MathNotations

Never too late in the school year to review percents, right? Well, even if you don't agree, here goes...

First a problem similar to the one I posted on Twitter the other day.

Middle School Level?

The cost of a meal including a 10% tip was 13ドル.75. What was the tip, in dollars?
Ans: 1ドル.25

SAT-type (Higher level of difficulty)

The cost of a meal is $M. With an x% tip included, the bill came to $T. Which of the following is an expression for x in terms of M and T? (A) T/M (B) (T-M)/M (C) 100T/M (D) 100(T-M)/M (E) (T-M)/(100M)
Ans: D


Thoughts and Questions...

What % of your middle school students could handle the first question? For that matter, what % of your secondary students would solve it?

Can you predict which of your students would be able to solve the first question mentally or with some quick trial-and-error (ok, G-T-R), using their calculators. I chose 10% to make this possible. Do you get upset when students do this? Should you?

What do you predict would be the difficulties your algebra students might confront in the 2nd problem?

Is it easy to eliminate some of the answer choices and to make an educated guess from the rest?
(NOTE: I composed the question and the answer choices and I know some of you could improve upon my efforts!)

NOTE: The 2nd question is representative of the harder problems on the SATs and there are many of these in my new Challenge Math Problem/Quiz Book mentioned below.









If interested in purchasing my NEW 2012 Math Challenge Problem/Quiz book, click on BUY NOW at top of right sidebar. 175 problems divided into 35 quizzes with answers at back. Suitable for SAT I, Math I/II Subject Tests, Math Contest practice and Daily/Weekly Problems of the Day. Includes multiple choice, case I/II/III type and constructed response items. Price is 9ドル.95. Secured pdf will be emailed when purchase is verified. DON'T FORGET TO SEND ME AN EMAIL (dmarain "at gmail dot com") FIRST SO THAT I CAN SEND THE ATTACHMENT!

Posted by Dave Marain at 11:09 AM 0 comments

Labels: percent, percent word problem, SAT-type problems

Posted by Dave Marain at 7:31 AM 0 comments

Labels: percent, percent change, percent word problem

Have you forgotten to register for MathNotation's Third FREE Online Math Contest coming in mid-October? We already have several schools (from around the world!) registered. For details, link here or check the first item in the right sidebar!!

Before tackling a more challenging problem in the classroom, I would typically begin with one or more simpler examples. My objective was to review essential concepts and skills and demonstrate key ideas in the harder problem. This incremental approach (sometimes referred to as scaffolding) enabled some students to solve the problem or at least get started. Usually within each group of 3-4 students, there was at least one who could help the others. Some groups or classes might still not be ready after one example, so more would be needed. I never felt that this expense of time was too costly since my goal was to develop both skill and understanding.


SIMPLER EXAMPLE
Consider the following two statements about positive numbers A and B:

(1) A is 80% of B.
(2) A is 20% less than B .

Are these equivalent, that is, if values of A and B satisfy (1), will they also hold true for (2) and conversely?

How would you get this idea across to your students?

Again, depending on the students, I would often allow them to discuss it first in small groups for two minutes, then open up the discussion.

Note: If the group lacks the skills, confidence or background (note that I left ability out, intentionally!), I might first start with concrete values before giving them the 2 statements above: E.g., What is 80% of 100?

How would I summarize the methods of solution to this question. Here's what I attempted to do in each lesson. I didn't reach everyone but I found from further questioning and subsequent assessment that this multi-pronged approach was more successful than previous methods I had used. Most of these methods came from the students themselves!

INSTRUCTIONAL STRATEGIES
I. Choose a particular value for one of the numbers, say B = 100. Ask WHY it makes sense to start with B first and why does it make sense to use 100. Calculate the value of A and discuss.

II. Draw a pie chart (circle graph) showing the relationship between A and B. Stress that B would represent the whole or 100%.

III. Write out the sentence:
80% of B is the same as 100% of B - 20% of B
In other words:
80% of B is the same as 20% less than B.

IV. Express algebraically (as appropriate):
0.8B = 1B - 0.2B

Numerical (concrete values)
Visual (Pie chart)
Verbal (using natural language)
Symbolic (algebra)

Yes, it's Multiple Representations! The Rule of Four!

To me, it's all about accessing different modes of how students process. Call it learning styles, brain-based learning, etc., it still comes down to:
RARELY DOES ONE METHOD OF EXPLANATION, NO MATTER HOW CLEAR OR STRUCTURED, REACH A MAJORITY OF STUDENTS. YOUR FAVORITE EXPLANATION WILL MAKE THE MOST SENSE TO THE STUDENTS WHO THINK LIKE YOU!!


Now for today's challenge.
(Assume all variables represent positive numbers)

M is x% less than P and N is x% less than Q. If MN is 36% less than PQ, what is the value of x?

Can you think of several methods?
I will suggest one of the favorite of many successful students on standardized assessments:

Choose P = 10, Q = 10. Then...

Click on More (subscribers do not need to do this) to see the answer without details.




Answer: x = 20

Posted by Dave Marain at 6:02 AM 1 comments

Labels: conceptual understanding, instructional strategies, more, percent, percent word problem, SAT strategies, SAT-type problems

Example I
40% of the the Freshman Calculus class at Turing University withdrew. If 240 students left, how many were in the class to start?

Solution without explanation or discussion:

0.4x = 240 ⇒ x = 600


Example II
40% of the the Freshman Calculus class at Turing University withdrew. If 240 students were left, how many were in the class to start?

Solution without explanation or discussion:

0.6x = 240 ⇒ x = 400


Thinking that the issues in the problems above are more language-dependent than based on learning key mathematics principles or effective methods? I would expect that many would say that using the word "left" in both problems was unnecessarily devious and that clearer language should be used to demonstrate the mathematics here. Perhaps, but when I taught these types of problems I would frequently juxtapose these types of questions and intentionally use such ambiguous language to generate discussion - creating disequilibrium so to speak. If nothing else, the students may become more critical readers! Further, the idea of using similar but contrasting questions is an important heuristic IMO.

Even though I've been a strong advocate for a standardized math curriculum across the grades, I fully understand that the methods used to present this curriculum are even more crucial. Instructional methods and strategies are often unpopular topics because they seem to infringe on individual teacher's style and creativity. BUT we also know that some methods are simply more effective than others in reaching the maximum number of students (who are actually listening and participating!). I firmly believe there are some basic pedagogical principles of teaching math, most of which are already known to and being used by experienced teachers.

Percent word problems are easy for a few and confusing to many because of the wide variety of different types.

Here are brief descriptions of some methods I've developed and used in nearly four decades in the classroom.

I. (See diagram at top of page)
The Pie Chart builds a strong visual model to represent the relationships between the parts and the whole and the "whole equals 100%" concept. How many of you use this or a similar model ? Please share! There's more to teaching this than drawing a picture but some students have told me that the image stays longer in their brain. I learn differently myself but I came to learn the importance of Multiple Representations to reach the maximum number of students.

II. "IS OVER OF" vs. "OF MEANS TIMES"
The latter is generally more powerful once the student is in Prealgebra but, of course, the word "OF" does not appear in every percent so many different variations must be given to students and practiced practiced practiced practiced over time. The first method can be modified as a shortcut in my opinion to find a missing percent and that may be its greatest value. However many middle schoolers use proportions for solving ALL percent problems. I personally do NOT recommend this!

Well, I could expound on each of these methods ad nauseam and bore most of you, but I think I will stop here and open the dialg for anyone who has strong emotions about teaching/learning per cents...

Posted by Dave Marain at 6:52 AM 9 comments

Labels: heuristics, instructional strategies, middle school, pedagogy, percent, percent word problem, SAT strategies, SAT-type problems

Version I (Level of difficulty 3 - medium)

With a special promotion, Al received a 60% discount on a new stereo system and paid $x. Sylvia bought the same system (same original price) but only received a 20% discount. In terms of x, how much (in dollars), did Sylvia pay? Assume x> 0 and disregard sales tax.

(A) 4x (B) 3x (C) 2x (D) 4x/3 (E) x/3


Version II (Level of Difficulty 4 - medium/hard)
Grid-In Type

Maury purchased a new electronic game system with a 25% off coupon. His friend bought the same system (same original price) with a 40% off coupon. If his friend paid 45ドル less for the system, how much did Maury pay (disregard sales tax)?



For the answers, suggested solutions, strategies and discussion, click Read more...


Level I problem
Answer: (C) 2x

Possible Solutions:
Method I ("Plug-in" SAT Strategy - Student-preferred?)
Let original price = 100ドル.
Then Al's discount was 60,ドル so he paid 40ドル. Thus x = 40.
Sylvia's discount was 20,ドル so she paid 80,ドル which means she paid twice Al's price or 2x.

Method II (conceptual)
Al paid 40% of the original price, Sylvia paid 80%, therefore Sylvia paid twice as much as Al.

Method III (traditional - Algebraic)
Reasoning as in Method I, Al paid 40% and Sylvia paid 80% of the original price.
Let y = original price (before discounts).
Then Al paid 0.4y = x. Solving, y = 2.5x.
Sylvia paid (0.8)(2.5x) = 2x.

Level II Problem
Answer: 225ドル
Methods???
Note: There is a mental math method which will be discussed later.


FOOD FOR THOUGHT

• What are some of the factors which might make the second problem more difficult?
  
• Should these types of percent questions be included in prealgebra texts for middle schoolers?
• The first question appears to be fairly straightforward. Which part of the problem would cause the most difficulty for students in your opinion?
  
• Note that in the ever-popular 'plug in a number method', the ever-popular 100ドル was chosen for the original price. Can you think of an occasion where you might not want to substitute 100ドル in a discount problem?
• Students also need to recognize that I plugged in 100ドル for the original price rather than replace x by 100ドル. They often feel they must substitute a number for the variable given, which, in this case would lead to more work.
• There are other approaches here. Please share different methods you have used!
• How would we assess their learning of the concepts here? Include one of these on the next quiz or test? Use these as warmups occasionally? Develop a worksheet of similar problems for homework?
  

Posted by Dave Marain at 8:29 AM 11 comments

Labels: more, percent, percent word problem, SAT strategies, SAT-type problems

More of the same...


In a certain group:
The ratio of males to females is 4:5.
The ratio of left-handed people to right-handed is 1:11 (assume no one is ambidextrous!).
64% of of the left-handed people are males.

(a) What % of the males are left-handed?
(b) What % of the females are left-handed?


Comments;

• If students or any of us see enough variations of these, will they become almost mechanical or does one have to decide which method/model/representation is needed for each problem?
• How many models should students be shown for these? Usually students or our readers will find a method or model no one else imagined!
• Is algebra the most powerful method? The most efficient? How many variables? Is it usually best to use one variable and let it represent the total number of people in the group?
  
• Anyone ever use a matrix/spreadsheet/table/Punnett square model to represent the data for these kinds of relationship problems? Specifically, problems in which the entire group (the universe) is divided into either groups A and B or groups C and D. This will become less cryptic as the discussion unfolds.
• Do you think most students today would feel more comfortable working in %, decimal or fraction form? What about rest of us out there?
• Too challenging for middle schoolers or not? Math contest problem or just a challenge to develop facility with ratio thinking? How would most algebra students fare with this?
  

Posted by Dave Marain at 6:21 AM 6 comments

Labels: algebra, middle school math, percent word problem, ratios

Have you submitted your vote yet in the MathNotations poll in the sidebar?

Target audience for this investigation: Our readers and algebra students (advanced prealgebra students can sometimes find a clever way to solve these).

Let's resurrect for the moment those ever popular rate/time/distance classics. Hang in there -- there's a more interesting purpose here!

We'll start by using fictitious presidential candidates running in a 'race.' Any resemblance to actual candidates is purely coincidental.

R and J are running on a huge circular track. J can run a lap in one month whereas it takes R twelve months to run the same lap. To be nice, J gives R a 3-month head start. After how many months will J 'catch up' to (overtake) R?

Are those of us who were trained to solve these feeling a bit nostalgic? Do you believe that our current generation of students has had the same exposure to these kinds of 'motion' problems or have most of these been relegated to the scrap heap of non-real world problems that serve no useful purpose. Well, they still appear on the SATs, a weak excuse for teaching them, perhaps, but I can certainly see other benefits from solving these. Can you?

Ok, there are many approaches to the problem above. Scroll down a ways to see a couple of methods (don't look at these yet if you want to try it on your own):







Method I: Standard Approach (using chart)

..............RATE ...x.......TIME .....=.....DISTANCE
............(laps/mo).....(months)............(laps)

R.........1/12....................t........................t/12

J................1....................t...........................t

Equation Model (verbal): At the instant when J 'catches up' to R:
Distance (laps) covered by J = Head Start + Distance covered by R

Equation: t = 1/4 + t/12 [Note: The 1/4 comes from the fact that R covers 1/4 of a lap in 3 months]

Solving: 12t = 3 +t --> 11t = 3 --> t = 3/11 months.

Check:
In 3/11 months, J covers 3/11 of a lap.
In the same time, R covers (1/12) (3/11) = 1/44 lap. Adding the extra 1/4 lap, we have 1/4 + 1/44 = 12/44 = 3/11. Check!

[Of course, we all know these fractions would present as much difficulty for students as the setup of the problem, but we won't go there, will we!]

Method II: Relativity Approach
Ever notice when you're zipping along at 65 mph and the car in the next lane is going the same speed, it appears from your vehicle that the other car is not moving, that is, its speed relative to yours is zero! However, if you're traveling at 65 mph and the vehicle in front is going 75 mph, the distance between the 2 cars is ever increasing. In fact, the speedier vehicle will gain 10 miles each hour! This 75-65 calculation is really a vector calculation of course, but, in relativity terms, one can think of it this way:
From the point of view of a passenger in the the slower vehicle, that person is not moving (speed is zero) and the faster vehicle is going 10 miles per hour. We can say the relative speeds are 0 and 10 mph.

Ok, let's apply that to the 'race':

If R's relative speed is regarded as zero, then J's relative speed will be 1 - 1/12 = 11/12 laps/month.
Since R is not 'moving', J only needs to cover the head-start distance to catch up:
(11/12)t = 1/4 --> t = (1/4)(12/11) = 3/11 months. Check!
[Note: Like any higher-order abstract approach, some students will latch on to this immediately and others will have that glazed look in their eyes. It may take some time for the ideas to set in. This method is just an option...]

There are other methods one could devise, particularly if we change the units (e.g., working in degrees rather than laps). Have you figured out how all of this will be related to those famous clock problems? Helping students make connections is not an easy task. One has to plan for this as opposed to hoping it will happen fortuitously.

Here is the analogous problem for clocks:

At exactly what time between 3:00 and 4:00, will the hour and minute hands of a clock be together?

Notes:
(1) I will not post an answer or solution at this time. I'm sure the correct answers and alternate methods will soon appear in the comments.
(2) A single problem like this does not an investigation make. How might one extend or generalize this question? Again, these are well-known problems and I'm sure many of you have seen numerous variations on clock problems. Share your favorites!
(3) Isn't it nice that analog watches have come back into fashion so we can recycle these wonderful word problems!
(4) For many problem-solvers, part of the difficulty with clock problems is deciding what units to use for distance (rotations, minute-spaces, some measure of arc length, degrees, etc.). This is a critical issue and some time is needed to explore different choices here.

Posted by Dave Marain at 6:20 AM 6 comments

Labels: algebra, clock problems, percent word problem, rate-time-distance problems

[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?

Posted by Dave Marain at 5:51 AM 13 comments

Labels: algebra, middle school, percent increase problem, percent word problem

Since there has been some discussion about the complications in the previous post on percents, how do we as educators deal with adversity and transform it into a teachable moment. We're explaining a difficult problem that students are struggling with, so we try to explain it again, but to no avail. What do experienced educators do?
(A) Abandon the problem - it was simply too hard or they're not ready for it yet. Perhaps assign it for extra credit?
(B) Make the question simpler by removing some of the complexity. Consider a scaffolding approach?
(C) Re-think what prerequisite skills were needed?

In this case, let's re-work the problem as follows:

In the senior class, there are 20% more girls than boys. If there are 180 girls, how many more girls than boys are there among the seniors?

What do you think the results will be now that we've concretized the problem?
Do you think some students will make the classic error of taking 20% of 180 and subtract to obtain 144 boys? You betcha!! There's no getting around the issue of recognizing the correct BASE for the 20% in my opinion. Whether you consider the boys to be 100% and the girls to be 120% or you let x = number of boys and 1.2x = number of girls, the central issue is recognizing that you cannot take 20% of the girls! Sometimes students need to just use algebra as a tool - it's a great one! By the way, one student used algebra with ratios to solve this:
Boys = x
Girls = 1.2x
Difference = 0.2x
Therefore, the difference/girls = 0.2x/1.2x = 1/6 and 180 divided by 6 equals 30!

How would you have reworded the question to make it more accessible?

Posted by Dave Marain at 10:21 AM 9 comments

Labels: percent, percent word problem

PLS READ THE COMMENTS. VERY ASTUTE OBSERVATIONS FROM OUR REGULAR CONTRIBUTORS TO WHICH I RESPONDED IN GREAT DETAIL. THIS MAY CONTINUE...

For now, I'll just leave the title as the problem to be discussed. Please consider how middle and high school students would approach this. How many might incorrectly guess that 60% of the seniors are girls and 40% are boys. If the problem were rephrased as a multiple-choice question, this type of error would occur frequently from my experience. What causes the confusion? Is it just the wording of the question or is there also an underlying issue regarding conceptual understanding of percents? Could it be that the question is asking for a percent and hasn't provided any actual numbers of students? What are the most effective instructional methods and strategies to help students overcome these issues? Certainly, algebraic methods would be a direct approach, but what foundation skills and concepts should middle schoolers develop even before setting up algebraic expressions?

Posted by Dave Marain at 6:49 AM 14 comments

Labels: algebra, middle school, percent, percent word problem, ratios