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12 August 2011

Mathematics Workshop. for P3 and P4 parents. Problem Solving Heuristics. 12 August 2011. Programme. 5.45 – 6.00 – Registration 6.15 – 6.30 – Overview of our Primary Mathematics Curriculum 6.30 – 7.00 Heuristic 1 7.00 – 7.30 Dinner

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12 August 2011

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  1. Mathematics Workshop for P3 and P4 parents Problem Solving Heuristics 12 August 2011

  2. Programme 5.45 – 6.00 – Registration 6.15 – 6.30 – Overview of our Primary Mathematics Curriculum 6.30 – 7.00 Heuristic 1 7.00 – 7.30 Dinner 7.30 - 9.15 Heuristic 2 9.15 – 9.30 Q n A

  3. Objectives To familiarise parents with the more challenging heuristics from “Conquer Math”. To partner with parents so that they can help their children with the approaches the school uses to each problem solving.

  4. Mathematics Curriculum • Concepts • Basic mathematical knowledge needed for • solving mathematical problems. • - Numerical concepts • - Geometrical concepts • - Algebraic concepts • - Statistical concepts

  5. Mathematics Curriculum 2. Skills Topic-related manipulative skills that pupils are expected to perform when solving problems. - Estimation & approximation - Mental calculation - communication - use of mathematical tools - arithmetic & algebraic manipulation - handling data - use of calculator

  6. Mathematics Curriculum 2. Processes The knowledge skills ( process skills) involved in the process of acquiring and applying mathematical knowledge : - reasoning - communication and connection - thinking skills and heuristics - application and modelling

  7. Problem Solving Heuristics • To give a representation • - draw a diagram (eg. model) , make a list, • use equation • To make a calculated guess • - guess and check, look for patterns, make • supposition • To go through the process • - act it out , work backwards, before-after • To change the problem • - restate the problem, simplify the problem, • solve part of the problem

  8. Working Backwards Sharon had some sweets at first. After she gave away 5 sweets to her sister and ate another 3 sweets, she went to buy another 48 sweets. In the end she had thrice as many sweets as before. How many sweets did she have at first ? (P3 - pg 20) ? Extra At First 1 unit End 1 unit Extra 48 – 3 – 5 = 40 2 units 40 1 unit 40 ÷2 = 20 (She had 20 sweets at first.)

  9. Equal Concept Model (Internal Transfer) Dinesh and Melissa shared a sum of money equally. After Dinesh gave $36 to Melissa, Melissa had twice as much money as Dinesh. How much money did each of them receive at first ? (P3 – pg 15) 1 u Dinesh -$36 Melissa +$36 2 u 1 u $36 + $36 = $72 Amt. of money each received at first $72 +$36 = $108

  10. Equal Stage There was an equal number of boys and girls at a gathering at first. After 60 boys and 12 girls had left the gathering, there were twice as many girls as boys that remained behind. Find the number of boys at first ? (P4 – pg 30) left Boys 1u 60 Girls 1u12 Left 1u = 60 – 12 = 48 48 + 60 = 108 There were 108 boys

  11. Equal Stage Type 1 (Beginning) Boys Girls 1 unit  60 – 12 = 48 48 + 60 = 108 60 12 P4-page 30

  12. Equal Stage Raymond and Rauf had a total of 224 marbles at first. After Raymond had bought another 12 marbles and Rauf had given away 30 of his marbles, both had the same number of marbles left. How many marbles did Raymond have at first ? (P4 – pg 34)

  13. Equal Stage Type 2 (End) Ray Rauf 224 – 30 = 194 194 – 12 = 182 182 ÷ 2 = 91 224 12 30 P4-page 34

  14. Equal Stage There was an equal number of boys and girls at an exhibition. After of the boys and 42 girls had left the exhibition, there were 12 more boys than girls at the exhibition. How many students were there at the exhibition at first ? (P4 – pg 96)

  15. Equal Stage - Fraction (Type 1) Boys Girls 3u  42 – 12 = 30 1u  30 ÷ 3 = 10 10u  10 x 10 = 100 12 42 P4-pge 96

  16. Equal Stage There is a total of 620 apples and oranges at a stall. of the oranges is equal to of the apples. How many apples are there at the stall ? (P4 – pg 105)

  17. Equal Stage - Fraction (Type 3) Oranges Apples Oranges  16u Apples  15u Total  31u  620 1u  620 ÷ 31 = 20 15u  20 x 15 = 300 P4- Page 105

  18. Equal Stage – Fractions (Type 3) Unitary Method: 32 8 5 66 16 15 Oranges  16u Apples  15u Total  31u  620 1u  620 ÷ 31 = 20 15u  20 x 15 = 300

  19. Equal Stage At the graduation ceremony, there were 30 less boys than girls. After of the boys and of the girls had left the hall for recess, there was an equal number of boy and girls that remained behind. Find the number of boys who were present at the graduation ceremony (P4 – pg 106)

  20. Equal Stage – Fractions (Type 3) Unitary Method: 2 3 5 8 66 15 16 Boys  15u Girls  16u Difference  16u – 15u = 1u 1u  30 15u  30 x 15 = 450 P4- Page 106

  21. Internal Transfer Two classes ,3A and 3B , had an equal number of pupils. After 12 pupils left class 3A to join class 3B, there were twice as many pupils in class 3B as in class 3A. Find the number of pupils in each class at first ? (P3 – pg 23) 3A 1u 12 3B 1u 1u = 24 24 + 12 = 36 There were 36 pupils in each class

  22. Internal Transfer Esther has 86 more stickers than Jane. How many stickers must Esther give to Jane so that Esther will have 40 more stickers than Jane ? (P3 – pg 26) • 86 - 40 = 46 • 2 =23 • She gives 23 stickers Esther 86 Jane Esther 40

  23. Equal Stage (Internal Transfer) David had 5 times as many marbles as Serene. After David had given 32 of his marbles to Serene, they had an equal number of marbles in the end. Find the number of marbles each of them had at first . (P4 – pg 38)

  24. Equal Stage Type 3 (Internal Transfer) 5u David Serene 4u  32 + 32 = 64 1u  64 ÷ 4 = 16 Serene: 16 David: 16 x 5 = 80 32 32 4u P4- Page 38

  25. Linear Formation Trees were planted along each side of a square garden so that an equal number of trees are formed along each side of the garden. If there were 60 trees planted, how many trees were there along each side of the garden ? 60 ÷ 4 = 15 (because of 4 sides)..Nope.. Let’s pick a smaller number - 12. So 12 ÷ 4 = 3 trees per side. Right? WRONG! As you can see, when sets of 3 were put along the sides, some trees be part of ANOTHER side. It is 3 + 1 = 4 trees on each side of the garden. So, it is 15 + 1 = 16 trees on each side of the garden.

  26. Linear Formation Trees were planted along each side of a square garden so that an equal number of trees are formed along each side of the garden. If there were 60 trees planted, how many trees were there along each side of the garden ? Since the 4 corners are repeated, we take 60 – 4 = 56 Then 56 divided equally by 4 sides = 14 per side 14 + 2 ( 2 corners ) and we get 16 trees per side.

  27. Internal Transfer There are 25 more passengers on Bus A than Bus B. If 7 passengers transferred from Bus A to Bus B, how many more passengers will there be in Bus A than in Bus B ? (P3 – pg 25) 25 A B 7 7 25 – 7 – 7 = 11 There will be 11 more passengers in Bus A

  28. Regular Gaps Fig 1 Fig 2 Fig 3 4 dots 7 dots 10 dots Fig 1 – 4 dots Fig 2 – 4 + 3 Fig 3 – 4 + 3 + 3 Fig 5 – 4 + 3 + 3 + 3 + 3 = 16 Fig 20 – 4 + 19 x 3 = 61 109 dots -- 109 – 4 = 105 105 3 = 35 35 + 1 = 36 figure (P3 – pg 37)

  29. Equal Distribution without Remainder Mrs Singh has a certain number of sweets to distribute to her class. If she gives each of them 5 sweets, she will have 54 extra sweets. If she gives each of them 8 sweets, she would need another 33 more. How many pupils does she have in her class? (P3 – pg 45) 5 5 5 5 54 extra 5 5 5 5 54 extra 33 needed 3 3 3 54 + 33 = 87 sweets needed to form extra sets of 3s. Hence, 87 ÷ 3 = 29 pupils in her class. Sets of 5 is the same (number of students fixed)

  30. More than / Less than (External Unchanged) Joshua had 45 more stickers than Melvin at first. After Melvin had used up 15 of his stickers, Joshua had thrice as many stickers as Melvin in the end. Find the number of stickers Joshua had at first. (P4 – pg 21)

  31. More / Less than (External Unchanged) Joshua Melvin 2u  15 + 45 = 60 1u  60 ÷ 2 = 30 3u  30 x 3 = 90 45 15 2u P4 – Page 21

  32. More than / Less than (External Unchanged) There were thrice as many swimmers as non- swimmers at a swimming carnival. After 40 non-swimmers had left the carnival ,there were 5 times as many swimmers as non-swimmers at the carnival. How many swimmers were there at the carnival ? (P4 – pg 24)

  33. More/ Less than (External Unchanged) 5u Swim Non-swim 2u  40 x 3 = 120 1u  120 ÷ 2 = 60 5u  60 x 5 = 300 1u 40 1u 40 1u 40 1u 40 P4 – Page 24

  34. Internal Transfer Ryan and Aaron had 156 bottle caps. After Ryan gave 40 bottle caps to Aaron, Ryan had twice as many bottle caps as Aaron. How many bottle caps did Ryan at first ? (P3 – pg 69) 40 40 Ryan Aaron 40 3u = 156 – 120 = 36 1u = 12 24 + 120 = 144 156

  35. External Change Concept Mrs Tan and Mrs Lim had an equal number of pies at first. After Mrs Tan bought another 12 pies and Mrs Lim gave away 4 pies, Mrs Tan had twice as many pies as Mrs Lim. How many pies did each of them have at first ? (P3 – pg 77) At first Mrs Tan 12 bought Mrs Lim 4 : from equal amount, decrease by 4 (gave away) 12 + 4 = 16 ---- forms 1 unit Each of them had ---16 + 4 = 20 pies at first.

  36. External Unchanged There were as many men as women at a concert. Halfway during the concert, some of the men left the concert and the number of men became of the number of women. Given that there were 80 more women than men at first, find the number of men who left the concert halfway ? (P4 – pg 110) left 5u = 80 1u = 16 3u = 48 Men women 48 men left the concert

  37. External Unchanged During a class gathering were boys and the rest were girls. Halfway, 22 more girls joined in the gathering and there were as many boys as girls. Find the number of students at the class gathering in the end. (P4 – pg 113) 20u – 9u = 11u 11u = 22 1u = 2 32u = 64 + 22 Boys Girls 3u 3u 3u 3u 3u 3u 3u 4u 4u 4u 4u 4u 4u 4u 4u There are 64 students in the end.

  38. External Unchanged Johnson had as many red marbles as blue marbles. If he bought another 12 red marbles, he would be left with as many red marbles as blue marbles. How many marbles did he have altogether ? (P4 – pg 114) 6u 1u = 12 20u = 240 He had 240 marbles. 5u Red Blue + 12 5u 5u 5u 3u 3u 3u 3u 3u

  39. Repeated Identity The mass of a container with 7 glasses is 2 500 g The same container with 3 glasses is 1 500g. What is the mass of the container ? (P3 – pg 101) (Container+7 glasses) – (Container+3 glasses) = (No container + 4 glasses) 2 500 – 1 500 = 1 000 1 000 ÷ 4 = 250  1 glass 250 x 3 = 750  3 glasses 1 500 – 750 = 750  container

  40. Repeated Identity The mass of a container with 7 glasses is 2 500 g The same container with 3 glasses is 1 500g. What is the mass of the container ? (P3 – pg 101) 1500 2 500 4u = 2500 – 1500 = 1000 1 u = 250 3 u = 750 OR 7u = 1750 1500 – 750 = 750 (g) 2500 – 1750 = 750 (g)

  41. Repeated Identity Serene had as many chocolates as Melvin. Melvin had as many chocolates as Esther. If Esther had 38 more chocolates than Serene, find the total number of chocolates that was shared among these 3 children at first. (P4 – pg 117) 4u 4u 4u 4u Serene Melvin Esther 35 – 16 = 19 19u = 38 1u = 2 79u = 158 4u 4u 4u 4u 4u 4u 4u 7u 7u 7u 7u 35u There are 158 chocolates.

  42. Repeated Identity Inside a bag, there were twice as many red balloons as blue balloons and 15 more blue balloons than green balloons. If there was a total of 141 balloons of these colours, how many green balloons were there in the bag at first ? (P4 – pg 56) 4u = 141 – (3 x 15 ) = 96 1u = 24 There were 24 green balloons. Red Blue Green 15

  43. Internal Transfer (No change in total) Ryan and Aaron had 156 bottle caps. After Ryan gave 40 bottle caps to Aaron, Ryan had twice as many bottle caps as Aaron. How many bottle caps did Ryan at first ? (P3 – pg 69) ? Ryan 40 156 Aaron 3 units 156 1 unit 156 ÷3 = 52 No. of bottle caps Ryan had at first 52 x 2 + 40 = 144

  44. Repeated Identity There are 180 blue and green balls in bag A. There are 320 green and red balls in bag B. Given that there are thrice as many red balls as blue balls, and the number of green balls in both bags is the same, find the total number of green balls in both bags ? (P4 – pg 59) Green blue 2u = 320 -140 = 140 1u = 70 180 – 70 = 110 110 x 2 = 220 A B 180 320 Red There were 220 green balls.

  45. Repeated Identity Serene and Tommy have a total of 130 sweets. Tommy and Clara have a total of 141 sweets. Serene and Clara have a total of 99 sweets. How many sweets does Serene have ? (P4 – pg 60) S + T = 130 T + C = 141 S + C = 99 S + T + C = (130 + 141 + 99) = 185 2 185 – 141 = 44

  46. Fraction (Part-Whole Relationship) At a swimming event, of the students were boys. Given that of the girls and of the boys did not know how to swim, how many students were there altogether if there were 154 swimmers at the event ? (P4 – pg 88) Boys girls 11 u = 54 1u = 14 20u = 280 5u 5u 5u 5u 2u 3u 6u 9u

  47. Parents’ Help with Child At Home Math in real life • Talk to your children about mathematics and • help them to see the usefulness of math in our • daily life. • Be observant to improve their visualisation. • Real life application of measurements

  48. Reference • Conquer Problem Sums (Primary 3 and Primary 4 ) • Published by onSponge Pte Ltd • Mathematics Syllabus Primary 2007

  49. Q & A Kathleen_lee_mui_hwa@moe.edu.sg

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