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This was held in Henry Park Primar
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Problem Solving in
PSLE Mathematics
Yeap Ban Har Marshall Cavendish Institute
Singapore
Slides are available at
www.banhar.blogspot.com
Paper 1 (50 min) Paper 2 (1 hr 40 min)
Type Mark Value
Number
MCQ 1 mark 10 (10%)
MCQ 2 marks 5 (10%)
SAQ 1 mark 10 (10%)
SAQ 2 marks 5 (10%)
Type Mark Value
Number
SAQ 2 marks 5 (10%)
LAQ
3 marks 4 marks 5 marks
13 (50%)
Paper 1 (1 hr) Paper 2 (1 hr 15 min)
Type Mark Value
Number
MCQ 1 mark 10 (10%)
MCQ 2 marks 10 (20%)
SAQ 2 marks 10 (20%)
Type Mark Value
Number
SAQ 2 marks 10 (20%)
LAQ
3 marks 4 marks 5 marks
8 (30%)
The rationale of teaching mathematics is that it is “a good vehicle for the development and improvement of a
person’s intellectual competence”.
Ministry of Education 2006
Find the value of 12.2 ÷ 4 .
Example 1
Answer : 3.05 [B1]
12.20 4
3
12
0.20
.05
0.20
0
12.20
12 20 hundredths
Number Bond Method
Long Division Method
A show started at 10.55 a.m. and ended
at 1.30 p.m. How long was the show in
hours and minutes?
Example 2
11 a.m. 1.30 p.m.
2 h 30 min
Answer : 2 h 35 min [B1]
Find <y in the figure below.
360o – 210o = 150o
70 o
70 o
70 o
y
Example 3
The height of the classroom door is about __.
(1) 1 m
(2) 2 m
(3) 10 m
(4) 20 m
Example 4
Ministry of Education 2006
Cup cakes are sold at 40 cents each.
What is the greatest number of cup cakes that can be bought with $95?
$95 ÷ 40 cents = 237.5
Answer: 237 cupcakes
Example 5
From January to August last year, Mr
Tang sold an average of 4.5 cars per
month, He did not sell any car in the
next 4 months. On average, how many
cars did he sell per month last year?
Example 6
4.5 x 8 =
36 ÷ 12 = 3
Answer: 3 cars / month
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?
Example 7
$767.40 – 3 x $155 = $302.40
$302.40 ÷ 60 cents per km = 504 km
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?
Example 7
767.40 – 3 x 155 = 302.40
302.40 ÷ 0.60 = 504
He travelled 504 km.
Ministry of Education 2006
Ministry of Education 2006
Parents Up In Arms
Over PSLE
Mathematics Paper TODAY’S 10 OCT 2009
SINGAPORE: The first thing her son did when he came out from
the Primary School Leaving Examination (PSLE) maths paper on
Thursday this week was to gesture as if he was "slitting his
throat".
"One look at his face and I thought 'oh no'. I could see that he felt
he was condemned," said Mrs Karen Sng. "When he was telling
me about how he couldn't answer some of the questions, he got
very emotional and started crying. He said his hopes of getting
(an) A* are dashed."
Not for the first time, parents are up in arms over the PSLE
Mathematics paper, which some have described as "unbelievably
tough" this year. As recently as two years ago, the PSLE
Mathematics paper had also caused a similar uproar.
The reason for Thursday's tough paper, opined the seven parents
whom MediaCorp spoke to, was because Primary 6 students were
allowed to use calculators while solving Paper 2 for the first time.
…
Said Mrs Vivian Weng: "I think the setters
feel it'll be faster for them to compute with a
calculator. So the problems they set are much
more complex; there are more values, more
steps. But it's unfair because this is the first
time they can do so and they do not know
what to expect!"
…
"The introduction of the use of calculators
does not have any bearing on the difficulty of
paper. The use of calculators has been
introduced into the primary maths curriculum
so as to enhance the teaching and learning of
maths by expanding the repertoire of learning
activities, to achieve a better balance between
the time and effort spent developing problem
solving skills and computation skills.
Calculators can also help to reduce
computational errors."
…
Another common gripe: There was not
enough time for them to complete the paper.
A private tutor, who declined to be named,
told MediaCorp she concurred with parents'
opinions. "This year's paper demanded more
from students. It required them to read and
understand more complex questions, and go
through more steps, so time constraints would
have been a concern," the 28-year-old said.
Students in the highest international benchmark are able
to apply their knowledge in a variety of situations and able to explain themselves.
Ministry of Education 2006
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
49 50 51 52 53 54 55 56
Problem 4
Table 1 consists of numbers from 1 to 56. Kay and Lin are given a plastic
frame that covers exactly 9 squares of Table 1 with the centre square
darkened.
(a) Kay puts the frame on 9 squares as shown in the figure below.
3 4 5
11 13
19 20 21
What is the average of the 8 numbers that can
be seen in the frame?
Table 1 consists of numbers from 1 to 56. Kay and Lin are given a plastic
frame that covers exactly 9 squares of Table 1 with the centre square
darkened.
(a) Kay puts the frame on 9 squares as shown in the figure below.
3 4 5
11 13
19 20 21
What is the average of the 8 numbers that can
be seen in the frame?
Alternate Method
4 x 24 = 96
96 ÷ 8 = 12
3+4+5+11+13+19+20 = 96
96 ÷ 8 = 12
(b) Lin puts the frame on some other 9 squares.
The sum of the 8 numbers that can be seen in the frame is 272.
What is the largest number that can be seen in the frame?
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
49 50 51 52 53 54 55 56
34
A figure is formed by arranging equilateral triangles pieces of sides 3 cm in a line. The figure has a perimeter of 93 cm. How many pieces of the equilateral triangles are used?
93 cm ÷ 3 cm = 31
31 – 2 = 29
29 pieces are used. Problem 2
Problem 3
40 cm x 30 cm x 60 cm = 72 000 cm3
72 000 cm3 ÷ 5 x 3 = 43 200 cm3
43 200 cm3 ÷ 1800 cm2 = 24 cm
Problem 3
40 cm x 30 cm x 60 cm = 72 000 cm3
72 000 cm3 ÷ 5 x 2 = 28 800 cm3
28 800 cm3 ÷ 1200 cm2 = 24 cm
Rena used stickers of four different shapes
to make a pattern. The first 12 stickers are
shown below. What was the shape of the
47th sticker?
………? 1st 12th 47th
Problem 4
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?
Problem 5
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?
Problem 5
182 7 = 20 + 6 = 26
182 x 2 + 26 = 364 + 26 = 390
There were 390 coins altogether.
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?
$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.
Visualization
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?
Problem 5
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?
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?
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?
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?
19 cm x 5 = 95 cm
150 cm – 95 cm = 55 cm
55 cm was left.
Problem 7
Problem 7
9
2
4
1
Number Sense
Patterns
Visualization
Communication
Metacognition
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.