# Is there a more intuitive/easier way to solve these combined rate problems?

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 $t_S$.

• How about changing the units to "houses/hour" and "lawns/hour"? No need to muck around with meters Nov 28, 2016 at 19:24
• @Omnomnomnom Yeah I tought about that but personally I found it more intuitive to think of $m^2/hour$ when combining rates especially
– Ovi
Nov 28, 2016 at 19:44

## 1 Answer

Om$\overline{\mbox{nom}}$ is right. The usual way to work such problems to work out how much of the job is done in one hour (or day or whatever.) If Karen can do the job in $k$ hours, then she paints $1/k$ of the house in one hour. Betty paints $1/6$ of the house in one hour. So here's the good part: Together they paint $1/k + 1/6$ of the house in one hour.

On the other hand, we know it takes $2k/3$ hours for both of them to paint the house, so that's $1/(2k/3)$ of the house per hour. Now you have an equation in one variable:

$$\frac{1}{k}+\frac{1}{6} = \frac{1}{2k/3}.$$

Multiply through by $6k$ to clear fractions:

$$6+k=9.$$

So $k=6$.

The philosophy here is that math lives and dies on equal signs. So you need things that are equal, so you can stick a $=$ between them. The great equalizer on these problems is expressing everything in one unit: "How much of the job is done in ONE hour?"

For the second problem, if Hilda takes $h$ minutes then Stan takes $2h$ minutes. In one minute they do $1/h+1/2h$ of the job. But the problems says this is $1/40$ of the job, and there's your equal sign.

We did both problems with one variable. Your way, I think, overcomplicates things.