A classic "work" problem involves one pipe (or hose, or faucet) pouring water into a pool (or tub or sink), while another (or the drain) empties it out.
In such a case, the pipes are working against each other. This means that their labors are subtractive, rather than additive. Otherwise, though, the process for setting up and solving this sort of problem remains the same.
It runs something like this:
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Okay, yeah; in "real life" you'd go find the main water shut-off valve and turn off the water to the whole house, but this is math, not real life.
Thinking about this problem, I see that the drain can empty of the tub per minute. The faucet can fill of the tub per minute. Then, working together, they can empty of the tub per minute.
Note: The subtraction indicates that the faucet is actually working against the drain.
I'll let "t " indicate how long it takes to drain the whole tub, with time being counted in minutes. Then of the tub is drained per minute. Setting things up in the usual way, I get:
minutes to complete job:
faucet: 20 minutes to fill
drain: 15 minutes to empty
together: t minutes to empty
completed per minute:
faucet: filled
drain: emptied
together: emptied
adding (or, in this case, subtracting) their labors:
60 = t
Recalling that the time "t " is defined in minutes, I know that "60" means "60 minutes", or "1 hour".
It will take one hour for the tub to drain.
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A related technique for "work" word problems uses the measuring unit called "man-hours". A man-hour is the labor done by one worker over the period of an hour. If one person works for three hours, this is three man-hours. If one person works for one hour and another works for two hours, this is also three man-hours. If both work for three hours, then this is six man-hours. By multiplying the number of workers by the number of hours worked (or by summing the hours of each employee), we can find the total man-hours devoted to a particular task.
When using man-hours, we have to assume that every worker's productivity is pretty much the same. That is, contrary to the previous examples, we have to assume that the inputs (pipes, contruction workers, typists, whatever) are interchangeable. So, for instance, rather than having one person complete a job in two hours while another takes five hours, we have to assume that every person on the job site can complete the job in two hours.
Let's do an example:
They haven't told me that various employees work at varying levels of productivity. Instead, all the employees are being treated as being equally productive. So I'll need to work in terms of man-hours.
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In the usual course of these projects, nine people each work eight hours a day, for a total of seventy-two man-hours per day. They do this over a period of twenty-five days, for a grand total of 72 × 25 = 1,800 man-hours for the entire project.
For this particular location, though, the crew is being given only fifteen days to complete the built. How many man-hours will they need to work during each of these days? I'll divide to find out:
1,800 ÷ 15 = 120
(The above represents [man-hours for the whole project] divided by [days for the total project], which equals [man-hours per day]. If you're not sure about how these units work, please take a quick detour to this lesson on cancelling units.)
So, for each of the fifteen days, the workers will need to put in one hundred twenty man-hours. There will be twelve workers each day. The 120 man-hours will be evenly spread amongst the twelve workers. Then each worker will be putting in:
120 ÷ 12 = 10 hours
In other words, the workers will be on the job for an extra two hours a day.
They will need to work ten hours each day.
In four hours, the twenty men will have put in eighty man-hours. In those eighty man-hours, they'll have felled thirty trees. How many trees is that, per man-hour? I'll divide to find out:
30 ÷ 80 = 3/8
So three-eighths of a tree is felled per one man-hour.
If four guys leave, then there will be sixteen workers remaining. If they work for six hours, then they'll have put in 16 × 6 = 96 man-hours. Since 3/8 of a tree is felled each man-hour, this gives a total of:
(3/8) × 96 = 36
They will fell thirty-six trees.
There is a variant of this sort of exercise, which is designed to drive students batty. You can see an assorment of them here. I'll do an example:
Nasty, right? But it is do-able. I'll just need to be clear in my steps.
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I'll take "a hen and a half" to mean "1.5 chickens". I'll take "a day and a half" to mean "1.5 days". I'll take "an egg and a half" to mean "1.5 eggs". Clearly, these values don't make practical sense, since "half a chicken" is, well, a dead chicken. But this'll make the math work out okay.
So 1.5 chickens work for 1.5 days. This is 2.25 chicken-days. (I don't think I can say "chicken-days" aloud without giggling, but... whatever.) In that time, they will produce 1.5 eggs. This then gives me:
(1.5 eggs) ÷ (2.25 chicken-days)
= 2/3 eggs per chicken-day
Then one chicken lays two-thirds of an egg per day (or, in practical terms, two eggs every three days).
They want me to figure out how many eggs five hens will lay in six days; in other words, how many eggs will be produced in thirty chicken-days. Since two-thirds of an egg are laid per chicken-day, then:
(2/3) × 30 = 20
The hens will lay twenty eggs.
This last exercise was rather strange, but it worked out in exactly the same way as the man-hour examples. Any time you're faced with workers (or producers, or livestock) that are being viewed as being interchangeable, you can use this man-hour methodology.
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