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"Work" Word Problems (page 2 of 4)


A classic "work" problem involves one pipe pouring water into a pool, while another drains it out. It runs something like this:

  • When the tub faucet is on full, it can fill the tub to overflowing in 20 minutes (we'll ignore the existence of the overflow drain). The drain can empty the tub in 15 minutes. Your four-year-old has managed to turn the faucet on full, and the drain was closed. Just as the tub starts to overflow, you run in and discover the mess. You grab the faucet handle, and it comes off in your hand, leaving the water running at full power. You yank the drain open, and run for towels to clean up the overflow. How long will it take for the tub to empty, with the faucet still on but the drain now open?

    Okay, yeah; in "real life" you'd go find the shut-off valve and turn off the water to the whole house, but this is math, not real life. Thinking about this problem, we see that the drain can empty 1/15 of the tub per minute. The faucet can fill 1/20 of the tub per minute. Then, working together, they can empty 1/15 1/20 of the tub per minute. The subtraction indicates that the faucet is actually working against the drain. Copyright Elizabeth Stapel 1999-2011 All Rights Reserved

    Again, let "t" indicate how long it takes to drain the whole tub. Then 1/t is drained per minute. Then 1/15 1/20 = 1/t, so 1/60 = 1/t, and t = 60. That is:

      minutes to complete job:
        faucet:
      20 minutes to fill
        drain:
      15 minutes to empty
        together:
      t minutes to empty

      completed per minute:
        faucet:
      1/20 filled
        drain
      : 1/15 emptied
        together:
      1/t emptied

      adding (or, in this case, subtracting) their labor:

        1/15 1/20 = 1/t

        1/60 = 1/t

        60 = t

    Recall that the time "t" is defined in minutes, so "60" is "60 minutes", or "1 hour".

      It will take an hour to drain the tub.

Another "typical" work problem is the "one guy did part of the job" type. It might run something like this:

  • Two mechanics were working on your car. One can complete the given job in six hours, but the new guy takes eight hours. They worked together for the first two hours, but then the first guy left to help another mechanic on a different job. How long will it take the new guy to finish your car?

    The first guy can do 1/6 per hour. The new guy can do 1/8 per hour. Together, they can do 1/6 + 1/8 = 7/24 per hour.  That is:

     

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      hours to complete job:
        first guy:
      6
        new guy:
      8
        together:
      t

      completed per hour:
        first guy:
      1/6
        new guy:
      1/8
        together:
      1/t

      adding their labor:

        1/6 + 1/8 = 1/t

        7/24 = 1/t

    They worked for two hours, so they got 2( 7/24 ) = 7/12 of the job done. That leaves 5/12 of the job to do.

(I calculated this last bit by using the fact that (fraction of work) = (rate per unit) (number of units). That is, I multiplied how much they can do per hour by the number of hours they worked, to find the fraction of the entire job that they had completed.)

    The new guy can do 1/8 of the job per hour. So the real problem here is figuring out how long the new guy will take to do 5/12 of the job. Let "h" indicate the number of hours he needs. Then the equation is ( 1/8 of the job per hour) times (however many hours) equals ( 5/12 of the job done). That is, I need to set up the equation so the units cancel:

      ( 1/8 job / hour) (h hours) = 5/12 job

      h/8 job = 5/12  job

      h/8 = 5/12

      h = ( 5/12) ( 8/1) = 10/3 = 3 1/3

    It takes the new guy another 3 hours and twenty minutes to finish fixing your car.

By the way, this means that they took a total of [two hours with two guys] plus [three hours and twenty minutes with one guy] equals [seven-and-a-third man-hours] to fix your car.

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Cite this article as:

Stapel, Elizabeth. "'Work' Word Problems." Purplemath. Available from
    http://www.purplemath.com/modules/workprob2.htm. Accessed
 

 



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