Transcript SECONDARY MATHEMATICS WORKSHOP

SECONDARY
MATHEMATICS
WORKSHOP
ENGLISH LANGUAGE
ENGLISH LANGUAGE
LEARNERS
IN THE
MATHEMATICS
CLASSROOM
SAMPLE QUESTION 1
Questions to help students rely on their own understanding,
ask the following :
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DO YOU THINK THAT IS TRUE? WHY?
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DOES THAT MAKE SENSE TO YOU?
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HOW DID YOU GET YOUR ANSWER?
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DO YOU AGREE WITH THE EXPLANATION?
SAMPLE QUESTION : 2
To promote problem solving, ask the following :
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WHAT DO YOU NEED TO FIND OUT?
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WHAT INFORMATION DO YOU HAVE?
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WILL A DIAGRAM OR NUMBER LINE HELP YOU?
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WHAT TECHNIQUE COULD YOU USE?
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WHAT DO YOU THINK THE ANSWER WILL BE
SAMPLE QUESTION : 3
Questions to encourage students to speak out, ask the following
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What do you think about what ……… said?
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Do you agree what I have said?
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Why?
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Or why not?
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Does anyone have the same answer but a different way to
explain it?
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Do you understand what …… ?
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Are you confuse?
SAMPLE QUESTION : 4
Question to check the students progress, ask the following:
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What have you found out so far?
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What do you notice about?
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What other things that you need to do?
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What other information you need to find out?
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Have you though of another way to solve the questions?
SAMPLE QUESTION : 5
Question to help students when they get stuck,ask the following
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What have you done so far?
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What do you need to figure out next?
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How would you say the questions in your own words?
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Could you try it the other way round?
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Have you compared your work with anyone else?
SAMPLE QUESTION : 6
Question to make connection among ideas and application,
Ask the following:
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What other problem does this remind you of?
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Can you give me an example of ?
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Can you write down the objective or aim?
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Can you write down the formulae?
EXAMPLE TO COMMUNICATE
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CAN YOU REPEAT THAT PLEASE?
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HOW DO YOU SPELL________?
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WHAT DOES ____MEAN?
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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
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MORE THAN
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LESS THAN
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ALTOGETHER
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AT FIRST
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SUM
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DIFFERENT
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COMPARE
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DIGITS
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FIND THE LENGTH /MASS
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PLACE VALUE
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WHOLE NUMBER
KNOW YOUR KEY WORDS
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ORDINAL NUMBER
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SUBTRACT
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SUBTRACT 2 FROM 5
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GREATER THAN
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LESS THAN
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SHORT/SHORTER/SHORTEST
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TALL/TALLER/TALLEST
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ARRANGE THE NUMBER FROM THE GREATEST TO THE
SMALLEST
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ARRANGE THE STRINGS FROM THE SHORTEST TO THE
LONGERST
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READ THE QUESTIONS CAREFULLY
KNOW YOUR KEY WORDS
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LABEL THE FOLLOWING
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EVALUATE
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HEAVY/HEAVIER/HEAVIEST
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NUMBER SEQUENCE
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HOW MUCH MONEY I LEFT?
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1 MORE THAN 10
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3 LESS THAN 10
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HOW MANY MARBLE HAD SHE LEFT?
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HOW MUCH MORE MONEY JOHN HAVE THAN MARY?
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PRODUCT
KNOW YOUR KEY WORDS
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FACTORS
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MULTIPLES OF 2, 3
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NUMBER LINES
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POSITIVE NUMBER
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NEGATIVE NUMBER
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INTEGERS
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3 TO THE POWER OF 2
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PRIME NUMBER
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VENN DIAGRAM
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INEQUALITIES
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MULTIPLY
KNOW YOUR KEY WORDS
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DIVIDE
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ADD TWO NUMBER UP TO THREE DIGITS
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FACTION
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MIXED NUMBER
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IMPROPER FRACTION
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CONVERT THE FOLLOWING FRACTION TO DECIMALS
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EQUILATERAL
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ISOSCELES
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RIGHT ANGLE TRIANGLE
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NUMBERATOR
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DENOMINATOR
FACTORS AND MULTIPLES
1. We can write a whole number greater than 1 as a
product of two whole numbers.
E.g. 18
18
18
= 1 x 18
= 2 x 9
= 3 x 6
Therefore, 1, 2, 3, 6, 9 and 18 are called factors of 18.
Tip : Note that 18 is divisible by each of its factors.
2. Factors of a number are whole numbers which multiply
to give that number.
3. The common factors of two numbers are the factors
that the numbers have in common.
E.g. Factors of 12: 1, 2, 3, 4, 6, 12
Factors 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
2x3=6
3x3=9
4 x 3 = 12
5 x 3 = 15
Multiples
of 3
1x5=5
2 x 5 = 10
3 x 5 = 15
4 x 5 = 20
5 x 5 = 25
Multiples
of 5
Therefore, the multiples of 3 are 3, 6, 9, 12, 15, … and
the multiples of 5 are 5, 10, 15, 20, 25, …
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, ...
The first three common multiples of 4 and 6 are 12, 24
and 36.
PRIME NUMBERS,
PRIME FACTORISATION
1. 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.
2. The numbers 2, 3, 5, 7, 11, 13, 17, … are prime
numbers.
3. A composite number is a whole number greater than 1
that has more than 2 different factors.
E.g. 6 = 1 x 6
4 Factors
6=2x3
Therefore, 6 is a composite number.
PRIME NUMBERS,
PRIME FACTORISATION
4. 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.
5. Prime factors are factors of a number that are also
prime.
E.g. The factors of 18 are 1, 2, 3, 6, 9, and 18.
The prime factors of 18 are 2 and 3.
6. The process of expressing a composite number as the
product of prime factors is called prime factorisation.
7. We can use either the factor tree or repeated division
to express a composite number as a product of its prime
factors.
PRIME NUMBERS,
PRIME FACTORISATION
WORKED EXAMPLE 1:
Express 180 as a product of prime factors.
SOLUTION:
Method I (Using the Factor Tree)
180
Steps:
2 x 90
2 x 2 x 45
2 x 2 x 3 x 15
2x2x3x3x5
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.
Therefore, 180 = 2 x 2 x 3 x 3 x 5
= 22 x 32 x 5
A quicker and more concise way to write the
product is using index notation.
PRIME NUMBERS,
PRIME FACTORISATION
WORKED EXAMPLE 1:
Express 180 as a product of prime factors.
SOLUTION:
Method II (Using Repeated Division)
2
2
3
3
5
180
90
45
15
5
1
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.
Therefore, 180 = 2 x 2 x 3 x 3 x 5
= 22 x 32 x 5
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
35 is read as ‘3 to the power of 5’
base
35
index
In index notation, 3 is called the base and the number at the
top, 5 is called the index.
E.g. 2 x 2 x 2 x 5 x 5 = 23 x 52
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.
The common factors of 12 and 18 are 1, 2, 3 and 6.
The highest common factor (HCF) of 12 and 18 is 6.
2. Another method to find the HCF of two or more numbers
is by using prime factorisation which is the more
efficient way.
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:
225 =
32 x 5 2
Find the prime factorisation of each number first.
270 = 2 x 33 x 5
HCF =
32 x 5 = 45
To get the HCF, multiple the lowest power
of each common prime factor of the given
numbers.
Therefore, the HCF of 225 and 270 is 45.
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, ...
The common multiples of 8 and 12 are 24, 48, ...
The lowest common multiple (LCM) of 8 and 12 is 24.
2. Another method to find the LCM of two or more numbers
is by using prime factorisation which is the more
efficient way.
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:
24
= 23 x 3
90
= 2 x 32 x 5
LCM = 23 x 32 x 5 = 360
To get the LCM, multiple the highest
power of each set of common prime
factors. Also include any uncommon
factors
Therefore, the LCM of 24 and 90 is 360.
SQUARES AND SQUARE ROOTS
1. When a number is multiplied by itself, the product is
called the square of the number
E.g. 5 x 5 = 25
or
52 = 25
2. 5 is the positive square root of 25.
E.g. √25 = 5
3. The numbers whose square roots are whole numbers
are called perfect squares.
E.g. 1, 4, 9, 16, 25, ... are perfect squares.
Tip :
22 = 4
and
√4=2
32 = 9
and
√9=3
42 = 16
and
√16 = 4
SQUARES AND SQUARE ROOTS
WORKED EXAMPLE 1:
Using prime factorisation, find the square root of 5184.
SOLUTION:
5184 = 26 x 34
√5184 = √26 x 34
= 23 x 32
= 8x9
= 72
2
2
2
2
2
2
3
3
3
3
5184
2592
1296
648
324
162
81
27
9
3
1
CUBES AND CUBE ROOTS
1. When a number is multiplied by itself thrice, the product
is called the cube of the number
E.g. 5 x 5 x 5 = 125
or
53 = 125
2. 125 is the cube of 5 and 5 is the cube root of 125.
E.g. ∛125 = 5
3. The numbers whose cube roots are whole numbers are
called perfect cubes.
E.g. 1, 8, 27, 64, 125, ... are perfect cubes.
Tip :
23 = 8
and
∛8 =2
33 = 27
and
∛27 = 3
43 = 64
and
∛64 = 4
CUBES AND CUBE ROOTS
WORKED EXAMPLE 1:
Using prime factorisation, find the cube root of 1728.
SOLUTION:
1728 = 26 x 33
∛1728 = ∛26 x 33
= 22 x 3
= 4x3
= 12
2
2
2
2
2
2
3
3
3
1728
864
432
216
108
54
27
9
3
1
REAL NUMBERS
1. Numbers with the ‘negative sign’ (‘ - ’) are called
negative numbers.
E.g. -1, -2, -3, -4, -5, ...
2. Integers refer to whole numbers and negative numbers.
E.g. ..., -3, -2, -1, 0, 1, 2, 3, 4, ... are integers.
3. Positive integers are whole numbers that are greater
than zero.
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, ...
5. Zero is an integer that is neither positive nor negative.
REAL NUMBERS
6. A number line showing integers is shown below:
-5 -4 -3 -2 -1 0 1
Negative Integers
2 3 4
5
Positive Integers
7. The arrows on both ends of the number line show that
the line can be extended on both ends.
8. Every number on the number line is greater than any
number to its left.
-5 -4 -3 -2 -1 0 1
2 3 4
5
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
REAL NUMBERS
9. >, <, > and < are called inequality signs.
> means ‘is greater than’
< means ‘is smaller than’
> means ‘is greater than or equal to’
< means ‘ is smaller than or equal to’
10. 1, 2, 3, 4, 5, 6, 7, ... are called natural numbers. The
natural numbers are also called positive integers.
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.
E.g. |2| = 2, |0| = 0, |-2| = 2
ADDITION OF INTEGERS
Rules for adding two integers:
Sign of numbers
Both numbers have the
same signs
(+a) + (+b) = +(a + b)
(-a) + (-b) = -(a + b)
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
Method
1. Add the numbers while ignoring
their signs.
2. Write the sum using their
common sign.
E.g. (+3) + (+8) = +11 = 11
E.g. (-3) + (-8) = -(3 + 8) = -11
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 = 8
E.g. 5 + (-11) = -(11 – 5) = -6
E.g. -8 + 3 = -(8 – 3) = -5
E.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
(+a) x (+b)
(- a) x (- b)
(+a) x (- b)
(- a) x (+b)
= +(a x b)
= +(a x b)
= - (a x b)
= - (a x b)
Examples
3 x 4 = 12
(-5) x (-6) = +(5 x 6) = 30
8 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
(+a) ÷ (+b)
(- a) ÷ (- b)
(+a) ÷ (- b)
(- a) ÷ (+b)
= +(a ÷ b)
= +(a ÷ b)
= - (a ÷ b)
= - (a ÷ b)
Examples
16 ÷ 2 = 8
(-20) ÷ (-5) = +(20 ÷ 5) = 4
36 ÷ (-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
INTEGERS
1. Addition and multiplication of integers obey the
Commutative Law.
Commutative Law of Addition of Integers:
a+b=b+a
Commutative Law of Multiplication of Integers:
axb=bxa
E.g. 1 2 + (-10) = (-10) + 2 = -8
E.g. 2 2 x (-10) = (-10) x 2 = -20
2. Addition and multiplication of integers obey the
Associative Law.
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
E.g.2
[3 + (-5)] + 8 = 3 + [(-5) + 8] = 6
[3 x (-5)] x 8 = 3 x [(-5) x 8] = -120
RULES FOR OPERATING ON
INTEGERS
3. Multiplication of integers is distributive over
a) addition
b) subtraction
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)
E.g. 1 -2 x (-3 + 5) = -2 x (-3) = (-2) x 5 = -4
E.g. 2 -2 x (-8 + 6) = -2 x (-8) = (-2) x 6 = 28
4. The order of operation on integers is the same as those
for whole numbers
Order of operations
1.Simplify expressions within the brackets first.
2.Working from left to right, perform multiplication or
division before addition or subtraction.
RULES FOR OPERATING ON
INTEGERS
WORKED 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
INTEGERS
SOLUTION:
b)(-10) – (-6) + (-9) ÷ 3
= (-10) – (-6) + (-3)
= -10 + 6 – 3
= -7
c) {-15 – [15 + (-9)]2} ÷ (-3)
= [-15 – (15 – 9)2] (-3)
= (-15 – 62) ÷ (-3)
= (-15 – 36) ÷ (-3)
= (-51) ÷ (-3)
= 17
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
INTRODUCTION TO ALGEBRA
Using Letters to Represent Numbers
1. In algebra, we use letters (e.g. x, y, z, a, b, P, Q, …) to
represent numbers.
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.
2. A variable is a letter that is used to represent some
unknown numbers/quantity .
E.g. x, y, z, a, b, P, Q, … are variables
INTRODUCTION TO ALGEBRA
Using Letters to Represent Numbers
3. An algebraic expression is a collection of terms
connected by the signs ‘+’, ‘-‘, ‘x’, ‘÷’.
E.g. 3x + y, a2 – ab, 2x2 + 3x – 4.
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 – c
Note: d – c ≠ c – d
Multiple e by f
Product = e x f
=fxe
= 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
Add 3 to the product of p and q
The total cost of x books and y
magazines if each book cost $4
and each magazine costs $5.
Algebraic Expression
pq + 3
Cost of x books
= $4 x x
= $4x
Cost of y magazines
= $5 x y
= $5y
Total 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
index
base
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’)
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
b) 2a x 3b x a
= 6a2b
= 12xy ÷ 6z
= 12xy/6z
=
2xy/
z
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
b) Product of c and d
= 5a + 4b
= c x d = cd
Required expression
Cube of e = e x e x e = e3
= 5a + 4b – 3 (ans)
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
SOLUTION:
b) a(2b – c) – 3b2 = 4[2(2) – (-3)] – 3(2)2
= 4(4+3) – 3(4)
= 4(7) – 12
= 28 – 12
= 16
c)
a/
b
– (a+b)/ac
= 4/2 – (4+2)/4(-3)
= 2 – (6/-12)
=2+½
= 2½
ALGEBRAIC EXPRESSIONS
a) Find the total cost of m cups and n plates if each cup
cost $3 and each plate costs $4.
1 cup = $3
m cups = m x $3
= $3m
1 plate = $4
n plates = n x $4
= $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 EXPRESSIONS
c) 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 NUMBERS
1. 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.
A rational number can be written in the form a/b where a
and b are integers and b ≠ 0
E.g. -3/5, ½, 5/3, 12/3, … are rational numbers.
2. All integers are rational numbers since each integer, n
can be written as n/1.
E.g. 3 and -6 are rational nos. since 3 = 3/1
and -6 = -6/1
3. Most decimals can be expressed as rational numbers too.
E.g. 0.5 and 3.2 are rational nos. since 0.5 = 5/10
and 3.2 = 32/10
RATIONAL NUMBERS
Recall:
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.
32/
16/ = 31/
10 =
5
5
16/ is called an improper
5
fraction. The numerator is
greater than or equal to the denominator in an
improper fraction.
31/5 is called a mixed number. It represents the sum of
whole number and a proper fraction.
A proper fraction has its numerator smaller than its
denominator.
E.g. ½ , 3/7, and 5/9 are proper fractions.
ADDITION AND SUBTRACTION OF
RATIONAL NUMBERS
1. 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:
SOLUTION:
a) 61/6 – 23/4
b) (-51/4) + (-12/5) + (-½)
= (5 – 2) + (11/6 – ¾)
= - (51/4 + 12/5 + ½)
= 3 + (7/6 – ¾)
= - (55/20 + 48/20 + 10/20)
= 3 + (14/12 – 9/12)
= - (563/20)
= 3 + 5/12
= - (5 + 33/20)
= 35/12
= -83/20
MULTIPLICATION AND DIVISION OF
RATIONAL NUMBERS
1.
a)
b)
c)
d)
a/
b
x c/d = a x c/b x d , where a, b, c, d are integers and b ≠ 0, d ≠ 0
2.
a)
b)
c)
d)
a/
b
To multiply two rational numbers:
Convert all mixed numbers to improper fractions first.
Simplify the fractions first by crossing out the common
factors of the numerators and denominators.
Multiply the numerators, then the denominators.
Reduce answer to its simplest form.
To divide a rational number by another number :
Convert all mixed numbers to improper fractions first .
Invert the second fraction by interchanging its
numerator and denominator .
Multiply the numerators, then the denominators .
Reduce answer to its simplest form.
÷ 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:
SOLUTION:
a) (-21/2) x 31/5
c) (3/15 – 5/15) ÷ (-1/4 x 2/9)
b)
= - 5/2 x 16/5
= - 2/15 ÷ (- 1/18)
= -8
= - 2/15 x (- 18/1)
(-2/11)
=-
÷
2/
11
= 3/5
x
(-10/33)
(-33/10)
= 36/15
= 22/5
TECHNIQUE & STRATEGIES IN
SOLVING MATHEMATICS WORD
PROBLEM SUM
1/2
SHARING HOW TO GO ABOUT
TEACHING
SYNOPSIS
In 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.
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.
Through-out this seminar, I will share with you
the various types of strategies used to solve
word problems sum.
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.




Guess and Check
Making A Table
Make a List
Draw a Picture
 Find a Pattern
 Working Backwards
 Model Drawing
• A nswer check
Answer must be check to be able
to satisfy the condition of the
question.
Strategy 5 : Find a Pattern
Q1.
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.
STEP 2 : I find the patterns present in the data to complete the table.
No. of Children
1
2
3
4
5
Multiple of 5 and add 4
9
14
19
24
29
+5
Multiple of 6 and add 1
7
+5
13
+6
+5
19
+6
+5
25
+6
31
+6
From the table, I can see that number 19 satisfies both
conditions of the question.
Checking:
( 3 x 5 ) + 4 = 19
( 3 x 6 ) + 1 = 19 ------- Correct
Ans: There were 3 children sharing the chicken
wings. There were 19 chicken wings.
Strategy 5 : Find a Pattern
Q2.
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.
STEP 2 : I find the patterns present in the data to complete the table.
No. of Students
1
2
3
4
5
Multiple of 5 and add 4
9
14
19
24
29
+5
Multiple of 6 and add 1
7
+5
13
+7
+7
+5
19
+5
25
+7
+7
From the table, I can see that number 19 satisfies both
conditions of the question.
Checking:
( 3 x 5 ) + 4 = 19
( 3 x 6 ) + 1 = 19 ------- Correct
Ans: There were 3 students sharing the stamps.
There were 19 stamps.
31
Strategy 5 : Find a Pattern
Q3.
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.
STEP 2 : I find the patterns present in the data to complete the table.
Red (every 2 mins)
8.00
8.02 8.04
8.06
8.08
8.10 8.12
Yellow (every 3 mins)
8.00
8.03 8.06
8.09
8.12
8.15 8.18
Blue (every 4 mins)
8.00
8.04 8.08
8.12
8.16
8.20 8.24
Looking at the pattern above, the starting time is 8.00pm.
Next time the bulbs will flash together 8pm + 12mins
Ans: The next time all the bulbs will flash together is
at 8.12pm.
Strategy 6 : Working Backwards
Q1.
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.
Amount I
started with
?
Step 1
Withdrew $100
from bank
+ $100
Step 2
Spent $40 on a
pair of shoes
÷2
- $40
Next, I work backward by undoing ( +
order.
Step 3
Spent half of the
money on T-Shirt
-) (x
Amount
left
$65
÷ ) each step in reverse
Strategy 6 : Working Backwards
Q1.
con’t
Amount I
started with
?
Step 1
Withdrew $100
from bank
Step 2
Spent $40 on a
pair of shoes
$70
Undo Amount
Step 1 before Step 2
- $100
÷2
- $40
+ $100
Next, I work backward by undoing ( +
order.
Amount I
started with
Step 3
Spent half of the
money on T-Shirt
$170
-) (x
$65
÷ ) each step in reverse
Undo Amount
Step 2 before Step 3
+ $40
Amount
left
$130
Ans: I had $70 before I went to the bank to withdraw
the money.
Undo
Step 3
Amount left
x2
$65
Strategy 6 : Working Backwards
Q2.
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 ?
+ 1.2ℓ
1.3ℓ
÷2
2.5ℓ
Amount left in
the jug
Ans: John brought 0.55 ℓ of drink.
- 0.7ℓ
1.25ℓ
0.55ℓ
Strategy 6 : Working Backwards
Q3.
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?
x3
14
+ 24
42
Amount left in
the jar
Ans: She baked 110 chocolate cookies.
+ 44
66
110
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 : 2
Q1.
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?
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.
1 Unit
Brother
$150
Jim
$50
Model Drawing : 2
Con’t
Q1.
Brother
1 Unit
$50
$150
Jim
$50
$150 - $50 = $100
$100 ÷ 2 = $50
1 Unit (
) / Brother
Step 4 : After finding your 1 unit, look back at the MODEL
DRAWING. The one that has 1 unit is Brother.
Therefore, the brother get $50
Step 5 : As for Jim, look at the model drawing it consist
of 1 unit + $50
$50 + $50 = $100
Ans: Therefore, Jim gets $100 and his brother gets
$50.
Model Drawing : 2
Q2.
Peter and David share a total of 300 sweets.
If David gets 60 more than Peter.
How many sweets do Peter and David get?
Step 1 : Identified KEY PERSON that has only 1 UNIT (
CAN YOU FIGURE IT OUT?
Step 2 : DRAW THE MODEL
1 Unit
Peter
120
300
David
300 – 60 = 240
240 ÷ 2 = 120
120 + 60 = 180
60
1 unit (
David
) / Peter
Ans: Peter gets 120 sweets and David gets 180
sweets.
)
Hand-On : Exercise 2
Q1.
$400 is to be shared between Susan and Mary.
If Susan gets $20 more than Mary.
How much money do Susan and Mary get?
1 Unit
Mary
$190
$400
Susan
$400 – $20 = $380
$380 ÷ 2 = $190
$20
1 unit (
) / Mary
Step 4 : After finding your 1 unit, look back at the MODEL
DRAWING. The one that has 1 unit is Mary.
Therefore, Mary gets $190
Step 5 : As for Susan, look at the model drawing it
consist of 1 unit + $20
$190 + $20 = $210
Susan
Ans: Therefore, Mary gets $190 and Susan gets
$210.
Hand-On : Exercise 2
Q2.
Amy and John have 240 stickers altogether. If Amy has 80 stickers
more than John. How many stickers does John have?
1 Unit
John
80
240
Amy
240 – 80 = 160
160 ÷ 2 = 80
80
1 unit (
) / John
Step 4 : After finding your 1 unit, look back at the MODEL
DRAWING. The one that has 1 unit is John.
Therefore, John gets 80
Step 5 : As for Amy, look at the model drawing it consist
of 1 unit + 80
80 + 80 = 160
Amy
Ans: Therefore, John gets 80 and Amy gets 160.
Model Drawing : 3
Q1.
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?
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.
1 Unit
Sharon
105
180
Janet
30
180 + 30 = 210
210 ÷ 2 = 105
1 unit (
105 - 30 = 75
Janet
) / Sharon
Ans: Therefore, Sharon gets 105 and Janet gets 75.
Hand-On : Exercise 3
Q1.
$150 is shared between Judy and Susan.
If Susan gets $30 less than Judy.
How much money do Judy and Susan get?
1 Unit
Judy
$90
$150
Susan
$30
$150 + $30 = $180
$180 ÷ 2 = $90
1 unit (
$90 - $30 = $60
Susan
) / Judy
Ans: Therefore, Judy gets $90 and Susan gets $60.
Hand-On : Exercise 3
Q2.
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?
1 Unit
David
200
360
John
40
360 + 40 = 400
400 ÷ 2 = 200
1 unit (
200 - 40 = 160
John
) / David
Ans: Therefore, David gets 200 and John gets 160.
Model Drawing : 4
Q1.
$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?
1 Unit
Jason
$100
$300
Kevin
$300 ÷ 3 = $100
1 unit (
$100 x 2 = $200
Kevin
) / Jason
Ans: Jason and Kevin each get $100 & $200
respectively.
Model Drawing : 4
Q2.
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?
1 Unit
Joseph
$50
$250
Ken
$250 ÷ 5 = $50
1 unit (
$50 x 4 = $200
Ken
) / Joseph
Ans: Joseph and Ken each get $50 & $200
respectively.
Model Drawing : 4
Q3.
Aaron has 4 times as many stamps as Jimmy. If he has 24 stamps
more than Jimmy. How many stamps does Aaron have?
Step 1 : Who is the KEY PERSON?
Step 2 : Draw the model drawing of that person first.
1 Unit
Jimmy
8
24
Aaron
24 ÷ 3 = 8
1 unit (
8 x 4 = 32
Aaron
) / Jimmy
Ans: Jimmy and Aaron each get 8 & 32 stamps
respectively.
Model Drawing : 4
Q4.
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
25
50
Sarah
50 ÷ 2 = 25
1 unit (
25 x 3 = 75
Sarah
) / Jenny
Ans: Jenny and Sarah each get 25 & 75 stickers
respectively.