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.

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.

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.

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.