Is there a more intuitive way to solve combined rates of work problems?

Question

I am helping my brother study for the GRE and we have come across some problems like this in my old precalculus textbook:

1) Karen and Betty have been hired to pain a house. Working together, they can paint the house in $\dfrac 23$ the time it would take Karen working alone. Betty could paint the house by herself in $6$ hours. How long would it take Karen to paint the house working alone?

2) Stan and Hilda can mown the lawn in 40 minutes if they work together. If Hilda works twice as fast as Stan, how long would it take Stan to mown the lawn alone?

The way I advised my brother to do these is to write the "$d=rt$" equation for every case: each person working alone, and both people working together; introduce as many variables as you need. From then, I just told him to try to manipulate the equations to solve for the unknown variable. However, he found this approach difficult. He said that somebody really smart could understand what's going on better and not have to rely on algebraic manipulations. Is there a better way to do these?

This is how I solved them:

1) Suppose the house has $x$ square meters of surface area that needs to be painted.

$r_k =$ rate of Karen (in $m^2$/hour)

$r_B =$ rate of Betty (in $m^2$/hour)

$t_k =$ time it takes Karen to paint the house alone (in hours)

$r_k = 6$ hours (given)

The three $d=rt$ equations are

$$x = r_k \cdot t_k$$

$$x = 6 \cdot r_B$$

$$x = \dfrac 23 \cdot t_k \cdot (r_B + r_k)$$

We are trying to solve for $t_k$. Transform the second equation into $r_B = \dfrac x6$ and substitute in for $r_B$. Distribute $\dfrac 23 t_k$ as well:

$$x = \dfrac 23 \cdot t_k \cdot \dfrac x6 + \dfrac 23 t_k r_k$$

Substitute $r_k \cdot t_k = x$

$$x = \dfrac 23 \cdot t_k \cdot \dfrac x6 + \dfrac 23 x$$

Now you can cancel the $x's$ and solve for $t_k$.

2) Using similar definitions as above (and $x$ is $m^2$ of lawn now)

$$x = r_S \cdot t_S$$

$$x = (2 r_S) \cdot t_H$$

$$x = (r_S + 2r_S) \cdot 40$$

Setting the first and third equations equal to each other,

$$r_S \cdot t_S = (3r_S) \cdot 40$$

Now you can cancel $r_S$ and solve for $t_S$.

Answer 1

By throwing in all of the $r$, $t$, and $d$, not to mention $x$, you're overly complicating things, especially if your brother is algebra-averse. There's really only one unknown, and you can call it $k$. It's the number of hours it would take Karen to paint $1$ house by herself. Everything else can be expressed in terms of $k$. Don't complicate things with $x$ and units of $m^2$; call your units "houses" instead, and then you can talk about $1$ $house$ instead of $x$ $m^2$.

For example, the simplest way I can think to do this one is as follows:

• Betty paints $1$ house in $6$ hours. Therefore, Betty paints at a rate of $\frac16$ houses per hour.
• We are looking for "How long would it take Karen to paint the house working alone". Let's call this $k$ hours. That is, Karen paints $1$ house in $k$ hours. Therefore, Karen paints at a rate of $\frac1k$ houses per hour.

At this point, you can see that it's important to understand the inverse relationship between "houses per hour" and "hours per house", and how to manipulate each quantity to obtain the other.

Continuing:

• Together, assuming they don't get in each other's way, they can paint $\frac1k + \frac16$ houses per hour, so it will take them $\frac{1}{\frac1k + \frac16}$ hours to paint $1$ house. But that latter figure, $\frac{1}{\frac1k + \frac16}$ hours, is $\frac23\space *$ "the time it would take Karen working alone" or $\frac23 * k$.

So we have $$\frac{1}{\frac1k + \frac16} = \frac23 * k = \frac{2k}{3}$$ $$\frac1k + \frac16 = \frac{3}{2k}$$ $$\frac16 = \frac{3}{2k} - \frac{2}{2k} = \frac{1}{2k}$$ $$6 = 2k$$ $$3 = k$$ $$k = 3\space hours$$

Checking:

Karen paints a house in $3$ hours, so she paints $\frac13$ of a house per hour. Betty paints a house in $6$ hours, so she paints $\frac16$ of a house per hour. Together they paint $\frac16 + \frac13 = \frac12$ of a house per hour. Therefore, together they paint a house in $2$ hours. $2$ hours = $\frac23$ of $3$ hours, check.

You can do the lawn-mowing and tub-filling problems similarly, by simplifying to talking about $1$ lawn and $1$ tub.

Important concepts:

• Simplify to a single variable if possible
• Inverse relationships between "things per time unit" and "time units per thing"
• Fractional manipulation (add, subtract, multiply, divide) with variables
• Check your work at the end

Answer 2

I don't have an answer for whether there is a "royal road" to solving such problems (I don't think so), but want to comment on why such problems are not so easy to think about intuitively.

In Gelfand and Shen's Algebra, there is the following nice puzzle:

Problem 130. A swimming pool is divided into two equal sections. Each section has its own water supply pipe. To fill one section (using its pipe) you need $a$ hours. To fill the other section you need $b$ hours. How many hours would you need if you turn on both pipes and remove the wall dividing the pool into sections?

This involves the so-called harmonic mean, which is not a particularly well-known or used mean among high schoolers. If we had the same intuition for that as the regular (additive/arithmetic) mean, maybe this kind of problem would be obvious.

In the context of your problems, consider this version:

Betty could paint the house by herself in 6 hours. Karen could paint it by herself in 5 hours. How long would it take them together?

This already requires thinking of a problem like $$x(1/6+1/5)=1$$ $$x=\frac{1}{1/6+1/5}=\frac{30}{11}$$ Now the problems you actually state require one to solve a version of this where the individual rates aren't both given. So unless one had a very intuitive grasp of the harmonic mean and used at least a little algebra, I don't feel like there would be a nice way for a high schooler to do these (without the methods you or other commenters mention).

However, I bet that with time and thought one could come up with a nice algorithm that avoided algebra to solve them, and likely such problems are of great enough antiquity that people did come up with them. The problem then is that you have to memorize some non-algebraic algorithm; the algebraic advantage is that you are directly translating everything and so one can recover the problem from the algebra.

Answer 3

I don't know much about math education, but I can give my thoughts on how to solve this problem. I think the main issue is whether to express the rate of work as a amount of work per unit time or its reciprocal, the amount of time per unit work. When you have two people working together, it is the amount of work per unit time that combines additively, so that is the natural thing to work with. (In another situation where you have one person painting two houses of different sizes, and you were asked how long it would take, you would want to work with the amount of time per unit work.)

To see how thinking of it in terms of the reciprocal can help, let's do the two problems.

If they can paint the house in 2/3 the time it takes Karen to paint the house, then they can paint 3/2 houses in the time it takes Karen to paint one house. Therefore, when Karen has painted one house, Betty has only painted a half a house. This means Betty is half as fast as Karen. So if it takes Betty 6 hours to paint a house, it takes Karen only 3 hours. (Here taking the reciprocal was the key step that allowed us to compare how many houses each was able to do in a fixed amount of time, and thereby compare how quickly they work.)

Stan and Hilda mowed a lawn, with Hilda doing twice as much as Stan. This means Stan only mowed 1/3 lawn in forty minutes, so it takes him 3*40 minutes to mow a whole lawn. (Again we decomposed the rate 1 lawn / 40 minutes into the sum of two contributions: 1/3 lawn in 40 minutes and 2/3 lawn in 40 minutes. Notice these are both amounts of work per unit time.)