Saturday, March 31, 2012

If a hen and a half can lay...

Share your teaching methods for the classic problem in the title:

If a hen and a half can lay an egg and a half in a day and a half, how many eggs can 3 hens lay in 3 days?

Would you ask students for an immediate intuitive guess and expect many to say 3?
The answer is 6 so the purpose of this post is reflection on sharing instructional strategies.

What are the BIG IDEAS here? Do you use one basic strategy in teaching all ratio problems?  Does dimensional analysis work for middle schoolers?  Should students always reduce everything to a single unit ratio like 1 hen per day? How many ways could your students devise if we tell them the answer is not 3?

Not much sharing going on like in the old days of this blog but maybe it's time..


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4 comments:

Joshua Zucker said...

I would point out that there are TWO things that have changed, and encourage them to deal with the changes one at a time.

What happens when you double the number of hens?
What happens when you double the number of days?

Dave Marain said...

That's wonderful, Joshua. It is not algorithmic, appeals to intuition and is mathematically powercul and generalizes to many variables. You and I know how iften students rely on procedures which causes a loss in confidencr in trusting one's instincts. I tend to use a TABLE approach with 3 columns for those who need a visual ir more structure.

Even some younger children are capable of reasoning:
Hey, there are twice as many chickens and twice the time so they can lay 2•2 =4 times as many eggs

OrangeMath said...

I would teach it as "we're multiplying by hens and days, and get eggs. Set up the ratios appropriately to cancel units." I'm not saying this is smart, but it's my general method!

Dave Marain said...

I think it is smart. The deeper issue is why does it make mathematical sense to multiply those units:

The fundamental rate which we're assuming is constant here is eggs per hen per day or (E/H)/D = E/(H•D). Thus E = k(HD) We used to say that E varies JOINTLY as H and D. Do texts still do that?

To determine k, we sub in values:
3/2 = k(3/2)(3/2) or k = 2/3. Now we can replace H and D by 3 to obtain E = (2/3)(3)(3)=6. Voila!