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Problem Solving in
PSLE Mathematics

      Yeap Ban Har
  Marshall Cavendish Institute
          Singapore

 banhar.yeap@pathlight.org.sg




 Slides are available at
www.banhar.blogspot.com
Type     Mark     Number      Type     Mark     Number
          Value                         Value
 MCQ     1 mark    10 (10%)    SAQ     2 marks   5 (10%)
 MCQ     2 marks   5 (10%)             3 marks
 SAQ     1 mark    10 (10%)    LAQ     4 marks   13 (50%)
                                       5 marks
 SAQ     2 marks   5 (10%)




Paper 1 (50 min)              Paper 2 (1 hr 40 min)
Type     Mark     Number      Type     Mark     Number
          Value                         Value
 MCQ      1 mark   10 (10%)    SAQ     2 marks   10 (20%)
 MCQ     2 marks   10 (20%)            3 marks
 SAQ     2 marks   10 (20%)    LAQ     4 marks   8 (30%)
                                       5 marks




Paper 1 (1 hr)                Paper 2 (1 hr 15 min)
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 .




                           Answer : 3.05 [B1]




                                    Example 1
3 .05
     12.20           4   12.20
                         12
  12    20 hundredths
                           0.20
Number Bond Method         0.20
                              0
                     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?



                       2 h 30 min


           11 a.m.                  1.30 p.m.



                     Answer : 2 h 35 min [B1]

                                        Example 2
   Find <y in the figure below.

                        70 o
                     70 o      y
                       70 o




   360o – 210o = 150o




                                   Example 3
The height of the classroom door is about __.

(1)    1m
(2)    2m
(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?
4.5 x 8 = 

36 ÷ 12 = 3

              Answer: 3 cars / month
                              Example 6
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?

    $767.40 – 3 x $155 = $302.40

$302.40 ÷ 60 cents per km = 504 km


                               Example 7
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?

    767.40 – 3 x 155 = 302.40

    302.40 ÷ 0.60 = 504
    He travelled 504 km.

                               Example 7
Ministry of Education 2006
Ministry of Education 2006
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 = 96
                             3       4    5       96 ÷ 8 = 12

                            11            13        Alternate Method
                                                    4 x 24 = 96
                            19      20    21        96 ÷ 8 = 12


                                 What is the average of the 8 numbers that can
                                 be seen in the frame?
(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
   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
                                           Problem 2
                             29 pieces are used.
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

                                     Problem 3
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
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?


   19 cm x 5 = 95 cm

   150 cm – 95 cm = 55 cm

   55 cm was left.
Problem 7
1 2
 
4 9



Problem 7
   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.

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Problem Solving Strategies in PSLE Mathematics

  • 1. Problem Solving in PSLE Mathematics Yeap Ban Har Marshall Cavendish Institute Singapore banhar.yeap@pathlight.org.sg Slides are available at www.banhar.blogspot.com
  • 2.
  • 3.
  • 4. Type Mark Number Type Mark Number Value Value MCQ 1 mark 10 (10%) SAQ 2 marks 5 (10%) MCQ 2 marks 5 (10%) 3 marks SAQ 1 mark 10 (10%) LAQ 4 marks 13 (50%) 5 marks SAQ 2 marks 5 (10%) Paper 1 (50 min) Paper 2 (1 hr 40 min)
  • 5. Type Mark Number Type Mark Number Value Value MCQ 1 mark 10 (10%) SAQ 2 marks 10 (20%) MCQ 2 marks 10 (20%) 3 marks SAQ 2 marks 10 (20%) LAQ 4 marks 8 (30%) 5 marks Paper 1 (1 hr) Paper 2 (1 hr 15 min)
  • 6.
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  • 10. The rationale of teaching mathematics is that it is “a good vehicle for the development and improvement of a person’s intellectual competence”.
  • 12. Find the value of 12.2 ÷ 4 . Answer : 3.05 [B1] Example 1
  • 13. 3 .05 12.20 4 12.20 12 12 20 hundredths 0.20 Number Bond Method 0.20 0 Long Division Method
  • 14. A show started at 10.55 a.m. and ended at 1.30 p.m. How long was the show in hours and minutes? 2 h 30 min 11 a.m. 1.30 p.m. Answer : 2 h 35 min [B1] Example 2
  • 15. Find <y in the figure below. 70 o 70 o y 70 o  360o – 210o = 150o Example 3
  • 16. The height of the classroom door is about __. (1) 1m (2) 2m (3) 10 m (4) 20 m Example 4
  • 18. 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
  • 19. 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? 4.5 x 8 =  36 ÷ 12 = 3 Answer: 3 cars / month Example 6
  • 20. 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? $767.40 – 3 x $155 = $302.40 $302.40 ÷ 60 cents per km = 504 km Example 7
  • 21. 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? 767.40 – 3 x 155 = 302.40 302.40 ÷ 0.60 = 504 He travelled 504 km. Example 7
  • 24. Students in the highest international benchmark are able to apply their knowledge in a variety of situations and able to explain themselves.
  • 25.
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  • 30.
  • 31.
  • 32.
  • 34. 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
  • 35. 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?
  • 36. 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 = 96 3 4 5 96 ÷ 8 = 12 11 13 Alternate Method 4 x 24 = 96 19 20 21 96 ÷ 8 = 12 What is the average of the 8 numbers that can be seen in the frame?
  • 37. (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
  • 38. 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 Problem 2 29 pieces are used.
  • 39. 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. 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 Problem 3
  • 41. 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
  • 42. 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
  • 43. 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?
  • 44. 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?
  • 45. 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.
  • 47. 1 2  4 9 Problem 7
  • 48. Number Sense  Patterns  Visualization  Communication  Metacognition
  • 49. 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.