
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by theCodeToGMAT » Wed Oct 02, 2013 5:02 am
resolehtmai wrote:i hoped to get just 1 method clear.
instead i got so many approaches.
thanks members. :)
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. :)
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