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.
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.
Here are some pointers I would share with my students at the start of a secondary math olympiad programme. In Singapore, most mathletes attending these enrichment classes are usually selected by the form or math teacher, who tends to choose the best three math students from each class to form a small group of 15 to 20 participants. They would then graduate to represent the school after attending a six-, eight-, or ten-session training programme, depending on the mathematical needs and wants of the school.
• Take your time! Very few contestants can solve all given problems within the time limit. For instance, in the Singapore Mathematics Olympiad (SMO), both at the junior and senior levels, [unconfirmed] feedback based on different schools’ results hints to the fact that those who can win a medal hover around five percent.
Interestingly but disturbingly, an SMO mathlete who can get six or seven out of 35 questions correct may still win a bronze medal, revealing how unmoderated olympiad math papers had been in recent years, going by the abnormally high rate of failures among the participants. This is primarily due to the fact that few, if any, faculty members who set these competition papers, are familiar with what elementary and middle school teachers are covering in local schools.
• Try the “easier” questions first. The questions aren’t set in ascending order of difficulty. It’s not uncommon to see easier questions in the second half of the paper.
• Olympiad questions don’t “crack” immediately. Be patient. Try various approaches. Experiment with simple cases. Working backwards from the desired result in some cases is helpful.
• If you’re using a contests book, and you’re stuck, glance at the “Hints” section. Sometimes a problem requires an unusual idea or technique.
• Even if you can solve a problem, read the hints and solutions. The hints may contain some ideas or insights that didn’t occur in your solution, and they may discuss intuitive, strategic, or tactical approaches that can be used elsewhere.
Remember that modeled or elegant solutions often conceal the torturous or tedious process of investigation, false starts, inspiration and attention to detail that led to them. Be aware of the behind-the-scenes hours-long dirty mathematical work! When you read the modeled solutions, try to reconstruct the thinking that went into them. Ask yourself, “What were the key ideas?” “How can I apply these ideas further?”
• Go back to the original problem later, and see if you can solve it in a different way, or in a different context. When all else fails, remember the reliable old friend, the guess-and-check strategy (or heuristic, as it’s being arguably called in Singapore)—for instance, substituting the optional answers-numbers given in an MCQ into some given equation or expression may yield the answer sooner than later.
Meaningful or creative problem solving takes practice, with insightful or elegant solutions not being the norm. Don’t get discouraged if you don’t seem to make any headway at first. The key isn’t to give up; come back to the question after a day or a week. Stickability and perseverance are two long-time buddies for full-time problem solvers.
Stealing the idea from Nancy Vu’s “Just Little Things,” I tried to reflect on some mathematical things that give us delight and pleasure as we go about our hectic 24/7/365 lives.
Here are some of my personal mathematical pleasures.
It’s now your turn to reflect on some aha! mathematical moments to share with the rest of us.
References
Wu, N. (2013). Just little things: A celebration of life’s simple pleasures. New York: A Perigee Book.
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?
ANumerical 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
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.
Since the number of candies remains the same in both cases, we have
3 units + 16 = 5 units – 6
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.
The difference in the number of candies is 5 – 3 = 2.
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.
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.
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.
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?
