Units: “How many days…”

This is an example of a common sort of story problem encountered in standardized tests: “1. A team of five professionals can do a certain job in nineteen days; a team of nine apprentices can do the same job in the same amount of time. Assuming all professionals work at the same rate and all apprentices work at the same rate, how long would a team of three professionals and six apprentices, working together, to do the job?”

This is more complicated than the typical version, which is usually more along the lines of, “2. Adam can paint a room in three hours and Bob can paint a room in six hours. How long will it take them if they work together?” There are other variants as well, such as, “3. Adam can paint a room in eight hours and Bob can paint a room in seven hours. Adam starts painting before Bob shows up. Bob arrives at some point, and they finish the job in five hours. How long had Adam been painting by the time Bob started?”

These sorts of problems create issues for students, particularly those who are used to the story problem strategy of taking the numbers given and applying some basic operator to them. In these problems, that will generally yield an incorrect, even an illogical, answer.

A way in which we math teachers get in the way of successful solution of these problems is in not stressing the importance of units enough. While these are mathematics problems in the sense that numbers are operated upon, the key to understanding them is understanding how to work with the units (which is still a mathematical task, but not one that people tend to think of as “mathematics”).

Let’s start with problem 2: Adam can paint a room in three hours and Bob can paint a room in six hours. How long will it take them if they work together?

The key is to think in terms of hours. Each hour, Adam paints 1/3 of a room, while Bob paints 1/6 of an hour. Working together, each hour, they paint 1/3 + 1/6 = 1/2 of the room. So it will take them two hours to finish the entire room. In this example, a student using the strategy of just applying some operator to the numbers in the problem might get “two hours” by 6/3, but that’s a coincidence.

Using units more formally, let A and B be the rates of the two painters. C is the rate of the two painters working together. So \[A = \frac{1\ room}{3\ hour} = \frac{1}{3}\ room/hour \\ B = \frac{1\ room}{6\ hour} = \frac{1}{6}\ room/hour \\ C = A + B = \left(\frac{1}{3} + \frac{1}{6}\right)\ room/hour = \frac{1}{2}\ room/hour\]

We can then calculate how long it takes to paint a room at that speed: \(1\ room \div \frac{1}{2}\ room/hour = 2\ hour\).

For problem 3, here are the relevant rates: \[A = \frac{1}{8}\ room/hour \\ B = \frac{1}{7}\ room/hour \\ C = A + B = \left(\frac{1}{8} + \frac{1}{7}\right)\ room/hour = \frac{15}{56}\ room/hour\]

However, the problem now requires some algebra, since \(aA + cC = 1\), where \(a\) represents the number of hours that Adam painted alone and \(c\) the number of hours Adam and Bob worked together. The unit of \(a\) and \(c\) is hours and that of \(A\) and \(C\) is rooms per hour, so the unit of \(aA\) and \(cC\) is hours, which is what we want (we want to paint one room, as indicated on the right hand side of the equation).

Since \(a = 5 – c\), we can substitute and simplify: \[(5 – c)A + cC = 5A – cA + cC = 5A + c(C – A) = 1 \\ c(C – A) = 1 – 5A) \Rightarrow c = \frac{1 – 5A}{C – A}\] Replacing \(A\) and \(C\) with the respective rates gives us: \[c = \frac{1 – \frac{5}{8}}{\frac{15}{56} – \frac{1}{8}} =\frac{56 – 35}{15 – 7} = \frac{21}{8} = 2.625\]

This means that the two worked together for 2.625 hours, and Adam worked by himself for 2.375 hours.

Going back to the original problem: “1. A team of five professionals can do a certain job in nineteen days; a team of nine apprentices can do the same job in the same amount of time. Assuming all professionals work at the same rate and all apprentices work at the same rate, how long would a team of three professionals and six apprentices, working together, to do the job?”

The second problem assumes there is one Adam and one Bob. This problem states that there are three professionals and six apprentices. The fractions are more challenging, but otherwise the problems are the same.

We can create two units: Professional rate (P) and apprentice rate (A). However, we aren’t given the rate for a single professional or a single apprentice; make sure to set the formulas up correctly. We have: \[5P = \frac{1\ job}{19\ day} \\ \Rightarrow P = \frac{1}{95}\ job/day \\ 9A = \frac{1\ job}{19\ day} \Rightarrow A = \frac{1}{171}\ job/day\]

We develop a team rate (T) based on multiplying each person’s work rate by the number of people in that category: \[T = 3P + 6A = 3 \times \left(\frac{1}{95}\right) + 6 \times \left(\frac{1}{171}\right) \\ T = \frac{3}{95} + \frac{6}{171} = \frac{27 + 30}{855} = \frac{57}{855} = \frac{1}{15}\] The unit of T is the same as that of P and A, that is, \(job/day\). We want the number of days to do one job, so we take the inverse. So it will take 15 days for this team to do one job.

These problems can be difficult if a student tries to rush through them, and particularly if a student has a weak understanding of how units work. This problem is certainly more challenging that the second problem is, even with this understanding, but each step should be straightforward if (a) you lay out the meaning of each value, including the units and (b) you carefully step through the problem, either keeping the units present or at least articulating (as above) how the units are affected by each step.

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