"Work" Word Problems
Purplemath
It's often easier to do these "work" word problems when you're going through the chapter, section by section, in the textbook. They'll cover a particular technique in a given section, and then ask only those types of questions in that section's homework. Then, in the next section, they'll cover another technique and you'll do exercises that use only that method. Then, when you get to the chapter review (or the next test), the questions aren't grouped according to the method that works best. And everything falls apart.
But all of these exercises have some basic commonalities. And, if you work logically and clearly, you can usually think your way to the answer.
Below are a few examples, so you can see what that "thinking your way to the answer" process looks like in practice.
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Working together, Bill and Tom painted a fence in 8 hours. Last year, Tom painted the fence by himself. The year before, Bill painted it by himself, but took 12 hours less than Tom took. How long did Bill and Tom take, when each was painting alone?
Bill's time for completion is stated in terms of Tom's time for completion. So I'll pick a variable for Tom's time; I'll use T for the number of hours Tom needs. Then Bill's time will be T − 12. So I can set up the completion-time info as:
completion time, in hours:
Tom: T
Bill: T − 12
together: 8
Inverting, I can tabulate their per-hour completion rates:
completed per hour:
Tom:
Bill:
together:
This works out to be a simple additive-labor exercise. So I'll add their individual completion rates, and set this equal to the "together" completion rate:
I can multiply through by the common denominator of 8T(T − 12):
8(T − 12) + 8T = T(T − 12)
8T − 96 + 8T = T 2 − 12T
0 = T 2 − 28T + 96
0 = (T − 4)(T − 24)
T = 4, 24
(Review how to factor quadratics, if you're not sure how I just got to that last line above.)
I've derived two solution values. But if I say that Tom takes four hours to complete the job by himself, (1) this won't make any sense in light of the two of them together needing eight hours (it's not possible that one does it in half the time as the two of them together), and (2) this wouldn't make any sense in light of Bill's time being twelve hours less than Tom's (since Bill can't work for a negative number of hours). That means that this "T = 4" solution is "extraneous" (pronounced "ekk-STRAY-nee-uss"), meaning "valid mathematically, but pointless as far as our situation is concerned".
So I can ignore the T = 4 solution. Instead, I'll use the other solution, which says that Tom takes twenty-four hours to paint the whole house himself. Then:
Tom takes twenty-four hours, and Bill takes twelve hours.
This next example is stated a bit differently:
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Ben takes 2 hours to wash 500 dishes, and Frank takes 3 hours to wash 450 dishes. How long will they take, working together, to wash 1000 dishes?
For this exercise, we are given how many items in a job can be done in one time unit, rather than how much of a job can be completed in one time unit. But the thinking process is otherwise the same.
Ben can wash 500 dishes in 2 hours, so, by dividing, I find that he can do 250 dishes per hour. Similarly, Frank can wash 450 dishes in 3 hours, so he can do 150 dishes per hour. Working together, they can do 250 + 150 = 400 dishes an hour. That is:
dishes per hour:
Ben: 500 ÷ 2 = 250
Frank: 450 ÷ 3 = 150
together: 250 + 150 = 400
They can do 400 dishes each hour. I need to find how many hours it takes for the two of them to wash 1000 dishes. To find out, I ask myself, how many sets of 400 dishes are there in 1000 dishes? I'll divide to get the value:
1000 ÷ 400 = 10 ÷ 4 = 2.5
In other words, in two hours, they'll wash two sets of 400, or 800 dishes; in the additional half of an hour, they'll wash an additional half of another set of 400 dishes, which is 200 dishes. This gives me the required total of 1000 dishes.
They'll take 2.5 hours.
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If six men can do a job in fourteen days, how many fewer men would be needed if they were allowed twenty-one days for the job?
The men are being treated as interchangeable; there is no "faster" or "slower" here. So I'll start by converting this to man-hours — well, in this case, man-days. If it takes six guys fourteen days to finish the job, then my total is:
6 × 14 = 84
That is, the entire job requires 84 man-days.
This exercise is asking me to expand the time allowed from fourteen days to twenty-one days. Obviously, if they're giving my guys more time, then I'll need fewer guys. But how many fewer guys, exactly?
However many guys I end up needing, I'll still need them to provide me with 84 man-days of work; those man-days will be spread out over 21 working days. This means:
84 ÷ 21 = 4
I've divided the number of man-days by the number of days, which leaves me the number of men; it appears that I'll only need four guys. Does this check? Well, if I put four guys to work for each of the twenty-one days, then I'll end up with 4 × 21 = 84 man-days, which is exactly what I need.
Originally, I needed six guys. Now I'll need only four. So:
I'll need two fewer guys.
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If Bob can dig out twenty cubic feet of dirt in an hour, and Carl can dig out twenty-four cubic feet of dirt in an hour, how much dirt will be in the hole, if they dig together?
This is a trick question. If they've dug the dirt out of the hole then, by definition, there will be no dirt in the hole. Holes are empty. Har. De. Har.
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Jill, Karen, and Lisa are painting a house. Working together, they can paint the house is 6 hours. Working alone, Jill can paint the house in five hours faster than can either of Karen or Lisa. How long would it take Jill to paint the house working alone?
Okay, there are a lot of unknowns here. But I can start with the usual hourly-rate stuff, and I'll pick variables that make sense. First, I'll note that, since Jill takes five hours less than either of Karen or Lisa, then Karen and Lisa take the same amount of time. So I can set up the hours for completion for each:
Karen: K
Jill: K − 5
Lisa: K
together: 6
Taking the reciprocals, I can find their hourly rates:
Karen:
Jill:
Lisa:
together:
Since I'm assuming their labors are additive, I can add their individual per-hour accomplishments, and set this equal to their combined accomplishment:
Multiplying through by the common denominator of 6K(K − 5), I get:
6(K − 5) + 6K + 6(K − 5) = K(K − 5)
6K − 30 + 6K + 6K − 30 = K2 − 5K
18K − 60 = K 2 − 5K
0 = K 2 − 23K + 60
0 = (K − 20)(K − 3)
K = 20, K = 3
But if Karen takes only three hours, then Jill would be into negative time, in order to be five hours faster. So I can discard the "K = 3" solution as extraneous. The only valid solution is "K = 20". Since Karen takes twenty hours, and since Jill is five hours faster, then:
Jill can paint the house in fifteen hours.
You may have noticed that each of these problems used some form of the "how much can be done per time unit" construction, but other than that, each problem was done differently. That's how "work" problems often are. You'll have to be alert and clever to do these. But as you saw above, if you label things neatly and do your work clearly and logically, you should find your way to the solution.
By the way, I've mentioned once or twice that these exercises often assume that people are equal in their productivity, that chickens can come in fractional units, and that rates of completion (such as how much a person gets done in an hour) are additive. I'm sure you can think of somebody that isn't working as hard as you are in your class, and the chicken thing is kind of obvious. The classic counter-example for labor being additive is the idea that, if one woman takes nine months to have one baby, then nine women can have one baby in one month.
So definitely use the reasoning during this part of your course, but keep in mind that it isn't generally reflective of "real life".
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