They do what they do.
Thinking about schools and peers and parent-child attachments....I came across one of my favorite posts .
I don't like brute force problems. A brute force problem is a problem that I can't solve in some nice way, and simply have to crank through the work. On a test like an SAT, I said you should know how to solve the problem by the time you've finished reading it. In this case, the way I knew to solve the problem quickly was the brute force method, and thinking for 30 more seconds hadn't produced any other idea. My brute force method was 3 equations with 3 unknowns, solve for x by row reduction.And apparently I'm lazy now, because my idea of brute force is "I have to write something down."
I hate that I keep getting these right but I don't set the problem up I just sort of logic it out. That is not helpful when the numbers are big or complicated.
I saw a slightly cleaner approach, Allison. If I subtracted the number of students in only one club from the sum of the total students, then the answer was 1/2 of the remaining students (since they were double counted). Of course, that's really kind of the same as three equations, three unknowns, but I'm not quick with row reduction.
These kind of questions crop up far more often than is statistically likely. Allison is right to say that it's three equations and three unknowns, but those equations are a+b=15, b+c=12 and a+c=13. Adding all three equations (ChemProf's double counted idea) gives 2a+2b+2c=40, from which there are 20 total students (a+b+c=20) and so our answer is 7 (I think, the question has already changed!!).Like I said, systems of linear equations where each variable is summated the same number of times crop up pretty often. The best question I've seen that has the same idea isA list of six positive integers p, q, r, s, t, u satisfies p < q < r < s < t < u. There are exactly 15 pairs of numbers that can be formed by choosing two different numbers from this list. The sums of these 15 pairs of numbers are:25, 30, 38, 41, 49, 52, 54, 63, 68, 76, 79, 90, 95, 103, 117:Which sum equals r + s?(A) 52 (B) 54 (C) 63 (D) 68 (E) 76That questions was taken from the Canadian Gauss Contest 2009. Other contests can be found here.Richard I
When I followed that link, I got an easy fill-in-the-blank vocabulary question. Which I don't think was intended. :-)Looks like their "problem of the day" changes from day to day. Does that site offer a permalink for the problem in question? Because now I'm curious!
About a month ago on a math question, none of the provided answers was correct but one at least had the right cents so I clicked that.The answer explanation had changed a dollar amount in the question to fit the provided answers.I contacted the College Board and after going through several levels, they couldn't understand why changing the question was a perfectly good fix.And no they had no interest in telling the Problem of the Day subscriber base that they had screwed up.
Yesterday when it was a math problem, I had 15 yr. old math kid read it. He did, and immediately answered, "Oh, seven," before walking off. I didn't get a chance to ask him if it was fun.SusanS
Today it was an easy vocabulary question but I was appalled that only 61% of the people who had attempted the question answered it correctly.
Post a Comment