Thursday, November 15, 2007

Percent Increase vs. Decrease Redux: A Rich Investigation for Middle Schoolers and Algebra Students

[As always, don't forget to give proper attribution when using the following in the classroom or elsewhere as indicated in the sidebar]


While we are contemplating tc's rectangles inscribed in circles problem (and we will post some solutions in a couple of days as needed), here's a change of pace. Awhile back, there was considerable interest in a percent problem posted on MathNotations involving an artificial scenario in which there were 20% more girls than boys in a group. Remember the heated discussion about the semantics of that problem?


Well, here's another scenario for you to challenge your middle school students or yourself...

STUDENT ACTIVITY or CHALLENGE FOR OUR READERS

Final score in the basketball game: Central 90, Eastside 60.
Jay, who played on Eastside, thought to himself after the game: "If we had scored 15% more points and they had scored 15% fewer points, we would have tied."

Now, how's that for a real-world application of percents. I'm sure you know hundreds of students who would think like that after the big game. Well, it's my blog and Jay is my invention and that's how he was thinking, so there! Of course we know that Jay was confusing increase and decrease of points with % increase and % decrease. But, solve the following:

(a) Increase 60 by 15% and decrease 90 by 15% to show numerically that Jay's reasoning was incorrect.

(b) Increase 60 by 20% and decrease 90 by 20% to show that 20% is the percent Jay had intended.

(c) Determine algebraically that the correct answer is 20%, again starting with scores of 60 and 90.
Note: While parts (a) and (b) can be handled by most middle school students in prealgebra, this question should prove more difficult, even for Algebra I students. However, some will get it and, with guidance, the rest can too!

(d) You surely didn't think we would let you off that easily, did you? Of course, we will now ask you to generalize the result:

Suppose A and B are positive numbers and A is less than B.
If A is increased by X% and B is decreased by X%, the results are the same. Determine an expression for X in terms of A and B.
Note: If you come up with the formula, think about why it makes sense. Any thoughts?

13 comments:

Totally_clueless said...

Interesting!

Let us consider the number of points scored in each case.

(1) If the student thought: If we had scored X more points, and the opponents had scored X fewer points, we would have been equal, For the right value of X, the points scored by each team would have been the Arithmetic mean of A&B.

(2) If the student thought (as in this problem), that if we had scored X% more points, and the opponents had scored X% fewer points, then we would have been equal. For the right value of X, the points scored by each team would be the Harmonic mean of A&B.

I am wondering - 'What is the similar thought to get the geometric mean?' I can write it as an equation, but that does not seem to lend itself to a cogent thought.


TC.

Dave Marain said...

Beautiful generalization, tc! How many hs students never get to see the harmonic mean! I'll have to play with the geometric mean also. I'm sure there is an interesting interpretation we can make based on 60:x = x:90.

Totally_clueless said...

OK, One thought seems to be : If we had scored X times higher, and they had scored X times lower, we would be equal.' For the right value of X, we get the geometric mean.

TC

Totally_clueless said...

An offshoot problem: Find two positive integers whose arithmetic, geometric and harmonic means are all integers. For bonus points, find the two integers for which the sum of the three integer means is lowest.

Dave Marain said...

tc--
I think your extra problem should be entitled, "By All Means!" (sorry, I couldn't resist that).
This is a nice problem and, in its more general form, perhaps could stand alone as a "tc challenge!"

For those who might have forgotten the definition of the harmonic mean of A and B, the formula is 2AB/(A+B). I realize that these days we go right to Wikipedia or MathWorld to get this background!
In general, the HM is defined as the reciprocal of the Arithmetic Mean of the reciprocals of A and B (or, in general, for n numbers).

My analysis is fairly unsophisticated and I suspect, tc, you have a more penetrating analysis. Well, first of all, I assume you intended the problem to read "unequal integers", otherwise we have trivial solutions like A=1, B=1. I will refer to this as the trivial case below. The rest of the discussion below generally assumes A and B are not equal.

My initial analysis was fairly basic. A and B had to have the same parity from the Arithmetic Mean condition. I decided to work from the Geometric Mean condition and restrict attention to perfect squares, i.e., AB = C^2. I concluded that C^2 cannot be the square of an odd prime (which produces only the trivial solution A=B) or a power of 2, so I was able to get to 225 = 15^2 fairly quickly. This led to the factors A=5, B=45, which satisfy all 3 conditions. You can tell me if this satisfies your bonus in the non-trivial case.

My instinct is that a general analysis of conditions insuring that the harmonic mean is an integer is non-trivial.

Totally_clueless said...

Hi Dave,

I realized very soon that I had not explicitly ruled out the trivial case, but trusted that our readers can be discerning enough to eschew this.

Yes, 5 & 45 is the bonus answer. In my typical rambling fashion, I first found 10 & 40, and then found 5 & 45.

You realize very soon that both numbers may have to be multiples of 5.

The extra challenge is to find two such integers (with all three integer means) such that both are not multiples of 5. Alternatively, prove that they both have to be multiples of 5.

TC

Totally_clueless said...

If someone can actually prove that in order for the arithmetic, geometric and harmonic means of two integers to be integers, the two numbers have to be multiples of five, I would really like to see that proof (since I can then have the challenge of finding the flaw in the proof:-))

TC

Dave Marain said...

tc--
Is the multiple of 5 idea a conjecture based on observation or do you have an algebraic argument? This is impressive if it turns out to be a necessary condition. Certainly, we can generate an infinitude of solutions from A=5, B=45 by taking integer multiples, i.e., A=5m, B=45m will also satisfy all 3 conditions. Proving necessity in general seems more formidable.

Totally_clueless said...

HI Dave,

The point of my last comment was that there was bound to be a flaw in the proof because the statement is wrong.

Some other doubles include (13,325), (34, 544), (104,234).

While trying to generalize these, I arrived at a couple of other conjectures:
For any integer m
(1) 7 does not divide m^2+1
(2) 11 does not divide m^2+1

TC

Jayant said...

What if we add the condition that, find A and B such that abs(A - B ) is minimum ?! Then clearly we can do better than A = 45, B = 5...

Matt Badley said...

Wow, I know I'm really late to this party, but I just now was directed here. I did a little investigation on this and found that each pair of unequal integers are related to Pythagorean triples. The 45 & 5 and the 10 & 40 are from a 3-4-5. The arithmetic mean is always the square on the hypotenuse.
I was hoping to explore this more in depth, if anyone is interested...

Matt Badley said...

Oops, forgot to mention that the harmonic mean is the square on one of the legs.

Dave Marain said...

Sorry, Matt, that I didn't acknowledge you earlier...

Nice observations and I would definitely like to see more of your thinking on this. Could you email me details at dmarain "at" gmail dot com?