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Pumps A and B

Expert replies
by heshamelaziry » Sat Nov 28, 2009 12:35 pm
Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will take to empty the pool?

OA[spoiler] A(B + 1)/ A + B[/spoiler]
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Source: — Problem Solving |

by aspirant1 » Sat Nov 28, 2009 9:28 pm
rate pump A = 1/A
rate pump B = 1/B

Pump A worked for 1 minute alone, so work done = 1/A (amount of water emptied by pump A for 1 minute)

remaining water = 1-(1/A)

Both pumps working together = 1+ [(1- (1/A)) / ((1/A)+(1/B))] = A(1+B)/(A+B)

good one!
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by ssuarezo » Mon Nov 30, 2009 12:29 pm
aspirant1 wrote:rate pump A = 1/A
rate pump B = 1/B

Pump A worked for 1 minute alone, so work done = 1/A (amount of water emptied by pump A for 1 minute)

remaining water = 1-(1/A)

Both pumps working together = = A(1+B)/(A+B)

good one!
Hi Aspirant:

You have (remaining water/sum of rates) and you add 1 because of the minute already gone ! .. I didn't get it ... Thank you.

Silvia

Silvia.
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by aspirant1 » Thu Dec 03, 2009 9:03 pm
yeah, pump A has already worked for 1 minute
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