Tuesday, August 4, 2009

Another 'Average' Problem for Standardized Tests and Conceptual Understanding

After 4 tests, Barry's average score was 5 points higher than Michelle's. After the 5th test, Michelle's overall average was 5 points higher than Barry's. Michelle's score on the 5th test was how many points higher than Barry's?

Can you find at least three methods for solving this?
Algebraic, "plug-in", conceptual, etc...

As teachers we need to have a deep understanding of these kinds of problems and familiarity with several approaches. Of course, our students will show us a variety of methods, both right and wrong, when we open up the dialog!


Comments
Students from middle school on see many problems relating to means. However, they need to see a variety of problems of increasing difficulty. This question is certainly not a highly challenging math contest problem but I believe it demonstrates some important principles of averages and can be used to review different problem-solving strategies. Middle schoolers would struggle with the algebraic approach (a system of two equations), however they should be thoroughly comfortable with the underlying ideas.

Since the focus is on concept and method, I will give the answer: 45


















9 comments:

watchmath said...

An attempt to avoid algebra:
After 4 test the difference point between Michael and Barry is -4*5=-20(it is minus since Barry is higher)

After 5 test the difference point between Michael and Barry is 5*5=25.

Which means that the difference between Michael and Barry on the 5th test is 25-(-20)=45.

Dave Marain said...

Nice, watchmath!
A "verbal" way of expressing what you did is to say that Michelle was 20 points "down" and ended 25 points "ahead". Same as yours but language also can play a key role in conceptual understanding.

Well, you found the conceptual method based on the principle that SUM = N ⋅AVG. How many of your students do you think would be able to "see" this?

We still have the algebraic approach and the "plug-in" method to go...

Totally_clueless said...

Given the above data, it may also be an interesting exercise to ask whose median score is higher.

TC

Dave Marain said...

Agreed, tc!
Issues of how medians are related usually require more logic and consideration of cases.

Your question also made me wonder about replacing 'mean' by 'median' throughout the question. This requires somewhat more analysis since the relationship between their 5th scores cannot be determined.

Example:
B's scores: 60,60,90,90
Median = 75
If his 5th score is, say, 75, then his median remains constant.

M's scores: 60,60,80,80
Median = 70
If her 5th score is greater than or equal to 80, her new median of course would become 80.

Eric Jablow said...

Let's bring up a non-mathematical point: how do you choose the proper nouns you use in your questions? I presume you chose Barry because 'Barack' would have been jarring.

There's the traditional Alice and Bob. My, they really get around.

Legal questions often involve John (or Jane) Doe, or Richard Roe.

You can take your names from one television show or another. One can often gain great insight into a teacher's mindset by looking at his choices. Moe/Curly/Larry? Mary/Lou/Sue Ann/Ted? Ginger/Mary Ann? Homer/Marge?

Similarly, companies also need names. I was always traditional; I preferred to use Acme.

Dave Marain said...

Eric,
My blanket disclaimer:

Any resemblance to any real person or people
Is purely fortuitous!

Thanks for reminding me of other names I can use!

letsplaymath said...

The algebra method is definitely beyond my current middle school students.

"Plug in" can be very easy: set Michelle's original average at 0 and Barry's at 5, then let him score another 5 on the last test. The hardest part about this method would be to make sure my students see why it gives a correct answer no matter what their scores had really been.

My favorite method for average problems like this is to create a rough bar diagram. The kids need to understand that average is the amount the scores would be "if they were spread out evenly"---so we can show them as bars the same height. Sort of like liquid from 4 different glasses that was all mixed together and then poured back out so everybody (or in this case, every test) had the same amount. It doesn't matter how much liquid was in any single glass at first, only what is there after it's been evened out.

With bars, it's easy to see the relationship that watchmath pointed out: That we would need 5 points to bring each of Michelle's test scores up to match Barry's average, and then another 5 per test to raise her higher. So she needs 40 points to lift up the tests she's already done, plus she will need to beat Barry's score by 5 on the last test, too. That's a total of 45 points she will need to "catch up" to the situation at the end of the story.

The principle that SUM = N ⋅ AVG is key to solving many MathCounts puzzles. My students don't "get" it intuitively, but I try to push it enough that by the time they leave my class, it comes almost automatically. So that whenever they see "average" in a problem, they automatically think, "average-->sum."

Anonymous said...

b1=Barry average after 4 tests
m1=Michelle average after 4 tests

Using Sum=n x avg

(4b1)/4=(4m1)/4 + 5
So, 4b1=4m1+20

b5=Barry's score on 5th test
m5=Michelle's score on 5th test

(4b1 + b5)/5 + 5=(4m1 + m5)/5
So, 4b1 + b5 + 25=4m1 + m5

Using substitution:
4m1 + 20 + b5 + 25=4m1 + m5

Eliminate 4m1 from both sides:
20 + b5 + 25=m5
45=m5-b5, the difference in the scores.

Not sure my students could get this, however. I really like the analogy with the level of the glasses!

Dave Marain said...

Nice algebraic work, Anon...

And, yes, I was awed by the ingenuity of Denise's mixing of glasses model. I believe that finding such analogies is a major part of the art and science of teaching concepts. The greatest teachers I've had seemed to be able to come up with these types of illustrations spontaneously, whereas I usually needed to think deeply about them beforehand.