surviving math! 3
TRANSCRIPT
Surviving Math! 3 presented by Gold 90.5 FM
Dr Yeap Ban Har
Marshall Cavendish Institute
Singapore
Slides are available at www.facebook.com/MCISingapore
Da Qiao Primary School, Singapore
Mathematics is “an excellent vehicle for
the development and improvement of a
person’s intellectual competence”.
Ministry of Education, Singapore (2006)
thinkingschools learningnation
Ministry of Education, Singapore (1991, 2000, 2006, 2012)
Reflection of the Shifts in the Test Questions
When we compare the tests from the past with the present, we see that:
• Questions from previous tests were simpler, one or two steps, or were heavily scaffolded. The new questions will requires multiple steps involving the interpretation of operations.
• Questions from the past were heavy on pure fluency in isolation. The new questions require conceptual understanding and fluency in order to complete test questions.
• Questions from past tests isolated the math. The new problems are in a real world problem context.
• Questions of old relied more on the rote use of a standard algorithm for finding answers to problems. The new questions require students to do things like decompose numbers and/or shapes, apply properties of numbers, and with the information given in the problem reach an answer. Relying solely on algorithms will not be sufficient.
Department of Education
New York State (2013)
5
What is Happening Around
the World?
Bringing Up Children Who Are Ready for the Global, Technological World
Da Qiao Primary School, Singapore
17 – 3 = 17 – 8 =
Primary 1 Singapore
7
Da Qiao Primary School, Singapore
Primary 1 Singapore
8
Learning by Doing Learning by Interacting Learning by Exploring
Gardner’s Theory of Intelligences Bruner’s Theory of Representations Dienes’ Theory of Learning Stages
St Edward School, Florida
Grade 2 USA
9
St Edward School, Florida
Grade 2 USA
10
Archipelschool De Tweemaster – Kameleon, The Netherlands
Grade 5 The Netherlands
11
Archipelschool De Tweemaster – Kameleon, The Netherlands
Grade 5 The Netherlands
12
Archipelschool De Tweemaster – Kameleon, The Netherlands
Grade 5 The Netherlands
13
mathematics
14
Students in Advanced Benchmark can
• apply their understanding and
knowledge in a variety of relatively
complex situations
• and explain their reasoning.
King Solomon Academy, London
Year 7 England
15
16
The sum of the two numbers is 88.
The greater number is 6 x (88 11) = 48.
The other number is 5 x (88 11) = 40.
High School Attached to Tsukuba University, Japan
Draw a polygon with no dots inside it. Investigate.
A polygon has 4 dots on the perimeter. Find an expression for its area.
Grade 9 Japan
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What Do Children Learn in
School Mathematics? And How You Can Coach Them
Students who have mastered the basic skills which include basic one-step and two-step problems are ready to handle at least the least demanding of the secondary courses.
Jay
Sam
34.7 kg
34.7 kg x 2 = (68 + 1.4) kg
34.7 kg x 2 = 69.4 kg Sam’s mass is 69.4 kg.
19
4. Find the value of 1000 – 724 . 5. Find the value of 12.2 4 .
20
999 – 724 = 275
1000 – 724 = 276
12.20 4 = 3.05
12.2 4 =
12 2 tenths = 20 hundredths
12.2 4
12
2 0 2 0
0
3.05
What Are the Challenging
Aspects of Mathematics? And How Children Develop Competencies to Handle Them
Problem 1
Cup cakes are sold at 40 cents each.
What is the greatest number of cup cakes that
can be bought with $95?
Answer:_____________
22
$95 40 cents = 237.5
237
Problem 1
Cup cakes are sold at 40 cents each.
What is the greatest number of cup cakes that
can be bought with $95?
Answer:_____________
23
237
Problem 2
Mr Tan rented a car for 3 days. He was
charged $155 per day and 60 cents for
every km that he travelled. He paid
$767.40. What was the total distance that
he travelled for the 3 days?
24
$155 x 3 = $465
$767.40 - $465 = $302.40
$302.40 60 cents / km = 504 km
He travelled 504 km.
Problem 2
Mr Tan rented a car for 3 days. He was
charged $155 per day and 60 cents for
every km that he travelled. He paid
$767.40. What was the total distance that
he travelled for the 3 days?
25
(767.40 - 155 x 3) 0.60 = 504
He travelled 504 km.
26
(25 + 2) 3 = 9
9 + 1 = 10
10 x 8 + 25 = 105
105
27
11m + 6 = 8(m + 1) + 25
3m = 27
m = 9
105 11m + 6 = 99 + 6 = 105
28
Number of Girls 11 sweets 6 sweets
105
2 11 + 6 12 + 25
3 22 + 6 18 + 25
4 33 + 6 24 + 25
29
men
women
There were 4 x 30 = 120 men and women at first.
After
30
2 fifths of the remainder were 38
3 fifths of the remainder were 19 x 3 = …
There were 19 x 5 pears and peaches.
1 −1
4−
1
3= …
5 twelfths of the fruits = 19 x 5 fruits
So, there were 19 x 12 fruits altogether.
Answer: 228 fruits
1
4
1
3
38
2 units = 38 5 units = 19 x 5 = 95
1
4+
1
3=
7
12
5
12 → 95
12
12 → 95 ÷ 5 12 = 19 12 = 190 + 38 = 228
There were 228 fruits altogether.
31
32
0 + 1 + 2 + 3 = 2 x 3 = 6
6 x $3 = $18
$100 - $18 = $82
$82 : 4 = $20.50
$20.50 + $9 = $29.50
Problem 7 Mr Lim packed 387 apples.
Each apple had a mass of about 24g.
He put them into three different baskets.
The mass of the apples in Basket A was 3 times that of the apples in Basket C.
The mass of the apples in Basket B is twice that of the apples in Basket C.
The mass of the empty Basket C was 140g.
What was the total mass of Basket C and the apples in it?
Source: Sent by a Parent to Gold 90.5FM
33
C
A
B
The total mass of the apples is about 387 x 24g = 9 288g
6 units = 9 288 g
1 units = 1 548 g
Basket C : 140 g + 1 548 g = 1 688 g = 1.688 kg
m
m
m
3
Problem 8
34
Problem 9 John had 1.5 m of
copper wire. He cut
some of the wire to
bend into the shape
shown in the figure
below. In the figure,
there are 6
equilateral triangles
and the length of XY
is 19 cm. How much
of the copper wire
was left? 5 x 19 cm = 95 cm
150 cm – 95 cm = …
35
a
b
c
Problem 10
36
1
4
2
9
(a) 41 is under M (b) 101 is under S (c) 2011 is under T …. Really? How do you know?
Problem 11
Problem 12
Weiyang started a savings plan by putting 2 coins in a money box every day. Each coin was either a 20-cent or 50-cent coin. His mother also puts in a $1 coin in the box every 7 days. The total value of the coins after 182 days was $133.90.
(a) How many coins were there altogether?
(b) How many of the coins were 50-cent coins?
182 7 = …
2 x 182 + 26 = …
$133.90 - $26 = $107.90
There were 50-cent coins.
50-cent 20-cent
Suppose each day he put in one 20-cent and one
50-cent coins, the total is $127.40
But he only put in $107.90 ..
to reduce this by $19.50, exchange 50-cent for
20-cent coins
$19.50 $0.30 = 65
There were 182 – 65 = 117 fifty-cent coins.
• Number Sense
• Patterns
• Visualization
• Communication
• Metacognition
Five Core Competencies
Try to do as you read the problems. Do not wait till the end of the question to try to do something.
Try to draw when you do not get what the question is getting at. Diagrams such as models are very useful.
Do more mental computation when practising Paper 1.
Surviving Math! 3 presented by Gold 90.5 FM
Dr Yeap Ban Har
Marshall Cavendish Institute
Singapore
Slides are available at www.facebook.com/MCISingapore
Da Qiao Primary School, Singapore
Surviving Math! 3 presented by Gold 90.5 FM
Dr Yeap Ban Har
Marshall Cavendish Institute
Singapore
Some Singapore data on how Primary 4 students performed on
some mathematics problems. This was the data for TIMSS2011.
Da Qiao Primary School, Singapore
Trends in International Mathematics
and Science Study TIMSS
