A popular kind of word problem tells us how many people (or cats, or hens, …) it takes to make some number of houses (or kill some number of mice, or lay some number of eggs) in some amount of time, and then asks us to fill in one of the blanks for a different scenario. This week we’ll look at several problems of this sort, each asking a different question (how many, how much, how long?), and each solved by several different methods. (This post was inspired by Peter Murphy, who recently asked a question of this type, which I answered by providing these links.)

## Cats, rats, and minutes: How many cats?

To introduce the topic, consider this question from 2009, where we’ll see four different ways to solve the problem:

Four Methods of Solving Classic Chicken and Egg ProblemsIf 6 cats can kill 6 rats in 6 minutes, how many cats will it take to kill 100 rats in 50 minutes?I think the answer is 50 cats but I really can't tell. I just don't understand how to solve it.

Doctor Greenie answered:

Hi, Mikayla -- A more common problem of this type involves chickens laying eggs instead of cats killing rats. But the idea of the problem is the same. You can find links to several pages in the Dr. Math archives where this type of problem is discussed by searching the archives using the keywords "chicken egg". Following is my quick summary of some possible approaches to the problem.

We’ll be seeing those chicken-and-egg problems next week, because they have an extra little challenge. (The search isn’t available any more, either.)

### Changing two numbers at a time

The first method uses natural reasoning rather than formal math:

(1) Start with the given information andvary two of the three numbers at a time, keeping the third one unchanged. For example,twice as many ratsand the same amount of time requirestwice as many cats; ortwice as many catsand the same number of rats requireshalf as much time. The following table shows one of many different paths to the answer to your problem using this method: cats rats minutes --------------------- 6 6 6 [given] 6 1 1 [same number of cats; 1/6 as much time means 1/6 as many rats] 6 100 100 [same number of cats; 100 times as many rats (to give us the number of rats we want) means 100 times as many minutes] 12 100 50 [same number of rats; half as many minutes (to give us the number of minutes we want) means twice as many cats] It takes 12 cats to kill 100 rats in 50 minutes.

Reducing at least one number to 1 is often a good place to start.

### Two ratios

Having thought about which numbers change in the same direction, we can now think about ratios:

(2) Here is a variation of the above method, getting directly to the answer by usingratios. It takes 6 cats to kill 6 rats in 6 minutes. We want to kill 100 rats instead of 6; that will require(100/6) times as many cats. We have 50 minutes instead of 6; that will require(6/50) times as many cats. So the number of cats we need is (6)(100/6)(6/50) = 12

This is the same sort of thinking we needed in the first method; the main difference is that rather than sticking with simple multipliers, we have let ourselves use uglier fractions, because we held off the actual calculations until last, when we could cancel.

### Cat-minutes

(3) Another approach is to find the number of "cat-minutes" it takes to kill each rat. 6 cats take 6 minutes to kill 6 rats, so the number ofcat-minutesit takes to kill a rat is (6)(6) cat-minutes ------------------ = "6 cat-minutes per rat" (6) rats We are supposed to find the number of cats it takes to kill 100 rats in 50 minutes. 100 rats, at 6 cat-minutes per rat, means 600 cat-minutes. If we have 50 minutes, then the number of cats we need is 600/50 = 12

This is a classic method in business, where we can use, say, “man-hours” (or, sometimes, “person-hours” or “worker-hours” to avoid implying gender) as a unit of work: how much work one person does in one hour. Of course, this assumes that each person’s productivity rate remains the same under different circumstances, which may not always be true of real people (or cats), who might perhaps either be distracted by others, or help one another work faster. And, of course, the rate of rat-catching really depends on how many rats there are!

The concept is similar to units in physics like the “kilowatt-hour” or “kilogram-meter per second” or “foot-pound”, where the hyphen represents multiplication.

### Joint variation

“Cat-minutes” encodes a proportional relationship that we can use explicitly instead:

(4) A formal mathematical approach that is very much like the preceding method usesdirect variation(actually, if you have studied this topic,"joint" variation). The idea is that the number of rats killed increases with either increased numbers of cats or increased numbers of minutes. So we can write a joint variation equation r = kcm where r is the number of rats,k is a constant of variationto be determined (it is related to the "6 cat-minutes per rat from the previous method), c is the number of cats, and m is the number of minutes. We can use the given information to find the value of our constant of variation. r=6 when c=6 and m=6; so 6 = k(6)(6) --> k = 1/6 Now we use this constant of variation and the new numbers of rats (100) and minutes (50) in our equation to find the new number of cats required: r = kcm 100 = (1/6)(c)(50) --> c = 12 Try understanding each of the above methods and pick the one that you find easiest to understand and use.

The constant *k* here is actually the number of rats per cat-minute.

## Men, work, and days: How many days?

The next question, from 2011, is initially about a simpler problem, with only two numbers given (or, if you prefer, with the amount done being 1):

Combining Rates of Work, Revealing Constants of Word Problems I saw this question in an old textbook on arithmetic:If 15 men can do a piece of work in 7 days, in how many days can 21 men do the same work?Now, the authors did not explain what mathematical logic is used to determine the answer, and they explain proportions in another chapter later in the book -- so it was not their intention to use proportional reasoning here. But how would you determine the answerwithout using an indirect proportion? I cannot think of a logical, mathematical approach that does not in some way require setting up this proportion: 15 men x 7 days ------------------- = 21 men 5 x 7 days ------------- = 5 days 7 But I just resorted to trial and error to assign 21 men as the denominator, and the product of 15 men and 7 days as the numerator. No other relationship used the information in a way that made the units work out and still led to the common sense conclusion that -- with more labor -- the job must take less than 7 days. My method does not use adequate mathematical thinking. There must be a better one that still does not use an indirect proportion, but does lead to the solution, above. What do you suggest? I thank you for your reply and explanations.

All the methods we’ll be seeing are in some way representations of the proportion; but the least obviously so is this:

### Man-days

Doctor Jerry answered:

Hello Kenneth, Thanks for writing to Dr. Math. If 15 men can do a piece of work in 7 days, in how many days can 21 men do the same work? If 15 men can do a piece of work in 7 days, then it takes15*7 man-daysto do the job. If you have 21 men, then you would need x days to produce 15*7 man-days. So, 21*x = 15*7 x = 15*7/21 = 5

In other words, the job required \(15\times7=105\) man-days, so 21 times the number of men has to equal 105. Division does the job.

## Men, houses, and days: How many houses?

A week later, Kenneth wrote back with the kind of problem we’re focusing on here:

Hello Doctor Jerry: I want to thank you for the reply and answer to my first question! Can the answer to the following be determined in the same way that the answer to my first question was determined?Four carpenters can build eight houses in 10 days. Two carpenters can build how many houses in 15 days?I set up these proportions: 4/2 = 8/? = 10/15 Here, 4/2 represents the carpenter ratio 8/? represents the house ratio 10/15 represents the day ratio. I thank you for your reply and assistance!

Kenneth this time tried to make a proportion as he did for the simpler problem, but it isn’t true that all three numbers are proportional separately. We’ll see a different way to deal with the proportion in a moment; but this could in fact be solved using the man-day (or carpenter-day) approach. We find that 8 houses require \(4\times10=40\) man-days; so 1 house requires \(40\div8=5\) man-days. Two men working for 15 days would do \(2\times15=30\) man-days; this is \(30\div5=6\) times the work for one house, so they can make 6 houses.

… or not. It’s conceivable that some parts of the work would require three men working together, so two men simply couldn’t build a house at all! All the calculations we do here assume that the variables are truly proportional. If the workers don’t work independently, each at a constant rate, then this can’t really be assumed. I used to give a bonus question on tests that was something like this: “If it takes one man 10 hours to plant a tree, how long will it take ten men to plant the tree? Why is the method we’ve learned not applicable to this problem?” The answer is that, with only one tree, the men *can’t* be working separately, each at a consistent rate; in fact, they would probably get in one another’s way!

### Direct proportion (joint variation)

Doctor Jerry answered again:

Hello Kenneth, Thanks for writing to Dr. Math. Ever since I took 9th grade algebra, these "combining rates of work" problems have always bugged me: I can solve them, but the reason I have relied on has always seemed a bit ad hoc -- similar to what you wrote about earlier, with your "trial and error" method guided by common sense and dimensional analysis. Your questions have moved me to think about this kind of problem from a slightly different viewpoint. Implicit in these problems is an assumption that there is a "direct proportion" involved. Specifically, the number H of houses that c carpenters can build in d days is ... H = K*c*d, ... where K is some constant, at the moment unknown. The problems provide information that will determine K.

Proportion and variation are essentially the same idea; this form is more properly called joint variation, as we saw above, which allows us to combine multiple variables in one relationship. In this case, it is equivalent to saying that the ratio \(K=\frac{H}{cd}\) is constant; in fact, this is the rate in houses per carpenter-day.

In the specific problem you raise, we know that 8 = K*4*10 So, K = 8/40 = 1/5 We now can write H = (1/5)*c*d We were asked how many houses can two carpenters build in 15 days. So, H = (1/5)*2*15 = 6 This seems more straight-forward to me.

What makes this straightforward is that it is a routine solution method for a routine kind of problem. The only trick is to be sure that the proportion is valid.

## People, walls, and minutes: How many minutes?

Our final question comes from 2013:

Combining Rates of Work: From Unit Rates to a General Model So I have this problem:Seven people can paint 4 walls in 41 minutes. How many minutes will it take for 8 people to paint 7 walls? Round to the nearest minute.I know the formulas and I have the answer already, but I can't get my head around the concept of rate in this context. Theformulais: time = walls / rate x people The rate is: rate = walls / time x people If I divide walls by people or time, I get the number of walls by one person or in one minute; that's clear. Butwhat is the product of time and people?What does "walls divided by time multiplied by people" mean? What exactly am I getting when I multiply time by people? I tried toincrease or decrease the values of each one equallybetween the others, but that didn't work out as I expected. This whole problem is very confusing for me. I understand the simpler rate-time-distance problem, but this one with the extra variable seems way harder. I don't understandhow to make sense of it. Can anybody explain this problem to me?

This could be explained using person-minutes as we have above; or, as Doctor Jerry did above, we can rethink the problem to use a method we are more comfortable with. Danilo appears to have tried using a step-by-step method, changing pairs of numbers, and that’s what Doctor Ian will demonstrate as his preferred method, using a different example:

### Incremental adjustments (changing two at a time)

Hi Danilo, There are formulas for this, but I can't ever remember them, so I always makeincremental adjustments. For example, 7 people can paint 4 walls in 41 minutes. If they only paint1/4 as many walls, they would need1/4 as long, so 7 people can paint 1 wall in 41/4 minutes. If there are1/7 as many people, they will need7 times as long, so 1 person can paint 1 wall in (7*41)/4 minutes. So far, so good? Now we can just go back the other way. We want to have 8 people working, so they can do the job in1/8 the time: 8 people can paint 1 wall in (7*41)/(4*8) minutes. And if they want to paint 7 walls? That will take7 times as long: 8 people can paint 7 walls in (7*7*41)/(4*8) minutes. Does this make sense?

This is less formulaic, but requires thinking at each step how changing one quantity will affect another. And thinking more about the meaning of the problem is a good thing!

But routine can be good, too:

### Direct and inverse proportion (variation)

Now, that's how I do it if I'm just doing it once in a while, so I don't have to memorize anything (and so I don't have to think about it in terms of rate, which I -- like you -- find confusing here). But if I were going to do it a bunch of times, I'd want to be a little more efficient, and for that I'd want to set up a proportion. To do that, I'd reason about which items aredirectly and inversely proportionalto each other. For example, if Idecreasethe number of people, Iincreasethe amount of time. That means people and time areinversely proportional: k' = people * time If Iincreasethe number of people, Iincreasethe number of jobs I can do; similarly for time. That means people and jobs aredirectly proportional, and so are time and jobs: k'' = people/jobs k''' = time/jobs Put it all together, and I get people * time k = --------------- jobs That's my constant. Now I can take my original situation, and set up a proportion with any other situation where I know two of these quantities.

This version of variation is like what I said above about looking for a constant (the rate). Direct proportion means that the ratio of two quantities is constant; inverse (or indirect) proportion means that their product is constant. He has done that for each pair of variables (taking the other as constant) and then combined the constants into one.

Let's check it against the problem we just did: 7 people * 41 minutes 8 people * ? minutes ----------------------- = ---------------------- 4 walls 7 walls Cross-multiplying to isolate the unknown, we get 7 people * 41 minutes * 7 walls --------------------------------- = ? minutes 8 people * 4 walls This gives us what we got doing it the other way. Once I've figured this out, I can generate all kinds of questions ... and answer questions that other people have generated. The template is a people * b units of time d people * e units of time ---------------------------- = ----------------------------- c units of work f units of work Pick any 5 values, and that gives you a "working together" problem. Solve for the remaining value, and that gives you the answer.

This is not a formula I would want to memorize long-term, but this way of building it helps to make a formula when you need one for a while.

Next week, we’ll see similar problems, all about hens and eggs, involving half hens and half eggs, which make it a little weirder. But we’ll see all the same methods in this odder setting.

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