Friday, January 19, 2007

Problems for 1-22-07

I am trying to post these problems before 3 PM each day, Mon-Fri. Again, I depend on your feedback to gauge whether these problems are useful for your students (grade-level dependent of course) or engaging even if you're not a frontline math teacher. I'm trying to choose problems I have written that address current math standards in most states or topics I feel are underrepresented. Moreover, I try to select questions that are nonroutine requiring more time and analysis than students usually give to their homework. Do they fit naturally anywhere in your curriculum? Should our textbooks have some questions like these throughout or just in some ancillary workbook for enrichment? Can you guess my preference!

For Monday, #1 and #3 are Algebra I/II and #2 is for Gr 7-12:

1. If n2+n = x, then, in terms of x, (2n+1)2 = ?
(A) x+1 (B) 4x (C) 4x+1 (D) 4x2 (E) 4x2+1

2. If a number is chosen randomly from the set of 2-digit positive odd integers, what is the probability that the product of its digits is even?

3. For the system x+y > -3 and x-y <5, which of the following must be true?
(A) y >-4 (B) y<-4 (C) x>1 (D) x<1 (E) none of the preceding answers is correct


Anonymous said...

For my practice, these questions fit best into the opener time slot. I wouldn't use #2, for example, to assess my students' knowledge of probability. If a kid failed it, it would be hard to tell if he failed because he doesn't understand probability or because he can't parse the multi-layered math terminology.

That said, I think #2 does a great job forcing kids to think hard about what each modifier means. Two-digit ... okay, 1 - 9 are out ... positive ... got that ... odd ... well that complicates things a bit.

All the two-digit numbers with an even number in the ten's place are good. No matter the unit's digit, the result will be even. So that right there is 40 numbers.

For the 10's, 30's, 50's, 70's, and 90's, there are 5 numbers each (the even ones) which give an even product. That's 25.

Total desired outcomes is 65. Total outcomes is 90. That gives us about 72%. (Would you bet for or against getting an even product, I'd ask, always putting some mental stakes on the problem for them.)

Good problem, thanks.

Anonymous said...

I like (1) and (3) since they cannot be solved by just going through the possibilities (maybe (1), but it is a lot more tedious than by just doing it). You have to actually do the problem.

(2) is quite nice. Part of getting a probability problem is to correctly identify the entire sample space and the chosen set. This can be done by explicit counting, where possible, or by determining the number of possiblilties. Maybe you can show both ways if someone goes wrong.

My answer, anyway, is 4/9.

Anonymous said...


I teach an elective "combinatorics" course. #2 is just the kind of question I like to use a lot in class. (except more likely as a "how many?" than as a combinatorial probability quesiton. On a test that would be the last of a group of (more straight-forward) questions.

On today's (today's!) final, I asked (after two simpler questions) How many 3 digit numbers have exactly two repeated digits?

For your number 2, a twist. Expand the set to all 2-digit positive integers.

Dave Marain said...

thanks for the support dan et anon!
#2 requires careful reading with max info compressed into a few words - favorite type on hard SAT problems or math contests but reasonable to give to middle schoolers; i'm a great believer in using these kinds of problems to develop reasoning and problem-solving prowess while reinforcing basic math principles
dan, check the wording again for the sample space on which this problem is based -- it was intentionally nasty!

i definitely agree that it wouldn't be useful to assess knowledge of prob concepts but repeated exposure to these without the threat of a grade penalty can only help students improve; here's my naive philosophy: you can only improve at solving hard problems by doing hard problems! i never improved my chess rating by playing opponents with lower ratings!
btw, replacing n by 1 in #1 gives a value of x = 2; then (2n+1)^2 = 9; now test choices replacing x by 2; definitely not more desirable than doing the algebra but an interesting functional approach since many would be tempted to choose a value for x first; remember some students will always find such methods far more seductive even if the algebra is easier -- facts of life!

druin said...

I guess I need some clarification on #2 because from how I read it, you are only interested in the odd numbers (11, 13, 15, 17, 19, 21, etc), so I would have 45 possibilities, not the 90 that was mentioned. Thanks

Dave Marain said...

1) C 2) 20/45 = 4/9 3) A
#2 is definitely my favorite because you have to be very alert (hard when you're getting 3-4 hrs sleep!) to realize, as anon and mrs temple did, that your sample space is just the set of odd numbers; jonathan suggested the variation of using all 2-digit numbers (as dan did) which is also a nice combinatorial problem. However, most students usually mess up the one I gave so I don't need to make it more complicated!
Personally, I have an affinity for combinatorial questions since they require the most thought. Students want a simple procedure or formula for each problem because they do not want to take the time to think; after all, the object is to get through all of their homework ASAP to have time for important things! Anyway, I do encourage students to consider using the multiplication principle for most (not all) digit problems:
# of ODD 2-digit integers: (9)(5) = 45
# of ODD 2-digit integers with EVEN product:
(4)(5) = 20

I also like #3 because it isn't asking for the SOLUTION of the system. It's asking for a conclusion, i.e., if both inequalities are true, then what. Again the reason I love rich problems (other than for my own personal challenge) is that it can lead to such fruitful discussion (if one values this). The instructor can demonstrate rewriting this in a form for the graphing calculator and using the shade feature. The solution of the system shows that (A) is always true and the other choices are not, although the graphing calculator does not make clear whether y could equal -4, but that's not needed here. Some students will try to plug numbers in to eliminate choices but it's not so obvious when there are 2 variables!
Here's another approach:
Multiply the 2nd inequality by (-1) obtaining:
-x+y > -5.
Since both inequalities are now 'greater than', you can ADD them to obtain (A)! Does this really prove that (A) MUST be true? Yes! But not every ordered pair with y > -4 is a solution. The logic is one-way only. Pretty subtle stuff...

No more questions until Mon afternoon! I promised my wife and kids! But then I'll lose the audience and I may have withdrawal symptoms... Is there a 12-step program for blog addicts!
Have a great weekend! [Of course, if you write any comments I will respond in less than 30 nanoseconds so don't ever stop.]

Just a question here. I feel like the most important stuff in this blog is in your comments and my replies. Does it make sense to refer readers to these in the original post or does everyone realize this?

Anonymous said...

I've been had.

Anonymous said...

Hey, Dave, you have any good open-ended final-exam-worthy first-semester Geometry questions?

We're through constructions, inductive reasoning, triangle properties/proofs, polygon properties, and circle properties.

I know this isn't Total Request Live or anything, but will you be my hero, Dave?

Dave Marain said...

dan, let me look for you...
if i can't find any i might be able to locate some online - no promises - I'm feeling a bit overwhelmed but I'm sure you can relate!
if you send me a personal email i can reply directly with attachments

Anonymous said...

I just started to read this blog, and work through some of the problems. I'm lost on #1. Could you show the work on how to get solution to this problem?