## Thursday, May 13, 2010

### If a hen and a half can lay an egg and a half in a day and a half...

The full version in one of its many many variations:

If a hen and a half can lay an egg and a half in a day and a half, how many eggs can three hens lay in three days? Assume that all hens are a-laying at the same rate.

Putting aside the silliness of the riddle, there really is some serious mathematics going in these kinds of rate/ratio/proportion problems. Rather than solve the "hen" problem for you, I'll leave it to my readers to solve it by their own favorite methods. By the way, the answer to this riddle is in the description of the video below on my YouTube channel. Sorry 'bout that!!

Instead, the video below, which appears on my YouTube channel, MathNotationsVids, presents a developmental approach to a more complicated ratio problem for middle schoolers and beyond. I'm far more interested in your thoughts about the teaching strategies than I am about the problem itself. Please understand, further, that I am not suggesting the method shown in the video is efficient nor would it make much sense for the upper level math or science student. See comments below the video for further discussion of this.

The Problem in the Video Below:

If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days? Assume all workers build at the same rate.

(1) We assume from the "constant rate" assumption in  the problem that the number of houses (H) which can be built varies jointly as the number of workers (W) and the number of days (D).
Thus, H = kWD.

Substituting, H=3, W=10 and D=60, we obtain:
3 = k(10)(60) or k = 1/200. Note that the units of k are Houses/(Workers x Days).
We can interpret k to mean that 1/200 of a house can be built by 1 worker in 1 day. Thus, k is not only a constant but actually represents a rate. Another way of expressing this rate is
(1 House)/(200 Worker-Days) or the reciprocal version:
(200 Worker⋅Days)/(1 House)

Substituting the new set of values into the relationship H = (1/200)WD, we obtain:
5 = (1/200)(W)(40) or W = 25 workers.

(2) This can be made even more efficient using the "factor-label" (dimensional analysis, etc.) format:

(200 Worker⋅Days)/(1 House)) x (5 Houses)/(40 Days) = 25 Workers!

(3)  I could also exploit the inverse variation between W and D, but that's for my readers to bring up or for another video!

I see these efficient methods as "black box" methods for some students. Developing a deeper understanding of direct and inverse variation is far more important for the younger student.

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"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860)

letsplaymath.net said...

For middle school, I would have added one more layer to the chart. The key tool in this sort of problem is to "Think 1," which is a simple enough concept that the students should be able to remember for reasoning through the steps:

You did...
Houses/Workers/Days
3/10/60
Think 1 house:
1/10/?
1/10/20
Then 2 houses are:
2/10/?
2/10/40
All keeping the number of workers the same. Now you've got the number of days where you want them. It's time to "Think 1" again...

Think 1 house:
1/?/40
1/5/40
And from there, you can easily move to 5 houses:
5/?/40
5/25/40

The proportion mushes the last two steps together, but I think for most middle school students it would be a "trick method" that they apply without understanding. On the other hand, "Think 1" seems logically intuitive.

Or perhaps your middle schoolers reason more abstractly than mine...

Denise

jd2718 said...

How many worker-days does it take to build a house?

The rest gets sort of easy.

10 workers, 60 days each - that's 600 work days for 3 houses, or 200 worker days per house.

That means 200 workers would take 1 day. Or 1 worker would take 200 days.

So you want 5 houses, that's 5*200 or 1000 worker-days? And we have 40 days to do it?

Jonathan

Dave Marain said...

Denise--
Excellent simplification! I was torn between that and leaving one proportion to be solved as a demonstration of its use. For the younger student, yours is superior.

Jonathan,
I had a hard time deciding about using both direct and inverse ratios in this video. Originally, your method was in the video. Ultimately, I decided to abandon the inverse relationship between workers and days to keep the instruction at a more basic level. However, this problem lends itself naturally to a discussion of inverse ratio concepts:
"If we have fewer days, boys and girls, do we need fewer workers or more workers to complete the same job?"

In the end, H = kWD is the essence of this problem. This equation can be written in many ways to demonstrate the direct or inverse relationship between any two of the quantities. Jonathan, is "joint variation" stressed as much these days in our algebra curriculum? Do we say that H varies jointly as W and D meaning H varies directly as the product of W and D?

By the way, a recent SAT problem assessed a conceptual understanding of inverse ratio, that is:
y varies inversely as x is equivalent to
y varies directly as 1/x. I was happy to see this kind of question.