#### (New question of the week)

A conversation last week went through a number of interesting questions, starting with a couple on percentages, and moving into some that I would call rate questions. I will extract these, which I think will be useful for others. (The rest could, too, but there was just too much there for one post!)

## Two problems

Picking up in the middle of the discussion:

These questions remind me of a similar one I worked out a few days ago, which works on the same principles, but easier to miss. I thought you’d like it, so here it is:

Lily, Betty, and Louisa race each other in a 100 meters race.

All of them run at a constant speed throughout the race.

Lily beats Betty by 20 meters.

Betty beats Louisa by 20 meters.How many meters does Lily beat Louisa by?

(I did this, though it gave me more of a challenge)

Then, there’s this which is even harder (for me).

The grass in a field grows in a constant rate. If 240 sheep were to graze on it, it will become bare in 5 weeks. If 200 sheep were to graze on it, it will become bare in 7 weeks. How many weeks would it take 170 sheep to graze, before the field becomes bare?

For this, I just figured out that the grass will keep on growing during the 5/7 or however weeks it takes, and 240*5=1200 and 200*7=1400, the difference is of 200 in 2 weeks, which is 100/week. For 170 sheep, it’s 100+ … that’s the part I haven’t figured out.

## Solution to 100-meter race

I replied, giving my answer directly for the problem Sarah had already solved, and expanding on what she said about the other, which was a little cryptic:

Yes, these are nice challenges that can probably be solved in many ways, some of which are surely nicer than what I’ll do.

For the first, Lily runs 100 m in the time Betty runs 80 m, so Lily’s speed is 100/80 times Betty’s speed. Similarly, Betty’s speed is 100/80 times Louisa’s speed, and therefore Lily’s speed is 100/80 * 100/80 = 100/64 times Louisa’s. In the time Lily runs 100 m, Louisa runs 64 m; so Lily beats Louisa by 36 m. (Not, as one might initially think, 40 m. But you and I know better!)

## Thoughts about the grazing sheep

The second is harder. You made a very nice start.

One way to finish would be to repeat what you did, comparing the 170 sheep case with one of the others, perhaps using a variable.

It may help to label the quantities, so you can see better what you are doing. Your 1200 and 1400 are measured in

sheep-weeks(number of sheep times number of weeks); this can be thought of as a measure of the quantity of grass (how much a sheep eats in a week). So the grass grows at a rate of 100 sheep-weeks per week. That is, 100 sheep could graze in the field indefinitely, without using up any grass, because they are eating only the amount that grows; any additional sheep will start overgrazing, decreasing the amount of grass. So if you reduce the number of sheep from 200 to 170, how much longer can they graze? This becomes just another proportion problem,if you look at it right.I’d usually just do this with algebra, though that may get pretty complicated, in order to make sure I was thinking right. Variables might be the rate at which each sheep eats grass (say,

apounds per week), the rate at which grass grows (bpounds per week), the amount of grass initially in the field (cpounds), and what we need to solve for, the number of weeks with 170 sheep,w. This works, but is a little ugly.

Sarah answered:

As for the “ugly” algebra, I think I’d prefer that to simple proportion, because I don’t think I’m seeing it right.

Simple proportion, maybe you do this:

240 sheep = 5 weeks = 1200 sheep-weeks – (5*100) = 1200 – 500 = 700

So 240 sheep = 700 sheep-weeks, 170 sheep = ?

When you say simple proportion, we have to use inverse proportion here, right?

I’m not quite seeing what to do with algebra, but we know b, right?

Thanks!

We had previously discussed the need to write carefully what you mean; the string of equalities above is a little confusing to me, because these quantities are not really *equal*. I think she meant something like this:

240 sheep for 5 weeks = 1200 sheep-weeks eaten

grass growing for 5 weeks = 5*100 = 500 sheep-weeks added

1200 – 500 = 700 sheep-weeks used up

So 240 sheep use 700 sheep-weeks, 170 sheep = ?

This is all good thinking, but doesn’t lead yet to the answer. I’ll come back later to an insight that could go straight from here to the answer, but frankly, I didn’t notice at the time how close she was.

## Solution by proportion

I replied, actually finishing the work by the method I had described, because it struck me as too complicated to communicate without just showing it:

The proportion is pretty subtle — enough so that I, too, would be a lot more confident with the algebra. But here’s the thinking: since 100 sheep would be just enough to keep the grass constant (eating just as much each week as grew), the time to use up the grass will be inversely proportional to the

excess over 100 sheep. Since 200 sheep use up the grass in 7 weeks, 170 use up the grass inwweeks:(200-100):(170-100) =

w:7100:70 =

w:7

w= 10As for algebra, I suggested the variables might be the rate at which each sheep eats grass (say,

), the rate at which grass grows (apounds per week), the amount of grass initially in the field (bpounds per week), and what we need to solve for, the number of weeks with 170 sheep,cpounds.wWe know none of these things initially; and we know nothing about pounds (or any other actual unit of quantity), so we will never actually solve for

a,b, orc! (We could if I had used sheep-weeks instead of pounds, but I probably wouldn’t think of that until later.)From the first given fact, we know that \(5b – 240\cdot 5a = c\) (amount grown – amount eaten = amount in field). Similarly, \(7b – 200\cdot 7a = c\). Those two equations can be solved to find the

ratiosof the three variables, though not their actualvalues; this is in effect what you already did. Then we have to solve \(wb – 170\cdot wa = c\) for w, This is what I had in mind when I said you could finish by (in some sense) doing the same thing but with a variable.

## Solution by algebra

Sarah was able to do the algebra as far as she got on her own previously:

I follow most of what you said. I don’t think I would have managed the proportion.

You say w = 10 weeks, but

how can you checkto confirm it’s correct?And if you use ratios with variables as your second method above, here is what I have:

5b – 1200a = 7b – 1400a

-1200a + 1400a = 7b – 5b

200a = 2b

100a = b

which we had already found. I don’t see how that helps us finish it off.

I just gave a nudge:

Now use the third equation!

As I said, “Those two equations can be solve to find the

ratiosof the three variables, though not their actualvalues;this is in effect what you already did. Then we have to solve \(wb – 170\cdot wa = c\) forw.” Actually, it’s notjustthat one equation; I’dreplacewith the expression from one of the first two equations, and alsocreplacewith 100ba. Then solve forw.As for checking an answer, I’d probably just plug my answer into the equations. In this case you don’t have actual values for

a,b, andc, but you could express them all in terms ofa, plug those expressions in, and see what happens. If I hadn’t used algebra, I might work through the statement of the problem in terms of sheep-weeks, showing that my answer worked.

After a false start, she got it:

Combining the first and third equations,

wb – 170*wa = 7b – 200*7a

w(100a) – 170*wa = 7 (100a)- 200*7a

100aw – 170aw = 700a – 1400a

Dividing by a

100w – 170w = 700 – 1400

-70w = -700

w = -700/-70 = 10 weeks

Let me try with the first equation:

wb – 170*wa = 5b – 240*5a

w (100a) – 170wa = 5 (100a) – 1200a

100aw – 170aw = 500a – 1200a

100w – 170w = 500 – 1200

-70w = -700

w = -700/-70

w = 10 weeks which is correct!

(I used both equations because I made a mistake in my first one and whilst doing it with the second equation I found it)

Now for the checking:

100aw – 170aw = 500a – 1200a

100a (10) – 170a (10) = -700a

1000a – 1700a = -700a

-700a = -700a

I don’t see how to do it using sheep weeks.

## Checking by sheep-weeks

Again, I just showed the check because it would be too hard to give a mere hint; and I wasn’t sure what it would look like until I did it:

Let’s see if I can check it with sheep-weeks. (I hadn’t tried.) The problem was

The grass in a field grows in a constant rate. If 240 sheep were to graze on it, it will become bare in 5 weeks. If 200 sheep were to graze on it, it will become bare in 7 weeks. How many weeks would it take 170 sheep to graze, before the field becomes bare?

We claim the answer is

If 170 sheep graze, it will be

10 weeksbefore the field becomes bare.We also found that the rate of growth is

100 sheep-weeks per week.In 5 weeks, 500 sheep-weeks of grass grows; and 240 sheep each 1200 sheep-weeks of grass. That tells us that the field starts out with

700 sheep-weeksof grass.Now we have numbers for everything, so we can check that the other two scenarios also work:

In 7 weeks, 700 sheep-weeks of grass grows, and 200 sheep eat 1400 sheep-weeks of grass. Starting with 700, adding 700, and removing 1400 leaves the field bare.

In 10 weeks, 1000 sheep-weeks of grass grows, and 170 sheep eat 1700 sheep-weeks of grass. Starting with 700, adding 1000, and removing 1700 leaves the field bare.

So everything checks.

That sort of check takes more work (because we had to find all the relevant numbers, and check without ready-made equations), but it’s more satisfying to me, because it confirms that the equations made sense.

Notice where I found that the field holds 700 sheep-weeks of grass. Now look back at Sarah’s attempt at solving with this approach; she ended with, “So 240 sheep use 700 sheep-weeks, 170 sheep = ?” So she’d found how much grass there was in the field! She might have continued like this:

With 170 sheep, we are 70 sheep above the number that eat just the amount that grows. In how many weeks will these 70 extra sheep eat up the 700 sheep-weeks of grass to use it up? 10 weeks!

Which is better: the algebra, or all the ad-hoc thinking about proportions and sheep-weeks and sustainable grazing? I’m not sure. I like solving a tricky problem in two very different ways (and also checking), in order to gain confidence.

Thank you so much for all your help. I appreciate your time. It makes even more sense when you see it like that, and yes, I agree it’s more satisfying. I wasn’t expecting you to do all of it again, thanks! I feel more confident doing these kinds of problems now 🙂

While writing this, I found that we have previously answered a question of this type; the initial answer from 1996 was incomplete, but a reader later wrote in with an algebraic solution:

Feeding Oxen

In this more complicated problem, the solution involves growth rates measured in ox-weeks per acre.