It's that time of year folks! Here's a problem for you or your students to think about as a warm-up for this week or for the SATs or for just developing logical thinking. It might also deepen student thinking about the distribution of data. You might find this question trivial but don't be too quick to judge this until your students try it! I would allow a calculator to be used. Observe how students approach this: Guess-Test, consideration of the 'mean', etc...
In a certain high school election, there were 12 candidates for President of the Student Council. If 1600 votes were cast and Denise received more votes than any of the others (i.e., she received a plurality), what is the least number of votes she could have received?
Sunday, October 19, 2008
A 'Mean' Problem to Ponder
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11 comments:
When a kid tells me that something is easy, I ask them to show me.
And here, I glanced, it was easy. And then I stopped to answer. I like the twist. It took me a moment longer than I thought at first blush.
Jonathan
I missed the "twist" at first and might not have bothered to even look at the given numbers until JD mentioned it.
I asked my youngest the question (before noticing the twist) with easier numbers -- 1000 voters, 10 candidates, and he answered quickly (and correctly). When I asked him about 14 voters and 3 candidates, he decided the problem was "boring" (too hard) and left the room. ;-)
Thanks jonathan and mathmom--
It appears that the 'twist' made the problem more interesting than the main logic!
Ok, the answer to this question is 135, not 134 as most students would conjecture (until they actually check to see if it works!).
Because of insanity, I actually did this problem algebraically in addition to incorrectly 'guessing' that the answer was 1 more than the mean. Further, I solved the problem in more general terms using T for the total number of votes cast and N for the number of candidates. If K represent the fewest number of votes Denise could have earned, K must satisfy:
K + (N-1)(K-1) ≥ T
Solving for K we obtain
K ≥ T/N + 1 - 1/N.
Note that T/N is the mean.
Of course K has to be an integer, so K would be the least integer greater than or equal to the expression on the right side of the above inequality. (This can be described using the 'ceiling' function).
Now, here's the rub. Depending on the remainder when T is divided by N, the value of this integer is either [Mean] + 1 or [Mean] + 2, where [...] represents the greatest integer (aka 'floor' function).
Can you determine the rule!
Hint: For T = 1600 and N = 12, the remainder is 4.
Funny, this problem scratched the "pigeonhole principle" part of my brain, not the "arithmetic mean" part. But either way I think I did the same thing - divided, tested, and adjusted.
Funny, Kate, but I had that same itch too! I see it both ways, distribution about the mean and pigeons flying into their coops! Worth exploring further...
Do you think your students would struggle at all with the logic? the adjustment? Let me know if you try it.
I didn't really think about it in terms of the "mean" just thought about dividing the votes evenly, and then adjusting slightly.
mathmom--
I think that's how the 'problem solvers' in the class would see it too. However, it is my experience that many students struggle "seeing" what appears somewhat "obvious" to others, i.e., those who have developed number sense vs. those who operate more mechanically and procedurally. This type of question tends to discriminate between these two groups in my experience. However, the brain is strengthened by doing many of these kinds of problems. This tends to close the gap somewhat between the two groups. But I'm preaching to the choir here!
There's a similar problem accessible to 10-year-olds.
P.S. 279 wishes to send its 70 3rd graders on a field trip to the Rose Planetarium. The school buses have seats for 30 people, but one teacher and one parent volunteer will be on each bus. How many buses will the school need?
Many students will answer 2½, no many times one tells them that ½ a school bus isn't very useful.
Dave, I think for kids who don't "see" the logic of it right away, it would be good to take the "solve a simpler problem" approach to see if they can start to see how it "works".
Eric, I think your problem is accessible to younger kids as well (2nd grade). And if you ask them that young, they won't tell you 2½ ;-)
mathmom--
'Making it simpler' is always my favorite.
Eric--
There's an additional subtlety in the 'tweaking' with the question I posted, since, depending on the total number of votes, one adds ONE or TWO to the greatest integer value of the mean (see my comment above). So rounding up may not be enough here.
Nah I don't think the logic of the pigeonhole principle is too hard for even young kids. Start with socks in a drawer and build it up. There are lots of good Math Olympiad type resources of questions of varying difficulty. Older kids can think about Ramsey numbers...fun!
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