## Introduction to Singapore Math

INTRODUCTION TO SINGAPORE MATH

Welcome to Singapore Math! The math curriculum in Singapore has been recognized worldwide for its excellence in producing students highly skilled in mathematics. Students in Singapore have ranked at the top in the world in mathematics on the Trends in International Mathematics and Science Study (TIMSS) in 1993, 1995, 2003, and 2008. Because of this, Singapore Math has gained in interest and popularity in the United States.

Singapore Math curriculum aims to help students develop the necessary math concepts and process skills for everyday life and to provide students with the ability to formulate, apply, and solve problems. Mathematics in the Singapore Primary (Elementary) Curriculum cover fewer topics but in greater depth. Key math concepts are introduced and built-on to reinforce various mathematical ideas and thinking. Students in Singapore are typically one grade level ahead of students in the United States.

The following pages provide examples of the various math problem types and skill sets taught in Singapore.

At an elementary level, some simple mathematical skills can help students understand mathematical principles. These skills are the counting-on, counting- back, and crossing-out methods. Note that these methods are most useful when the numbers are small.

1. The Counting-On Method

Used for addition of two numbers. Count on in 1s with the help of a picture or number line.

7 + 4 = 11
+1 +1 +1 +1

7 8 9 10 11

2. The Counting-Back Method

Used for subtraction of two numbers. Count back in 1s with the help of a picture or number line.

16 - 3 = 13
-1 -1 -1

13 14 15 16

3. The Crossing-Out Method

Used for subtraction of two numbers. Cross out the number of items to be taken away. Count the remaining ones to find the answer.

20 - 12 = 8

A number bond shows the relationship in a simple addition or subtraction problem. The number bond is based on the concept "part-part-whole." This concept is useful in teaching simple addition and subtraction to young children.

whole

3. Addition Number Bond (double and single digits)

2 + 15
5 10

To find a whole, students must add the two parts.
To find a part, students must subtract the other part from the whole.

The different types of number bonds are illustrated below.

1. Number Bond (single digits)

9 36

3 (part) + 6 (part) = 9 (whole) 9 (whole) - 3 (part) = 6 (part) 9 (whole) - 6 (part) = 3 (part)

2. Addition Number Bond (single digits)

9+5

1.

2.

The Adding-Without-Regrouping Method

= 9 + 1 + 4 = 10 + 4
= 14

Make a ten first.

+1 5 3

6 4 5

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part

part

Students should understand that multiplication is repeated addition and that division is the grouping of all items into equal sets.

1. Repeated Addition (Multiplication)

Mackenzie eats 2 rolls a day. How many rolls does she eat in 5 days? 2 + 2 + 2 + 2 + 2 = 10

5  2 = 10

She eats 10 rolls in 5 days.

2. The Grouping Method (Division)

Mrs. Lee makes 14 sandwiches. She gives all the sandwiches equally to 7 friends. How many sandwiches does each friend receive?

14  7 = 2 Each friend receives 2 sandwiches.

One of the basic but essential math skills students should acquire is to perform the 4 operations of whole numbers and fractions. Each of these methods is illustrated below.

14 HTO 14 9 2

O: Ones
T: Tens
H: Hundreds

= 2 + 5 + 10 = 7 + 10
= 17

Regroup15 into 5 and 10.

4. Subtraction Number Bond (double and single digits)

12 - 7

2 10 10 - 7 = 3

3+ 2= 5
5. Subtraction Number Bond (double digits)

H

3 + 5 8

T O 2 1 6 8 8 9

O: Ones
T: Tens
H: Hundreds

10

20 - 15

10 10 5

10 - 5 = 5 10 - 10 = 0 5+ 0= 5

Since no regrouping is required, add the digits in each place value accordingly.

The Adding-by-Regrouping Method

In this example, regroup 14 tens into 1 hundred 4 tens.

3. The Adding-by-Regrouping-Twice Method

11. The Addition-of-Fractions Method

H 12

+3

6

T O 18 6 6 5 5 1

_1 × 2

_1 × 3 2 3 _5_ × 3 = ___ + ___ =

H T O

73 9 -32 5 41 4

O: Ones
T: Tens
H: Hundreds

O: Ones
T: Tens
H: Hundreds

× 2 +

4 12 12 12
Always remember to make the denominators common before adding the

fractions.

12. The Subtraction-of-Fractions Method

_1_ × 5 _1_ × 2 _5__ _2__ _3_ 2 × 5 - 5 × 2 = 10 - 10 = 10

Always remembers to make the denominators common before subtracting the fractions.

13. The Multiplication-of-Fractions Method

1 _3_ _1_ _1_ 5 ×39 = 15

When the numerator and the denominator have a common multiple, reduce them to their lowest fractions.

14. The Division-of-Fractions Method

6

Regroup twice in this example.
First, regroup 11 ones into 1 ten 1 one. Second, regroup 15 tens into 1 hundred 5 tens.

4. The Subtracting-Without-Regrouping Method

Since no regrouping is required, subtract the digits in each place value accordingly.

5. The Subtracting-by-Regrouping Method

HTO

5 78111 - 2 4 7 334

O: Ones
T: Tens
H: Hundreds

2
_7_ ÷ _1_ = _7_ × _6_ = _1_4_ = 4 _2

In this example, students cannot subtract 7 ones from 1 one. So, regroup the tens and ones. Regroup 8 tens 1 one into 7 tens 11 ones.

6. The Subtracting-by-Regrouping-Twice Method

When dividing fractions, first change the division sign (÷) to the multiplication sign (×). Then, switch the numerator and denominator of the fraction on the right hand side. Multiply the fractions in the usual way.

Model drawing is an effective strategy used to solve math word problems. It is a visual representation of the information in word problems using bar units. By drawing the models, students will know of the variables given in the problem, the variables to find, and even the methods used to solve the problem.

Drawing models is also a versatile strategy. It can be applied to simple word problems involving addition, subtraction, multiplication, and division. It can also be applied to word problems related to fractions, decimals, percentage, and ratio.

The use of models also trains students to think in an algebraic manner, which uses symbols for representation.

The different types of bar models used to solve word problems are illustrated below.

HTO 78 90100

-593

207

O: Ones
T: Tens
H: Hundreds

9 6 39 1 3 3

In this example, students cannot subtract 3 ones from 0 ones and 9 tens from 0 tens. So, regroup the hundreds, tens, and ones. Regroup 8 hundreds into 7 hundreds 9 tens 10 ones.

7. The Multiplying-Without-Regrouping Method

TO 2 4

1.

2.

3.

4.

The model that involves addition

Melissa has 50 blue beads and 20 red beads. How many beads does she

O: Ones 2 T: Tens

48

have altogether?

?

Since no regrouping is required, multiply the digit in each place value by the multiplier accordingly.

8. The Multiplying-With-Regrouping Method

50 + 20 = 70 The model that involves subtraction

Ben and Andy have 90 toy cars. Andy has 60 toy cars. How many toy cars

 50 20

H T O

O: Ones
T: Tens
H: Hundreds

13 24 ×

1,04

9 3 7

does Ben have?

90

In this example, regroup 27 ones into 2 tens 7 ones, and 14 tens into 1 hundred 4 tens.

9. The Dividing-Without-Regrouping Method

241

2482 -4

8 -8

2 -2 0

Since no regrouping is required, divide the digit in each place value by the divisor accordingly.

10. The Dividing-With-Regrouping Method

166

5830 -5

33 -30

30 -30 0

In this example, regroup 3 hundreds into 30 tens and add 3 tens to make 33 tens. Regroup 3 tens into 30 ones.

90 - 60 = 30 The model that involves comparison

Mr. Simons has 150 magazines and 110 books in his study. How many more magazines than books does he have?

 60 ?

Magazines Books

?

150 - 110 = 40
The model that involves two items with a difference

A pair of shoes costs \$109. A leather bag costs \$241 more than the pair of shoes. How much is the leather bag?

150

110

 \$241

Bag Shoes

?

\$109 + \$241 = \$350

\$109

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1. The model that involves multiples

Mrs. Drew buys 12 apples. She buys 3 times as many oranges as apples. She also buys 3 times as many cherries as oranges. How many pieces of fruit does she buy altogether?

Apples
Oranges ? Cherries

13 × 12 = 156

2. The model that involves multiples and difference

There are 15 students in Class A. There are 5 more students in Class B than in Class A. There are 3 times as many students in Class C than in Class A. How many students are there altogether in the three classes?

?

Tie Belt

\$539

Jack's height is _2_ of Leslie's height. Leslie's height is _3_ of Lindsay's height. If 34

Lindsay is 160 cm tall, find Jack's height and Leslie's height. ?

12

? \$539 ÷ 7 = \$77

Tie (2 units) → 2 x \$77 = \$154 Belt (5 units) → 5 x \$77 = \$385

11. The model that involves comparison of fractions

Class A Class B Class C

(5 × 15) + 5 = 80
7. The model that involves creating a whole

Jack

Leslie Lindsay

?

15

 5

?

Ellen, Giselle, and Brenda bake 111 muffins. Giselle bakes twice as many muffins as Brenda. Ellen bakes 9 fewer muffins than Giselle. How many muffins does Ellen bake?

1 unit → 160 ÷ 4 = 40 cm
Jack's height (2 units) → 2 × 40 = 80 cm

Thinking skills and strategies are important in mathematical problem solving. These skills are applied when students think through the math problems to solve them. Below are some commonly used thinking skills and strategies applied in mathematical problem solving.

1. Comparing

Comparing is a form of thinking skill that students can apply to identify similarities and differences.

When comparing numbers, look carefully at each digit before deciding if a number is greater or less than the other. Students might also use a number line for comparison when there are more numbers.

Example:

012345678

3 is greater than 2 but smaller than 7.

2. Sequencing

A sequence shows the order of a series of numbers. Sequencing is a form of thinking skill that requires students to place numbers in a particular order. There are many terms in a sequence. The terms refer to the numbers in a sequence.

To place numbers in a correct order, students must first find a rule that generates the sequence. In a simple math sequence, students can either add or subtract to find the unknown terms in the sequence.

Example: Find the 7th term in the sequence below.
1, 4, 7, 10, 13, 16 ?

160 cm Leslie's height (3 units) → 3 × 40 = 120 cm

 ? 9

Ellen Giselle Brenda

111 + 9

8. The model that involves sharing

(111 + 9) ÷ 5 = 24 (2 × 24) - 9 = 39

There are 183 tennis balls in Basket A and 97 tennis balls in Basket B. How many tennis balls must be transferred from Basket A to Basket B so that both baskets contain the same number of tennis balls?

 ?

Basket A Basket B

183

183 - 97 = 86 86 ÷ 2 = 43

 97

9. The model that involves fractions

George had 355 marbles. He lost _1_ of the marbles and gave _1_ of the remaining 54

marbles to his brother. How many marbles did he have left? 355

? 5 parts → 355 marbles

1 part → 355 ÷ 5 = 71 marbles 3 parts → 3 × 71 = 213 marbles

10. The model that involves ratio

L: Lost
B: Brother
R: Remaining

1st term

Step 1: Step 2:

Step 3:

2nd 3rd 4th 5th 6th 7th term term term term term term

This sequence is in an increasing order.

4 - 1 = 3 7 - 4 = 3
The difference between two consecutive terms is 3.

16 + 3 = 19
The 7th term is 19.

 L B R R R

Aaron buys a tie and a belt. The prices of the tie and belt are in the ratio 2 : 5. If both items cost \$539,

1. (a)  what is the price of the tie?

2. (b)  what is the price of the belt?

3. Visualization

Visualization is a problem solving strategy that can help students visualize a problem through the use of physical objects. Students will play a more active role in solving the problem by manipulating these objects.

The main advantage of using this strategy is the mobility of information in the process of solving the problem. When students make a wrong step in the process, they can retrace the step without erasing or canceling it.

The other advantage is that this strategy helps develop a better understanding of the problem or solution through visual objects or images. In this way, students will be better able to remember how to solve these types of problems.

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Some of the commonly used objects for this strategy are toothpicks, straws, cards, strings, water, sand, pencils, paper, and dice.

4. Look for a Pattern

This strategy requires the use of observational and analytical skills. Students have to observe the given data to find a pattern in order to solve the problem. Math word problems that involve the use of this strategy usually have repeated numbers or patterns.

Example: Find the sum of all the numbers from 1 to 100.

assumptions will eliminate some possibilities and simplifies the word problems by providing a boundary of values to work within.

Example: Mrs. Jackson bought 100 pieces of candy for all the students in her class. How many pieces of candy would each student receive if there were 25 students in her class?

In the above word problem, assume that each student received the same number of pieces. This eliminates the possibilities that some students would receive more than others due to good behaviour, better results, or any other reason.

1. Representation of Problem

In problem solving, students often use representations in the solutions to show their understanding of the problems. Using representations also allow students to understand the mathematical concepts and relationships as well as to manipulate the information presented in the problems. Examples of representations are diagrams and lists or tables.

Diagrams allow students to consolidate or organize the information given in the problems. By drawing a diagram, students can see the problem clearly and solve it effectively.

A list or table can help students organize information that is useful for analysis. After analyzing, students can then see a pattern, which can be used to solve the problem.

2. Guess and Check

One of the most important and effective problem-solving techniques is Guess and Check. It is also known as Trial and Error. As the name suggests, students have to guess the answer to a problem and check if that guess is correct. If the guess is wrong, students will make another guess. This will continue until the guess is correct.

It is beneficial to keep a record of all the guesses and checks in a table. In addition, a Comments column can be included. This will enable students to analyze their guess (if it is too high or too low) and improve on the next guess. Be careful; this problem-solving technique can be tiresome without systematic or logical guesses.

Example: Jessica had 15 coins. Some of them were 10-cent coins and the rest were 5-cent coins. The total amount added up to \$1.25. How many coins of each kind were there?

Use the guess-and-check method.

There were ten 10-cent coins and five 5-cent coins.

10. Restate the Problem

When solving challenging math problems, conventional methods may not be workable. Instead, restating the problem will enable students to see some challenging problems in a different light so that they can better understand them.

The strategy of restating the problem is to "say" the problem in a different and clearer way. However, students have to ensure that the main idea of the problem is not altered.

How do students restate a math problem?

First, read and understand the problem. Gather the given facts and unknowns. Note any condition(s) that have to be satisfied.

Next, restate the problem. Imagine narrating this problem to a friend. Present the given facts, unknown(s), and condition(s). Students may want to write the "revised" problem. Once the "revised" problem is analyzed, students should be able to think of an appropriate strategy to solve it.

11. Simplify the Problem

One of the commonly used strategies in mathematical problem solving is simplification of the problem. When a problem is simplified, it can be "broken down" into two or more smaller parts. Students can then solve the parts systematically to get to the final answer.

Step 1: Step 2:

Step 3:

Step 4:

Simplify the problem.

Find the sum of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

Look for a pattern.

1 + 10 = 11 4 + 7 = 11

2 + 9 = 11 5 + 6 = 11

3 + 8 = 11

Describe the pattern.
When finding the sum of 1 to 10, add the first and last numbers to get a result of 11. Then, add the second and second last numbers to get the same result. The pattern continues until all the numbers from 1 to 10 are added. There will be 5 pairs of such results. Since each addition equals 11, the answer is then 5 × 11 = 55.

Use the pattern to find the answer. Sincethereare5pairsinthesumof1to10,thereshouldbe(10×5= 50 pairs) in the sum of 1 to 100.

Note that the addition for each pair is not equal to 11 now. The addition for each pair is now (1 + 100 = 101).

50 × 101 = 5050
The sum of all the numbers from 1 to 100 is 5,050.

1. Working Backward

The strategy of working backward applies only to a specific type of math word problem. These word problems state the end result, and students are required to find the total number. In order to solve these word problems, students have to work backward by thinking through the correct sequence of events. The strategy of working backward allows students to use their logical reasoning and sequencing to find the answers.

Example: Sarah has a piece of ribbon. She cuts the ribbon into 4 equal parts. Each part is then cut into 3 smaller equal parts. If the length of each small part is 35 cm, how long is the piece of ribbon?

3 × 35 = 105 cm 4 × 105 = 420 cm

The piece of ribbon is 420 cm.

2. The Before-After Concept

The Before-After concept lists all the relevant data before and after an event. Students can then compare the differences and eventually solve the problems. Usually, the Before-After concept and the mathematical model go hand in hand to solve math word problems. Note that the Before-After concept can be applied only to a certain type of math word problem, which trains students to think sequentially.

Example: Kelly has 4 times as much money as Joey. After Kelly uses some money to buy a tennis racquet, and Joey uses \$30 to buy a pair of pants, Kelly has twice as much money as Joey. If Joey has \$98 in the beginning,

1. (a)  how much money does Kelly have in the end?

2. (b)  how much money does Kelly spend on the tennis racquet?

 Number of 10¢ Coins Value Number of 5¢ Coins Value Total Number of Coins Total Value 7 7 × 10¢ = 70¢ 8 8 × 5¢ = 40¢ 7 + 8 = 15 70¢ + 40¢ = 110¢ = \$1.10 8 8 × 10¢ = 80¢ 7 7 × 5¢ = 35¢ 8 + 7 = 15 80¢ + 35¢ = 115¢ = \$1.15 10 10 × 10¢ = 100¢ 5 5 × 5¢ = 25¢ 10 + 5 = 15 100¢ + 25¢ = 125¢ = \$1.25

Before

Kelly

Joey

After

Kelly Joey

??

1. (a)  \$98 - \$30 = \$68 2 × \$68 = \$136

Kelly has \$136 in the end.

2. (b)  4 × \$98 = \$392
\$392 - \$136 = \$256
Kelly spends \$256 on the tennis racquet.

7. Making Supposition

Making supposition is commonly known as "making an assumption." Students can use this strategy to solve certain types of math word problems. Making

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70 Must-Know Word Problems Level 3

\$98

\$30