## Wednesday, September 10, 2008

### This Logic Challenge is 'Par for the Course'!

Don't forget our MathAnagram for Aug-Sept. Thus far we have received a couple of correct responses. You are encouraged to make a conjecture!
Look here for directions. Here is the anagram again:

PRINCE? NAH! E-ROI!

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A former student sent me a wonderful reasoning problem involving mean, median, and mode, so it is accessible to middle schoolers. The question came from his teacher so I decided to revise it, put it in a different context but preserve the essence of the logic. The student will need to know some basics of scoring in golf but most of it should be clear. If not,
this may help.

This kind of question will frustrate some but reasonable frustration can often lead to 'pearls of wisdom.' Clear thinking and careful attention to detail is necessary. Certainly basic knowledge of measures of central tendency is a prerequisite, but this question can also serve to review these ideas.

Have fun with it yourself and, if you can, try it as a 5-minute warmup in class, preferably with students working in pairs. Let us know if they make a 'hole in one'! Again, thanks to my student and his teacher for the original source of this challenge.

Alex played 18 holes of golf and we know the following information:
His maximum score on any hole was a '5' and he shot this on six holes.
His median score on the 18 holes was 4.
The mode was 3.
What was the lowest possible mean score he could have achieved on the 18 holes?

mathmom said...

Nice! I am definitely adding this to my list of mean/median/mode/range questions.

mathmom said...

Here's a more advanced problem of this type, from 1995 AHSME: "A list of five positive integers has mean 12 and range 18. The mode and median are both 8. How many different values are possible for the second largest element in the list?"

Dave Marain said...

Mathmom--
There's nothing like a meaty AHSME (now termed AMC) question. I love this question. Although the 'range'info makes this a bit different from the golf question, there's just as much challenge to it, if not more. I found myself falling into an algebraic representation for the terms: x,8,8,y,x+18. From there the solution became fairly evident since y has to be even and greater than 8 and the sum has to be 60. I would imagine most students would use a guess-test approach. THANK YOU FOR SHARING THIS ONE!

Many educators realize that questions like these are learning instruments, not just ways of determining who 'gets it' and who doesn't. Unfortunately, many others may feel there is too much overhead (time 'wasted') in presenting these kinds of challenges in the classroom. These kinds of questions are often seen as only appropriate for those math team g----. I guess you know how I feel about this! In fact, I would give these kinds of problems to my 'fundamentals/skills' classes. I would review what was needed, give clues only after they struggled for awhile and was always gratified to see some youngsters make a breakthrough. More important is how they felt about themselves when I told them this was difficult math contest problem!

mathmom said...

Dave, I think it is tougher than the golf one. I came up with exactly the same algebraic representation, although I think that it is possible that the "line up" could end up looking like x,y,8,8,x+18 (where the second-highest number turns out to be 8) although there is no value of x (for x<= 8 which it must be) where it actually turns out that is the case. I made a chart an listed all the x values from 1 through 8, to check whether x+18 turned out to be bigger than y or not, and to see if y was ever going to be less than 8 (because, if there were multiple such values, they wouldn't each contribute a new "second largest element").

I do contest math with all my kids, and I know well that look of triumph when they solve one and start to realize that they really are good at math!

Dave Marain said...

mathmom--
I left out some details in my explanation. I had concluded that the 2nd highest number could not be 8, since the extreme case 8,8,8,8,26 would not sum to 60 (therefore x,y,8,8,x+18 would produce an even smaller sum. Once I obtained 2x+y=26, I realized x could range from 1 to 8 inclusive. Any value of x greater than 8 would lead to a y value less than x, which is impossible. Thus there are 8 possible values for both x and y.

You're probably right that this is more challenging than the other problem, but it's a close call, IMO.

I'd be interested in how students would compare these. Since the golf problem does not require an algebraic set-up, they'd probably find it easier, but we'd have to try it out...

mathmom said...

Dave, I don't think there are 8 values. Because if x=1, then x+18=19, but to sum to 60 we'd need y=24, which ruins the range constraint. Same with x=2.

Dave Marain said...

My error, mathmom. You're absolutely right. I didn't check carefully enough. Shame on me! I should have added the constraint:
x+18 > 26-2x or 3x > 8, which eliminates x =1 and 2. Yup, this is definitely a nice question!

Anonymous said...

I like this (now that I finally found it)

No algebra for me, just some blanks and a list. It went quick.

Jonathan

Anonymous said...

Alex played 18 holes of golf and we know the following information:

His maximum score on any hole was a '5' and he shot this on six holes.
His median score on the 18 holes was 4.
The mode was 3.
What was the lowest possible mean score he could have achieved on the 18 holes?

Solution:

Given: Maximum score = 5
Median score = 4
Mode = 3

Hole Score
1 -----> 3

2 -----> 3

3 -----> 3

4 -----> 3

5 -----> 3

6 -----> 3

7 -----> 3

8 -----> 3

9

(9.5) -----> 4

10

11 -----> 3

12 -----> 3

13 -----> 5

14 -----> 5

15 -----> 5

16 -----> 5

17 -----> 5

18 -----> 5

Mean should be 3.56

Dave Marain said...

Anon--
My individual results are a bit different from yours. I obtained 11/3 or 3.67 rounded as follows:
(These scores are listed in increasing order, not by holes)

1-1
7-3's
6-4's
4-5's

In sequence form:
1,3,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5
Note that the 9th and 10th scores need to be 4's in order to insure a median of 4 (using 3 and 5 would force more 5's, therefore a higher mean). Where I differ from you is on the 11th and 12th scores. Once the scores are listed in order, 3's are not an option after the 8th score.
My logic may be flawed but let me know if this makes sense.

David said...

I get the following
one hole in 1
seven 3s (since there are 6 5s and 3 is the mode)
four 4s
six 5s

Giving a total of 76, with a mean average of 4.22.

I think that's right, this is a great question. I'm putting together a collection of weekly problems and this is a great one to include!

Dave Marain said...

David,
Look at my solution above yours. I interchanged the number of 4's and 5's. By having 6 fours and 4 fives, I was able to produce a lower total, namely 66. This dropped the mean score to 3.67. Please check my logic and see if you agree.

Also, thanks for the kind words. I believe this problem would force youngsters to think quite a bit and perhaps deepen their understanding of mean, median and mode.
Dave

David said...

But you can't do that! The problem as stated says that there are 6 5s

Dave Marain said...

David,
You're absolutely right! Shame on me -- I was solving my own problem from memory without looking at the details. Thanks for the correction.
I'd better get some more sleep...
Dave

Dave Marain said...

David,
I should know better than to post anything in the evening. Thanks again for reminding me to read my own problem more carefully!
One more thing, either I'm going loony or your scores add up to 68, not 76, which would give a mean of 3.78 rounded. I'm doing this in my head at this moment but I think that's right.