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word problems bar model method of solving mathematical Yeap Ban Har National Institute of Education Nanyang Technological University Singapore [email protected] Slides are available for download from www.banhar.blogspot.com or www.mmepdpm.pbworks.com SEMINAR at RYE CITY SCHOOL DISTRICT
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Rye Bar Model

May 10, 2015

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Education

Jimmy Keng

This presentation on bar modeling was for Grades Four and Five teachers from elementary and middle schools in Rye City School District in New York
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Page 1: Rye Bar Model

word problemsbar model method

of solving mathematical

Yeap Ban HarNational Institute of Education

Nanyang Technological UniversitySingapore

[email protected]

Slides are available for download fromwww.banhar.blogspot.com or www.mmepdpm.pbworks.com

SEMINAR at RYE CITY SCHOOL DISTRICT

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intellectual competenceintroduction

singapore curriculum

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curriculum framework

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Wellington Primary School, Singapore

Ministry of Education Singapore 2006

an excellent vehicle for the development and improvement of

a person’s intellectual competence

educationmathematics

intellectual competence

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Move 3 sticks to get two squares.

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Wellington Primary School, Singapore

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Wellington Primary School, Singapore

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Wellington Primary School, Singapore

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Wellington Primary School, Singapore

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Wellington Primary School, Singapore

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Move 3 sticks to get two squares.

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Move 3 sticks to get two squares.

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Wellington Primary School, Singapore

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VisualizationPatterning

Number Sense

“…development and improvement of a person’s

intellectual competencies...” Singapore Ministry of Education 2006

 

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PSLE ItemJohn 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 = 105 cm

105 cm of the copper wire was left.

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Primary Mathematics Standards Edition

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coaching techniquesmodel method

pre-algebra

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Siti

Rahim

29 kg

11 kg

Siti packs her clothes into a suitcase and it weighs 29 kg. Rahim packs his clothes into an identical suitcase and it weighs 11 kg. Siti’s clothes are three times as heavy as Rahim’s clothes.

What is the mass of Rahim’s clothes?What is the mass of the suitcase?

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Siti packs her clothes into a suitcase and it weighs 29 kg. Rahim packs his clothes into an identical suitcase and it weighs 11 kg. Siti’s clothes are three times as heavy as Rahim’s clothes.

What is the mass of Rahim’s clothes?What is the mass of the suitcase?

Siti

Rahim

29 kg

11 kg

11 kg 18 kg

2 units = 18 kg1 unit = 9 kg

Rahim’s clothes is 9 kg.The suitcase is 2 kg.

We can also find the mass of Siti’s clothes (27 kg) if required.

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Siti

Rahim x y

x y y yx + y = 11

x + 3y = 29

2y = 29 – 11 = 18

y = 18 ÷ 2 = 9

Siti packs her clothes into a suitcase and it weighs 29 kg. Rahim packs his clothes into an identical suitcase and it weighs 11 kg. Siti’s clothes are three times as heavy as Rahim’s clothes.

What is the mass of Rahim’s clothes?What is the mass of the suitcase?

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and primary onemodel method

in kindergarten

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Ali has 3 sweets. Billy has 5 sweets.How many sweets do they have altogether?

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mental computationsmodel method

alternate methods

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Cheryl has $20 less than David.Cheryl and David have $148 altogether,Find the amount of money Cheryl has.

Cheryl

David 20

$148

Let the amount of money that Cheryl has be $y.

y + (y + 20) = 148

2y + 20 = 148

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Cheryl has $20 less than David.Cheryl and David have $148 altogether,Find the amount of money Cheryl has.

Cheryl

David 20

$148 - $20 = $128

$128 ÷ 2 = $64

Cheryl has $64.

How about David? $84

2y + 20 = 148

2y = 148 – 20 = 128

y = 128 ÷ 2 = 64

Cheryl has $64.

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Cheryl has $20 less than David.Cheryl and David have $148 altogether,Find the amount of money Cheryl has.

Cheryl

David 20

$148

Let the amount of money that David has be $y.

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Cheryl

David 20

$148 + $20 = $168

20$168 ÷ 2 = $84

David has $84.

Cheryl has $64.

Cheryl has $20 less than David.Cheryl and David have $148 altogether,Find the amount of money Cheryl has.

Let the amount of money that David has be $y.

y + (y – 20) = 148

2y – 20 = 148

2y = 148 + 20 = 168

y = 168 ÷ 2 = 84 Cheryl has $64.

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see abstract ideasmodel method

helps average learners

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Josh spent 2/5 of his savings to buy a gift and 1/6 of the remainder to buy a snack.Josh then has $7.50 left. Find the amount Josh spent on the gift.   

5 units = $7.50

1 unit = $1.50

4 units = $1.50 x 4 = $6

Josh spent $6 on the gift. How about the snack? $1.50How much is his savings? $7.50

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There were three times as many students in soccer as there were in basketball. After 12 students moved from soccer to basketball, there number of students in both sports became equal. Find the number of students in soccer at first.  

soccer

basketball 12

Soccer 12 x 3 = 36

There were 36 students in soccer .

How about basketball?

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There were four times as many students in soccer as there were in basketball. After 12 students moved from soccer to basketball, there number of students in both sports became equal. Find the number of students in soccer at first.  

soccer

basketball

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There were four times as many students in soccer as there were in basketball. After 12 students moved from soccer to basketball, there number of students in both sports became equal. Find the number of students in soccer at first.  

soccer

basketball

3 units = 12

1 unit = 4

8 units = 32

There were 32 students in soccer at first

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problem solvingmodel method

new situations

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88 children took part in a swimming competition. 1/3 of the boys and 3/7 of the girls wore swimming goggles. Altogether 34 children wore swimming goggles. How many girls wore swimming goggles on that day?

boys

girls

34 34

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88 children took part in a swimming competition. 1/3 of the boys and 3/7 of the girls wore swimming goggles. Altogether 34 children wore swimming goggles. How many girls wore swimming goggles on that day?

boys

girls

34 34

88 – 34 – 34 = 20

34

2 units = 34 – 20 = 14

1 unit = 7

7 x 3 = 21 21 girls wore goggles

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leaving examinationmodel method

primary school

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Jim bought some chocolates and gave half of them to Ken.  Ken bought some sweets and gave half of them to Jim. 

Jim ate 12 sweets and Ken ate 18 chocolates. After that, the number of sweets and chocolates Jim had were in the ratio 1 : 7 and the number of sweets and chocolates Ken had were in the ratio 1 : 4. 

How many sweets did Ken buy?PSLE 2009

chocolates

Jim

Ken

sweets

18

3 parts 12 + 12 + 12 + 12 + 18 = 661 part 22

Half of the sweets Ken bought = 22 + 12 = 34So Ken bought 68 sweets.`

12

12

12

12

12

12