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Re: GRE Math Challenge #27-Working alone jerry can complete [#permalink]
2
Working alone Jerry can complete a work in 6 minutes.Working alone ,Adam can complete a work in 8 minutes.Working together,Jerry leaves the work after 2 minutes.How long will it take Adam to complete the work?

LCM of 8 and 6 is 24
so W=24

Jerry:
W=RT
24=R6
R=4 per min

Adam:
W=RT
24=R8
R=3 per min

Total Rate = R1 + R2
R=3+4
R=7 per min

If they work for 2 mins together with the same rate they can complete:
W=RT
W=7*2
W=14

14 units they can complete together. So, 24-14 = 10 units will remain. Adam's rate is 3 therefore,
10=3*T
T=10/3 min = 3.33*60 = 200 seconds

Originally posted by Farina on 16 Jun 2020, 22:45.
Last edited by Farina on 19 Aug 2020, 22:42, edited 2 times in total.
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Re: GRE Math Challenge #27-Working alone jerry can complete [#permalink]
I feel the question is kinda misleading in the end. It should ask for how long did Adam work MORE than Jerry in the job together. Because I tried to add Adam's total time as '2 minutes with jerry' PLUS '200 seconds' individually and got an answer of 320 seconds.
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Re: GRE Math Challenge #27-Working alone jerry can complete [#permalink]
1
Let the work be to clean 48rooms since 48 is a multiple of 6 & 8.
So for Jerry, he cleans 48rooms in 6mins, that's 8rooms/min
Adam cleans 48rooms in 8mins, which is 6rooms/min.

Working together for 2 minutes before Jerry leaves, their combined respective cleaning rates per minute cleans a total of 18 rooms (Jerry 8*2 = 16 + Adam 6*2 = 12). So the new total of rooms to be cleaned by Adam alone is 48 - 28 = 20 rooms. But Adam cleans at 6rooms/min. Thus the time it'll take Adam to clean 20 rooms = 20/6 = 3.3mins or 200 seconds. Hope this helps
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Re: GRE Math Challenge #27-Working alone jerry can complete [#permalink]
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