Thursday, February 5, 2009

Work Problem 3 Solution

In this Algebra work problem solution we have a similar situation in dealing with hours and and figuring out how much of a job is done in 1 hour. The only difference is part of the job is being taken away -- negative --- by the leak. So just subtract this. That's the only difference between this work problem and the others.

So with Pump A in 1 hour it would 1/12th of a job Pump B would do 1/6th of a job and the leak would take away 1/24th of the job.

So algebraically we have:

x/12 + x/6 - x/24 = 1

multiply by 24

2x + 4x - x = 24

5x = 24

x = 24/5 or 4 and 4/5ths

4/5ths of an hour is 60*4/5 or 240/5 = 48 minutes

So the pool would be filled in 4 hours and 48 minutes.


  1. hey who put this equation here? its wrong. because if we compute for their filling rates without the leak, it'll be 4 hours. so you mean with the leak it would be faster to fill the pool? i'll post the equation later

  2. 4 hours w/o leak < 4 and 48 hrs w/ leak

  3. The given solution is right. However, the equation must be 1/x + 1/y - 1/z = 1/n ; where: x = no. of hours for pipe A; y = no. of hours for pipe B; z = no. of hours for the leak; and, n = no. of hours if they are opened at the same time


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