Sunday, October 14, 2007

Last minute PSAT Prep... A Challenge for the rest of us!

NOTE: In addition to the PSAT/SAT practice in this post, I strongly recommend that interested readers go to the topic index in the sidebar, scroll down to SAT-Type Problems, click on this and you will pull up about 50 different posts of practice SAT questions!


While I'm working on the Alec Klein interview, here are a few problems for you or your students to work on as the PSAT fast approaches. These paired questions are generally similar to the test but some are on a higher level. The (b) part of each question should be somewhat more difficult. It generally helps, performance-wise, to practice with more challenging problems. Encourage students to visit the MATHCOUNTS home page (and many other math contest sites) to find similar challenges!

1) (a)How many 2-digit positive integers have exactly one digit equal to 4?

1) (b) How many 3-digit positive integers have exactly one digit equal to 4?

2) (a) How many 3-digit positive integers have a sum of digits equal to 3?

2) (b) How many 3-digit positive integers have a sum of digits equal to 6?

3) (a) When expanded, 1020 - 1 contains how many digits equal to 9?

3) (b) If k is a positive integer, 102k+1 - 1 contains how many digits equal to 9? Express your answer in terms of k.

4) (a) In terms of k, what is the sum of all values of x for which (x-k)2 = x-k?

4) (b) If f(x) = x2 - 2007x, for how many integer values is f(x) less than zero?

Comments:
(i) 3(b) and 4(a) would of course be multiple-choice type questions.
(ii) A problem like 4(b) would probably not appear on the PSAT as it is more advanced. Something like it (but simpler) could definitely appear on the SAT.

10 comments:

mathmom said...

Unless I'm missing something, 4b seems like one of the easier problems to me. if we want x^2 - 2007x < 0, first we note that x must be positive (since if x were negative, then x^2 would be positive and -2007*x would also be positive). So, we need 2007x > x^2 or 2007 > x (we can divide through by x without worrying because we know it's positive). Checking x=0, we see that f(x) = 0. So 0<x<2007 which range contains 2006 positive integers. Even kids not used to working with inequalities, this seems pretty easy to "eyeball" to me. you want 2007*x to be bigger than x*x which will be true as long as 2007 is bigger than x.

Dave Marain said...

Mathmom--
You rarely if ever miss something! I agree that this is straightforward, however there are several points to be made here:
(1) Expressing the problem in function notation immediately raises the difficulty level since many high school students are still intimidated by this, even Algebra 2 students (experience talking). The function piece is what makes this unlikely on the PSAT.
(2) I really had an ulterior motive for selecting this problem. It encourages the use of multiple representations (aka, 'Rule of Four') by the instructor:
(a) Function table approach (with or without the graphing calculator)
(b) Verbal interpretation, i.e., x cannot be negative using your explanation)
(c) Graphical/visual approach: The parabola is only below the x-axis for x-values between the intercepts
(d) Algebraic solution using 'sign-charts' (or the myriad of names for this):
+++++++++|--------|++++++++
0 2007
I agree with all of your comments except for the fact that, in actual practice, student performance on these kinds of questions is surprisingly low. That's why it typically appears among the last 5 questions of a section on the exam. Remember, the Educational Testing Service has a pretty good idea of levels of difficulty (and therefore problem placement) even before they administer the real test! This is because these questions have been placed in an experimental section prior to the actual 'live' implementation (the 'operational' version). If, say, 75% of the students answered this question incorrectly on an experimental section, ETS places this question among the last 5 on the actual test. In some cases, questions are rejected based on the field-test statistics (reliability, p-value too low, etc.).

mathmom said...

Expressing the problem in function notation immediately raises the difficulty level since many high school students are still intimidated by this, even Algebra 2 students (experience talking).

Really? I would have guessed that having a real number in the problem (instead of k, as in part a) would make it easier, and the function notation is so peripheral there. But I don't have experience with Algebra II kids, so I'll take your word for it. :)

Jackie said...

I'm going to back Dave up on this one (not that he needs it). Function notation is confusing for kids.

f(3)=2x+4

Many (of my) students still think this means 3 = 2x +4.

mathmom said...

Wait a minute: what does f(3)=2x+4 mean? Are you using x as a constant there?

Totally_clueless said...

History repeats itself as to the methods that can be used. Thus, the answers to 2(a) and 2(b) are 4C2 and 7C2.

TC

Dave Marain said...

Hey tc!
I knew you wouldn't miss that opportunity! What's really interesting is that many students appreciate the pattern of triangular numbers for these types.
For sum of 3:
300; 210,201; 120,102,111
1+2+3 = 6 = 4C2 = (3)(2)/2
In general, the nth triangular number is nC2 = n(n-1)/2.
Works for a sum of 6 as well, of course.

Jackie said...

argh, that's what I get for trying to comment before 6:00 in the morning. I meant, they have trouble finding f(3), when f(x) = .....

Sorry.

Dave Marain said...

Jackie,
That's weird. All of my mistakes occur AFTER 6 AM!
Dave

Jackie said...

Before 6am, after 6am, I've stopped tracking when they occur. All I know is the number of occurrences is in one-to-one correspondence with the reals.

That's about the extent of the math I'll be able to add to the conversation at this point.