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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:
Okay, yeah; in "real life" you'd go find the shutoff 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 19992011 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: completed per minute: 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:
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:
hours to complete job: completed per hour: 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 [sevenandathird manhours] to fix your car. << Previous Top  1  2  3  4  Return to Index Next >>



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