Tag Archives: Sakamoto method

A Grade 5 Bicycles-and-Tricycles Problem

20140220-124202.jpg

In an earlier post, I shared about the following chickens-and-rabbits problem.

There are 100 chickens and rabbits altogether. The chickens have 80 more legs than the rabbits. How many chickens and how many rabbits are there?

Other than using a guess-and-guess strategy and an algebraic method, both of which offering little pedagogical or creative insight, let me repeat below one of the two intuitive methods I discussed then.

Since the chickens have 80 more legs than the rabbits, this represents 80 ÷ 2 = 40 chickens.

Among the remaining (100 – 40) = 60 chickens and rabbits, the number of chicken legs must be equal to the number of rabbit legs.

Since a rabbit has twice as many legs as a chicken, the number of chickens must be twice the number of rabbits in order for the total number of legs to be equal.

20140220-180550.jpg

From the model drawing,

3 units = 100 − 40 = 60
1 unit = 60 ÷ 3 = 20

Number of rabbits = 1 unit = 20
Number of chickens = 2 units + 40 = 2 × 20 + 40 = 80

The Bicycles-and-Tricycles Problem

Again, if we decided to ban any trial-and-error or algebraic method, how would you apply the intuitive method discussed above to solve a similar word problem on bicycles and tricycles?

There are 60 bicycles and tricycles altogether. The bicycles have 35 more wheels than the tricycles. How many bicycles and tricycles are there?

Go ahead and give it a try. What do you discover? Do you make any headway? In solving the bicycles-and-tricycles question, I find that there are no fewer of half a dozen methods or strategies, which could be introduced to elementary school students, three of which lend themselves easily to the model, or bar, method, excluding the Sakamoto method.

© Yan Kow Cheong, March 4, 2014.

The Singapore Excess-and-Shortage Problem

In Singapore, in grades four and five, there is one type of word problems that seldom fail to appear in most local problem-solving math books and school test papers, but almost inexistent in local textbooks and workbooks. This is another proof that most Singapore math textbooks ill-prepare local students to tackle non-routine questions, which are often used to filter the nerd from the herd, or at least stream the “better students” into the A-band classes.

Here are two examples of these “excess-and-shortage word problems.”

Some oranges are to be shared among a group of children. If each child gets 3 oranges, there will be 2 oranges left. If each child gets 4 oranges, there will be a shortage of 2 oranges. How many children are there in the group?

A math book costs $9 and a science book costs $7. If Steve spends all his money in the science books, he still has $6 left. However, if he buys the same number of math books, he needs another $8 more.
(a) How many books is Steve buying?
(b) How much money does he have?

A Numerical Recipe

Depicted below is a page from a grade 3/4 olympiad math book. It seems that the author preferred to give a quick-and-easy numerical recipe to solving these types of excess-and-shortage problems—it’s probably more convenient and less time-consuming to do so than to give a didactic exposition how one could logically or intuitively solve these questions with insight.

A page from Terry Chew’s “Maths Olympiad” (2007).

Strictly speaking, it’s incorrect to categorize these questions under the main heading of “Excess-and-Shortage Problems,” because it’s not uncommon to have situations, when the conditions may involve two cases of shortage, or two instances of excess.

In other words, these incorrectly called “excess-and-shortage” questions are made up of three types:
・Both conditions lead to an excess.
・Both conditions lead to a lack or shortage.
・One condition leads to an excess, the other to a shortage.

One Problem, Three [Non-Algebraic] Methods of Solution

Let’s consider one of these excess-and-shortage word problems, looking at how it would normally be solved by elementary math students, who have no training in formal algebra.

Jerry bought some candies for his students. If he gave each student 3 candies, he would have 16 candies left. If he gave each student 5 candies, he would be short of 6 candies.
(a) How many students are there?
(b) How many candies did Jerry buy?

If the above question were posed as a grade 7 math problem in Singapore, most students would solve it by algebra. However, in lower grades, a model (or intuitive) method is often presented. A survey of Singapore math assessment titles and test papers reveals that there are no fewer than half a dozen problem-solving strategies currently being used by teachers, tutors, and parents. Let’s look at three common methods of solution.

Method 1

20131026-231338.jpg

Difference in the number of candies = 5 – 3 = 2

The 16 extra candies are distributed among 16 ÷ 2 = 8 students, and the needed 6 candies among another 6 ÷ 2 = 3 students.

Total number of students = 8 + 3 = 11

(a) There are 11 students.

(b) Number of candies = 3 × 11 + 16 = 49 or  5 × 11 –  6 = 49

Jerry bought 49 candies.

Method 2

Let 1 unit represent the number of students.

20131027-213637.jpg

Since the number of candies remains the same in both cases, we have

3 units + 16  = 5 units – 6

20131026-231524.jpg

From the model,
2 units = 16 + 6 = 22
1 unit = 22 ÷ 2 = 11
3 units + 16 = 3 × 11 + 16 = 49

(a) There are 11 students.
(b) Jerry bought 49 candies.

Method 2 is similar to the Sakamoto method. Do you see why?

Method 3

The difference in the number of candies is 5 – 3 = 2.

20131026-231534.jpg

The extra 16 candies and the needed 6 candies give a total of 16 + 6 = 22 candies, which are then distributed, so that all students each received 2 extra candies.

The number of students is 22 ÷ 2 = 11.

The number of candies is 11 × 3 + 16 = 49, or 11 × 5 – 6 = 49.

Similar, Yet Different

Feedback from teachers, tutors, and parents suggests that even above-average students are often confused and challenged by the variety of these so-called shortage-and-excess problems, not including word problems that are set at a contest level. This is one main reason why a formulaic recipe may often do more harm than good in instilling confidence in students’ mathematical problem-solving skills.

Here are two grade 4 examples with a twist:

When a carton of apples were packed into bags of 4, there would be 3 apples left over. When the same number of apples were packed into bags of 6, there would still be 3 apples left over. What could be the least number of apples in the carton? (15)

Rose had some money to buy some plastic files. If she bought 12 files, she would need $8 more. If she bought 9 files, she would be left with $5. How much money did Rose have? ($44)

Conclusion

Exposing students of mixed abilities to various types of these excess-and-shortage word problems, and to different methods of solution, will help them gain confidence in, and sharpen, their problem-solving skills. Moreover, promoting non-algebraic (or intuitive) methods also allows these non-routine questions to be set in lower grades, whereby a diagram, or a model drawing, often lends itself easily to the solution.

References

Chew, T. (2008). Maths olympiad: Unleash the maths olympian in you — Intermediate (Pr 4 & 5, 10 – 12 years old). Singapore: Singapore Asian Publications.

Chew, T. (2007). Maths olympiad: Unleash the maths olympian in you — Beginner (Pr 3 & 4, 9 – 10 years old). Singapore: Singapore Asian Publications.

Yan, K. C. (2011). Primary mathematics challenging word problems. Singapore: Marshall Cavendish Education.

© Singapore Math, October 27, 2013.

PMCWP4-2
See Worked Example 2 on page 8; try questions 7-8 on page 12.

A Before-and-After Singapore Math Problem

Picture

A Singapore math primer for grades 4–6 students, teachers, and parents

In Model Drawing for Challenging Word Problems, one of the better Singapore math primers to have been written by a non-Singaporean author for an American audience in recent years, under “Whole Numbers,” Lorraine Walker exemplified the following before-and-after problem, as we commonly call it in Singapore.

Mary had served $117, but her sister Suzanne had saved only $36. After they both earned the same amount of money washing dishes one weekend, Mary noticed she had twice as much money as Suzanne. What was the combined total they earned by doing dishes?

The solution offered is as follows:

Picture

© 2010 Crystal Springs Books

The author shared that she did two things to make the model look much clearer:

• To add color in the “After” model;
• To slide the unit bars to the right.

This is fine if students have easy access to colored pens, and know which parts to shift, but in practice this may not always be too convenient or easy, especially if the question gets somewhat more complicated.Let me share a quick-and-dirty solution how most [elementary math] teachers and tutors in Singapore would most likely approach this before-and-after problem if they were in charge of a group of average or above-average grades 4–5 students.

Picture

From the model drawing,

1 unit = $117 – $36 = $81
1 unit – $36 = $81 – $36 = $45

2 × $45 = $90

They earned a total of $90 by doing dishes.

Analysis of the model method

Notice that the placement of the bars matters—whether a bar representing an unknown quantity is placed before or after another bar representing a known quantity.

In our model, had we placed the [shaded] bar representing the unknown unit on the right, it would have been harder to deduce the relationship straightaway; besides, no sliding or shifting is necessary. So, placing the bar correctly helps us to figure out the relationship between the unknown unit and the known quantities easier and faster.

In general, shading and dotting the bars are preferable to coloring and sliding them, especially when the problem gets harder, with more than two conditions being involved.

The Stack Method

This word problem also lends itself very well to the Stack Method. In fact, one can argue that it may even be a better method of solution than the bar model, especially among visually inclined below-average students.

Take a look at a quick-and-dirty stack solution below, which may look similar to the bar method, but conceptually they involve different thinking processes. To a novice, it may appear that the stack method is just the bar method being depicted vertically, but it’s not. Perhaps in this question, the contrast isn’t too obvious, but for harder problems, the stack method can be seen to be more advantageous, offering a more elegant solution than the traditional bar method.

Picture

From the stack model diagram, note that the difference $81(= $117 – $36) must stand for the extra unit belonging to Mary.

1 unit = $81
$36 + ▅ = $81
▅ = $81 – $36 = $45
2 ▅ = 2 × $45 = $90

So, they had a total of $90.

The Sakamoto Method

This before-and-after problem also lends itself pretty well to the Sakamoto method, if the students have already learned the topic on Ratio. Try it out!

Let me leave you with three practice questions I lifted up from a set of before-and-after grades 4–6 problems I plan to publish in a new title I’m currently working on, all of which encourage readers to apply both the bar and the stack methods (and the Sakamoto method, if they’re familiar with it) to solving them.

Practice

Use the model and the stack methods to solve these questions.

1. At first, Joseph had $900 and Ruth had $500. After buying the same watch, Joseph has now three times as much money as Ruth. How much did the watch cost?

2. Moses and Aaron went shopping with a total of $170. After Moses spent 3/7 of his money and Aaron spent $38, they had the same amount of money left. How much money had Aaron at first?

3. Paul and Ryan went on a holiday trip with a total of $280. After Paul had spent 4/7 of his money and Ryan had spent $52, the amount Paul had left was 1/4 of what Ryan had left. How much money did Ryan have at first?

Answers
1. $300 2. $86 3. $196

Reference
Walker, L. (2010). Model drawing for challenging word problems: Finding solutions the Singapore way. Peterborough, NH: Crystal Springs Books.

© Yan Kow Cheong, August 4, 2013.

Problem Solving Made Difficult

Picture

The US edition of a grade 5 Singapore math supplementary title.

Recently, while revising a grade 5 supplementary book I wrote for Marshall Cavendish, I saw that other than the answer, there was no solution or hint provided to the following question.

If Ann gave $2 to Beth, Beth would have twice as much as Ann.
If Beth gave $2 to Ann, they would have the same amount of money.
How much did each person have?

Most grade 7 Singapore math textbooks and assessment books would normally carry a few of these typical word problems, whereby students are expected to use an algebraic method to solve them. For instance, using algebra, students would form two linear equations in x and y, before solving them by the elimination, or substitution, method. A pretty standard application of solving a pair of simultaneous linear equations, by an analytic method.

However, it’s not uncommon to see these types of word problems appearing in lower-grade supplementary titles, whereby students could solve them, using the Singapore model, or bar, method; and the Sakamoto method. In other words, these grade 7 and 8 questions could be solved by grade 5 and 6 students, using a non-algebraic method.

Algebra versus Model Drawing

Conceptually speaking, I think a grade 6 or 7 student who can solve the above word problem, using a model drawing, appears to exhibit a higher level of mathematical maturity than one who simply uses two variables to represent the unknowns, before forming two simultaneous linear equations to solve them. Of course, because the numbers in this question are relatively small, it’s not surprising to catch a number of average students relying on the trial-and-error method to find the answer.

Try to solve the question, using both algebra and a model; then compare the two methods of solution. Which one do you think demands a deeper or higher level of reasoning or thinking skills?

Depicted below is a model drawing of the above grade 5 word problem.

Picture

From the model drawing,

1 unit = 2 + 2 + 2 + 2 = 8
1 unit + 2 = 10
1 unit + 6 = 14

Ann had $10.
Beth had $14.

Generalizing the Problem

A minor change in the question, by altering the “number of times” Beth would have as much money as Ann, reveals an interesting pattern: the model drawing remains unchanged, except for the varying number of units that represent the same quantity.Here are two modified versions of the original grade 5 question.

If Ann gave $2 to Beth, Beth would have three times as much as Ann.

If Beth gave $2 to Ann, they would have the same amount of money.
How much did each person have?

Answer: Ann–$6; Beth–$10.

If Ann gave $2 to Beth, Beth would have five times as much as Ann.
If Beth gave $2 to Ann, they would have the same amount of money.
How much did each person have?

Answer: Ann–$4; Beth–$8.

From Problem Solving to Problem Posing

The two modified questions could serve as good practice for students to become skilled in model drawing, and to help them deduce numerical relationships confidently from them. Besides, they provide a good opportunity to challenge students to pose similar questions, by altering the “number of times” Beth would have as much money as Ann. Which numerical values would work, and what ones wouldn’t, in order for the model drawing to make sense, or for the question to remain solvable?

Conclusion

Let me end, by tickling you with another grade 5 question, similar to the previous three word problems.

If Ann gave $2 to Beth, Beth would have three times as much as Ann.
If Beth gave $2 to Ann, they would have twice as much money as Beth.
How much did each person have?

Answer: Ann–$4.40; Beth–$5.20.

How do you still use the model method to solve this slightly modified ratio question? Test it on your better students or colleagues! It’s slightly harder, because any obvious result isn’t easily deduced from the model drawing, as compared to the ones posed earlier on. Besides, unlike the three previous word problems whose answers are integers, this last problem has a decimal answer—it just doesn’t lend itself well to the guess-and-check strategy.

Share with us how your students or colleagues fare on this last question. Remember: No algebra allowed!

© Yan Kow Cheong, July 12, 2013.

Is Singapore math frickin’ hard?

There is a millennium myth that Singapore math is hard or tough—that only geeks from some remote parts of Asia (or from some red little dot on the world map) should do it. The mass media (and math educators, too) have implicitly mythologized that those who are presently struggling with school math, should avoid Singapore math, in whatever form it’s being presented, totally. A grave mistake, indeed!

Is Singapore math a mere fad?

After reading dozens of tweets and blog posts on the pluses and minuses of Singapore math, it sounds as if Singapore math has a certain mystique around it—some kind of foreign math bestowed by some creatures from outer space to terrorize those who wished math were an optional subject in elementary school.

Not to say, folks who totally reject Singapore math on the basis that it’s just another fad in math education, or another marketing gimmick to promote an allegedly “better foreign curriculum” to math educators. They believe that “back-to-basics math,” whatever that phrase means to them, with all its memorizing and drill-and-kill exercises, is a necessary mathematical evil to get kids to learn arithmetic.

Singapore math OR/AND Everyday Math

Singapore math? Sure, no problem! It’s no big deal!

Well, it’s a big deal for traditional publishers, which may lose tons of money if more states and schools continue to embrace this “foreign brand” of math education.

Poor writing and teaching from a number of us could have indirectly contributed to the white lie that if you can’t cope with Everyday Math or Saxon Math, or whatever math textbook your school or state is currently using, Singapore math is worse! You might as well forget about it!

Picture

One of the first few Singapore supplementary titles to promote the model method in the mid-nineties.

The bar method as a powerful problem-solving strategy

Objectively speaking, Singapore math doesn’t come close to most pedagogical insights or creative ideas featured in journals and periodicals published by the MAA and the NTCM. Personally, I must admit that I become a better teacher, writer, and editor, thanks to these first-class publications. The Singapore model method, although a key component of the Singapore math curriculum, is just one of the problem-solving strategies we use every day, as part of our problem-solving toolkit.

On the other hand, for vested interests on the part of some publishers, little has been done to promote other problem-solving strategies, such as the Stack Method and the Sakamoto method, which are as powerful, if not more elegant, than the bar method—in fact, more and more local students and teachers are using them as they see their advantages over the model method in a number of problem situations.

Picture

Fabien Ng in his heyday was a household name in mathematics education in Singapore—it looks like he’s since almost disappeared from the local publishing scene.

Don’t throw out the mathematical baby yet!

You may not wish to adopt the Singapore math curriculum, or even part of it; but at least consider the model and stack methods, not to say, the Sakamoto method, as part of your arsenal of problem-solving strategies (or heuristics, as we call them here). Don’t let traditional publishers (or “math editors” with a limited repertoire of problem-solving strategies) prevent you from acquiring new mathematical tools to improve your mathematical problem-solving skills.

In Singapore, the model, or bar, method is formally taught until grade six to solve a number of word problems, because from grade seven onwards, we want the students to switch over to algebra. However, this doesn’t mean that we’d totally ban the use of the model method in higher grades, because in a number of cases, the model or stack method often offers a more elegant or intuitive method of solution than its algebraic counterpart.

© Yan Kow Cheong, May 27, 2013.

Picture

A booklet comprising of PSLE (grade 6) past exam papers, which cost me only 88 cents in the eighties.

The Dolls Problem à la Singapour

Following a request from a Linkedln friend to provide a solution that makes use of the Singapore model method to the question below—I couldn’t trace the origin of this word problem—here’s a quick-and-dirty sketch of a five-model-drawing solution.

Jazmine buys and sells antique dolls on the Internet. Yesterday, she focused on dolls from the Civil War period. She began the day by selling one-fourth of her dolls from that period. Then she sold six more. Just before lunch she sold one-fourth of the remaining Civil War dolls. After lunch, she bought some Civil War dolls, increasing her collection by one-sixth. Then she bought some more, doubling her collection. Just before she quit for the day, she sold two thirds of her Civil War dolls. After all that, she had fourteen of these dolls left. How many dolls did Jazmine have before she began trading yesterday?

It wouldn’t be surprising that this kind of brain-unfriendly word problem, set in a test or exam, might give some un-mathophobic grade five or six students sweaty palms, or goose pimples, if they started feeling clueless after attempting to solve it for some five odd minutes!

Picture

A quick-and-dirty solution that makes use of the model method.

Using the “work backwards” strategy repeatedly, the model drawings show that Jazmine had 40 dolls before she began trading yesterday.

If you’re an “algebraic freak,” by all means, use algebra to check your answer—I decided to give the algebraic approach a miss this time round.

Disproportionate parts or units

Notice that I’ve loosely used “units” and “parts” alternately to represent each model drawing. And I’ve also used each unit, or part, in a rather disproportionate manner, as compared to textbooks’ modeled solutions, which generally depict the bars (or rectangles) proportionately, based on their respective numerical values—which is secondary to the reasoning or thinking processes.

The above dolls problem is similar to a question I discussed in an earlier post, except that the present one is slightly harder; otherwise, it adopts the same problem-solving strategies for  its solution.

Sakamoto and Stack Methods

My next task is to check whether the Stack method or the Sakamoto method to the above word problem is conceptually “friendlier” than the model method. Are there intuitive or elegant solutions other than the one that embraces the bar method? Meanwhile, please send us your solution(s) to the dolls problem.

© Yan Kow Cheong, March 27, 2013.

Postscript: Although math was my favorite subject in school, I don’t recall solving questions similar to the above word problem. I doubt if I would be able to solve it when I was in grade five or six. It looks like this present younger generation has been given the shorter end of the mathematical stick—worse, if math happens not to be their cup of tea! It’s no surprise that strangers, young and old, angrily tell me of their negative mathematical experiences in school—how they disliked math (and their math teachers).