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.
[埋込みオブジェクト:http://www.youtube.com/v/P_VCYl0zdts&hl=en_US&fs=1&rel=0]
More Advanced and Efficient Algorithms
(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)
Posted by Dave Marain at 2:37 PM 3 comments
Labels: average rates, math videos, mathnotationsvids, middle school, proportions, ratios, work problems