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At their respective rates, pump A, B, and C can fulfill an

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At their respective rates, pump A, B, and C can fulfill an [#permalink]

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New post 29 Jul 2008, 09:22
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A
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C
D
E

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At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full 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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New post 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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New post 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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New post 29 Jul 2008, 16:22
arjtryarjtry wrote:
At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full 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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New post 03 Aug 2008, 10:41
arjtryarjtry wrote:
At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full 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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New post 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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New post 03 Aug 2008, 18:00
net effect of all these JOBLESS :lol: 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 :twisted:
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Re: work and time PS [#permalink]

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New post 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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New post 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   [#permalink] 05 Aug 2008, 15:56
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