Thursday, November 28, 2013

Turkey Day stuffing recipe...Have to double 3/4 cup...Three coins...

Happy Thanksgiving and Happy Hanukkah to all out there in cyberspace! The confluence of these two holidays provides plenty of grist for a math blogger but I'm not going there today...

Silly title, perhaps, but to all the experts who will tell us that money is the root of all evil when teaching fraction concepts/skills, uh whatever...

Are there any adults out there who don't want to mess up that Turkey Day recipe and will ask the 'math expert' in the family to verify what doubling three-fourths is?

Three-fourths cup...three quarters...75 cents...double...$1.50...1.5...1 1/2 cups

But, Dave, this won't work for multiplying 5/7 by 2 so this is worthless...

To quote a certain ad,
"Hey, if it works, it's not crazy!"

Seriously, from my perspective, money is as "real" as it gets. A child can be guided to discover the algorithm for multiplying a fraction by a whole number in an endless variety of ways ---
Concrete objects
Fraction pieces and all those other manipulatives
Bar models
Pictures
Simple numerical patterns

But I wouldn't be afraid to use money! Imagine making connections...

Wednesday, November 27, 2013

How (m^2)/(n^2)=(m/n)^2 is Fundamental to Geometry!

OVERVIEW
The Common Core stresses the importance of students developing a deeper understanding of fundamental concepts and to discover/uncover the interrelatedness of mathematics. The discussion below can be used to demonstrate how a basic law of exponents is tied to the geometry of similar figures.
THE PROBLEM/INVESTIGATION
1) If the sides of 2 squares are in the ratio 2:1, show that their areas are in the ratio 4:1
(a) visually
(b) numerically by examining particular cases
(b)  algebraically
2) If the sides are in the ratio 3:1, do you think the areas will be in the ratio 6:1 or 9:1? Now do parts a-c as in 1).
3) If the ratio of the sides is 3:2 show algebraically that the ratio of the areas is 9:4.
4) Show algebraically that if the ratio of the sides of 2 squares is m:n then the ratio of their areas is (m/n)^2.
Note: How does this result connect to the idea that the area of a square varies directly as the square of its side length?
4) If squares are replaced by circles using radii or diameters in place of "sides" show that the results of questions 1-4 are the same.
How does this result connect to the idea that the area of a circle varies directly as the square of its radius or diameter (or circumference)?
REFLECTIONS
• Squares and circles are of course special cases of similar figures. Beyond this investigation lies the BIG IDEA:
The areas of 2-dim similar figures are proportional to the squares of their linear dimensions.
Note: In 3 dim, we can replace 'area' by what?
• Do you see this as one of the fundamental theorems of Euclidean geometry? Is it sufficiently stressed in textbooks and in the standards? Of course you may not feel as I do about all this!
• So what is the geometry connection to
(m/n)^3 = (m^3)/(n^3)...
'.

Tuesday, November 26, 2013

Today's MathNotations # is 972...


OVERVIEW AND THE PROBLEM
As posted on Twitter (@dmarain), student's task is to list as many interesting properties/facts about 972 as possible, at least 10.  I posted a couple of possibilities but you may not want to give both of these away...
1) Prod of a perfect sq and a perf cube
2) a perf 5th power × a perf sq
...
REFLECTIONS
• Appropriate grade level? 3rd? 4th? Middle grades?
• 10+ properties way too much? Why do you think I chose that number? Depends on grade level? Abilities?
• What do you see as the principal
learning goals/benefits here? How highly would you rank "Expressing mathematical ideas/number properties in words?"
• We might naturally have students work in teams but I tended to respect that individual student who chose to work alone. Would you make this competitive to stimulate interest or that's not desirable?
• Beyond making a list of properties, what else should be expected here? How important is it for students to check their results? How about independent verification from another team member?
• Allow use of calculator throughout or after a little while?
• How is this different from MAA's "Today's Number is"?
• Anyone recall the last number I asked about a long time ago? [97 and 153!] You may want to click on the link...

Saturday, November 23, 2013

Six parking spots were assigned randomly. What is the probability that...

OVERVIEW
No matter how many of these appear on standardized tests, a large per cent of test takers continue to get these wrong. Teachers will teach that unit on combinations, permutations, Multiplication Principle and rules of probability. But learners will still struggle and even if they survive the chapter test, they will probably get the following problem wrong. But some of you may have overcome this...

THE PROBLEM

Six parking spots are assigned randomly to six employees. If Jake and Alex are 2 of these employees, what is the probability they will be assigned to the first 2 spaces?
(A) 1/60 (B) 1/30 (C) 1/15 (D) 1/6 (E) 1/3

Answer: (C)

REFLECTIONS

• It's so simple: (2/6) (1/5) = 1/15  Next...
Of course that would never happen in a classroom!
How much understanding of probability principles and practice is required to feel comfortable with this efficient approach? More importantly when would this approach fail or need to be revised?

• So is it combinations? Permutations! Do we also need the multiplication principle?
How much experience do students need before being able to write
(2P2) (4P4) ÷ (6P6)

• Those who have experience teaching these will have developed their favorite instructional strategies. Please share!

Friday, November 22, 2013

"LEAST number to select to be CERTAIN of getting at LEAST..." -- Those annoying logic problems!

OVERVIEW

You've seen these on SATs, GREs, etc..
Where exactly does this fit into the Common Core?

Should the underlying Pigeonhole Principle and/or specific strategies/methods be taught to middle or high schoolers?

You can guess my feelings about this...

THE PROBLEM
There are thirty solid-colored scarves in 30 identical unmarked closed boxes, one per box. The 30 scarves consist of:
6 red
7 blue
8 yellow
9 green

What is the LEAST number of boxes that need to be opened to be CERTAIN of getting at LEAST

(A) 2 of the same color [5]
(B) 4 of the same color [9]
(C) one of each color [25]
(D) 4 of each color [28]
(E) 2 green and 2 blue [25]

REFLECTIONS

• Oh those nasty but critical keywords which I wrote in uppercase. Initial instruction must focus on developing meaning for these.

• Those who have had experience teaching this may want to share their strategies. Here are some I've used...

* Act it out with coins in a bag or any concrete objects. Let's say the goal is to get at least one of each color and the student or you selects 4 boxes or objects. I might say, "Do you want to stop? This is surely the LEAST number!" Were hoping students will respond with, "But we can't be sure!"
OR I might grab all of the boxes and say, "Ok, now we are CERTAIN!" Hopefully someone will respond...

* Without getting into the Pigeon Hole Principle I have used what I call the WORST POSSIBLE LUCK (CASE) approach. For example, if the goal is to get one of each color the WORST CASE would be getting the same color repeatedly. Don't like that approach? This is subtle and requires patience, time and several examples. Some learners will grasp it way before others but acting it out with objects and making a game out of it really has made a difference for my students.

Your thoughts?

Tuesday, November 19, 2013

When Mom was 40, Son was twice as old as Daughter... Age Problems and Singapore Bar Models

THE PROBLEM
When mom was 40, her son was twice as old as her daughter. Now her daughter is 28 and mom is twice as old as her son. How old is mom now?

Answer: 64

REFLECTIONS
Not exactly an authentic real world assessment but there's probably a reason why these are "Problems for the Ages"!! I don't believe harm is done by having students develop these kinds of relationships...

Of course for years I used to teach this using a chart and setting it up algebraically.

For younger students I would encourage a Guess-Test-Revise approach. Say, girls was 3 so boy was 6. Now it's 28-3 = 25 years later, so boy would be 6+25=31 and mom would be 40+25=65 which is a little more than twice 31. Revise to girl was 4, son was 8. 24 years later, girl would be 28, boy would be 32 and mom would be 40+24=64, Bingo!

Ah but children in Singapore perhaps as early as Grade 4 are representing these relationships using unit lengths and bar models. I simply don't have enough experience doing this so those out there with more knowledge please improve upon this so I can learn!

Then
Mom |-----40------|
Son |----||----|
Daughter |----| (some unit length)

Note: [//////] below represents the additional years from Then to Now

Now
Daughter |----| [//////] = [--28--]
Son |----||----| [//////]
Mom |----40---| [//////] = |----||----| [//////] |----||----| [//////] = |----||----||----| [//////] [--28--]
So 40 = 3u + 28 or u = 4 yrs, etc.

I'm limited by my graphics but I believe there are different,  better and more efficient models used in Singapore Math. If you're thinking I retrofitted an algebraic solution to make it look like a bar model you're probably right! I need wiser and more experienced 'bar modelers' to help me!

Sunday, November 17, 2013

Mean Equals Median Problem

THE PROBLEM
5, 11, 19, 22, x
If the mean and median of these 5 numbers are equal, determine all possible values of x.
ANSWER
-2, 14.25, 38
REFLECTIONS
• If the question asked for one possible value for x, this could be an SAT Grid-In question of above-average difficulty. I chose to ask for ALL possible values not only to make it more challenging but also to encourage students to probe more deeply.
• Thinking of many strategies how YOU would solve this? But that's not the focus of this blog. After WE figure out how to solve the problem ourselves and, yes, we should try to find more than one method, we should be asking ourselves as educators:
What questions need to be asked to enable our students to solve it for themselves?
What learning environment best facilitates this?
• Do you believe some of your students would reason that the median could only be 11, 19, or x itself? Listen to their discussions. Give them time but if no one is thinking that way we might ask:
Which numbers could NOT be the median? Why can't 5 be the median? 22?
• Beyond solving this problem is an even bigger question:
What must be true about a set of data values if their mean (avg) equals their median? We know there are sufficient conditions like:
The data is normally distributed about the mean, in which case, the mean, median and mode coincide.
But this is NOT NECESSARY!
In the above example, you might want to ask students to examine the case where x=14.25:
"What is the mean of the numbers if x is NOT in the set? Why does this make sense?
[If you have an 80 average and one of your scores was exactly 80, what would happen to the mean if you removed that score?]
So what does it 'mean' if the mean and median of a set of data are equal?
Now that's an open-ended assessment question!

Wednesday, November 13, 2013

'Half as many as' vs 'Twice as many as' Revisited

OVERVIEW
Hundreds of math posts over the last 6+ years and my most popular post is still by far "There are twice as many girls as boys...".
http://mathnotations.blogspot.com/2008/08/there-are-twice-as-many-girls-as-boys.html

Still hard to do word problems without coping with the vagaries of language and grammar in particular.

Contrast phrases such as

'Half OF the girls' and 'Half AS MANY Girls as' OR

'Twice as many X's as Y's'

These are confusing enough for native English speakers. Imagine the terror if you're not! Phrases like these don't respect anyone!

THE PROBLEM

Attendance at a stadium broke down as follows...
Of the children, half as many girls as boys
Of the adults, half as many women as men
The number of females was what fractional part of the total attendance?

(A) 1/2  (B) 1/3  (C) 1/4  (D) 1/6  (E) cannot be determined from info given

Answer: B

REFLECTIONS

1. Although the title of the post suggests the focus was on interpreting the phrase "as many as", there are some significant underlying ratio concepts here.

2.  As I discussed in my earlier post, I've observed that students do better with the semantics if they first decide which is the larger of two quantities. Thus "half as many girls as boys" hopefully suggests that there are fewer girls but my experience is that some are simply blocked by the sentence structure and will guess randomly or say nothing. Remember we can always go back to concrete numerical relationships:

"Ok say there are 12 boys.
If there are half as many girls, then how many girls?"

Note how I not only used numbers but I also inverted the problem by giving the number of boys first. Yes we are also teachers of reading! Will students do this on their own? If trained!!

3. From this point there are many solution paths and I would definitely allow upper  elementary or middle school students to play with this for a few minutes in their groups. There's no rushing this process.

We can simply explain our method to them but this is only a part of their learning. Of course that's my opinion and there are many out there who would cringe at this. The eternal battle between The Direct Instructionists and the Constructivists! Those are just labels about which I care little. Whatever works... Since I can speak from 40 yrs of experience I know what worked for my students...
Besides I've already debated this ad infinitum and ad nauseam with some of the best. No one ever changes their mind!

4.  There are some subtle part:whole ratio concepts embedded here. Isn't it tempting to pick choice (E) here because we're not given the Adult:Child ratio. But the question asked for the ratio of the combined female to the total. It is very instructive to see this algebraically.

5. 'Plugging in' convenient numbers for the subgroups in this problem makes it accessible to 4th-5th graders. Organization is very helpful. I use a tree model to represent this kind of data but most do not do this.  Say there are 2 girls, 4 boys; 5 women, 10 men. Then the number of females (7) is still one-third of the total (21)!

6. Hopefully my readers will suggest more efficient ratio methods, Singapore models, other algebraic representations, etc... OR no comments at all!