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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
The Rivervale – School of Excellence, Individuals of Character
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)
The Rivervale – School of Excellence, Individuals of Character
Achievement Bands, Grades / Mark Range:
P5, P6:
The Rivervale – School of Excellence, Individuals of Character
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
The Rivervale – School of Excellence, Individuals of Character
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
The Rivervale – School of Excellence, Individuals of Character
List of approved calculators for PSLE:
http://www.seab.gov.sg/content/calcula
tor/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
The Rivervale – School of Excellence, Individuals of Character
Assessments:P6 PSLE:
• T-score
• Aggregate score
• Performance is relative
The Rivervale – School of Excellence, Individuals of Character
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
The Rivervale – School of Excellence, Individuals of Character
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
The Rivervale – School of Excellence, Individuals of Character
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
?
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
The Rivervale – School of Excellence, Individuals of Character
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
PSLE Maths 2015 P2 Q16
(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
PSLE Maths 2015 P2 Q17
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
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