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'Work' problems usually involve situations such as two people working together to paint a house. You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together.
Teaches every lesson in the Saxon Algebra 1, 3rd Edition Textbook from a Biblical foundation.Teaching 3 to 5 practice problems, he review previous concepts and connects them to previous ones, giving students the big picture. During the lectures, students take notes and work problems along with Dr. Mathematics—3 years. Must include first-year algebra and two additional years of college-preparatory mathematics, including Algebra II (or equivalent) or higher. An advanced mathematics course is highly recommended in your senior year. One year of either algebra or geometry taken prior to ninth grade are acceptable. Science—3 years.
Objective To identify common features of cases with novel coronavirus disease (COVID-19) so as to better understand what factors promote secondary transmission including superspreading events. Methods A total of 110 cases were examined among eleven clusters and sporadic cases, and investigated who acquired infection from whom. The clusters included four in Tokyo and one each in Aichi, Fukuoka. Each student will be given a textbook at the beginning of the year. If a student is in need of an online version of the textbook they can follow the link below.
Many of these problems are not terribly realistic — since when can two laser printers work together on printing one report? — but it's the technique that they want you to learn, not the applicability to 'real life'.
The method of solution for 'work' problems is not obvious, so don't feel bad if you're totally lost at the moment. There is a 'trick' to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time. For instance:
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Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours to paint a similarly-sized house. How long would it take the two painters together to paint the house?
If the first painter can do the entire job in twelve hours and the second painter can do it in eight hours, then (this here is the trick!) the first guy can do 1/12 of the job per hour, and the second guy can do 1/8per hour. The question then becomes, how much then can they do per hour if they work together?
To find out how much they can do together per hour, I make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner), and I add together what they can do individually per hour. So, per hour, their labors are:
1/12 + 1/8 = 5/24
In other words, they can do 5/24 of the job per hour.
But the exercise didn't ask me how much they can do per hour; it asked me how long they'll take to finish one whole job, working togets. So now I'll pick the variable 't' to stand for how long they take (that is, the time they take) to do the job together. Then they can do:
1/t per hour
This gives me an expression for their combined hourly rate. I already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get:
5/24 = 1/t
I can solve by flipping the equation; I get:
t = 24/5 = 4.8 hours
An hour has sixty minutes, so 0.8 of an hour has forty-eight minutes. Then:
They can complete the job together in 4 hours and 48 minutes.
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The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate.
Each person took a certain number of hours to complete the task:
hours to complete job:
Since the unit for completion was 'hours', I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for 'hours to complete job':
completed per hour:
first painter: 1/12
second painter: 1/8
together: 1/t
Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the 'together' rate:
adding their labor:
1/12 + 1/8 = 1/t
5/24 = 1/t
24/5 = t
As you can see in the above example, 'work' problems commonly create rational equations. But the equations themselves are usually pretty simple to solve.
One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?
My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the 'together' time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times.
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You might wonder how to work with that '1.25 times as fast' part. If you're not sure, try doing similar computations with simpler numbers. For instance, if the faster pipe filled the pool 2 times as fast as the second pipe, then it would take one-half as long to fill the pool. If the faster pipe filled the pool 3 times as fast as the second pipe, then it would take one-third as long to fill the pool. Following this same reasoning, since the faster pipe fills the pool 1.25 = 5/4 as fast as the second pipe, then it takes 4/5 as long as does the second pipe to fill the pool.
Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time:
hours to complete job:
slow pipe: s
fast pipe: 4/5s
together: 5
Next, I'll convert all of the completion times to per-hour rates:
completed per hour:
slow pipe: 1/s
fast pipe: 5/(4s)
together: 1/5
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Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate:
adding their labor:
1/s + 5/(4s) = 1/5
multiplying through by 20s (being the lowest common denominator of all the fractional terms):
They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is:
The slower pipe takes 11.25 hours.
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Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been 'f ' for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been '1.25f '.
URL: https://www.purplemath.com/modules/workprob.htm
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8-8 Function Junction
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