Surviving Math! 3 presented by Gold 90.5 FM

Dr Yeap Ban Har

Marshall Cavendish Institute

Singapore

banhar@sg.marshallcavendish.com

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

17

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

banhar@sg.marshallcavendish.com

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