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29 Mar 2019, 01:37
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (03:02) correct
61%
(03:15)
wrong
based on 59
sessions
History
Date
Time
Result
Not Attempted Yet
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?
(A) 1/16
(B) 1/8
(C) 3/16
(D) 1/4
(E) 5/16
(A) 1/16
(B) 1/8
(C) 3/16
(D) 1/4
(E) 5/16
03 Apr 2019, 04:03
Kudos
Bookmarks
Bunuel
Key points
1. Rachel and Robert start from the same line at the same time.
2. Rachel completes a lap every 90 seconds.
3. Robert completes a lap every 80 seconds.
4. 1/4th of the track centered on the starting line is captured by the photographer.
Inference: So, 1/8th of the track on either side of the starting line is captured in the photo.
5. The photographer captures it sometime in the 60 second window between 10 minutes and 11 minutes i.e., 600 seconds and 660 seconds
Approach: Let us find the time interval between 600 seconds and 660 seconds in which Rachel and Robert will be running in the region that is 1/4th of the length of the track centered on the starting line. i.e., 1/8th of the track length on either side of the starting line.
Robert completes 1 lap in 80 seconds. So, Robert will be at the starting line at the 640th second (after 8 laps). Robert will take 10 seconds to run 1/8th of a lap. So from 630 seconds to 650 seconds Robert will be in the region of the ground that is captured by the photographer.
Rachel completes 1 lap in 90 seconds. So, Rachel will be at the starting line at the 630th second (after 7 laps). Rachel will take 90/8 seconds to run 1/8th of a lap. So, from (630 - \(\frac{90}{8}\)) seconds to (630 + \(\frac{90}{8}\)) seconds Rachel will be in the region of the ground that is captured by the photographer.
So, between 630 seconds and 630 + \(\frac{90}{8}\) seconds both Rachel and Robert are in the region of the ground that is captured by the photographer.
i.e., for a duration of \(\frac{90}{8}\) seconds out of the 60 seconds both of them are in the frame captured by the photographer.
Required probability = {time window in which both Rachel and Robert are in the favorable zone}/{time window in which the photographer captures the picture}
= {90/8}/60 = \(\frac{3}{16}\)
03 Apr 2019, 05:12
Kudos
Bookmarks
what does this mean "picture that shows one-fourth of the track, centered on the starting line"
I tried to visualies it like this , but got a wrong answer :
Rachel will cover 2/3 lap every minute
Robert : 3/4 lap every minute ,
I tried to know their positions after 10 and 11 minutes :
after 10 minutes : Rachel will cover 20/3 laps , that means 6 laps + 2/3
Robert will cover 30/4 laps , that means 7 laps +1/2
now the next minute :Robert wil cover 2/3 lap , we want him to be in the 1/4 area during this minute,
so the probability is (1/4) / (2/3 )
the same for Robert : (1/4) / ( 3/4 )
p = both in the 1/4 area , so p=(1/4) / (2/3 ) * (1/4) / ( 3/4 ) =1/8
can anyone let me know why my approach is wrong
Thanks ,
I tried to visualies it like this , but got a wrong answer :
Rachel will cover 2/3 lap every minute
Robert : 3/4 lap every minute ,
I tried to know their positions after 10 and 11 minutes :
after 10 minutes : Rachel will cover 20/3 laps , that means 6 laps + 2/3
Robert will cover 30/4 laps , that means 7 laps +1/2
now the next minute :Robert wil cover 2/3 lap , we want him to be in the 1/4 area during this minute,
so the probability is (1/4) / (2/3 )
the same for Robert : (1/4) / ( 3/4 )
p = both in the 1/4 area , so p=(1/4) / (2/3 ) * (1/4) / ( 3/4 ) =1/8
can anyone let me know why my approach is wrong
Thanks ,
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