Lesson HOW TO Solve Rate of Work (painting, pool filling, etc) Problems

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Word Problems: Rate of work, PAINTING, Pool Filling

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Source code of 'HOW TO Solve Rate of Work (painting, pool filling, etc) Problems'

This Lesson (HOW TO Solve Rate of Work (painting, pool filling, etc) Problems) was created by by mbarugel(146)

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About mbarugel: Passionate about Math :) I'm currently running OnlineMathAnswers.com

I've seen many questions in this site from students that are struggling with this kind of problems. These problems usually go along the lines of:

<ul>
<li>"Worker A can finish a job in 3 hours. When working at the same time as Worker B, they can finish the job in 2 hours. How long does it take for Worker B to finish the job if he works alone?"</li>
<li>"Painters A and B can paint a wall in 10 hours when working at the same time. Painter B works twice as fast as A. How long would it take to each of them to paint it if they worked alone?"</li>
<li>"A 10,000 litre pool is filled by two pipes: A and B. Pipe A delivers 1,000 litres per hour. When pipe A and B are both on, they can fill an empty pool in 4 hours. How many litres per hour can pipe B deliver?"</li>
</ul>

Many times, students don't know how to begin to deal with these problems. Actually, they are quite simple once you know how to set up the appropiate equation or system of equations. I'll give you now the single most important formula that will help you solve these problems.

<h3>Basic Formula</h3><table border="1" align="center">
<tr><td align="center">Let's assume we have two workers (pipes, machines, etc): A and B.<br>Worker A can finish a job in X hours when working alone.<br>Worker B can finish a job in Y hours when working alone<br>The number of hours they need to complete the job<br>when they're both <i>working at the same time</i> is given by</td></tr><tr><td align="center">{{{Time=1/(1/X + 1/Y)}}}</td></tr></table>

Once you're "armed" with this formula, solving these problems will be very easy. We'll see a few examples below; but first, let me show where this formula comes from.

<h3>Where does the formula come from?</h3>We started the problem with the following assumptions. There are two workers (pipes, machines, etc), A and B. Worker A can finish a job in X hours when working alone; Worker B can finish the same job in Y hours when working alone.

Now consider this: if A can finish the job in X hours, then what's his <i>rate of work per hour</i>? That is, how many jobs can he complete in one hour? Clearly, if he takes X hours to complete a job, the he can complete {{{1/X}}} jobs per hour. For example, if he needs 10 hours to finish a job, then he can complete 1/10 (one tenth) of the job in 1 hour. If he needs 30 minutes (0.50 hours), then he can complete 1/0.50 = 2 jobs in one hour. The same reasoning applies to worker B.

What happens if they're working together? In this case we have to add their <i>rate of work</i>. Many people make the mistake of assuming that if A takes X hours, and B takes Y hours, then they both take X+Y hours. This is clearly wrong, as it would imply that they need more time when working together than when working by themselves. So we need to sum their rate of work. If A completes {{{1/X}}} jobs per hour, and B completes {{{1/Y}}} jobs per hours, then, when working together, they can complete {{{1/X+1/Y}}} jobs per hour.
Finally, given that they can complete {{{1/X+1/Y}}} jobs per hour; then how many hours do they take to complete one job? This is simply the inverse of the above formula. So they can complete 1 job in {{{1/(1/X+1/Y)}}} hours.

<h3>What does it mean that a worker works N times faster than another one?</h3>
There's also some confusion when the problem states, for example, "Worker B is twice (N=2 times) as fast as Worker A". Notice that if someone works twice as fast as someone else then he needs <strong>half</strong> the time to finish the job; if he works 4 times as fast, then he needs 1/4 the time to finish the job, and so on. Thereofore, if X is the number of hours worker A needs to finish a job, and Y is the number of hours worker B needs to finish a job, then the statement "Worker A works N times faster than worker B" is "translated" as {{{X = Y/N}}}. For example, if he works twice as fast as B, then {{{X = Y/2}}}

<h3>Sample Problems</h3>
Let's solve the 3 problems I started with using what we learned here.

<i>1. Worker A can finish a job in 3 hours. When working at the same time as Worker B, they can finish the job in 2 hours. How long does it take for Worker B to finish the job if he works alone?</i>

Always start by defining the variables. Let's call X to the number of hours worker A needs to finish the job, and Y to the number of hours worker B needs to finish the job. We already know that X = 3. We also know that when working at the same time, they need 2 hours. So, using the formula I gave you before:

{{{Time=1/(1/X + 1/Y)}}}
{{{  2 = 1/(1/3 + 1/Y)}}}
Rearranging terms, we get:
{{{1/3 + 1/Y = 1/2}}}
{{{1/Y = 1/2 - 1/3 = 1/6}}}
{{{1/Y = 1/6}}}
{{{Y = 6}}}

We conclude that B needs 6 hours to complete the job when working alone.

<i>2. Painters A and B can paint a wall in 10 hours when working at the same time. Painter B works twice as fast as A. How long would it take to each of them to paint it if they worked alone?</i>

Let's call X to the number of hours painter A needs to finish the job, and Y to the number of hours painter B needs to finish the job. We know that "painters A and B can paint a wall in 10 hours when working at the same time". Using the formula I gave above, this means that we have the equation:
{{{10 = 1/(1/X + 1/Y)}}}

We also know that "Painter B works twice as fast as A". So the other equation is:
{{{Y = X/2}}}
[remember, the fact that he works twice as fast implies that he need half the time to do the same job]

So we have the system of equations:
{{{system(10 = 1/(1/X + 1/Y), Y = X/2)}}}

Rearranging the 1st equation:
{{{1/X + 1/Y = 1/10}}}

Substituting the 2nd equation into this one:

{{{1/X + 1/(X/2) = 1/10}}}
{{{1/X + 2/X = 1/10}}}
{{{3/X = 1/10}}}
{{{X = 10*3= 30}}}

So painter A would need 30 hours to finish the wall if he worked alone. Since painter B is twice as fast, he would need 15 hours if worked alone.

<i>3. A 10,000 litre pool is filled by two pipes: A and B. Pipe A delivers 1,000 litres per hour. When pipe A and B are both on, they can fill this pool in 4 hours. How many litres per hour can pipe B deliver?</i>

Notice that there is an important difference between the wording of this problem and the other two. In this case, we're given the rate of work of pipe A rather than the time it needs to complete an activity. I'm now going to give a solution that sets the variables a little differently but uses the same principle.

Let's call X to the number of litres per hour pipe A delivers and Y to the number of litres per hour pipe B delivers. Notice the difference with problems 1 and 2: I'm setting the variables to be rates of work instead of "time to finish a job".

When both pipes are working, they can deliver {{{X+Y}}} litres per hour. Notice that we sum rates of work, just as we did with {{{1/X + 1/Y}}} in the previous problems. We should now use the information that says that "When pipe A and B are both on, they can fill this pool in 4 hours". Since the pool has 10,000 litres, the fact that they can fill it in 4 hours implies that when they are both turned on, they can deliver 10,000 litres every 4 hours; or 10,000/4 = 2,500 litres per hour. So we get the equation:

{{{X+Y = 2500}}}

Since we already know that X = 1000, then we conclude that <i>pipe B can deliver 1,500 litres per hour</i>.

As you might have guessed if you've had trouble with this kind of problems before; their main difficulty is to correctly define the variables and understand how rates of work are added. I hope that this lesson has helped you gain a better understading of these problems.

Martín
Private tutor at <a href=http://www.onlinemathanswers.com>Online Math Answers.com</a>