POWERMATH© SERIES PSLE Math - Simon Eio Learning Lab · 5 PSLE MATH BOOK 1 P6 POWERMATH© SERIES...

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PSLE MATH BOOK 1 P6 POWERMATH© SERIES

SIMON EIO LEARNING LAB PTE LTD ©

UPDATED 2019

PRIMARY POWERMATH© SERIES

PSLE Math Concepts You Must Master!

BOOK 1

SIMON EIO LEARNING LAB PTE LTD Blk 603 Clementi West St 1, #01-34 Reach For The Stars Singapore 120603

Tel/Fax: 67742157 Website: www.selearninglab.com.sg Email: [email protected]

BOOKLET IS TO BE USED EXCLUSIVELY DURING LESSONS AT SIMON EIO LEARING LAB PTE LTD. ALL RIGHTS RESERVED. WRITTEN PERIMISSION MUST BE SECURED FROM THE COMPANY TO USE OR REPRODUCE THIS BOOK.

NOT FOR SALE

BOOK 1 E-BOOK VERSION

http://www.selearninglab.com.sg/

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UPDATED 2019

Why some students don’t do well

Dear parents and students,

We wrote this short and simple E-book to aid parents and students in the

learning and mastery of mathematics. The subject can be frustrating to learn and

grasp for some.

One of the problems and concerns that we want to point towards to is – some

pupils may put in their 100% effort and time to do and practice math, but yet, do

not score well.

Here’s the reason:

Majority of students are taught to solve each question using only 1 method: A

“One-size fits all” methodology. Students may only be equipped with one single

problem solving strategy such as model drawing.

The rationale is – every child is unique and different. Every child learns

differently.

Your child may be a:

1. Numbers inclined student: unitary method, listing or algebraic

approach will suit him or her better.

2. Visually inclined – diagrams, shapes and pictorial examples helps him

or her see things better: modelling and guess and check approach

We have to discover and unveil a pupil’s learning preference and get him or her

to focus on this area of strength.

In this E-book, there are questions that we list down one or more methods. One

method or the other method may be your child’s preferred way of solving

mathematics.

We hope this will be an eye-opening experience for you and hopefully change

the way we approach teaching and educating our children on the subject of

mathematics.

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UPDATED 2019

For any enquiries on concepts, strategies or any other relevant questions, you

may reach us at [email protected] or whatsapp us directly at

97116072.

Regards

Mathematics Curriculum Team

Simon Eio Learning Lab

mailto:[email protected]

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UPDATED 2019

PSLE Math Concepts You Must Know! (Book 1)

Concepts Included in this Book:

Concept 1: Equal Fractions

Concept 2: Constant Difference

Concept 3: Double Total with Common Item

Concept 4: Unchanged Item

Concept 5: Unchanged Total

Concept 6: Repeated Identity

Concept 7: Repeated Identity (with diagrams)

Concept 8: Number x Value (Grouping)

Concept 9: Extra and Shortfall

Concept 10: Assumption

Concept 11: Remainder Concept (Type B)

Concept 12: Remainder Concept (Type C)

Concept 13: Remainder Concept (3 Branches)

Concept 14: Before-After Model

Concept 15: Simultaneous Relations

Concept 16: Internal Transfer

Concept 17: Internal Transfer (Cartoon)

Concept 18: Part-Whole (Cut by Parts)

Concept 19: Stacking Model

Concept 20: Working Backwards (Unitary)

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Level 1

In a class gathering of 39 pupils,

1

4 of the boys is equal to

2

5 of the girls. How many

more boys than girls are there at the gathering?

Method 1 – Model Drawing

Boys

Girls

Total units = 8 units + 5 units 8 units – 5 units = 3 units

= 13 units

13 units = 39 3u = 3 x 3

1 unit = 39 ÷ 13 = 9 more boys than girls

= 3

Method 2 – Unitary Method

1

4 of boys =

2

5 of girls

* In equal fractions, make the numerator EQUAL *

2

8 of boys =

2

5 of girls

Boys = 8 units

Girls = 5 units

Total units = 8 units + 5 units 8 units – 5 units = 3 units

= 13 units

13 units = 39 3u = 3 x 3

1 unit = 39 ÷ 13 = 9 more boys than girls

= 3

1 unit of boys

2 unit of girls

39

x 2

x 2

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Method 3 – Algebraic Approach (for the abler pupils)

Boys Girls

x y

x + y = 39 …. (1)

1

4 x =

2

5 y

x 20

5x = 8y …. (2)

(1) x 5 5x + 5y = 195 …. (3)

Replace ‘5x’ in (3) by 8y in (2)

So: 8y + 5y = 195

13y = 195

y = 195 ÷ 13 = 15

So there are 15 girls

x = 39 – 15 = 24 (number of boys)

24 – 15 = 9 more boys than girls

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Level 2

There are 792 people at a Community Club for a sport event. After 38

of the adults

and 14

of the children left the club half-way, the number of adults and children that

remained behind were equal. How many adults and children were left in the Club?

3

8 (left)

1

4 (left)

Adult Children

5

8 (remained)

3

4 (remained)

5

8 of adults =

3

4 of children

15

24 of adults =

15

20 of children (remained behind)

Adults : Children = 24 : 20 (at first)

24 + 20 = 44 units 792

1 unit 792 ÷ 44 = 18

15 u + 15 u = 30 units 30 x 18 = 540 left behind

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Concept 2: Constant Difference

Level 1 – Age based questions Joyce is 3 years old and her mother is 40 years older than Joyce. In how many

years’ time will Joyce’s mother be 5 times as old as Joyce?

Difference between Joyce and mum = 40 years (difference doesn’t change)

In the future…

Mum

Joyce

Difference between Joyce and her mum will still be 40 no matter how many

years have passed.

4 units = 40

1 unit = 10 (Joyce is 10 years old when her mum is 5 times as old as her)

10 – 3 = 7 years time

40

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Concept 2: Constant Difference (with equal increase or equal decrease)

Level 2

On a coach to Kuala Lumpur, there were

2

5 as many adults as children. After 14

children and 14 adults boarded the bus at Johor Bahru, there are now

3

4 as many

adults as children. How many adults were there in the bus at first?

Cool Note: If 14 children and 14 adults boarded the bus , the difference

between adults and children will still be the same.

Adults Children Difference

At first 2 5 3

At end 3 x 3 4 x 3 1 x 3

9 12 3

From the above, we see that both the adults and children increased by 7

units after 14 children and 14 adults boarded.

9 units – 2 units = 7 units 14

1 unit 14 ÷ 7 = 2

Adults at first = 2 x 2

= 4

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Concept 3: Double Total with Common Item When a box is filled with chocolates, its total mass is 488 g. When the same box is

filled with sweets, its total mass becomes 194 g. If the mass of the cholates is 7

times the mass of the sweets, what is the mass of the empty box?

Note: The box is the common item

Box + Choco

Box + Sweet

488 – 194 = 294

6 units = 294

1 unit = 294 ÷ 6

= 49

Mass of Box = 194 – 49

= 145 g

194

488

294

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Concept 4: Unchanged Item

On a MRT train, there were

3

5 as many children as adults. After 125 adults alighted

from the train, the number of children became

2

3 the number of adults. How many

children were there in the train?

Note: Adults changed, the ratio part of children should remain the same!

Adults Children

At first 5 x 2 3 x 2

10 6

At end 3 x 3 2 x 3

9 6

From the above, we see that the units of adults decreased from 10 units to 9

units.

10 units – 9 units = 1 unit 125

Children = 6 units 125 x 6 = 750

Make the ratio of

the children equal

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Concept 5: Unchanged Total Level 1 – Internal Transfer The ratio of Annabel’s ribbons to Clara’s ribbons is 1 : 3. After Clara gave Annabel 9

ribbons, the ratio becomes 2 : 3. Find the number of ribbons Annabel has at first.

Cool Note: Clara gave Annabel, their total will still be the same

Annabel Clara Total

At first 1 x 5 3 x 5 4 x 5

5 15 20

At end 2 x 4 3 x 4 5 x 4

8 12 20

From the above, we see that Clara’s ribbons decreased by 3 units and

Annabel’s ribbons increased by 3 units.


3 unit 9 ÷ 3 = 3

Annabel at first = 5 x 3

= 15

Make the total of

the ratio parts the

same

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Concept 5: Unchanged Total Level 2

Lena’s monthly expenditure was 3

10 of the total of Maria’s and Nora’s monthly

expenditure. Maria’s monthly expenditure was 1

5 of the total of Lena’s and Nora’s

monthly expenditure. The difference between Lena’s and Nora’s monthly

expenditure was $1682. How much was the total monthly expenditure of the three

women?

Cool Note: The total of Lena, Maria and Nora must be the same

Lena Maria + Nora Total

3 x 6 10 x 6 13 x 6

18 60 78

Maria Lena + Nora Total

1 x 13 5 x 13 6 x 13

13 65 78

Lena = 18 units

Maria = 13 units

Nora = 65 – 18 = 47 units


1 unit 1682 ÷ 29 = 58

Total expenditure = 78 units x 58

= $4524

Make the ratios of

the “totals” the

same!

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Concept 5: Unchanged Total

Level 3

Jack and Kate had some money. If Jack gave Kate $20, the ratio of Jack’s money to

Kate’s money became 3 : 5. If Jack gave Kate $70, the ratio of Jack’s money to

Kate’s money became 1 : 3. How much did each of them have?

Cool Note: Jack gave Kate in both cases, the total of the two person Is

unchanged.

Jack Kate Jack Kate

# * # *

-20 +20 -70 +70

3 : 5 1 : 3

2 : 6

Case 1: Jack gave away $20

Case 2: Jack gave away $70

So in case 2, Jack is $50 poorer.

Thus; 3u – 2u = 1u $50

3u $150

5u $250

Working Backwards:

Jack (#) : $150 + $20 = $170

Kate (*) : $250 - $20 = $230

x 2

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Concept 6: Repeated Identity Level 1

Siti had

2

7 as many sweets as Joel. Melvin had

3

4 as many sweets as Joel. If

Melvin had 39 more sweets than Siti, find the total number of sweets that

were shared among these 3 children at first.

Cool Note: Joel is the repeated person; his ratio / units must be the same

Siti Joel Melvin

2 x4 : 7 x4

4 x7 : 3 x7

8u 28u 21u

21 units – 8 units = 11 units

11 units 39

1 unit 39 ÷ 11 = 3

Total units = 21u +28u + 8u = 57u 57 x 3

= 171

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Concept 6: Repeated Identity Level 2

1

5 of the audience of a musical show were adults and the rest were children. The

number of girls was 3

8 of the number of girls. If there was an audience of 2500

people, find the number of girls at the musical show.

Cool Note: The children (boys and girls) are the repeated item, their

units must be the same

Number of Adults : Number of Children

1 x 2 : 4 x 2

2 : 8

Number of girls : Number of boys Total children

3 : 5 8

Total number of units = 2 + 8 = 10 units

10 units 2500

1 unit 2500 ÷ 10 = 250

3 units 250 x 3 = 750 girls

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Concept 7: Repeated Identity (Diagrams)

The figure is made up of a rectangle and a square. The area of the square is

2

5 the area of the rectangle. Given that

3

4 of the square is shaded and the

area of the unshaded part 72 cm2, what is the area of the square?

Solution:

Area of square : Area of rectangle

2 x 2 : 5 x2

4 10

Shaded of square : Unshaded of square Total

3 : 1 4

Shaded of rectangle : Unshaded of rectangle Total

3 : 7 10

Total unshaded area = 7u + 1u = 8u

8 units = 72

1 unit = 9

Total area = 1u + 3u + 7u (refer to diagram above)

= 10u x 9

= 90 cm2

2 units x 2 = 4u

5 units x2 = 10u

3

1

7

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Concept 8: Units x Value (Grouping) Level 1

Mrs Lim had a total of $360, consisting of $10 and $2 notes. Given that she had four

times as many $2 notes as $10 notes, how many notes did she have?

Method 1 – Grouping

$2 $2 $2 $2

$10

(5 notes per set)

Value of 1 Group = $2 x 4 + $10 = $18

$360 ÷ $18 = 20 groups

20 groups x 5 = 100 notes

Method 2 – Units multiply by value

Number of $2-notes Number of $10-notes

4u 1u

$8 units $10 units

$8 units + $10 units = $18 units $360

1 unit $360 ÷ $18 = 20 groups

5 unit 20 x 5 = 100 notes

There are 4 times as

many $2 notes as $10

notes

x $2 x $10

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Concept 9: Extra and Shortfall Mrs Lee bought some lollipops to give to some children. If she gave them 5 lollipops

each, she would have 4 left. If she gave them 7 lollipops each, she would be short

of 12. How many lollipops did Mrs Lee buy?

Method 1 – Make a List

Number of Children 1 2 3 4 5 6 7 8 9

Multiples of 5 :

+4

Number of sweets

5, 10, 15, 20, 25, 30, 35, 40 45

9, 14, 19, 24, 29, 34, 39 44 49

Multiples of 7

-12

Number of sweets

7 14 21 28 35 42 49 56 63

X 2, 9, 16, 23, 30, 37, 44, 51,

Robin has 44 lollipop

Method 2 – Draw a Model

Let number of children be 1 unit

If she gives out 5 sweets … 5u

If he gives out 7 sweets … 7u

From the model, we can see that:

2u 4 + 12 = 16

1u 8

To find total number of sweets = 5 x 8 + 4 = 44

Mrs. Lee has 44 sweets

No. of sweets

5u

7u

4

12

Case 1

Case 2

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Method 3 – Algebraic Approach

Let the number of children be x

Case 1 No. of sweets available = 5x + 4

Case 2 No. of sweets available = 7x – 12

So, logically:

7x – 12 = 5x + 4

+ 12 + 12

7x = 5x + 16

-5x -5x

2x = 16

x = 8

There are 8 children

Total number of sweets available = 5 x 8 + 4 = 44

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Level 1

Jaya had some twenty-cent and fifty-cent coins in his piggy bank. There are a

total of 280 coins in the piggy bank. If the total value of the coins was $119,

how many more fifty-cent coins than twenty-cent coins were there?

Assume all coins are 50-cent coins

280 x $0.50 = $140

$140 - $119 = $21

Cool note: What does $21 represent? $21 is the total difference in value

of the 20-cent and 50-cent coins.

$0.50 - $0.20 = $0.30

(difference in the value of 50-cent and 20-cent coins)

$21 ÷ $0.30 = 70 (number of 20-cent coins)

280 – 70 = 210 (number of 50-cent coins)

210 – 70 = 140 more fifty cent coins

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Level 2

Mindy seats for a math quiz with 60 questions. For every question that she

answered correctly, she was awarded 5 marks. For every question that she

answered incorrectly, 2 marks will be deducted from her marks. She scored a

total of 181 marks. How many questions did she answered correctly?

Assume all questions were answered correctly

60 x 5 = 300

300 - 181 = 119

5 + 2 = 7 marks (difference between correct and incorrect)

119 ÷ 7 = 17 (number of incorrect)

60 – 17 = 43 correct questions

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If you’d like to learn more concepts and strategies like the one you had in this E-Book, do email us at [email protected] or drop us a Whatsapp at 97116072

mailto:[email protected]

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