SECONDARYMATHEMATICS

WORKSHOP

ENGLISH LANGUAGE

ENGLISH LANGUAGE

LEARNERS

IN THE

MATHEMATICS

CLASSROOM

ENGLISH LANGUAGE

SAMPLE QUESTION 1

Questions to help students rely on their own understanding,

ask the following :

• DO YOU THINK THAT IS TRUE? WHY?

• DOES THAT MAKE SENSE TO YOU?

• HOW DID YOU GET YOUR ANSWER?

• DO YOU AGREE WITH THE EXPLANATION?

SAMPLE QUESTION : 2

To promote problem solving, ask the following :

• WHAT DO YOU NEED TO FIND OUT?

• WHAT INFORMATION DO YOU HAVE?

• WILL A DIAGRAM OR NUMBER LINE HELP YOU?

• WHAT TECHNIQUE COULD YOU USE?

• WHAT DO YOU THINK THE ANSWER WILL BE

SAMPLE QUESTION : 3

Questions to encourage students to speak out, ask the following

• What do you think about what ……… said?

• Do you agree what I have said?

• Why?

• Or why not?

• Does anyone have the same answer but a different way to explain it?

• Do you understand what …… ?

• Are you confuse?

SAMPLE QUESTION : 4

Question to check the students progress, ask the following:

• What have you found out so far?

• What do you notice about?

• What other things that you need to do?

• What other information you need to find out?

• Have you though of another way to solve the questions?

SAMPLE QUESTION : 5

Question to help students when they get stuck,ask the following

• What have you done so far?

• What do you need to figure out next?

• How would you say the questions in your own words?

• Could you try it the other way round?

• Have you compared your work with anyone else?

SAMPLE QUESTION : 6

Question to make connection among ideas and application,

Ask the following:

• What other problem does this remind you of?

• Can you give me an example of ?

• Can you write down the objective or aim?

• Can you write down the formulae?

EXAMPLE TO COMMUNICATE

• CAN YOU REPEAT THAT PLEASE?

• HOW DO YOU SPELL________?

• WHAT DOES ____MEAN?

• CAN YOU GIVE ME AN EXAMPLE?

Teacher : I am reading a book about amphibians

Students : Can you repeat that please?

Teacher : I said : “I’m reading a book on amphibians”

Students : How do you spell amphibians?

Teacher : A-M-P-H-I-B-I-A-N-S

Students : What does amphibians mean?

Teacher : It is an animal that is born in water but can live on land

Student : Can you give me an example?

Teacher : A frog

KNOW YOUR KEY WORDS

• MORE THAN

• LESS THAN

• ALTOGETHER

• AT FIRST

• SUM

• DIFFERENT

• COMPARE

• DIGITS

• FIND THE LENGTH /MASS

• PLACE VALUE

• WHOLE NUMBER

KNOW YOUR KEY WORDS

• ORDINAL NUMBER

• SUBTRACT

• SUBTRACT 2 FROM 5

• GREATER THAN

• LESS THAN

• SHORT/SHORTER/SHORTEST

• TALL/TALLER/TALLEST

• ARRANGE THE NUMBER FROM THE GREATEST TO THE SMALLEST

• ARRANGE THE STRINGS FROM THE SHORTEST TO THE LONGERST

• READ THE QUESTIONS CAREFULLY

KNOW YOUR KEY WORDS

• LABEL THE FOLLOWING

• EVALUATE

• HEAVY/HEAVIER/HEAVIEST

• NUMBER SEQUENCE

• HOW MUCH MONEY I LEFT?

• 1 MORE THAN 10

• 3 LESS THAN 10

• HOW MANY MARBLE HAD SHE LEFT?

• HOW MUCH MORE MONEY JOHN HAVE THAN MARY?

• PRODUCT

KNOW YOUR KEY WORDS

• FACTORS

• MULTIPLES OF 2, 3

• NUMBER LINES

• POSITIVE NUMBER

• NEGATIVE NUMBER

• INTEGERS

• 3 TO THE POWER OF 2

• PRIME NUMBER

• VENN DIAGRAM

• INEQUALITIES

• MULTIPLY

KNOW YOUR KEY WORDS

• DIVIDE

• ADD TWO NUMBER UP TO THREE DIGITS

• FACTION

• MIXED NUMBER

• IMPROPER FRACTION

• CONVERT THE FOLLOWING FRACTION TO DECIMALS

• EQUILATERAL

• ISOSCELES

• RIGHT ANGLE TRIANGLE

• NUMBERATOR

• DENOMINATOR

FACTORS AND MULTIPLES

1. We can write a whole number greater than 1 as a product of two whole numbers.

E.g. 18 = 1 x 1818 = 2 x 918 = 3 x 6

Tip : Note that 18 is divisible by each of its factors.

Therefore, 1, 2, 3, 6, 9 and 18 are called factors of 18.

3. The common factors of two numbers are the factors that the numbers have in common.

2. Factors of a number are whole numbers which multiply to give that number.

E.g. Factors of 12: 1, 2, 3, 4, 6, 12Factors of 21: 1, 3, 7, 21

The common factors of 12 and 21 are 1 and 3.

FACTORS AND MULTIPLES

4. When we multiply a number by a non-zero whole number, we get a multiple of the number.

E.g. 1 x 3 = 3 1 x 5 = 5 2 x 3 = 6 2 x 5 = 10 3 x 3 = 9 Multiples 3 x 5 = 15 Multiples 4 x 3 = 12 of 3 4 x 5 = 20 of 5 5 x 3 = 15 5 x 5 = 25

Therefore, the multiples of 3 are 3, 6, 9, 12, 15, … andthe multiples of 5 are 5, 10, 15, 20, 25, …

The first three common multiples of 4 and 6 are 12, 24 and 36.

5. The common multiple of two numbers is a number that is a multiple of both numbers.

E.g. Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, ...

Multiples of 6 are 6, 12, 18, 24, 30, 36, ...

PRIME NUMBERS,PRIME FACTORISATION1. A prime number is a whole number greater than 1 that

has exactly two different factors, 1 and itself.

E.g. 5 = 1 x 5

Since 5 has no other whole number factors other than 1 and itself, it is a prime number.

3. A composite number is a whole number greater than 1 that has more than 2 different factors.

2. The numbers 2, 3, 5, 7, 11, 13, 17, … are prime numbers.

E.g. 6 = 1 x 66 = 2 x 3

Therefore, 6 is a composite number.

4 Factors

PRIME NUMBERS,PRIME FACTORISATION4. The numbers 4, 6, 8, 9, 10, 12, 14, 15, 16, … are

composite numbers. In other words, all whole numbers greater than 1 that are not prime numbers are composite numbers.

Tip: 0 and 1 are neither prime nor composite numbers.

E.g. The factors of 18 are 1, 2, 3, 6, 9, and 18. The prime factors of 18 are 2 and 3.

5. Prime factors are factors of a number that are also prime.

7. We can use either the factor tree or repeated division to express a composite number as a product of its prime factors.

6. The process of expressing a composite number as the product of prime factors is called prime factorisation.

PRIME NUMBERS, PRIME FACTORISATIONWORKED EXAMPLE 1:

Express 180 as a product of prime factors.

180

2 x 90

2 x 2 x 45

2 x 2 x 3 x 15

2 x 2 x 3 x 3 x 5

Therefore, 180 = 2 x 2 x 3 x 3 x 5= 22 x 32 x 5

SOLUTION:Method I (Using the Factor Tree)

Steps:1.Write the number to be factorised at the top of the tree.2.Express the number as a product of two numbers.3.Continue to factorise if any of the factors is not prime.4.Continue to factorise until the last row of the tree shows only prime factors.

A quicker and more concise way to write the product is using index notation.

PRIME NUMBERS, PRIME FACTORISATIONWORKED EXAMPLE 1:

Express 180 as a product of prime factors.

2 180

2 90

3 45

3 15

5 5

1

Therefore, 180 = 2 x 2 x 3 x 3 x 5= 22 x 32 x 5

SOLUTION:Method II (Using Repeated Division)

Steps:1.Start by dividing the number by the smallest prime number. Here, we begin with 2.2.Continue to divide using the same or other prime numbers until you get a quotient of 1.3.The product of the divisors gives the prime factorisation of 180.

INDEX NOTATION

If the factors appear more than once, we can use the index notation to represent the product.

E.g. 3 x 3 x 3 x 3 x 3 = 35

In index notation, 3 is called the base and the number at the top, 5 is called the index.

35 is read as ‘3 to the power of 5’

E.g. 2 x 2 x 2 x 5 x 5 = 23 x 52

35 indexbase

The answer is read as 2 to the power of 3 times 5 to the power of 2.

HIGHEST COMMON FACTOR (HCF)1. The largest common factor among the common factors

of two or more numbers is called the highest common factor (HCF) of the given numbers.

E.g. Factors of 12 are 1, 2, 3, 4, 6, and 12.

Factors of 18 are 1, 2, 3, 6, 9, and 18.

2. Another method to find the HCF of two or more numbers is by using prime factorisation which is the more efficient way.

The common factors of 12 and 18 are 1, 2, 3 and 6.

The highest common factor (HCF) of 12 and 18 is 6.

3. We can also repeatedly divide the numbers by prime factors to find the HCF.

HIGHEST COMMON FACTOR (HCF)WORKED EXAMPLE 1:

Find the highest common factor of 225 and 270.

SOLUTION:

Therefore, the HCF of 225 and 270 is 45.

225 = 32x 52

270 = 2 x 33 x 5

HCF = 32 x 5 = 45

Find the prime factorisation of each number first.

To get the HCF, multiple the lowest power of each common prime factor of the given numbers.

LOWEST COMMON MULTIPLE (LCM)1. The smallest common multiple among the common

multiples of two or more numbers is called the lowest common multiple (LCM) of the given numbers.

E.g. Multiples of 8 : 8, 16, 24, 32, 40, 48, ...

Multiples of 12 : 12, 24, 36, 48, 60, ...

2. Another method to find the LCM of two or more numbers is by using prime factorisation which is the more efficient way.

The common multiples of 8 and 12 are 24, 48, ...

The lowest common multiple (LCM) of 8 and 12 is 24.

3. We can also repeatedly divide the numbers by prime factors to find the LCM.

LOWEST COMMON MULTIPLE (LCM)WORKED EXAMPLE 1:

Find the lowest common multiple of 24 and 90.

SOLUTION:

Therefore, the LCM of 24 and 90 is 360.

24 = 23x 3

90 = 2 x 32 x 5

LCM = 23x 32 x 5 = 360

To get the LCM, multiple the highest power of each set of common prime factors. Also include any uncommon factors

SQUARES AND SQUARE ROOTS1. When a number is multiplied by itself, the product is

called the square of the number

3. The numbers whose square roots are whole numbers are called perfect squares.

2. 5 is the positive square root of 25.

Tip : 22 = 4 and √ 4 = 2

32 = 9 and √ 9 = 3

42 = 16 and √16 = 4

E.g. 5 x 5 = 25 or 52 = 25

E.g. √25 = 5

E.g. 1, 4, 9, 16, 25, ... are perfect squares.

SQUARES AND SQUARE ROOTSWORKED EXAMPLE 1:

Using prime factorisation, find the square root of 5184.

5184 = 26 x 34 2 5184

√5184 = √26 x 34 2 2592

= 23 x 32 2 1296

= 8 x 9 2 648

= 72 2 324

2 162

3 81

3 27

3 9

3 3

1

SOLUTION:

CUBES AND CUBE ROOTS

1. When a number is multiplied by itself thrice, the product is called the cube of the number

3. The numbers whose cube roots are whole numbers are called perfect cubes.

2. 125 is the cube of 5 and 5 is the cube root of 125.

Tip : 23 = 8 and 8 = 2∛

33 = 27 and 27 = 3∛

43 = 64 and 64 = 4∛

E.g. 5 x 5 x 5 = 125 or 53 = 125

E.g. 125 = 5∛

E.g. 1, 8, 27, 64, 125, ... are perfect cubes.

CUBES AND CUBE ROOTS

WORKED EXAMPLE 1:

Using prime factorisation, find the cube root of 1728.

1728 = 26 x 33 2 1728

∛1728 = 2∛ 6 x 33 2 864

= 22 x 3 2 432

= 4 x 3 2 216

= 12 2 108

2 54

3 27

3 9

3 3

1

SOLUTION:

REAL NUMBERS

1. Numbers with the ‘negative sign’ (‘ - ’) are called negative numbers.

E.g. -1, -2, -3, -4, -5, ...

3. Positive integers are whole numbers that are greater than zero.

5. Zero is an integer that is neither positive nor negative.

2. Integers refer to whole numbers and negative numbers.

E.g. ..., -3, -2, -1, 0, 1, 2, 3, 4, ... are integers.

E.g. 1, 2, 3, 4, 5, ...

4. Negative integers are whole numbers that are smaller than zero.

E.g. -1, -2, -3, -4, -5, ...

REAL NUMBERS

6. A number line showing integers is shown below:

8. Every number on the number line is greater than any number to its left.

7. The arrows on both ends of the number line show that the line can be extended on both ends.

E.g. 2 is greater than -3 and is denoted by 2 > -3

We can also write -3 is smaller than 2 and is denoted by -3< 2

-5 -4 -3 -2 -1 0 1 2 3 4 5

Negative Integers Positive Integers

-5 -4 -3 -2 -1 0 1 2 3 4 5

REAL NUMBERS

9. >, <, > and < are called inequality signs.

10. 1, 2, 3, 4, 5, 6, 7, ... are called natural numbers. The natural numbers are also called positive integers.

> means ‘is greater than’< means ‘is smaller than’> means ‘is greater than or equal to’< means ‘ is smaller than or equal to’

E.g. |2| = 2, |0| = 0, |-2| = 2

11. The numerical or absolute value of a number x, denoted by |x|, is its distance from zero on the number line.

Since distance can never be negative, the numerical or absolute value of a number is always positive.

ADDITION OF INTEGERS

Rules for adding two integers:

Sign of numbers Method

Both numbers have the same signs

(+a) + (+b) = +(a + b)(-a) + (-b) = -(a + b)

1. Add the numbers while ignoring their signs.

2. Write the sum using their common sign.E.g. (+3) + (+8) = +11 = 11E.g. (-3) + (-8) = -(3 + 8) = -11

Both numbers have different signs

(+a) + (-b) = +(a – b) if a>b(+a) + (-b) = - (b – a) if b>a(-a) + (+b) = - (a – b) if a>b(-a) + (+b) = +(b – a) if b>a

1. Subtract the numbers while ignoring their signs.

2. The answer has the same sign as the number having the larger numerical value.E.g. 12 + (-4) = 12 – 4 = 8E.g. 5 + (-11) = -(11 – 5) = -6E.g. -8 + 3 = -(8 – 3) = -5E.g. -9 + 15 = 15 – 9 = 6

SUBTRACTION OF INTEGERS

To subtract integers, change the sign of the integer being subtracted and add using the addition rules for integers.

a – b = a + (-b)

E.g. 8 – 15 = 8 + (-15) = -(15 – 8) = -7

-11 – 7 = -11 + (-7) = -(11 + 7) = -18

-6 – (-10) = -6 + 10 = 10 – 6 = 4

3 – (-13) = 3 + 13 = 16

MULTIPLICATION OF INTEGERS

Rules for multiplying integers:

Multiplication Examples

(+a) x (+b) = +(a x b)(- a) x (- b) = +(a x b)(+a) x (- b) = - (a x b)(- a) x (+b) = - (a x b)

3 x 4 = 12(-5) x (-6) = +(5 x 6) = 308 x (-3) = -(8 x 3) = -24(-12) x 4 = -(12 x 4) = -48

Rules for signs:

( + ) x ( + ) = ( + ) The product of two positive integers is a positive integer

( - ) x ( - ) = ( + ) The product of two negative integers is a positive integer

( + ) x ( - ) = ( - )( - ) x ( + ) = ( - )

The product of a positive and a negative integer is a negative integer.

DIVISION OF INTEGERS

Rules for dividing two integers:

Division Examples

(+a) ÷ (+b) = +(a ÷ b)(- a) ÷ (- b) = +(a ÷ b)(+a) ÷ (- b) = - (a ÷ b)(- a) ÷ (+b) = - (a ÷ b)

16 ÷ 2 = 8(-20) ÷ (-5) = +(20 ÷ 5) = 436 ÷ (-4) = -(36 ÷ 4) = -9(-24) ÷ 8 = -(24 ÷ 8) = -3

Rules for signs:

( + ) ÷ ( + ) = ( + ) The quotient of two positive integers is a positive integer

( - ) ÷ ( - ) = ( + ) The quotient of two negative integers is a positive integer

( + ) ÷ ( - ) = ( - )( - ) ÷ ( + ) = ( - )

The quotient of a positive and a negative integer is a negative integer.

RULES FOR OPERATING ON INTEGERS1. Addition and multiplication of integers obey the

Commutative Law.

2. Addition and multiplication of integers obey the Associative Law.

E.g. 1 2 + (-10) = (-10) + 2 = -8E.g. 2 2 x (-10) = (-10) x 2 = -20

Commutative Law of Addition of Integers:a + b = b + a

Commutative Law of Multiplication of Integers:a x b = b x a

Associative Law of Addition of Integers:(a + b) + c = a + (b + c)

Associative Law of Multiplication of Integers:(a x b) x c = a x (b x c)

E.g. 1 [3 + (-5)] + 8 = 3 + [(-5) + 8] = 6E.g.2 [3 x (-5)] x 8 = 3 x [(-5) x 8] = -120

RULES FOR OPERATING ON INTEGERS3. Multiplication of integers is distributive over

a) addition b) subtraction

4. The order of operation on integers is the same as those for whole numbers

E.g. 1 -2 x (-3 + 5) = -2 x (-3) = (-2) x 5 = -4E.g. 2 -2 x (-8 + 6) = -2 x (-8) = (-2) x 6 = 28

Distributive Law of Multiplication over Addition of integers:

a x (b + c) = (a x b) + (a x c)Distributive Law of Multiplication over Subtraction of Integers:

a x (b – c) = (a x b) – (a x c)

Order of operations1.Simplify expressions within the brackets first.2.Working from left to right, perform multiplication or division before addition or subtraction.

RULES FOR OPERATING ON INTEGERSWORKED EXAMPLE 1:

Evaluate each of the following.

a) 25 – 36 ÷ (-4) + (-11)

b) (-10) – (-6) + (-9) ÷ 3

c) {-15 – [15 + (-9)]2} ÷ (-3)

d) (3 – 5)3 x 4 + [(-18) + (-2)] ÷ (-3)2

SOLUTION:

a)25 – 36 ÷ (-4) + (-11)= 25 – (-9) + (-11)= 25 + 9 – 11= 23

RULES FOR OPERATING ON INTEGERSSOLUTION:

b)(-10) – (-6) + (-9) ÷ 3= (-10) – (-6) + (-3)= -10 + 6 – 3= -7

d) (3 – 5)3 x 4 + [(-18) + (-2)] ÷ (-3)2

= (-2)3 x 4 + (-18 – 2) ÷ (-2)2

= (-2)3 x 4 + (-20) ÷ (-2)2

= (-8) x 4 + (-20) 4= -32 + (-5)= -32 – 5= -37

c) {-15 – [15 + (-9)]2} ÷ (-3)= [-15 – (15 – 9)2] (-3)= (-15 – 62) ÷ (-3)= (-15 – 36) ÷ (-3)= (-51) ÷ (-3)= 17

INTRODUCTION TO ALGEBRA

Using Letters to Represent Numbers

2. A variable is a letter that is used to represent some unknown numbers/quantity .

E.g. There are n apples in a bag.If there are 5 bags, then the total number

of apples is 5 x n.5 x n can be any whole number value.It can be 5, 10, 15, … depending on the

value of n. i.e. n = 1, 2, 3, …Here, n is called the variable and 5 x n is

called the algebraic expression.

E.g. x, y, z, a, b, P, Q, … are variables

1. In algebra, we use letters (e.g. x, y, z, a, b, P, Q, …) to represent numbers.

INTRODUCTION TO ALGEBRA

Using Letters to Represent Numbers

E.g. 3x + y, a2 – ab, 2x2 + 3x – 4.

3. An algebraic expression is a collection of terms connected by the signs ‘+’, ‘-‘, ‘x’, ‘÷’.

Tip: An algebraic expression does not have an equal sign (=).An algebraic expression is different from an algebraic equation. An equation is a mathematical statement that says that two expressions are equal to each other

E.g. A = lb is an equation. A and lb are algebraic expressions.

ALGEBRAIC NOTATIONS

1. We use the signs ‘+’, ‘-‘, ‘x’, ‘÷’ and ‘=’ in Algebra the same way as Arithmetic.

2. The examples below show how we rewrite mathematical statements as algebraic expressions.

Words Algebraic Expression

Add a to b Sum = a + b = b + a

Subtract c from d Difference = d – cNote: d – c ≠ c – d

Multiple e by f Product = e x f = f x e = ef

Divide g by h(h ≠ 0)

Quotient = g ÷ h = g/h

ALGEBRAIC NOTATIONS

More examples below show how we rewrite mathematical statements as algebraic expressions.

Words Algebraic Expression

Add 3 to the product of p and q pq + 3

The total cost of x books and y magazines if each book cost $4 and each magazine costs $5.

Cost of x books= $4 x x= $4xCost of y magazines= $5 x y= $5yTotal Cost= $(4x + 5y)

ALGEBRAIC NOTATIONS

3. In Algebra, we use the same index notation as in Arithmetic.

Index Notation

Recall: 5 x 5 x 5 = 53

53 is read as ‘5 to the power of 3’

In Algebra,

a x a = a2 (read as ‘a squared’)

a x a x a = a3 (read as ‘a cubed’)

a x a x a x a x a = a5 (read as ‘a to the power of 5’)

base

index

ALGEBRAIC NOTATIONS

WORKED EXAMPLE 1:

a) 3x x 4y ÷ 6z

b) 2a x 3b x a

c) 5p ÷ 10q + 7s x 2

SOLUTION:

a) 3x x 4y ÷ 6z

= 3 x x x 4 x y ÷ 6z

= 12xy ÷ 6z

= 12xy/6z

= 2xy/z

b) 2a x 3b x a

= 6a2b

c) 5p ÷ 10q + 7s x 2

= 5p/10q + 14s

= p/2q + 14s

ALGEBRAIC NOTATIONS

WORKED EXAMPLE 2:

a) Subtract 3 from the sum of 5a and 4b.

b) Add the product of c and d to the cube of e.

c) Multiple 2 to the quotient of f divided by g.

SOLUTION:

a) Sum of 5a and 4b

= 5a + 4b

Required expression

= 5a + 4b – 3 (ans)

b) Product of c and d

= c x d = cd

Cube of e = e x e x e = e3

Required expression

= cd + e3 (ans)

c) Quotient of f divided by g = f/g

Required expression = 2 x f/g = 2f/g (ans)

EVALUATION OF ALGEBRAIC EXPRESSIONS AND FORMULA

1. To evaluate an algebraic express, we substitute a number for the variable and carry out the computation.

WORKED EXAMPLE 1:

a) 3a + 2b – 4c,

b) a(2b – c) – 3b2,

c) a/b – (a+b)/ac,

given that a = 4, b = 2, c = -3.

SOLUTION:

a) 3a + 2b – 4c = 3(4) + 2(2) – 4(-3)

= 12 + 4 + 12

= 28

EVALUATION OF ALGEBRAIC EXPRESSIONS AND FORMULA

c) a/b – (a+b)/ac = 4/2 – (4+2)/4(-3)

= 2 – (6/-12)

= 2 + ½

= 2½

SOLUTION:

b) a(2b – c) – 3b2 = 4[2(2) – (-3)] – 3(2)2

= 4(4+3) – 3(4)

= 4(7) – 12

= 28 – 12

= 16

ALGEBRAIC EXPRESSIONSa) Find the total cost of m cups and n plates if each cup

cost $3 and each plate costs $4.

1 cup = $3 1 plate = $4

m cups = m x $3 n plates = n x $4

= $3m = $4n

Total Cost = $3m + $4n = $(3m + 4n) (ans)

b) Find the total cost of 7 bars of wafers at p cents each and q packets of sweets at $1 each.

1 bar = p cents 1 packet = 100 cents

7 bars = 7 x p cents q packets = q x 100 cents

= 7p cents = 100q cents

Total Cost = 7p cents + 100q cents = (7p + 100q) cents

ALGEBRAIC EXPRESSIONSc) John has $100, He bought n comic books at $9 each.

How much money had he left?

1 book = $9

n books = n x $9

= $9n

Amt left = $100 - $9n = $(100 – 9n) (ans)

d) The cost of 3 caps is $x. Find the cost of 5 caps. Each cap costs the same.

3 caps = $x

1 cap = $x ÷ 3 = $x/3

5 caps = $x/3 x 5 = $5x/3 (ans)

RATIONAL NUMBERS1. A rational number is any number that can be written as

a ratio of two integers. In other words, a number is a rational number if it can be written as a fraction where both the numerator and denominator are integers.

E.g. 3 and -6 are rational nos. since 3 = 3/1 and -6 = -6/1

E.g. 0.5 and 3.2 are rational nos. since 0.5 = 5/10 and 3.2 = 32/10

E.g. -3/5, ½, 5/3, 12/3, … are rational numbers.

A rational number can be written in the form a/b where a and b are integers and b ≠ 0

2. All integers are rational numbers since each integer, n can be written as n/1.

3. Most decimals can be expressed as rational numbers too.

RATIONAL NUMBERSRecall:

5/10 = ½ 5/10 and ½ are equivalent fractions.

½ is said to be in its simplest form or in its lowest terms. Here, the numerator and denominator have no common factors.

31/5 is called a mixed number. It represents the sum of whole number and a proper fraction.

E.g. ½ , 3/7, and 5/9 are proper fractions.

32/10 = 16/5 = 31/5

16/5 is called an improper fraction. The numerator is greater than or equal to the denominator in an improper fraction.

A proper fraction has its numerator smaller than its denominator.

ADDITION AND SUBTRACTION OF RATIONAL NUMBERS1. To add or subtract rational numbers, express the

rational numbers as equivalent fractions in the same denominators first

WORKED EXAMPLE 1:

a) 61/6 – 23/4, b) (-51/4) + (-12/5) + (-½)

SOLUTION:

a) 61/6 – 23/4

= (5 – 2) + (11/6 – ¾)

= 3 + (7/6 – ¾)

= 3 + (14/12 – 9/12)

= 3 + 5/12

= 35/12

SOLUTION:

b) (-51/4) + (-12/5) + (-½)

= - (51/4 + 12/5 + ½)

= - (55/20 + 48/20 + 10/20)

= - (563/20)

= - (5 + 33/20)

= -83/20

MULTIPLICATION AND DIVISION OF RATIONAL NUMBERS1. To multiply two rational numbers:

a) Convert all mixed numbers to improper fractions first.b) Simplify the fractions first by crossing out the common factors of the numerators and denominators.c) Multiply the numerators, then the denominators.

d) Reduce answer to its simplest form.a/b x c/d = a x c/b x d , where a, b, c, d are integers and b ≠ 0, d ≠ 0

2. To divide a rational number by another number :a) Convert all mixed numbers to improper fractions first .b) Invert the second fraction by interchanging its

numerator and denominator .c) Multiply the numerators, then the denominators .

d) Reduce answer to its simplest form.a/b ÷ c/d = a/b x d/c = a x d/b x c , where a, b, c, d are integers and

b ≠ 0, c ≠ 0, d ≠ 0

MULTIPLICATION AND DIVISION OF RATIONAL NUMBERS

WORKED EXAMPLE 1:

a) (-21/2) x 31/5, c) (1/5 – 1/3) ÷ (-1/4 x 2/9)

b) b) (-2/11) ÷ (-10/33)

SOLUTION:

a) (-21/2) x 31/5

= - 5/2 x 16/5

= -8

b) (-2/11) ÷ (-10/33)

= - 2/11 x (-33/10)

= 3/5

SOLUTION:

c) (3/15 – 5/15) ÷ (-1/4 x 2/9)

= - 2/15 ÷ (- 1/18)

= - 2/15 x (- 18/1)

= 36/15

= 22/5

TECHNIQUE & STRATEGIES IN SOLVING MATHEMATICS

WORD PROBLEM SUM1/2

SHARING HOW TO GO ABOUT TEACHING

SYNOPSISSYNOPSISIn solving math problem sum at secondary school level, it is widely acknowledged that heuristics strategies play a major role. By using suitable heuristics, it could greatly enhance pupil’s problem solving performance. Heuristics, referred to the method or strategies of achieving a solution to a given problem sum

Model Drawing is just one of the methods that can be used. Of course there are also various strategies that can be used.

Through-out this seminar, I will share with you the various types of strategies used to solve word problems sum.

The reason why Model drawing is used is because it is one of the most common heuristics used to solve word problems in Mathematics. It is recognized internationally as an effective way for young children to solve word problems and to be exposed early to algebraic concepts.

TECHNIQUES AND OTHER HEURISTICS STRATEGIES:

• Guess & Check method

• Making a Table

• Make a List (Listing method)

• Draw a Picture

• Find a Pattern

• Working Backwards

• Model Drawing

COMMON DIFFICULTIES IN MATHEMATICAL PROBLEM SOLVING

• Inability to read the problem

• Lack of comprehension of the problem posed

• Lack of strategy knowledge

• Inappropriate strategy used

• Inability to translate the problem into a mathematical form

• Computational error

4 - Step in solving problem sum:

• I dentify the problem(what is the questions exactly asking for?)

• D evise a plan(model method)

• E xecute the plan(work it out)

• A nswer check(number sense)

4 - Step in solving problem sum:

• I dentify the problem

After reading the problem sum, what is the questions exactly asking for?

• D evise a plan

‘By drawing models, pupils can represent the mathematical relationships in a problem pictorially. This helps them understand the problem and plan the steps for the solution’

4 - Step in solving problem sum:

• E xecuting the plan

In your plan, you might required to use one or more of the strategies (heuristics) listed below to help you solve the words problem sum.

• A nswer check

Answer must be check to be able to satisfy the condition of the question.

Strategy 5 : Find a PatternQ1. A few children had to share a plate of chicken wings. If each of them took 5

chicken wings, there would be 4 left. In the end, they decided to take 6 chicken wings each, leaving 1 chicken wing on the plate. How many children shared the chicken wings? What was the original number of chicken wings on the plate?

STEP 1 : Draw a Table and determine the information that needs to be found.

No. of Children

Multiple of 5 and add 4

STEP 2 : I find the patterns present in the data to complete the table.

From the table, I can see that number 19 satisfies both conditions of the question.

Ans: There were 3 children sharing the chicken wings. There were 19 chicken wings.

1 2 3 4 5

Checking: ( 3 x 5 ) + 4 = 19( 3 x 6 ) + 1 = 19 ------- Correct

9 14 19 2429

Multiple of 6 and add 1 7 13 19 2531

+5 +5 +5 +5

+6 +6 +6 +6

Strategy 5 : Find a PatternQ2. Mr. Tom wanted to distribute his stamps equally among a few of his students. If he

were to give each of them 5 stamps, he would have 4 left. If he gave each of them 6 stamps, he would have 1 left. How many students and stamps did he have?

STEP 1 : Draw a Table and determine the information that needs to be found.

No. of Students

Multiple of 5 and add 4

STEP 2 : I find the patterns present in the data to complete the table.

From the table, I can see that number 19 satisfies both conditions of the question.

Ans: There were 3 students sharing the stamps. There were 19 stamps.

1 2 3 4 5

Checking: ( 3 x 5 ) + 4 = 19( 3 x 6 ) + 1 = 19 ------- Correct

9 14 19 2429

Multiple of 6 and add 1 7 13 19 2531

+5 +5 +5 +5

+7 +7 +7 +7

Strategy 5 : Find a PatternQ3. A Christmas tree was decorated with flashing light bulbs. The red bulbs flashed

every 2 minutes. The yellow bulbs flashed every 3 minutes and the blue bulbs flashed every 4 minutes. At 8pm, all the light flashed simultaneously. Figure out the next time when all the bulbs will flash together?

STEP 1 : Draw a Table and determine the information that needs to be found.

Red (every 2 mins)

Yellow (every 3 mins)

Blue (every 4 mins)

STEP 2 : I find the patterns present in the data to complete the table.

Looking at the pattern above, the starting time is 8.00pm.

Ans: The next time all the bulbs will flash together is at 8.12pm.

8.00 8.02 8.04 8.06 8.08 8.10 8.12

Next time the bulbs will flash together 8pm + 12mins

8.00 8.03 8.06 8.09 8.12 8.15 8.18

8.00 8.04 8.08 8.12 8.16 8.20 8.24

Strategy 6 : Working BackwardsQ1. On my way to the shopping centre, I found that I did not bring enough money for

my shopping. I then went to the bank to withdraw $100. Next , I bought a pair of shoes for $40. Later, I paid for a T-shirt with half of the money I had left. I was left with $65. How much money did I have before I visited the bank to withdraw the money?

To find out how much money I had at first, I have to work backward by starting at the end and undoing each step in reverse order.

You can draw a flow chart or an arrow to show what happened.

Step 1 Step 2 Step 3Amount I Withdrew $100 Spent $40 on a Spent half of the Amountstarted with from bank pair of shoes money on T-Shirt left

? + $100 - $40 ÷ 2 $65

Next, I work backward by undoing ( + - ) ( x ÷ ) each step in reverse order.

Strategy 6 : Working BackwardsQ1. con’t

Ans:I had $70 before I went to the bank to withdraw the money.

Step 1 Step 2 Step 3Amount I Withdrew $100 Spent $40 on a Spent half of the Amountstarted with from bank pair of shoes money on T-Shirt left

? + $100 - $40 ÷ 2 $65

Next, I work backward by undoing ( + - ) ( x ÷ ) each step in reverse order.

Amount I started with

Amount before Step 2

Amount before Step 3

Amount leftUndo Step 1

Undo Step 2

Undo Step 3

$65x 2$130+ $40$170- $100$70

Strategy 6 : Working BackwardsQ2. Alice, Billy and John each bought some drink. John poured his drink in a jug.

Alice then added 0.7ℓ of drink into the jug. After that, Billy added enough drink to double the amount in the jug. All of them drank 1.2ℓ of it, leaving 1.3ℓ in the jug. How much drink did John bring ?

Ans: John brought 0.55 ℓ of drink.

1.3ℓ 2.5ℓ 1.25ℓ 0.55ℓ+ 1.2ℓ ÷ 2 - 0.7ℓ

Amount left in the jug

Strategy 6 : Working BackwardsQ3. One evening, Lily baked some chocolate cookies. She put 44 chocolate cookies in

a bag and packed another 24 in a tin. Then she divided the remainder equally among herself and 2 of her friends. She kept her share of 14 chocolate cookies in a jar. How many chocolate cookies did she bake that evening?

Ans: She baked 110 chocolate cookies.

14 42 66 110x 3 + 24 + 44

Amount left in the jar

UNDERSTANDING THE ‘8’ DIFFERENTS MODEL. (MODEL DRAWING)

The model drawing/diagram is a very important strategy in secondary school mathematics. Using it correctly, a child will be able to solve many types of challenging problems sum easily.

The 8 different types of model drawing is very useful as it can be used a ‘Diagnostic Tool’.

The trainers or the teachers can straight away identified what kinds or types of problem sum a child has instead of spending time figuring out where is the weakness of the child.

Therefore with constant practice on the model drawing it not only reinforce the understanding of the questions it also develop skill and the process thinking skill in solving word problems.

The ‘8’ Different Models can be used in the following types of problem sum :

• ADDITION

• SUBTRACTION

• COMPARISION

• MULTIPLICATION

• DIVISION

• 1 – STEP PROBLEM SUM

• 2 – STEP PROBLEM SUM

• 3 – STEP PROBLEM SUM

• CHALLENGING PROBLEM SUM INVOLVING BEFORE AND AFTER MODEL CONCEPT

Model Drawing : 2Q1. Jim and his brother share a sum of $150.

If Jim gets $50 more than his brother.How much money do Jim and his brother get?

$150

1 Unit

$50

Step 1 : Look out for KEY PERSON (REFERENCE POINT) Key person or reference point has only 1 UNIT ( )

Step 2 : In this case, the key person is his brother.

Step 3 : Draw the MODEL drawing of his brother first as 1 unit.

Brother

Jim

Model Drawing : 2Q1. Con’t

Ans: Therefore, Jim gets $100 and his brother gets $50.

Step 4 :After finding your 1 unit, look back at the MODEL DRAWING. The one that has 1 unit is Brother.

Step 5 :As for Jim, look at the model drawing it consist of 1 unit + $50

$150

1 Unit

$50

Brother

Jim

$150 - $50 = $100

$100 ÷ 2 = $50 1 Unit ( ) / Brother

$50 + $50 = $100

Therefore, the brother get $50

$50

Model Drawing : 2Q2. Peter and David share a total of 300 sweets.

If David gets 60 more than Peter.How many sweets do Peter and David get?

300

1 Unit

60

Step 1 : Identified KEY PERSON that has only 1 UNIT ( )CAN YOU FIGURE IT OUT?

Step 2 : DRAW THE MODEL

Peter

David

300 – 60 = 240240 ÷ 2 = 120 1 unit ( ) / Peter 120 + 60 = 180 David

Ans: Peter gets 120 sweets and David gets 180 sweets.

120

Hand-On : Exercise 2Q1. $400 is to be shared between Susan and Mary.

If Susan gets $20 more than Mary.How much money do Susan and Mary get?

$400

1 Unit

$20

Step 4 :After finding your 1 unit, look back at the MODEL DRAWING. The one that has 1 unit is Mary.

Mary

Susan

$400 – $20 = $380$380 ÷ 2 = $190 1 unit ( ) / Mary

$190 + $20 = $210 Susan

$190

Therefore, Mary gets $190Step 5 : As for Susan, look at the model drawing it

consist of 1 unit + $20

Ans: Therefore, Mary gets $190 and Susan gets $210.

Hand-On : Exercise 2Q2. Amy and John have 240 stickers altogether. If Amy has 80 stickers

more than John. How many stickers does John have?

240

1 Unit

80

Step 4 :After finding your 1 unit, look back at the MODEL DRAWING. The one that has 1 unit is John.

John

Amy

240 – 80 = 160

160 ÷ 2 = 80 1 unit ( ) / John

80 + 80 = 160 Amy

80

Therefore, John gets 80

Step 5 : As for Amy, look at the model drawing it consist of 1 unit + 80

Ans: Therefore, John gets 80 and Amy gets 160.

Model Drawing : 3Q1. Sharon and Janet share a total of 180 beads.

If Janet gets 30 bead less than Sharon.How many beads do Sharon and Janet get?

180

1 Unit

30

Step 1 : Look out for KEY PERSON (REFERENCE POINT) Key person or reference point has only 1 UNIT ( )

Step 2 : In this case, the key person is Sharon.

Step 3 : Draw the MODEL drawing of Sharon as 1 unit.

Sharon

Janet

Ans: Therefore, Sharon gets 105 and Janet gets 75.

180 + 30 = 210

210 ÷ 2 = 105 1 unit ( ) / Sharon

105 - 30 = 75 Janet

105

Hand-On : Exercise 3Q1. $150 is shared between Judy and Susan.

If Susan gets $30 less than Judy.How much money do Judy and Susan get?

$150

1 Unit

$30

Judy

Susan

Ans: Therefore, Judy gets $90 and Susan gets $60.

$150 + $30 = $180

$180 ÷ 2 = $90 1 unit ( ) / Judy

$90 - $30 = $60 Susan

$90

Hand-On : Exercise 3Q2. David and John share a total of 360 stamps.

If John gets 40 stamps less than David.How many stamps do David and John get?

360

1 Unit

40

David

John

Ans: Therefore, David gets 200 and John gets 160.

360 + 40 = 400

400 ÷ 2 = 200 1 unit ( ) / David

200 - 40 = 160 John

200

Model Drawing : 4Q1. $300 is to be shared between Jason and Kevin. If Kevin gets twice

as much as Jason. How much money do Jason and Kevin get?

$300

1 Unit

Jason

Kevin

$300 ÷ 3 = $100 1 unit ( ) / Jason

$100 x 2 = $200 Kevin

$100

Ans: Jason and Kevin each get $100 & $200 respectively.

Model Drawing : 4Q2. Ken and Joseph share a sum of $250.

If Ken gets 4 times as much as Joseph.How much money do Ken and Joseph get?

$250

1 Unit

Joseph

Ken

$250 ÷ 5 = $50 1 unit ( ) / Joseph

$50 x 4 = $200 Ken

$50

Ans: Joseph and Ken each get $50 & $200 respectively.

Model Drawing : 4Q3. Aaron has 4 times as many stamps as Jimmy. If he has 24 stamps

more than Jimmy. How many stamps does Aaron have?

1 Unit

Jimmy

Aaron

24 ÷ 3 = 8 1 unit ( ) / Jimmy

8 x 4 = 32 Aaron

8

Ans: Jimmy and Aaron each get 8 & 32 stamps respectively.

Step 1 : Who is the KEY PERSON?

Step 2 : Draw the model drawing of that person first.

24

Model Drawing : 4Q4. Sarah has 50 more stickers than Jenny. If Sarah has thrice as many

stickers as Jenny. How many stickers does Sarah have?

1 Unit

Jenny

Sarah

50 ÷ 2 = 25 1 unit ( ) / Jenny

25 x 3 = 75 Sarah

25

Ans: Jenny and Sarah each get 25 & 75 stickers respectively.

50