Wednesday, January 17, 2007

Problems for 1-18-07

Some SAT-types for today (more reasonable)...

1. A rectangle is inscribed in a semicircle so that the base of the rectangle lies on the diameter. If the diameter is 4 and the base of the rectangle is twice its height, what is the area of the rectangle? (Sorry, no diagrams today!)

2. If 1 + 2 + 3 + ... + 2006 = x, then, in terms of x, 2 + 3 + 4 + ... + 2007 = ?
(A) x - 1 (B) x + 1 (C) 2x (D) x + 2006 (E) x + 2007

3. On a certain 20-question math contest, 12 points are awarded for each correct answer, 4 points are deducted for each incorrect answer and a blank receives zero points. What is the fewest number of right answers needed to score at least 100 points if no questions are skipped?

Note: Although many students will use guess-test here (which is fine), we're looking for an algebraic approach as well.


Jonathan said...

I had a horrible geometry year in middle school, and never learned to do these. Now I know, recognize a missing segment... That's number 1.

Number 2??? I hope that doesn't trip up too many kids.

The factoring problem from a day back? I told you I used it with sophomores (as a time killer). Today I shared it with some freshmen, at the end of a review class. I was surprised - they chucked it into the calculator, then argued about how to move the decimal. I think that problem is a keeper.

Mrs. Temple said...

I liked all three of these :) The first one could again be used for 45-45-90 triangles (if I did it right heheh).

I agree with jonathan on #2, I would hope kids would notice how to add/subtract and not get hung up on the sequence

#3 took me a few minutes to think through, but as long as kids can identify their variables, they should be okay.

Again, great job! :)

Darmok said...

All right, I don’t have any paper with me and I can’t visualize it in my head well enough, so I’m skipping number one.

2 should be D, right? I mean, you start with x, but subtract 1 to get rid of that term, and then add 2007 to get the final term, so x - 1 + 2007 = x + 2006. Or I suppose you could think of adding 1 to each of the 2006 terms of x, yielding x + 2006.

3, assuming I understand the problem correctly: if we get x problems correct, our score will be 12x-4(20-x)=100; 8x-80=100; 8x=20, x= 2.5 → 3. Hmm, that's obviously not right. Oh, obviously, -4(20-x) is -80 + 4x so it would have been 16x=20, wait, what is wrong with me? Ahhh, clearly I cannot handle negative signs and clearly I should be writing this on paper instead of solving it in a comment...12x-4(20-x)=100; 16x-80=100; 16x=180; x=umm, 11.25 → 12. Check 12*12=144, 8*(-4) = -32; 11*12 = 132, 9*(-4) = -36, so yes it must be 12. That should have been much simpler.

Darmok said...

OK, I can’t resist at least trying number 1. So let's see, first off, let's say the height of the rectangle is x, so the base would be 2x, so the area is 2x^2.

If I were to draw a radius from the center of the (semi)circle to one of the upper corners of the rectangle, it would form a right triangle with sides of x and, oh obviously, x, so it's xr2, and since it's a radius of the circle, it = 4, so x = 4 / r2 or 4r2/2=2r2 (I think that's the hardest part of the problem for me!) Area = 2(2r2)^2 which is...2(4*2)=16. That better be right, though I wasn't expecting it to be an integer. I'm actually really surprised by that.

Though it makes sense, I could cut out the two small right triangles using the radii and then place them on top to make a large square ("diamond") with side 4. Huh.

Dave Marain said...

Here are the 'official' answers (not that you need them!)
1) 4 2) D (x + 2006) 3) 12
Other than a careless error with the radius (equals 2 not 4), Darmok's analyses are wonderful. Doing these in your head is great mental gymnastics but I need pencil and paper since visualization is not my strong suit...
Joanathan, thanks for trying these in the class and reporting back. I'm not surprised some tried to enter the powers of 10 factoring problem into the calculator. Now while we might be taken aback, the fact is that it led to a nice discussion of how to interpret the result. This is an aspect of calculator use often overlooked. The technology raises new questions that we would never have considered!
BTW, #2, is a typical SAT question that, IMO, develops reasoning and is a good vehicle for demonstrating several strategies. I like the add/subtract method mentioned by mrs. temple and darmok, but the other method mentioned by darmok is very nice also. Since the ti-83,-84,-89 can do summations (using LIST operations), some resourceful student could actually find each sum. Then again someone might invent a method we couldn't anticipate. The dialogue is so powerful here!
Finally, I'm wondering if the 4 of you are the only ones trying these! (Yes, yes, I can see I have many visitors daily but I have no idea why they're visiting or if they exit rapidly when seeing these problems!). Am I going to eventually lose my readership with these or do you think a few problems and a few comments about my passionate belief in a standardized curriculum are enough? I really appreciate your comments. There is nothing more gratifying for an 'author' than that one good review, never mind 4!!

Totally_clueless said...

The wording of (3) is somewhat confusing. I can answer 9 questions correctly, choosing not to provide answers to the rest and still get 108 points. Does this mean I skipped the 11 questions? A wording I would have preferred would be 'What is the minim um number of correct responses needed if no question was left blank?'

Dave Marain said...

I totally agree with totally_clueless! I got careless with my wording about 'skipped' vs. 'left blank'! The intent was that no question was left blank. Thanks!! Ah, the joy of writing unambigous math problems. That's why math people choose to communicate in cryptic symbolic code that is unambiguous to them! And I pride myself in using clear language!!

Darmok said...

Ah, you like the semi–stream-of-consciousness answers?

One thing I was always good at was showing my work; especially in undergrad I often wouldn't be able to come up with a solution, but I'd write down everything I was thinking of so at least the professor could see what my thought processes were. Or to take pity on me when I make stupid mistakes with radii and diameters (I’m a radius man). But yes, I’m a visual person too, as you can see from the several sloppy mistakes I made when trying the "solve in the comment box" approach.

Regarding #2, the only other solution I can think of would be to actually sum the sequences. The first one would be 2007*(2006/2), and the second would, 2009*(2006/2), so the difference would be (2009 - 2007) * (2006/2) = 2006. → x + 2006.

Dave Marain said...

Nice! Some students will always know those formulas for the sum of an arithmetic series! My goal is to show students how to solve these kinds of relationship questions without using the formula but the formula is always there if you need it...
In school we focus on procedures and formulas so much that when confronted with problems in which there is no obvious formula or it's forgotten, the student tends to give up too easily!

Darmok said...

Oh, I don't have the formulas memorized; I just derived them on the fly. You can pair up the sequence into 1+2006, 2+2005, and so on; each sum is 2007, and there are 2006 terms so 2006/2 pairs. The second's similar but the pairs ard 2+2007 = 2009. Of course, if one understands these principles it'd be easier just to use one of the other methods if you don't care about the actual sum.