Tag Archives: Sakamoto math

A Grade 5 Bicycles-and-Tricycles Problem

Problem Solving, Singapore Math, Singapore Model Method, , , , , , , ,

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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.

Problem Solving Made Difficult

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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.

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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.

Algebra or Model Method

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On December 19, 2012, in her Confession from a Homeschool Mom: Singapore Math Stumped Me Today, Monise L. Seward, blogged that her 6th grader woke her up to ask her for help on a word problem in her Singapore Math book. The nonroutine question she shared was the following:

Mrs. Pappas had some apples. She sold 1/3 of the apples plus 5 more on the first day. She sold 1/3 of the remaining apples plus 5 more on the second day. She had 125 apples left in the end. How many apples did Mrs. Pappas have in the beginning?

If you use algebra to solve this problem, it’s unlikely that the working will arouse any excitement; in fact, you may find this method of solution to be somewhat uninteresting or boring. Yes, algebra does religiously solve the problem, but the solution is anything but elegant. Moreover, most average grade five or six students wouldn’t have acquired the maturity to solve it algebraically.

An algebraic check

Besides working backwards to check the answer, after solving the question, using the Singapore model method, I also checked it out by algebra.Looking at the symbolic clutter, I guessed then that even our Singapore grades 7 and 8 average students would likely be challenged to solve this problem by algebraic means.If we stick to working with only one variable, then we may end up with an unappetizing equation such as the following to solve.

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A mum’s solution

Without a good working knowledge with the Singapore model method, we can expect most teachers and parents (who still remember their school math) to solve the question in a way similar to the one sketched and tweeted by my Pinterest and Twitter friend at PragmaticMom.com, as shown below:

2/3x – 5 = 125

2/3x = 125 + 5

2/3x = 130

x = (130 x 3)/2

x = 195

You add the 5 because you have to subtract the 5 apples from the 2/3 calculation because they were added in additionally. Then you do it again …

2/3x – 5 = 195

2/3 x = 195 + 5

2/3 x = 200

x = (200 x 3)/2

x = 300

No algebra, please!

Pretend for a while that algebra is an alien language to you! And you can only use a non-algebraic explanation to communicate your solution to a ten-year-old child! How would you go about doing it? Can you think of some intuitive methods?

After coming across this grade six question via @pragmaticmom, I tweeted a quick-and-dirty solution to the above problem, using the model, or bar, method. Give it a try first, before comparing yours with mine.

Sakamoto math

I also remarked that we could also solve this question by the Sakamoto method, which I assumed most of you in the United States might not be familiar with; so I’ll skip presenting the Japanese method of solution here.

Working backwards

Had the above grade five or six question involved the fraction 1/2 instead of 1/3, then using the “work backwards” strategy, via a flow chart, would have yielded an equally decent method of solution as the model approach.

Mrs. Pappas had some apples. She sold 1/2 of the apples plus 5 more on the first day. She sold 1/2 of the remaining apples plus 5 more on the second day. She had 125 apples left in the end. How many apples did Mrs. Pappas have in the beginning?

See a sketch of a common method of solution below.

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Two bonus questions

Let me leave you with two grade 3/4 problems that lend themselves easily to the model method.1. A shop owner sold 2 more iPads than half the number of iPads in his stock. He then sold 2 fewer iPads than half of the remaining iPads. If he was left with 28 iPads, how many iPads did he have in his stock in the beginning?

2. Sarah used $8 to buy a book. She then used half of the remaining money to buy a bag. Lastly, she spent $1 more than half of what she had left on a meal. In the end, she had only $5 left. How much money had Sarah at first?
Answers: 1. 108 iPads; 2. $32

© Yan Kow Cheong, March 13, 2013.