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Pump A and Pump B

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by logitech » Sat Jan 10, 2009 7:39 am
SO in 1 minute pump a can empty the 1/A of the tank, leaving (1-1/A) in the tank

( TIME-1 )x(1/A+1/B) = (1-1/A)

(TIME-1) = (AB-B)/(A+B)

TIME = (AB+A)/(A+B)

TIME = A(B+1)/(A+B)
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by Brent@GMATPrepNow » Sat Jan 10, 2009 7:49 am
Nice work, Logitech.
Here's my full solution as well:

After 1 minute, the pool has been emptied by 1/A, or we can say that (A-1)/A remains to be emptied.
After 1 minute, A can empty 1/A of the pool and pump B can empty 1/B of the pool. Combined they can empty 1/A + 1/B of the pool or (A+B)/AB. This means that they can empty the entire pool in AB/(A+B) minutes
But, we don’t need to empty the entire pool. After 1 minute (when both pumps are working together), the pool is only (A-1)/A full, so the length of time to empty it from this point is [AB/(A+B)][(A-1)/A]
To this value, we add the initial 1 minute to get [AB/(A+B)][(A-1)/A] + 1
This simplifies to be B
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by acecoolan » Sat Jan 10, 2009 7:20 pm
Anything wrong doing it this way

A's rate = 1/A
B's rate = 1/B

Say A works for t mins
so B works for t-1 mins

Hence t/A + (t-1)/B = 1 (i.e. total work)

t/A + t/B = 1 + (1/B)
t( A + B)/AB = (B + 1) B

t = A(B+1)/(A+B)

-A
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by logitech » Sat Jan 10, 2009 7:44 pm
acecoolan wrote:Anything wrong doing it this way

A's rate = 1/A
B's rate = 1/B

Say A works for t mins
so B works for t-1 mins

Hence t/A + (t-1)/B = 1 (i.e. total work)

t/A + t/B = 1 + (1/B)
t( A + B)/AB = (B + 1) B

t = A(B+1)/(A+B)

-A
Math can get pretty sexy sometimes and this is one of them! Beautiful work!
LGTCH
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by Brent@GMATPrepNow » Sat Jan 10, 2009 8:01 pm
acecoolan wrote:Anything wrong doing it this way

A's rate = 1/A
B's rate = 1/B

Say A works for t mins
so B works for t-1 mins

Hence t/A + (t-1)/B = 1 (i.e. total work)

t/A + t/B = 1 + (1/B)
t( A + B)/AB = (B + 1) B

t = A(B+1)/(A+B)

-A
Beautiful solution - very eloquent!
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by Ian Stewart » Sat Jan 10, 2009 10:18 pm
I'd almost certainly have done this algebraically, but it's also quick to choose some extreme numbers. For example, if A does next to no work - say A = 1,000,000 - and B fills the pool in a minute (B = 1), then the job is going to take a total of just shy of two minutes, because A does next to nothing, and B doesn't start working until one minute is up. It's then quick to rule out all the answers besides B.
Last edited by Ian Stewart on Sun Jan 11, 2009 5:11 am, edited 1 time in total.
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by cramya » Sat Jan 10, 2009 10:25 pm
Ian, thats one smart thinking, seriously... Thanks for the alternate approach.

Regards,
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by logitech » Sun Jan 11, 2009 1:10 am
That's how Ian Stewart rolls! Great explanation!
LGTCH
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