Embed Size (px)
Citation preview
“Strengthening the Mastery and
Understanding of
P5 and P6 Mathematics”
Parents Workshop (Mathematics)
Overview
• Presentation of Mathematical Solutions
• Addressing Common Misconceptions
• Solving Non-routine Word Problems
Presentation of Mathematical Solutions
Presentation of Mathematical Solutions
Common errors in mathematical statement
• Use of approximate sign (≈)
• Equations involving order of operations and remainder
• Use of ‘→’ and ‘=’
• Use of the percentage sign (%)
Use of approximate sign ≈Find the value of 850 ÷ 8. Express your answer to one decimal place.
Presentation 1 Presentation 3
850 ÷ 8 ≈ 106.3 850 ÷ 8 = 106.25
= 106.3 (1 decimal place)
Presentation 2 Presentation 4
850 ÷ 8 = 106.25 850 ÷ 8 = 106.3
≈ 106.3
Calculator
Equations involving order of operations
Find the value of 7 x 9 – 4 + 12
Presentation 1
7 x 9 – 4 + 12 = 63 – 4 + 12
= 59 + 12
= 71
Presentation 2
7 x 9 – 4 + 12 = 63
= 63 – 4 + 12
Presentation 37 x 9 – 4 + 12 = 63 – 4
= 59 + 12= 71
Equations involving remainder
Mrs Phua bought 50 m of cloth to make dresses. She used 3 m of cloth to make each dress. How many such dresses can she make?
Presentation 150 ÷3 = 16 R 2
She can make 16 dresses.
Presentation 2
50 ÷3 = 16
She can make 16 dresses.
Use of ‘→’ and ‘=’
• ‘’ means ‘represents’ while ‘=‘ means ‘equal to’
Acceptable Unacceptable
Scenario 1
23 60
23
= 60
23
of the students = 60 girls
Scenario 2
4 units = 60 4 units 60 (Not recommended)
Scenario 3
5y = 10 5y 10 (Not recommended for algebra)
Some examples of the acceptable and unacceptable use of ‘’ and ‘=’
Use of ‘→’ and ‘=’
Mrs Tay had $360. She spent 30% of the money on food. How much money had she left?
Presentation 1
100 % 360
1% 360 ÷ 100
= 3.60
30% 3.60 x 30
= 108
Presentation 2
100 % = 360
1% = 360 ÷100 = 3.60
30% = 3.60 x 30 = 108
Note:
We use the equal sign when
the terms are equivalent.
However, note that 100% = 1
and hence we cannot say that
100% = 360
It is also not advisable for
students to keep using .
They need to learn to use the
equal sign and arrow in an
appropriate manner.
Use of the percentage sign (%)
12 out of 25 children are boys. What percentage of the children are boys?
Presentation 1
1225 x 100% = 48%
Note:1225
x 100 = 48 not 48%
Presentation 21225
x 100 = 48%
Addressing Common Misconceptions
Addressing Common Misconceptions
Fractions • Fraction word problems• Finding the remainder in fractions
Decimals
• Finding the remainder from a string of decimals
Area of Triangle
• Finding the height of a triangle
Fraction Word Problems
Question 1:
A jug contained 78
ℓ of water. Sam poured 25
ℓ of water from the jug into
a mug. How much water was left in the jug?
Question 2:
A jug contained 78
ℓ of water. Sam poured 25
of the water from the jug
into a mug. How much water was left in the jug?
What is different about
Question 1 and Question 2?
Fraction Word Problems
Question 1:
A jug contained 78
ℓ of water. Sam poured 25
ℓ of water from the jug into
a mug. How much water was left in the jug?
What does 25
ℓ of water represent?
25
ℓ of water refers to a specific quantity of water and can be represented
as 400 ml of water
Hence, we will need to subtract25
ℓ of water from78
ℓ of water.
78
ℓ ─25
ℓ =1940
ℓ
Ans:1940
ℓ
Fraction Word Problems Question 2:
A jug contained 78
ℓ of water. Sam poured 25
of the water from the jug
into a mug. How much water was left in the jug?
25
of the water in this question means 25
of 78
ℓ because the jug contained78
ℓ of water at first.
25
x 78
ℓ = 720
ℓ
78
ℓ ─7
20ℓ =
2140
ℓ
Ans:2140
ℓ
A jug contains 910
ℓ of water. Raymond poured 18
ℓ of the drink into each
identical glass .
(a) How many glasses can be filled completely with the water from the jug?
(b) How much drink did he have left?
Fraction: Finding the remainder
(a) 9
10÷
18
= 9
10x 8
=7 15
Ans: 7 glasses
A jug contains 9
10ℓ of water. Raymond poured
18
ℓ of the drink into each
identical glass .
(a) How many glasses can be filled completely with the water from the jug?
(b) How much drink did he have left?
Fraction: Finding the remainder
(b) 15
x 18
= 1
40
Ans: 1
40ℓ
Note
The amount of drink left is not 15
ℓ.
It is 15
of 18
ℓ .
Alternative Method for (b) which is slightly longer
Capacity of 7 glasses = 7 x 18
= 78
Drink left = 9
10─
78
= 1
40Ans:
140
ℓ
Decimals: Finding the remainderQuestion:
Siti had a rectangular piece of paper, 35 cm by 24.6 cm. She cut out as many squares as possible from the paper. The side of each square was 5 cm. What area of the paper was left?
35 ÷ 5 = 724.6 ÷ 5 = 4 R 4.6 cm35 x 4.6 = 161
Ans: 161 cm2
Common mistake 24.6 ÷ 5 = 4 R 0.92 cm 35 x 0.92 = 32.2
Ans: 32.2 cm2
Using the calculator, theanswer shows 4.92.To get the remainder,multiply 0.92 by 5:0.92 x 5 = 4.6
Area of Triangle
Question 1: Find the area of the shaded triangle.
1
2x 14 x 16 = 112
Ans: 112 m2
Common mistake 1
2x 14 x 20 = 140
(Identified the wrong height, height must be perpendicular to the base)
Area of triangle = 1
2x base x height
10 m
Area of Triangle
Question: Find the area of the shaded triangle.
1
2x 14 x 16 = 112
Ans: 112 m2
Common mistake 1
2x 14 x 10 = 70
(Identified the wrong height, height must be from the base to its opposite vertex)
10 m
opposite vertex
Area of triangle = 1
2x base x height
Area of Triangle
Question : Find the area of the shaded triangle.
1
2x 14 x 16 = 112
Ans: 112 m2
Common mistake 1
2x 26 x 16 = 208
(Base must be the side of the triangle)
10 m
Area of triangle = 1
2x base x height
Solving Non-RoutineWord Problems
Polya’s 4 Steps for Problem Solving
� Read
� Identify Key Information and Goal(s)
Choose a strategy
� Draw a Diagram/Model
� Act it Out
� Look for a Pattern
� Guess and Check
� Make a Table
� Work Backwards
3 Do 4 Check
1 Understand 2 Plan
O Carry out the Plan
O Solve the Problem
O Does my Solution Make Sense?
O Is my Solution Reasonable?
O Can I Solve it Another Way?
Overview
• Solving Non-Routine Word Problems
- Restate the problem
- Identify Patterns and Relationships
- Missing Scale in Bar Graph
2019 PSLE Paper 2 Q16(a) (2 marks)Topic: Area and Volume
The figure is formed by 5 identical semicircles. What is the diameter of each semicircle?
The figure is formed by 5 identical semicircles. What is the diameter of each semicircle?
22 + 16 + 22 = 6012 + 12 = 2460 – 24 = 36
Ans: 36 cm
Method 1 Heuristic: Restate the problem by moving the semicircles to the left.
12 cm12 cm
The figure is formed by 5 identical semicircles. What is the diameter of each semicircle?
The two lengths highlighted in red are equal.22 – 12 = 10 cmThe two lengths highlighted in blue are equal.22 – 12 = 10 cmDiameter = 10 + 16 + 10
= 36 cmAns: 36 cm
Method 2 : Identify Patterns and Relationships
10 cm 10 cm
Topic: Area and Volume
The figure is made up of 2 squares, P and Q and a rectangle R.
The area of the shaded part is 1
2of the area of the figure.
Find the length of x.
3.5 cm
QP
2 cm 5 cm
x
R
The figure is made up of 2 squares, P and Q and a rectangle R.
The area of the shaded part is 1
2of the area of the figure.
Find the length of x.
3.5 cm
QP
2 cm 5 cm
x
R
Heuristic: Restate this figure in another way by adding 2 more rectangles
3.5 cm
QP
2 cm 5 cm
x
R
The area of the shaded part and the area of the unshaded part are the same.
Length of rectangle A = 5 – 2
= 3 cm
Area of Rectangle A = 2 x 3
= 6 cm2
Area of Rectangle B = 6 cm2
Area of red triangle = Area of blue triangle
Area of the shaded part = Area of the unshaded partArea of rectangle A = Area of rectangle B
3.5 cm
QP
2 cm 5 cm
x
R
Rectangle B
Rectangle A
Breadth of Rectangle B = 5 – 3.5
= 1.5
6 ÷ 1.5 = 4
Ans: 4 cm
2019 PSLE Paper 2 Q16 (Looking For a Pattern) The first 4 figures of a pattern are shown below.
Figure Number 1 2 3 4 5
Number of white triangles 1 1 6 6
Number of grey triangles 0 3 3 10
The table shows the number of white and grey triangles used for each figure.
(a) Fill in the table for Figure 5.
(b) What is the total number of white and grey triangles in Figure 250?
(c) In Figure 250, what percentage of the triangles are grey?
Figure Number 1 2 3 4 5
Number of White Triangles 1 1 6 6 15
Number of grey Triangles 0 3 3 10 10
The table shows the number of white and grey triangles used for each figure.
(a) Fill in the table for Figure 5.
Add another row of 9 white triangles
below.
Figure Number 1 2 3 4 5
Number of white triangles 1 1 6 6 15
Number of grey triangles 0 3 3 10 10
Total number of white and grey triangles 1 4 9 16 25
(b) What is the total number of white and grey triangles in Figure 250?
Figure 1 : 1 x 1 = 1
Figure 2 : 2 x 2 = 4
Figure 3 : 3 x 3 = 9
Figure 4 : 4 x 4 = 16
Figure 250 : 250 x 250 = 62 500
Ans: 62 500
Figure Number 1 2 3 4 5 6 7 9
Number of white triangles 1 1 6 6 15 15 28 28
Number of grey triangles 0 3 3 10 10 21 21 36
(c) In Figure 250, what percentage of the triangles are grey?
Method 1
Number of grey triangles
Figure 2: 1 x 3 = 3
Figure 4: 2 x 5 = 10
Figure 6: 3 x 7 = 21
Figure 8: 4 x 9 = 36
Figure 250: 125 x 251 = 31 375
+ 5 + 9+ 7 + 11
+ 13
+ 15
Figure Number ÷ 2 Figure Number + 1
31 375
62 500x 100 % = 50.2 %
Ans: 50.2 %
(c) In Figure 250, what percentage of the triangles are grey?
Method 2
Difference in Figure 250 = 250
62 500 + 250 = 62 750
62 750 ÷ 2 = 31 375 (Number of grey triangles)
Figure Number
31 375
62 500x 100 % = 50.2 %
Ans: 50.2 %
Figure Number 1 2 3 4 5
Number of white triangles 1 1 6 6 15
Number of grey triangles 0 3 3 10 10
Total number of white and grey triangles 1 4 9 16 25
Difference between the number of white
triangles and the number of grey triangles
1 2 3 4 5
Grey
White250
62 500
2016 PSLE Paper 2 Q18 (Looking For a Pattern)
Jasmin uses triangles and circles to form figures that follow a pattern as shown below.
Figure 1 Figure 2 Figure 3 Figure 4
(a) The table shows the number of triangles and circles for the first four figures. Complete the table for Figure 5.
(b) A figure has a total of 100 circles and triangles. What is the Figure Number?
Figure Number 1 2 3 4 5
Number of triangles 2 5 9 14
Number of circles 2 4 7 11
Total number of triangles and
circles
4 9 16 25
(b) A figure has a total of 100 circles and triangles.
What is the Figure Number?
Figure 1: 2 x 2 = 4Figure 2: 3 x 3 = 9Figure 3: 4 x 4 = 16Figure 4: 5 x 5 = 25
10 x 10 = 10010 – 1 = 9Ans: 9
20
16
36
Figure Number +1
(a)
PSLE 2015 Paper 2 Q18Ishak uses rods to form figures that follow a pattern. The first four figures are shown below.
(a) The table below shows the number of rods used for each figure.Complete the table for Figure 5 and Figure 6.
(b) What is the difference in the number of rods Ishak would use for Figure 9 and Figure 11?(c) How many rods would he use for Figure 30?
Figure Number Number of rods used
1 10
2 15
3 18
4 23
5
6
PSLE 2015 Paper 2 Q18Ishak uses rods to form figures that follow a pattern. The first four figures are shown below.
(a) The table below shows the number of rods used for each figure.Complete the table for Figure 5 and Figure 6.
Figure Number Number of rods used
1 10
2 15
3 18
4 23
5
6
+ 5
+ 3
+ 5
+ 3
+ 526
31
50 – 42 = 8The difference is 8 rods.
(b) What is the difference in the number of rods Ishak would use for
Figure 9 and Figure 11?
Figure Number Number of rods used
1 10
2 15
3 18
4 23
5 26
6 31
7 34
8 39
9 42
10 47
11 50
Figure Number Number of rods used
1 10
2 15
3 18
4 23
5
6
26
31
+ 8
+ 8
(c) How many rods would he use for Figure 30?
(c) How many rods would he use for Figure 30?
Figure 2Figure 4 Figure 6
15
15 8
15 8 8
As the figure number increases by 2, the number of rods used increases by 8.
Number of groups of 8 rods for Figure 30= (30 – 2) ÷2 = 14
Number of rods in Figure 30= 14 x 8 + 15= 127
127 rods will be used for Figure 30.
In order to solve for greater figure numbers, we “restate the problem” to show a common difference of 8 rods between every 2 figures.
2019 PSLE Paper 2 Q7Topic: Bar GraphsThe bar graph shows the number of books read by Class 6A from January to April. The number of books read is not shown on the scale.
(a) What was the percentage increase in the number of books read from January to February?
(b) The average number of books read in a month from January to April was 75. How many books did Class 6A read in April?
Data is missing
(a) What was the percentage increase in the number of books read from January to February?
Let each part be 1 unit.
January = 10 unitsFebruary = 20 units
Increase = 20 – 10 = 10
10
10x 100 % = 100 %
Ans: 100 %
(b) The average number of books read in a month from January to April was 75. How many books did Class 6A read in April?
Let each part be 1 unit.
Total units = 10 + 20 + 8 + 22= 60
Total = 75 x 4 = 300
60 units = 300
1 unit = 300 ÷60 = 5
22 units = 22 x 5= 110
Ans: 110
Topic: Area Figure ABCD is made up of a rectangle ABCE and a triangle ECD.
AD is a straight line. The length of AE is twice the length of ED. The area of the shaded part is 172 cm2. Find the area of the figure ABCD.
E
A B
C
D
172 cm2
Lengths are not given.What is the relationship between rectangle ABCE and the shaded triangle?- Rectangle ABCE is the related rectangle of the shaded triangle.Reasons: 1) Base of the triangle = Breadth of the rectangle2) Height of the triangle = Length of the rectangle
Area of rectangle ABCE = 2 x area of shaded triangle= 2 x 172= 344
E
A B
C
D
172 cm2
2u 1uThe length of AE is twice the length of ED. Compare triangle ECD and the shaded triangle.
- Both triangles have the same base.
- Height of triangle ECD is 1
2of the height of the shaded triangle.
- Area of triangle ECD = 1
2x area of shaded triangle.
Area of triangle ECD = 1
2x 172
= 86
86 + 344 = 430
Ans: 430 cm2
1u
2u
