Rivervale Primary School Maths Workshop for P4 to P6 …rivervalepri.moe.edu.sg/qql/slot/u143/Communications/Rivervale... · Maths Workshop for P4 to P6 Parents ... Rivervale Primary

Maths Workshop for P4 to P6 Parents

15 April 2016, Friday3.00 – 4.30 pm

Rivervale Primary School

P4 – P6

Maths Assessment &

Exam Paper Format

Assessments:Formative (Assessment for Learning, AfL):

• Gather feedback about how pupils have learnt, e.g. multiplication of fractions

• Teacher to reinforce concepts, skills to close any learning gap / stretch pupils’ learning

• Workbooks, worksheets, supplementary books

The Rivervale – School of Excellence, Individuals of Character

Assessments:Summative (Assessment of Learning, AoL):

• Measure pupil achievement at a particular point in time

• Covers a range of topics, concepts and skills

• Used for banding, promotion, awards

• Exams


2016 Assessments:P4, P5:

• 1 Semestral Continual Assessment (Term 3)

• 2 Semestral Assessments (Terms 2, 4)

P6:

• 1 Semestral Continual Assessment (Term 1)

• 1 Semestral Assessment (Term 2)

• 1 Term Test (Term 3)

• 1 Preliminary Exam (Term 3)


Achievement Bands, Grades / Mark Range:

P5, P6:


Band Mark Range

1 85-100

2 70-84

3 50-69

4 Below 50

P4:

Grade Mark Range

A* 91-100

A 75-90

B 60-74

C 50-59

D 35-49

E 20-34

U (Ungraded) Below 20

Exam Formats:

P4: 1 h 45 min, 100 marks

Section A (MCQ): 15 2-marks

Section B (SAQ): 15 2-marks

Section C (LAQ): 10 3/4/5-marks


Exam Formats:P5, P6:

Paper 1: 50 min paper, 40 marks (calculator not allowed)

Booklet A (MCQ): 10 1-mark and 5 2-marks

Booklet B (SAQ): 10 1-mark and 5 2-marks

Paper 2: 1 h 40 min paper, 60 marks(calculator allowed)

SAQ: 5 2-marks

LAQ: 13 3/4/5-marks


List of approved calculators for PSLE:

http://www.seab.gov.sg/content/calcula

tor/GuidelinesCalculators.pdf

http://www.seab.gov.sg/content/calculator/GuidelinesCalculators.pdf

Assessments:P4 Subject-based Banding:

• Choice to take a combination of standard or foundation subjects

• Recognise pupils’ different learning abilities and provide greater flexibility to concentrate on the subjects they are good at

• Details: https://www.moe.gov.sg/docs/default-source/document/education/primary/files/subject-based-banding2015.pdf


Assessments:P6 PSLE:

• T-score

• Aggregate score

• Performance is relative


http://nurs8004.wikispaces.com/file/view/no

rmal-distribution-

curve.jpg/365012298/normal-distribution-

curve.jpg

P4 & P5 HeuristicsGuess-and-check

Supposition Method

Model Drawing (Before and After)

Gaps and Differences (Listing and Model drawing)

Grouping v.s Number × value


Overview of Heuristics from P2 – P5

P2 P3 P4 P5

Guess & Check

Supposition Mtd

(drawing)

Guess & Check

Supposition Mtd

Guess & Check

Revision

Supposition Mtd

Guess & Check Revision

Supposition Mtd Revision

Units x Values

Pattern Pattern Pattern Pattern (grouping)

Working

Backwards

Working

Backwards

Working Backwards

(Model drawing)

Working Backwards

(Model drawing & listing)

Equal Gaps Equal Gaps revision Equal Gaps extension

Listing Method

(integrated into

Fractions)

Listing Method (in

gaps & diff qn)

Listing Method & Model

Drawing (in gaps & diff

qn)

Guess-and-checkA farmer has 15 chickens and rabbits.

These animals have 40 feet altogether.

How many of each type of animals does the farmer have?

No. of rabbits

(4 feet)

No. of chickens

(2 feet)FIXED

C + R = 15

TARGET

Total feet = 40

Check

√ / x

7

7 x 4 = 28

8

8 x 2 = 16

7 + 8

= 15

28 + 16

= 44 x

When doing guess and check/trial and error method,it’s encouraged to guess the number of each animal

by dividing the total by 2 first. In this case, 15 ÷ 2 = 7 R 1

So 1 type of animal will be 7 while the other is 8.

From the 1st guess, we need to look at the number of feet to see if we should decrease

the number of rabbits or chickens. Since rabbits have more feet, we should decrease

the number of rabbits.

6

6 x 4 = 24

9

9 x 2 = 18

6 + 9

= 15

24 + 18

= 42 x

5

5 x 4 = 20

10

10 x 2 = 20

5 + 10

= 15

20 + 20

= 40 √

Ans: 5 rabbits and 10 chickens

Guess-and-checkGrace bought 20 blouses and skirts. There were 5 buttons on each

blouse and 2 buttons on each skirt. She counted a total of 64 buttons.

How many of each type of clothes did she buy?No. of blouses

(5 buttons)

No. of skirts

(2 buttons)FIXED

B + S = 20

TARGET

Total buttons = 64

Check

√ / x

10

10 x 5 = 50

10

10 x 2 = 20

10 + 10

= 20

50 + 20

= 70 x

9

9 x 5 = 45

11

11 x 2 = 22

9 + 11

= 20

45 + 22

= 67 x

8

8 x 5 = 40

12

12 x 2 = 24

8 + 12

= 20

40 + 24

= 64 √

Ans: 8 blouses and 12 skirts

Supposition MethodA farmer has 15 chickens and rabbits.

These animals have 40 feet altogether.

How many of each type of animals does the farmer have?

Determine which animal has fewer legs.

Suppose: 15 chickens

no. of feet (chickens) =___ ___= _____15 x 230

excess feet =____ ____= ____

40 – 301010

A rabbit has 2 more feet than a chicken.

no. of rabbits=___ ___= _____

÷ 25

10

no. of chickens=___ ___= ____

15 – 510

There are ___ rabbits and ______ chickens.

510

Supposition MethodGrace bought 20 blouses and skirts. There were 5 buttons on each

blouse and 2 buttons on each skirt. She counted a total of 64 buttons.

How many of each type of clothes did she buy?

Determine which piece of clothing has lesser buttons.

Suppose: 20 skirts

no. of buttons =___ ___= ___

20 x 240

excess buttons=____ ____= ____

64 – 402424

A blouse has 3 more buttons than a skirt.

no. of skirts =___ ___= _____

÷ 38

24

no. of blouses =___ ___= ____

20 – 812

There are ___ skirts and ______ blouses.

812

Before and after (equal stage)Daniel and Patrick had an equal number of stickers at first. Daniel then gave away 20

of his stickers to his friend Melvin, and Patrick bought another 12 stickers. In the end,

Patrick had thrice as many stickers as Daniel. Find the number of stickers Daniel had

at first.

3 units

1 unit

Summary

Before: D

1u

P

1u

–20

After: D

1u

P

3u

+12

D

Before

P

After

D 20

P 121u

1u

3 units

2 units

?

20

2 units 20 + 12= 32

1 unit 32 ÷ 2= 16

Daniel at 1st 16+ 20

= 36

Daniel had _________.36 stickers2020

Before and after (equal stage)Mrs Tan had a total of 115 oranges and apples. She sold half the apples and 25

oranges. In the end, she had an equal number of oranges and apples left. Find

the number of oranges Mrs Tan had at first.

Summary

Before: O A

115

–25

After: O

1u

A

1u

–half

O

After

A

Before

O 25

A

115

?

3 units 115– 25= 90

1 unit 90 ÷ 3= 30

Oranges at 1st30+ 25

= 55

She had ___________.55 oranges

Before and after (equal stage)4) Rakesh and Xijie had a total of 216 stickers at first. After Rakesh bought another 24

stickers and Xijie gave away 60 of his stickers to David, both of them had the same

number of stickers left. How many stickers did Rakesh have at first?

Summary

Before: R X

216

+24

After: R

1u

X

1u

–60

R

After

X

Before

R 24

X 6024

216

?2 units 216– 24 – 60

= 132

1 unit 132÷ 2

= 66

He had ___________.66 stickers

Gaps and Differences (Listing)

3

9

(x6,– 3):

A group of children had to share a bag of badges. When each

child took 4 badges, there were 5 badges left. When each child

took 6 badges instead, 3 more badges were needed. How many badges were there in the bag?

No. of children

x 4

(x4,+5):

x 6

1

4

6

9

13

2

8

12

15

17

3

12

18

21

21

4

16

24

Ans: 21 badges

Gaps and Differences (Model)

When looking at this type of question, we have to bear in mind

that in both scenarios, the bars of the model drawn should be of

equal length.(because the total amount of items for both cases should be the same).

A group of children had to share a bag of badges. When each

child took 4 badges, there were 5 badges left. When each child

took 6 badges instead, 3 more badges were needed. How many badges were there in the bag?

x4

x6

5

35

?

diff 6 – 4

= 2

gaps 5+ 3

= 8

The difference in badges between the 2 scenarios will always be 2.

So, we have to find how many sets of 2 we need to cover

the gaps (8).

No. of sets 8 ÷ 2

= 4 (children)

1 child 4

4 children 4 x 4= 16

total badges 16 + 5

= 21 Ans: 21 badges

Gaps and Differences (Model)1) Sue bought some pencils for her nephews. If she gave

them 8 pencils each, she would have 3 pencils left. If she

gave them 11 pencils each, she would need 9 more

pencils. How many pencils did Sue buy?

When looking at this type of question, we have to bear in mind

that in both scenarios, the bars of the model drawn should be of

equal length.(because the total amount of items for both cases should be the same).

x8

x11

3

93

?

diff 11 – 8

= 3

gaps 3+ 9

= 12

The difference in pencils between the 2 scenarios will always be 3.

So, we have to find how many sets of 3 we need to cover

the gaps (12).

No. of sets 12÷ 3

= 4 (nephews)

1 nephew 8

4 nephews 8 x 4= 32

total pencils 32 + 3

= 35 Ans: 35 pencils

Number × value

At a school funfair, every girl was given 3 packet drinks while every boy was given 5 packet drinks. There were thrice as many girls as boys at the funfair. Given that a total of 350 packet drinks were given out, how many boys were at the funfair?

3 units 1 unit

3 things to look out for:

1) no. of units for each item

2) amount/value of each item

3) total amount/value

No. of children(units) x

No. of drinks

Total unit (packet drinks)

girl 3 units x 3 9 packet units

boy 1 unit x 5 5 packet units

total amount 9 units + 5 units

= 14 units

14 units 350

1 unit 350 ÷ 14

= ____(boys)25

There were ________________.25 boys

Grouping with extra part5) 30 pupils shared a box of pencils. Each girl received 4 pencils and

each boy received 5 pencils. If the girls received 30 more pencils

than the boys, how many girls were there?

Girls(4 pencils each)

Boys(5 pencils each)

B + G(30)

(G – B)30 pencils /x

15 x 4 = 60 15 x 5 = 75 15 + 15 60- 75 x

17 x 4 = 68 13 x 5 = 65 17 + 13 68-65=3 x

19 x 4 = 76 11 x 5 = 55 19 + 11 76-55=21 x20 x 4 = 80 10 x 5 = 50 20 + 10 80-50=30

Ans: ______________20 girls

Mrs Lim had a total of $360, consisting of $10 and $2

notes. Given that she had 6 more $10-notes than $2

notes, how many notes did Mrs Lim have?

In a farm, there are twice as many chickens as horses.

Given that there are 288 legs altogether, how many

chickens are there in the farm?

A sum of $960 is to be divided amongst a group of adults

and children. Each adult will receive $7 and each child

will receive $4. Given that there are four times as many

adults as children, how many adults are there?

Grouping or Number × value?

Number x value

Grouping

Number x value

P6 Heuristics and

Exam Skills in

Mathematics Problem

Solving


Commonly Used

Heuristics

Draw a model

Look for patterns

Act it out

Guess and Check

Draw a diagram

Work backwards

Solve part of the problem

Before and After

Supposition method

Examination

Skills

Scan the whole question

Simplify the problem

Read the question a few

times

Underline the key points

Metacognitive questioning:

- “Which is the best method to use?”

- “Have I done a similar question

previously?”

- “How does this number help me to

solve the problem?”

- “Am I answering to the question?”

Working Backwards

Start with the end result and work backwards towards the beginning to

find a solution. Then check your answerby working forward.

Peter had some money . He spent half of the amount on a pair of jeans. He then spent half of the remainder on a pair of shoes. After that, he was left with $12. How much money did Peter have at first?

Solution

$12 × 2 = $24

$24 × 2 = $48

Ali had $48 at first.

Check $48 ÷ 2 = $24 (jeans), $48 − $24 = $24

$24 ÷ 2 = $12 (shoes), $24 − $2 = $12 (√ )

jeans

shoes $12

?

Draw a Model

Ahmad and Benny have a total of 66 toys. Of Ahmad’s

toys and of Benny’s toys are ‘Matchbox’ cars. If they have

an equal number of ‘Matchbox’ cars, how many toys does

each one of them have?

5

4

3

2

Ahmad

Benny66

11 units 66

1 unit 66 ÷ 11 = 6

5 units 6 x 5 = 30

6 units 6 x 6 = 36

Ahmad has 30 toys and Benny has 36 toys

Ahmad and Benny have a total of 66 toys. of Ahmad’s toys and of Benny’s

toys are ‘Matchbox’ cars. If they have an equal number of ‘Matchbox’ cars, how

many toys does each one of them have?

5

4

3

2

Ahmad

Ahmad has 30 toys and Benny has 36 toys

Alternative solution:

Benny =5

4

3

2

6

4

Total number of units = 5 + 6 =11

11 units 66

1 unit 6

5 units 5 x 6 = 30 (Ahmad)

6 units 6 x 6 = 36 (Benny)

Common mistakes

• Reading too quickly and missed out details

• Misread of questions or numbers • 355 as 335

• Number left instead of number sold

• Forget to write the units or the wrong units

• Incorrect standard unit conversion• 1m = 100cm

• 1 hour = 60 mins

Common mistakes

• Wrong mathematical statements :

• Wrong use of equal signs / approximation signs

2

5= 40

2

5 40

1

5= 40 ÷ 2 = 20

1

5 40 ÷ 2 = 20

Express 3 ÷ 7 as a percentage, rounded off to the nearest 1 decimal place.

3 ÷ 7 = 42.9% 3 ÷ 7 42.9%

Common mistakes

• Not putting given information into diagrams

• ABCD is a rhombus and AEFD is a parallelogram.

Answering PSLE

Questions &

PSLE Exam tips


Question 1

PSLE Maths 2015 P2 Q16

Peiyi and Jamal bought potted plants at the

prices shown below.

Large potted plants

2 for $15

Small potted plants

3 for $10

Question 1


(a) Peiyi bought an equal number of large and small

potted plants. She spent $175 more on the large

ones. How many potted plants did she buy

altogether?

(b) Jamal spent an equal amount of money on the

large and small potted plants. What fraction of the

potted plants he bought were large?

(a) Correct Solution

Cost of 6 large potted plants = $15 x 3= $45

Cost of 6 small potted plants = $10 x 2= $20

Difference in cost = $45 - $20= $25

No. of sets of 6 large potted plants she bought = $175 ÷ $25= 7

Number of potted plants she bought altogether = (7 x 6) x 2 = 42 x 2= 84

(b) Correct Solution

No. Of large potted plants he can buy for $30

= 2 × 2 = 4

No. Of small potted plants he can buy for $30

= 3 × 3 = 9

Fraction of potted plants he bought that are large

= 𝟒

𝟗+𝟒

= 𝟒

𝟏𝟑

Question 2


Three girls Amy, Beth and Cindy had the

same number of coins. Amy and Beth

each had a mix of fifty-cent and ten-cent

coins. Amy had 9 ten-cent coins while

Beth had 15 ten-cent coins. Cindy had

only fifty-cent coins.

(a) Of the three girls, who had the most money

and who had the least?

(b) What was the difference in the total value of

Amy and Beth’s coins?

(c) Beth used all her fifty-cent coins to buy some

food. She then had $10 less in coins than

Cindy. How many fifty-cent coins did Cindy

have?

(a) Cindy had the most money and Beth had the least money.

(b) Difference in value between a 10-cent and a 50-cent coin= $0.50 - $0.10= $0.40Difference in total value of Amy and Beth’s coins= 6 x $0.40= $2.40

(c) Original difference in total value of Beth and

Cindy’s coins

= 15 x $0.40

= $6

Number of fifty-cent coins Beth had

= ($10 - $6) ÷ $0.50

= 8

Number of fifty-cent coins Cindy had

= 15 + 8

= 23

Tips For Answering

PSLE Questions

1) Be open-minded

2) Think flexibly and divergently

3) Do not always narrow down to 1 method of

solution

Tips For Answering

PSLE Questions

4)Don’t skip too many questions --- attempt all the questions

5) Check your work means physically check, NOT just read

6) Complete the whole PSLE Mathematics book

Documents

Rivervale Primary School Maths Workshop for P4 to P6 …rivervalepri.moe.edu.sg/qql/slot/u143/Communications/Rivervale... · Maths Workshop for P4 to P6 Parents ... Rivervale Primary