hougang primary school mathematics department · 2016. 9. 18. · final note regular revision by...
TRANSCRIPT
Hougang Primary School
Mathematics Department
Parent Workshop 2016
Types of PSLE Questions
Computation
cccc
Example 1
1 + 2 + 3 + 4 + …….. + 94 + 95 + 96 + 97
When the first 97 whole numbers are
added up, what is the digit in the ones
place of this total?
(1) 1
(2) 2
(3) 3
(4) 8
1 + 2 + 3 + 4 + …….. + 94 + 95 + 96 + 97
1 + 2 + 3 + 4 + …….. + 94 + 95 + 96 + 97
1 + 2 + 3 + 4 + …….. + 94 + 95 + 96 + 97
1 + 2 + 3 + 4 +5 …+ 50+….. + 94 + 95 + 96 + 97
100
100
100
= 100 x 47 + 1 + 2 +50
= 4700 + 53
=4753
Example 1
1 + 2 + 3 + 4 + …….. + 94 + 95 + 96 + 97
When the first 97 whole numbers are
added up, what is the digit in the ones
place of this total?
(1) 1
(2) 2
(3) 3
(4) 8
Ans : (3)
Example 2
A repeated pattern is formed using the numbers
1 and 0. The first 18 numbers are shown below.
1 0 1 0 1 1 1 0 1 0 1 1 1 0 1 0 1 1
1st 19th
What is the sum of the first 100 numbers?
(1) 60
(2) 64
(3) 66
(4) 68 2013 PSLE
There are 6 numbers in 1 group.
There are altogether 100 numbers.
How many groups are there? What is the
remainder?
100 ÷ 6 = 16 R 4
.
R1 R2 R3 R4
1 0 1 0
1 0 1 0 1 1 1 0 1 0 1 1 1 0 1 0 1 1
1st 18th
Sum of each group ------------1 + 0 + 1 + 0 + 1+ 1 = 4
Sum of 100 numbers ---------- 4 x 16 + 1 + 1 =66
Example 3
Other computation questions Round off 70.171 to the nearest tenth.
Find the value of 4.8 + 0.29.
Express 1090 cm in m
A solid cuboid of height 10 cm has a square base of
side 3 cm. What is its volume?
Zaleha bought 1.2kg of grapes. 100 g of grapes cost
70 cent. How much did she pay?
Tips to nail the computation
questions
Firm foundation in number bonds.
Must know the multiplication tables very well.
Encourage children to use mental sum tricks
Example:
0.46 x 70 = 0.46 x 7 x 10
= 3.22 x 10
= 32.2
Types of PSLE Questions
Scale reading
Three containers with some water are shown below.
Which container has the most water?
160 ml 150 ml 200 ml
Container C contains the most water
50 20
=500ml
0.1
or
100
ml
Concepts tested along with
Procedures
A number when divided by 30 gives a remainder of 8.
Which of the following can be added to the number to
change it to a multiple of 6?
Spatial Visualisation
Question 1
In the figure below, ABDF and BCEF are rectangles and
CDE is a straight line. AB = 6 cm, AF = 8 cm and
BF = 10 cm. Find the length of BC
10cm
B
A
C
D
E
F
6 cm
8 cm
Question 2
MQN is an equilateral triangle
QPM is an isosceles triangle
MPQ = ( 180 – 30 ) ÷2 = 75
MPN = 75 x 2 = 150
60o
60o
60o
30o
Question 4
In the figure below, a rectangular piece of paper is folded
at 2 of its corners A and C as shown. Find ABC
Practical Questions
In the square grid below, PQ and QR are straight lines.
(a) Measure and write down the size of PQR
(b) PQ and QR form 2 sides of a trapezium PQRS. PS is
parallel to QR. PS is twice the length of QR. Complete
the drawing of the trapezium PQRS.
Word Problems
Understand
Plan/Devise
Do/Carry Out
Check
Understand • What am I given? (information , data)
• What am I asked to find?
• How can I make sense of the information
given to me? (Keywords, models, tables )
Plan/Devise • What strategy should I use?
• Have I solved similar problems before?
Strategies / Heuristics
Working backwards
Make a table
Systematic listing
Pattern Recognition
Assumption
Guess and Check
Model Drawing
Before and After concept
Do/Carry Out
• Are my steps accurate?
• Are there traps that I need to be alert of?
• Have I used all the information?
• Do my steps make sense?
Check • Does my answer make sense?
• Did I answer the question?
• Could this problem be solved in a simpler
way?
Question 1
Meng sold a total of 368 large and small durians at
the prices shown below and collected $2760. How
many large durians did Meng sell?
Durian for Sale
Large $9 Small $5
2 totals given
Plan :
Assumption
Guess & Check
Assumption
Assume all are small durians.
Cost of durians ----- 368 x $5 = $1840
Difference between actual cost and assumed cost
----- $2760 - $1840 = $920
$5 $5 $5
………
No of large durians ------- $920 ÷ $ 4 = 230
Question 2
Robin bought some pens for $42. He bought the same
number of rulers for $22. Each pen cost $2 more than
each ruler.
How many rulers did Robin buy?
Same number
of rulers
1 Pen is more
expensive than
1 ruler
Plan :
Differences
Since the number of rulers and pens are the
same
No of
pens:
No of
rulers :
$42
$22
$42 - $22 = $20
$20 ÷ $2 = 10
Robin bought 10 rulers.
Question 2
160 more
chickens than
duck
Plan : model
drawing
ducks
chickens
S
S
S
L
S S L L
160
80
S
80
L
3 units + 80 ------- 230
3 units -------- 230 – 80 = 150
1 units -------- 150 ÷ 3 = 50
There are 50 ducks left.
Question 3
The ratio of the number of books to the number of files
was 7 : 2.
When 20 books and 25 files were added, the ratio
became 2 : 1.
How many files were there at first?
2 different
types of units
Plan :
units and parts
Before and after
Units and Parts
Books Files
Before
After
7 units + 20 ----------- 2 parts
2 units + 25 ----------- 1 part
7 units 2 units
+20 books +25 files
2 parts 1 part
(
)
4 units + 50 ----------- 2 parts
7 units +20 ---------- 4 units + 50
X2
7 units +20 ---------- 4 units + 50
3 units ---------- 50 – 20 = 30
1 unit --------- 30 ÷ 3 = 10
2 units -------- 10 x 2 = 20
There were 20 files at first.
Question 5
Ishak uses rods to form figures that follow a pattern. The
first four figures are shown below.
How many rods would he use for Figure 30?
Figure Number Number of rods used
1 10
2 15
3 18
4 23
Numbers
increase at
regular
intervals
Figure Number Number of rods used
1 10
2 15
3 18
4 23
5 23 + 3 = 26
6 26 + 5 = 31
7 31 + 3 = 34
…… …….
30 127
+ 5
+ 3
Systematic Listing
Grouping
Figure Number Number of rods used
1 10
2 15
3 18
4 23
+2 +8
Number of figure from Figure 4 to Figure 30
= 30 – 4 = 26
Number of increases of 8 rods from Fig 4 to 30
= 26 ÷ 2 = 13
Number of rods in Figure 30
= 23 + 13 x 8 = 127
Final Note
Regular revision by re-doing questions that your child has
done wrongly.
Encourage your child to simplify the problem by drawing
models (pictures) or presenting the information in a
table.
Ensure adequate and productive practice.
Provide opportunities for time-based practices.
Ground the basics first before moving on to challenging
problems.
Remind your child to start with the simpler questions and
move on if he is stuck.