Distributive Law for Excess and Shortage Problems in PSLE Math
Ho Soo Thong
This article illustrates the use of Arithmetic Distributive Law as essential mathematics for problem solving at primary school level. It begins with a bar modelling approach to relate the mathematical situations in some word problems to the Distributive bLaw. A further step forward is to use the similar approach to complex Excess and Shortage problems.
A renowned American psychologist, Jerome S. Bruner (1915- ) proposed the concrete-picture-abstract approach and the strategy of spiral curriculum in which some basic mathematics and problem solving skills can be repeatedly applied to deal with simple and complex mathematical situations. These suggest a comprehensive curriculum with pictorial problem solving approaches to teaching mathematics.
This article attempts to follow the approach by focusing on some pictorial problem solving skills with the use of the Distributive Law as essential mathematics for a spiral curriculum involving Excess and Shortage (E&S) Problems.
Bar Modelling Approach with Distributive Law
Let us begin with a simple example.
Lydia made some cakes and packed all the cakes into a certain number of boxes, each box containing 12 cakes.
When she packed each box with 10 cakes, there were 14 cakes left.
How many boxes were there? How many cakes were made?
After packing each box with 10 cakes, there were 2 (= 12 − 10) cakes left. Since there were a total of 14 ( = 7×2 ) cakes left, there must be 7 boxes.
Total number of cakes was 84 (= 7× 12).
In Example 1, the scenarios and the related arithmetic expression 7×12 =7×10 + 7×2 can be depicted by a comparison bar model as shown in Figure 1.
The above arithmetic expression 7×12 =7×10 + 7×2 is the result of the Distributive Law
n(x + y) = nx + ny where n is a positive integer.
About 2400 years ago, Euclid used rectangular area models to illustrate a proposition of Distributive Law of multiplication over addition in his book Euclid’s Elements Book II.
Figure 2 shows a general comparison bar model for depicting the Distributive Law.
Next, we will see how the Distributive Law is applied to solve a simple speed problem in which travelling distances are represented by line segments. Line segments for mathematical quantities were used in the well-known Euclidean Algorithm.
Andy walked to school at a speed of 65 m/min and he was 2 minutes early.
At the same time, his brother Bob walked to school at a speed of 50 m/min and he was 4 minutes late.
What was the total time taken by Andy?
How far was the school from their home?
Suppose Andy took n minutes.
Since Bob took 6 ( = 4 + 2 ) minutes more than Andy as shown in the time model in Figure 3.
Their speed difference is 65 − 50 = 15 m/min. After Andy reached the school, Bob had to travel a further distance of n × 15 m as depicted in the distance bar models as shown in Figure 4.
Andy took 20 ( = n )minutes.The school was 1.3 km ( = n×65 m) from home.
Note that there is a ratio approach to this problem (See ).
The next example shows the bar modelling approach to a counting problem.
In a conference, chairs were arranged in rows such that there were exactly 9 chairs in each row.
On a certain day, a broken chair was removed and the remaining chairs were arranged with exactly 7 chairs in each row. There were 3 more rows than before.
How many chairs were there at first?
Suppose there were n rows of 9 chairs at first.
First, we construct a bar model for n rows of 9 chairs. Next, with one chair less, we add a bar model with n + 3 rows of 7 chairs and the distributive lines as shown in Figure 5.
Next, we solve a Finance problem.
Steve bought some curry puffs at $ 1.50 each and some chicken pies at $ 2.20 each.
He spent $ 1.10 less on curry puffs than chicken Pies, but had 3 more curry puffs than chicken pies.
How many curry puffs did Steve buy?
Suppose Steve bought n chicken pies.
Noting that there are 3 curry puffs more than chicken pies, we construct a comparison bar model for the Distributive Law over the addition $ 2.20 = $ 1.50 + $ 0.70 as shown in Figure 6.
Therefore Steve bought 11 ( = 8 + 3 ) curry puffs.
Tha above examples illustrate a problem solving approach with Distributive Law at PSLE level. Next, we will proceed to apply the Distibutive Law for more complex excess and shorage problems.
Excess and Shortage Scenarios
First, we begin with the mathematical situations of excess and shortage scenarios.
There are 38 candies to be distributed to 6 girls, A, B, C, D, E and F. Some possible scenarios are :
If each girl gets 8 candies , there is a shortage of 10 candies.
If each girl gets 7 candies , there is a shortage of 4 candies.
If each girl gets 6 candies , there is an excess of 2 candies.
If each girl gets 5 candies , there is an excess of 8 candies.
Figure 7 shows the pictorial view of the excess and shortage scenarios.
The mathematical description of above scenarios can be depicted in the following comparison bar models as shown in Figure 8.
In Figures 8 & 9, we draw a vertical line to indicate the number of candies for a supposedly even distribution in each scenario. We will call these lines as distributive lines.
A general rule for identifying the distributive lines is that each line
cuts off the excess or makes up the shortage
as shown in the following examples.
3. Excess and Shortage Problems
The first example is that of an excess-shortage Problem.
Some pirates were to share a collection of gold coins. When each was given 11 coins, there was a shortage of 15 coins. After that, each was given 8 coins, there was an excess of 6 coins.
(a)How many pirates were there?
(b)What was the total number of coins?
Suppose there were n pirates.
In Figure 9, the two scenarios are depicted in a comparison bar model with two distributive lines ( Each gets 11, Each gets 8) for the Distributive Law over the addition 11 = 8 + 3.
(a) There were 7 pirates.
(b) There was a total of 62 gold coins.
Another example is that of an excess-excess problem.
Jane have some sweets for her students.
If she gives 5 sweets each, there will be 32 sweets left. If she gives 8 sweets each, there will be 5 sweets left.
How many sweets are there altogether?
Suppose there are n students.
Figure 10 shows a comparison bar model for the two scenarios with the Distributive Law over the addition 8 = 5 + 3 and the corresponding distributive lines.
Therefore, the number of sweets is 77.
The next example is that of a shortage-shortage problem.
Some students are to share an amount as a prize money.
If each gets $ 8.0, there will be a shortage of $15.8. If each gets $ 6.5 , there will still be a shortage of $ 5.3.
(a) How many students are there?
(b) How much is the prize money?
Suppose there are n students.
Figure 11 shows a comparison bar model for the Distributive Law over the addition $8.0 = $6.5 + $1.5.
(a) Number of students, n =7
(b) Prize Money = $ 42.2.
A well-designed spiral curriculum with pictorial problem solving strategies based on essential mathematics will help a knowledgeable teacher to achieve in teaching. The author hopes that this article will serve as reference materials for teachers preparing their curriculum tailed to the needs of their pupils.
The use of the Distributive Law for problem solving is far beyond the E&S problems; further illustrations will be given in future articles.
 PSLE examination Questions 2005 – 2009, Educational Publishing House Pte Ltd
 Ho Soo Thong, Ho Shuyuan, Bar Model Method for PSLE and Beyond, AceMath, Singapore 2011
 Ho Soo Thong, Ho Shuyuan, Bar Model Method for Speed Problems, barmodelhost.com, 2014