12 august 2011

River Valley Primary School – Strive for the Best

12 August 2011


5.45 – 6.00 – Registration6.15 – 6.30 – Overview of our Primary Mathematics Curriculum6.30 – 7.00 Heuristic 17.00 – 7.30 Dinner7.30 - 9.15 Heuristic 29.15 – 9.30 Q n A

Programme


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.

Objectives


Mathematics Curriculum

1. Concepts Basic mathematical knowledge needed

for solving mathematical problems. - Numerical concepts - Geometrical concepts - Algebraic concepts - Statistical concepts


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


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


•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

Problem Solving Heuristics


Working BackwardsSharon 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)At FirstEnd

1 unit

1 unit

Extra

Extra 48 – 3 – 5 = 40

2 units 40

1 unit 40 ÷2 = 20 (She had 20 sweets at first.)

?


Equal Concept Model (Internal Transfer)

Dinesh

Melissa

1 u

2 u

-$36

+$36

1 u $36 + $36 = $72

Amt. of money each received at first $72 +$36 = $108

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)


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)

Boys 1u 60

Girls 1u 12 Left

left

1u = 60 – 12 = 48

48 + 60 = 108

There were 108 boys


Equal Stage Type 1 (Beginning)

Boys Girls

1 unit 60 – 12 = 4848 + 60 = 108

60

12

P4-page 30


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)


Equal Stage Type 2 (End)

Ray Rauf

224 – 30 = 194194 – 12 = 182182 ÷ 2 = 91

12

30

224

P4-page 34


Equal StageThere 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)

53


Equal Stage - Fraction (Type 1)

Boys Girls

3u 42 – 12 = 301u 30 ÷ 3 = 1010u 10 x 10 = 100

12

42

P4-pge 96


Equal StageThere 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)

83

52


Equal Stage - Fraction (Type 3)

Oranges Apples

Oranges 16uApples 15uTotal 31u 6201u 620 ÷ 31 = 2015u 20 x 15 = 300

P4- Page 105


Equal Stage – Fractions (Type 3)

Unitary Method:3 2 8 5 6 616 15Oranges 16uApples 15uTotal 31u 6201u 620 ÷ 31 = 2015u 20 x 15 = 300


Equal StageAt 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)

85

53


Equal Stage – Fractions (Type 3)

Unitary Method:2 3 5 8 6 615 16Boys 15uGirls 16uDifference 16u – 15u = 1u1u 3015u 30 x 15 = 450

P4- Page 106


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


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)

Esther 86

Jane

Esther 40

86 - 40 = 46

46 2 =23

She gives 23 stickers


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)


Equal Stage Type 3 (Internal Transfer)

David Serene

4u 32 + 32 = 641u 64 ÷ 4 = 16Serene: 16David: 16 x 5 = 80

32

32

5u

4u

P4- Page 38


Linear FormationTrees 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.


Linear FormationTrees 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.


Internal TransferThere 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)

A

B

257

7

25 – 7 – 7 = 11

There will be 11 more passengers in Bus A


Regular Gaps

Fig 1 Fig 2 Fig 3 4 dots 7 dots 10 dots

Fig 1 – 4 dotsFig 2 – 4 + 3Fig 3 – 4 + 3 + 3Fig 5 – 4 + 3 + 3 + 3 + 3 = 16Fig 20 – 4 + 19 x 3 = 61 109 dots -- 109 – 4 = 105 105 3 = 35 35 + 1 = 36 figure (P3 – pg 37)


Equal Distribution without RemainderMrs 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

333

33 needed

Sets of 5 is the same (number of students fixed)

54 + 33 = 87 sweets needed to form extra sets of 3s.

Hence, 87 ÷ 3 = 29 pupils in her class.

54 extra


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)


More / Less than (External Unchanged)

Joshua Melvin

2u 15 + 45 = 601u 60 ÷ 2 = 303u 30 x 3 = 90

45

15

2u

P4 – Page 21


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)


More/ Less than (External Unchanged)

Swim Non-swim

2u 40 x 3 = 1201u 120 ÷ 2 = 605u 60 x 5 = 300

5u

40

40 40 40

1u

1u1u1u

P4 – Page 24


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)

RyanAaron 40

40 40

156 3u = 156 – 120

= 36

1u = 12

24 + 120 = 144


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)

Mrs Tan

Mrs Lim

At first

12

4 : from equal amount, decrease by 4 (gave away)

bought

12 + 4 = 16 ---- forms 1 unit

Each of them had ---16 + 4 = 20 pies at first.


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)

127

31

Menwomen

5u = 80

1u = 16

3u = 4848 men left the concert

left


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)

74

53

Boys Girls

+ 22

3u 3u 3u 3u 3u 3u 3u 4u 4u 4u 4u 4u 4u 4u 4u

20u – 9u = 11u

11u = 22

1u = 2

32u = 64

There are 64 students in the end.


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)

31

52

Red

Blue

+ 125u 5u 5u

3u 3u 3u 3u 3u

6u

5u1u = 12

20u = 240

He had 240 marbles.


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


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)

4u = 2500 – 1500 = 1000

1 u = 250

3 u = 750 OR 7u = 1750

1500 – 750 = 750 (g) 2500 – 1750 = 750 (g)

2 500

1500


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)

74

54

Serene

Melvin

Esther

4u 4u 4u 4u 4u 4u 4u

4u 4u 4u 4u

7u 7u 7u 7u

35u

35 – 16 = 19

19u = 38

1u = 2

79u = 158There are 158 chocolates.


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)

Red

Blue

Green 15

4u = 141 – (3 x 15 )

= 96

1u = 24

There were 24 green balloons.


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

Aaron156

3 units 156

1 unit 156 ÷3 = 52

No. of bottle caps Ryan had at first 52 x 2 + 40 = 144

40

?


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)

A

B

Green blue

180

320

Red

2u = 320 -140

= 140

1u = 70

180 – 70 = 110

110 x 2 = 220There were 220 green balls.


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


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)

41

52

53

5u 5u 5u 5u

Boys girls

2u 3u 6u 9u

11 u = 54

1u = 14

20u = 280


Parents’ Help with Child At Home •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

Math in real life


Reference

Conquer Problem Sums (Primary 3 and Primary 4 )-Published by onSponge Pte Ltd

-Mathematics Syllabus Primary 2007


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12 august 2011

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