Category Archives: Singapore Model Method

The Fake Bar Model Method

Recently, I was peeping at some postings on the Facebook PSLE Parents group, and I came across the following question:

Philip had 6 times as many stickers as Rick. After Philip had given 75 stickers to Rick, he had thrice as many stickers as Rick. How many stickers did they have altogether?

Here are two solutions that caught my attention to the above primary or grade 6 word problem.

Solution contributed by Izam Marwasi
Solution by Izam Marwasi
Solution by Jenny Tan
Solution by Jenny Tan

Pseudo-Bar Model Method?

Arguably, the solution by the first problem solver offered to parents looks algebraic, to say the least. Some of you may point out that the first part uses the “unitary method,” but it’s the second part that uses algebra. Fair, I can accept this argument.

Since formal algebra, in particular the solving of algebraic equations, isn’t taught in primary or grade six, did the contributor “mistake” his solution for some form of bar model solution, although no diagram was provided? It’s not uncommon to see a number of pseudo-bar model solutions on social media or on the Websites of tuition centers, when in fact, they are algebraic, with or without any model drawings.

Many parents, secondary school teachers, or tutors, who aren’t versed with the bar model method, subconsciously use the algebraic method, with a bar model, which on closer look, reveals that the mental processes are indeed algebraic. No doubt this would create confusion in the young minds, who haven’t been exposed to formal algebra.

Does the Second Solution Pay Lip Service to Design Thinking?

What do you make of the second solution? Did you get it on first reading? Do you think an average grade five or six student would understand the logic behind the model drawing? From a pedagogical standpoint, the second solution is anything but algebraic. Although it makes use of the bar model method, I wonder what proportion of parents and their children could grasp the workings, without some frustration or struggle.

One common valid complaint by both parents and teachers is that in most assessment (or supplementary) math books that promote bar modeling, even with worked-out solutions to these oft-brain-unfriendly word problems, they’re often clueless how the problem solver knew in the first place that the bar model ought to be presented in a certain way—it’s almost as if the author knew the answer, then worked backwards to construct the model.

Indeed, as math educators, in particular, math writers, we haven’t done a good job in this area in trying to make explicit the mental processes involved in constructing the model drawings. Failure to make sense of the bar models has created more anxiety and fear in the minds of many otherwise above-average math students and their oft-kiasu parents.

Poor Presentation Isn’t an Option

Like in advanced mathematics, the poor excuse is that we shouldn’t be doing math like we’re writing essays! No one is asking the problem solver or math writer to write essays or long-winded explanations. We’re only asking them to make their logic clear: a good presentation forces them to make their thinking clearer to others, and that would help them to avoid ambiguity. Pedantry and ambiguity, no; clarity and simplicity, yes!

Clear Writing Is Clear Thinking

It’s hard work to write well, or to present one’s solution unambiguously. But that’s no excuse that we can afford to be a poor writer, and not a good thinker. As math educators or contributors, we’ve an obligation to our readers to make our presentation as clear as possible. It’s not enough to present a half-baked solution, on the basis that the emphasis in solving a math problem is to get the correct answer, and not waste the time to write grammatically correct sentences or explanations.

I Am Not a Textbook Math Author, Why Bother to Be Precise?

As teachers, we dread about grading students’ ill-written solutions, because most of us don’t want to give them a zero for an incorrect answer. However, if we’re convinced based on their argument that they do know what they’re doing, or show mathematical understanding or maturity of the concepts being tested, then we’d only minus a few marks for careless computation.

Poorly constructed or ill-presented arguments, mathematical or otherwise, don’t make us look professional. Articulating the thinking processes of our logical arguments helps us to develop our intellectual maturity; and last but not least, it makes us become a better thinker—and a better writer, too.

© Yan Kow Cheong, November 1, 2017.

A Grade 5 Bicycles-and-Tricycles Problem

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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.

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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

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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.

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

3 units + 16  = 5 units – 6

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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.

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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.

The Chickens-and-Rabbits Problem

In Singapore, the chickens-and-rabbits question was in vogue in the late nineties, when the Ministry of Education then wanted teachers to formally teach problem-solving strategies (or heuristics, as we commonly call them here). Two common methods of solution favored by local teachers are “guess and check” (for younger students) and “make a supposition.” And in recent years, as Sakamoto math strategies gain currency in more local and regional schools, we’ve been blessed with no fewer than three other methods of solution to solve this type of problems.

A Grade 5 Contest Problem

In math contests and competitions, it’s not uncommon to witness some variations of the chickens-and-rabbits problem, which often pose much difficulty even to students, who are fluent in the Singapore model method. Let’s look at a grade 5 chickens-and-rabbits question, with a slight twist.

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?

Give it a try first, before comparing your solution(s) with the ones I’ve exemplified below.

Method 1

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 for both their total number of legs to be equal.

Picture

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

A check shows that the answers do satisfy the conditions of the question.

Method 2

The equations resulting from the models for Methods 1 and 2 are the same, but conceptually this method is slightly different from the previous one.

The bar representing the number of chickens must be half the length of the bar representing the number of chicken legs. The bar representing the number of rabbits must be one quarter the length of the bar representing the number of rabbit legs.

Picture

From the model drawing,

3 units = 100 – 40 = 60
1 unit = 60 ÷ 3 = 20
2 units + 40 = 2 × 20 + 40 = 80

Therefore, the number of rabbits is 20, and the number of chickens is 80.

Let me leave you with another fertile chickens-and-legs problem, which should challenge most grade 5 or 6 students, not to say, their teachers and parents.

Mr. Yan has almost twice as many chickens as cows. The total number of legs and heads is 184. How many cows are there?

Could you use the bar method, or the Sakamoto method, to solve it?

© Yan Kow Cheong, March 3, 2013.