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# A swimming pool has three pumps, A, B and C. Working

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A swimming pool has three pumps, A, B and C. Working [#permalink]  14 Sep 2006, 13:22
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A swimming pool has three pumps, A, B and C. Working together at their constant rates, A and B can drain the swimming pool in 10 hours, whereas A and C can drain the pool in 12 hours. How long does it take A to drain the pool?

(1) B and C, working together, can drain the pool in 15 hours.
(2) The time it would take 3 B's to drain the pool is equal to the time it would take 4 C's to do so.

Last edited by kevincan on 14 Sep 2006, 13:42, edited 1 time in total.
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In statment (1), is it A C working together or B C ?
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Re: DS: Swimming Pool Drainage [#permalink]  14 Sep 2006, 13:56
kevincan wrote:
A swimming pool has three pumps, A, B and C. Working together at their constant rates, A and B can drain the swimming pool in 10 hours, whereas A and C can drain the pool in 12 hours. How long does it take A to drain the pool?

(1) B and C, working together, can drain the pool in 15 hours.
(2) The time it would take 3 B's to drain the pool is equal to the time it would take 4 C's to do so.

I will explain later. B gets tricky.
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Re: DS: Swimming Pool Drainage [#permalink]  14 Sep 2006, 17:44
kidderek wrote:
kevincan wrote:
A swimming pool has three pumps, A, B and C. Working together at their constant rates, A and B can drain the swimming pool in 10 hours, whereas A and C can drain the pool in 12 hours. How long does it take A to drain the pool?

(1) B and C, working together, can drain the pool in 15 hours.
(2) The time it would take 3 B's to drain the pool is equal to the time it would take 4 C's to do so.

I will explain later. B gets tricky.

Hmmm, I might be over thinking, but I am going to retract the D answer and get back to it.
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Given
Ra = 1/Ta
Rb = 1/Tb
Rc = 1/Tc

Where Ta, Tb & Tc are times to drain the pool by themselves.

Given
Ra + Rb = 1/10 .... (1)

Rb + Rc = 1/12 ... (2)

=> Rb -Rc = 1/60 .... (3)

S1:
Rb + Rc = 1/15

From (3) and above eqn we can find Rb.

From Rb and (1) we can find Ra.
Sufficient.

S2:
3Rb = 4Rc

3Rb - 4Rc = 0

From (2) multiplying by 4:

4Rb +4Rc = 1/3
3Rb -4Rc = 0

Can solve for Rb... and from (1) can solve for Ra.

Sufficient.

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Arrived at D, but didn't work it out as detailed as Haas_mba07.
Statement 1: Gives you B+C. Imagine 2As, 1 B and 1 C [(A+B) and (A+C)] draining the tank and two more pumps (B+C) pouring in water, and you effectively negate B and C and can figure out how long it takes A to drain the tank. So, sufficient.
Statement 2: Gives you B in terms of C or vice versa. Substitute appropriately to find out A.

Wow! Typing this out took much longer than figuring it out in my head. Hope it makes sense to most people at least.
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D..for this reason...

1 and 2 result in 3 equations w/ 3 distinct variables.
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Hass_Mba or Anyone,

Pardon my lack of knowledge ,but how did u get to #3 Higlighted below

--Snippet from solution earlier
Given
Ra = 1/Ta
Rb = 1/Tb
Rc = 1/Tc

Where Ta, Tb & Tc are times to drain the pool by themselves.

Given
Ra + Rb = 1/10 .... (1)

Rb + Rc = 1/12 ... (2)

=> Rb -Rc = 1/60 .... (3)

could u pls explain ?
thanks
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