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At their respective rates, pump A, B, and C can fulfill an [#permalink]
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29 Jul 2008, 09:22
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At their respective rates, pump A, B, and C can fulfill an empty tank, or pumpout the full tank in 2, 3, and 6 hours. If A and B are used to pumpout water from the halffull tank, while C is used to fill water into the tank, in how many hours, the tank will be empty? A. 2/3 B. 1 C. 3/4 D. 3/2 E. 2



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Re: work and time PS [#permalink]
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29 Jul 2008, 09:30
my answer is C.
I can explain if it is correct.



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Re: work and time PS [#permalink]
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29 Jul 2008, 09:31
my answer is C.
I can explain if it is correct.



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Re: work and time PS [#permalink]
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29 Jul 2008, 16:22
arjtryarjtry wrote: At their respective rates, pump A, B, and C can fulfill an empty tank, or pumpout the full tank in 2, 3, and 6 hours. If A and B are used to pumpout water from the halffull tank, while C is used to fill water into the tank, in how many hours, the tank will be empty? A. 2/3 B. 1 C. 3/4 D. 3/2 E. 2 Since the tank is already half full you have 2/x instead of 1/x; and always substract the small quantity from the large one. Hence in this case water out > in water 2/x = (1/2+1/3)  1/6 2/x = 4/6 we need to find out in how many hours will the tank be empty that was already half full so x/2 x/2 = 6/4 = 3/2 IMO D



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Re: work and time PS [#permalink]
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03 Aug 2008, 10:41
arjtryarjtry wrote: At their respective rates, pump A, B, and C can fulfill an empty tank, or pumpout the full tank in 2, 3, and 6 hours. If A and B are used to pumpout water from the halffull tank, while C is used to fill water into the tank, in how many hours, the tank will be empty? A. 2/3 B. 1 C. 3/4 D. 3/2 E. 2 Lets take work done in one hr: Wa=t/2 Wb=t/3 Wc=t/6 where t=total tank capacity now in 1 hr if C fills and A,B empty the tank effective workdone woll be Wa+Wb  Wc = t/2 +t/3 t/6 =4t/6=2t/3 => to empty complete tank t/2t/3 =3/2 hrs are required IMO D
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Re: work and time PS [#permalink]
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03 Aug 2008, 12:00
C
pump A & B together can empty 5/6 of the tank in 1 hour, or half the tank in 3/5 hour (36 minutes). During 3/5 hour, pump C fills 3/30 of the tank, which causes pump A & B to spend another 6.5 minutes to empty, during this 6.5 minutes ....



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Re: work and time PS [#permalink]
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03 Aug 2008, 18:00
net effect of all these JOBLESS guys working together in 1 hr=1/6(fill)(1/2+1/3)(empty)=2/3 WHAT DOES THIS 2/3 MEAN??? it means that in if these guys work together 1 hr,then the net work done is ve or there is net emptying of the tank. so the work this probelm gives us is 1/2( means that 1/2 of work should be done and ve means that net effect should be emptying) so , no. of hrs=(1/2)/(2/3). this is same as saying if u have to go 100kms and your rate is 50kmph, then u compelte in 100/50=2hrs. =3/4. i finished 1 full bottle in the last 1/2 hr .. so i hope this ans is ok. otherwise , i pray for forgiveness



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Re: work and time PS [#permalink]
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05 Aug 2008, 15:20
Is there an alternate explanation? Not quite sure if I complete follow what was given above. Thanks.



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Re: work and time PS [#permalink]
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05 Aug 2008, 15:56
Oops i didnt notice half filled silly mistake even i get 3/4 hrs option C I opted following method : say t is the capacity of tank A to empty the tank in 1 hr : t/2 B to empty the tank in 1 hr : t/3 C to empty the tank in 1 hr: t/6 now effective work done for 1 hr if C were to fill the tank,A,B were to empty the tank : t/2 + t/3 t/6=4t/6=2t/3 hence time taken to empty half filled tank : t/2 / 2t/3 =3/4 hrs
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Re: work and time PS
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