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MATHEMATICS GRADE

National Book Foundation as

Federal Textbook Board Islamabad

5

Textbook of

MATHEMATICS Grade

5

National Book Foundation as

Federal Textbook Board Islamabad

OUR MOTTO Standards

Outcomes

Access

Style

© 2016 National Book Foundation as Federal Textbook Board, Islamabad. All rights reserved. This volume may not be reproduced in whole or in part in any form (abridged, photo copy, electronic etc.) without prior written permission from the publisher.

Textbook of Mathematics Grade - 5

Content Authors Pedagogical Author Designing / Layout Editor Desk Officer, Curriculum Management of Supervision of

: : : : : : :

Dr. Saleem Ullah, Mr. Khalid Mahmood, Mr. Obaid-Ur-Rehman Mr. Abdur Rashid Hafiz Rafiuddin, Mr. Shahzad Ahmad Mr. Majeed Ur Rehman Malik Mr. Javaid Saleem Mr. Aftab Soomro Prof. Dr. Attash Durrani (T.I), Advisor FTBB (NBF), Member National Curriculum Council

First Edition : 2014 Qty : 27,000 New Developed Edition : Mar. 2015 Qty : 30,000 2nd Print : Aug. 2015 Qty : 5000 3rd Print : Jan. 2016 Qty : Price : Rs. 140/Code : STE-513 ISBN : 978-969-37-0773-7 Printer : for Information about other National Book Foundation Publications, visit our Web site http://www.nbf.org.pk or call 92-51-9261125 or Email us at: [email protected]

Preface Preface Mathematics Grade - 5 is developed according to the National Curriculum 2006 and National Style Guide. It is being published since 2014 and in 2015 it was presented under the new management and supervision of textbook development principles and guidelines with new design and layout. The lessons are designed as per innovations. A colorful presentation is made so that this description should look as interesting as any literary or social subject. This may lead to an interactive approach. Our efforts are to make textbooks teachable with quality, i.e. maintaining of standards. It is a continuous effort and we will get feedback of the yearly feasibility reports and redesign the textbook every year. Like before, the National Book Foundation has made specific endeavours to publish the text and illustrations in much effective form. The meticulous effort of the team is acknowledged. Quality of Standards, Pedagogical Outcomes, Taxonomy Access and Actualization of Style is our motto. With these elaborations this series of new development was presented for use. After educational feedback, research and revision it is published again.

Dr. Inam ul Haq Javeid (Pride of Performance)

Managing Director National Book Foundation

Contents Contents 1

Numbers and Arithemtic Operation

9

2

P.6

Information Handling

HCF and LCM

P.26

P.145

8

3

Perimeter and Area

Fractions

MATHEMATICS

P.138

P.38

5 7

4

Geometry

Decimals and Percentages

P.116

P.54

6

5

Unitary Method

Distance, Time and Temperature

P.102

P.80

Answers

Glossary

P.170

P.183

4

1 Numbers and Arithmetic Operations

1 NUMBERS AND ARITHMETIC OPERATIONS After completing this lesson, you will be able to:

This is a 21 days lesson

Read and write numbers up to one billion in numerals and in words. Add and subtract numbers of arbitrary size. Multiply and divide up to 6-digit numbers by 2-digit and 3-digit numbers. Recognize BODMAS rule, using parenthesis. Verify distributive law. Solve real life problems involving mixed operations.

Look A number is a mathematical object If you are good in Mathematics, then you will enjoy. used to count, label, and measure. In mathematics, the definition of number has been extended over the years to include new numbers. Many people used numbers to create fun by making different number patterns. Have you seen any number pattern?

Reading

1.1 NUMBERS UP TO ONE BILLION In grade IV, we have learnt numbers up to one hundred million. In grade V we will learn numbers up to one billion in numerals and words identifying place values of digits in ten.

5

1 Numbers and Arithmetic Operations

1.1.1 Read Numbers up to One Billion We know that 100,000,000 is a 9 digits number. We read it as one hundred million. We write the place value of each digit in this number as follows Milions

Thousands

Ones

H.M

T.M

M

H.Th

T.Th

Th

H

T

U

1

0

0

0

0

0

0

0

0

The greatest 9 digits number is 999,999,999. The number next to 999,999,999 is 1,000,000,000. It is 10 digits number and is read as “one billion”. The place value of each digit in this number is Bilion

Milions

Thousands

Ones

B

H.M

T.M

M

H.Th

T.Th

Th

H

T

U

1

0

0

0

0

0

0

0

0

0

Example 1: Identify the place values of digits and then read the following numbers. (i) 1,236,658,211 (ii) 9,024,579,320 (iii) 3,201,506,056 Solution: (i) 1,236,658,211 Bilion

Millions

Thousands

Ones

B

H.M

T.M

M

H.Th

T.Th

Th

H

T

U

1

2

3

6

6

5

8

2

1

1

We read it as “one billion, two hundred thirty six million, six hundred fifty eight thousand two hundred eleven”. (ii) 9,024,579,320 Bilion

Millions

Thousands

Ones

B

H.M

T.M

M

H.Th

T.Th

Th

H

T

U

9

0

2

4

5

7

9

3

2

0

6

1 Numbers and Arithmetic Operations

We read it as “nine billion, twenty four million, five hundred seventy nine thousand and three hundred twenty”. (iii) 3,201,506,056 Bilion

Millions

Thousands

Ones

B

H.M

T.M

M

H.Th

T.Th

Th

H

T

U

3

2

0

1

5

0

6

0

5

6

We read it as “three billion, two hundred one million, five hundred six thousand and fifty six”.

1.1.2 Write Numbers up to One Billion To write numbers, we can follow the idea of place values for better understanding. Example 2: Write the following numbers in figures. i. Eight billion, two hundred ninety million four hundred three thousand, six hundred forty five. ii. One billion, five hundred eighty eight million, seven hundred fifty three thousand forty two. iii. Three billion, nine hundred million, four hundred four thousand, one hundred five. Solution: (i) Bilion

Millions

Thousands

B

H.M

T.M

M

H.Th

T.Th

Th

H

T

U

8

2

9

0

4

0

3

6

4

5

We write the number as 8,290,403,645. (ii)

Ones

1,588,753,042

(iii) 3,900,404,105

1 Numbers and Arithmetic Operations

7

Exercise 1.1 1. Identify the place values of digits and then read the following numbers. (i)

4,276,898,236 (ii)

7,154,890,722 (iii) 3,038,134,405

(iv) 4,632,569,730 (v)

6,336,480,422 (vi) 8,402,981,358

(vii) 9,178,980,786 (viii) 5,103,663,870 (ix) 4,401,876,280 2. Write the following numbers in figures. i. Seven billion, four hundred five million, six hundred twenty two thousand three hundred forty six. ii. Five billion, six hundred seventy eight million, nine hundred sixty two thousand seventy three. iii. Two billion, one hundred million, five hundred three thousand six hundred five. iv. Three billion, five hundred fifty two million, three hundred two thousand, seven hundred sixty five. v. Eight billion, two hundred ninety nine million, one hundred fifty two thousand, two hundred forty eight. vi. Nine billion, six hundred million, three hundred three thousand, five hundred six.

Reading

1.2 ADDITION AND SUBTRACTION The basic operations with whole numbers are addition, subtraction, multiplication and division. These are also called fundamental operations.

8

1 Numbers and Arithmetic Operations

1.2.1 Addition i. For addition and subtraction we generally write the numbers in columns according to their place values. ii. We start addition and subtraction with the ones, tens, hundreds and so on. iii. The numbers to be added are called addends and the result is called the sum. For example 2613 +1374

Addend Addend

3987

Sum

Example 3: Add 5396273 and 3241628. Solution: 5396273

+ 3241628 8637901

1.2.2 Subtraction The numbers to be subtracted is known as Subtrahend and the number from which it is being subtracted is known as Minuend and the result is called difference. For example 8929 – 1374 7555 Example 4: Subtract 24905713 from 36728526. Solution:

Minuend Subtrahend Difference

36728526 –24905713

11822813

9

1 Numbers and Arithmetic Operations

Exercise 1.2 1. Solve: (i)

3624667

(ii)

+ 316746

(iii)

5156928

+ (v)

(vii)

+

713476

82139167

(iv)

1732594

1432574

103603

+32670570

6234192

+

43506789

(vi)

31347465

2957421

21341375

450589

+ 74896234

5214196

(viii)

33747025

4967329

16369378 +

+ 1478582

(ix)

6134482

(x)

8876246

24680468

2957728

98765432

+ 5513590

+ 12345678

10

1 Numbers and Arithmetic Operations

2. Solve: (i)

82754380

(ii)

–10817593

(iii)

27763508 –

(v) –

(vii)

–

(iv)

(vi)

(viii)

– 197825932

(ix)

927047659

–

87832158

447147907

31347465 – 21341375

97921921

218328767

2776359

–136869851

7219479

819358757

54792465

42458576 – 30326261

(x)

719058007

– 297026924

11

1 Numbers and Arithmetic Operations

Reading

1.3 MULTIPLICATION AND DIVISION 1.3.1 Multiplication We know that multiplication is a process of repeated addition. The number by which any number is multiplied, is known as multiplier. The number which is to be multiplied is known as multiplicand. The result of multiplication is known as product of the multiplier and the multiplicand. For example 265 × 3 795

Multiplicand Multiplier Product

1.3.2 Multiplying Numbers by 10,100 and 1000 To multiply a number by 10, we write one zero to the right of the number and similarly to multiply a number by 100 and 1000, we write two and three zeros to the right of the number respectively. For example 49 ×10 = 490, 49 × 100 = 4900, Similarly 583 × 10 = 5830 583 × 100 = 58300 583 × 1000 = 583000 These further imply that 83 × 30 = (83 × 3) ×10 = 249 × 10 = 2490 83 × 300 = (83 × 3) × 100 = 249 × 100 = 24900 83 × 3000 = (83 × 3) × 1000 = 249 × 1000 = 249000

49 × 1000 = 49000 It is very simple. I can do it.

In a school, the students of grade V were asked about the answer of 2 + 2 × 2. Some students answered as 8 and other answered as 6. What is your answer? 6 or 8….

12

1 Numbers and Arithmetic Operations

1.3.3 Multiplication of a Number by 2-digit and 3-digit Numbers Example 5: Multiply 4137 by 213. Solution: Horizontal Form 4137 × 213 = 4137 × (200 + 10 + 3) = 4137 × 200 + 4137 × 10 + 4137 × 3 = 827400 + 41370 + 12511 = 881281 Vertical Form 4137 ×

213

(2 0 0 + 1 0 + 3)

12411

4137×3

41370

4137×10

827400

4137×200

Locate the mistake in the solution.

881181 In general we omit the detail of the procedure followed. Example 6: Multiply 724392 by 24. Solution: 724392 ×

24

2897568 14487840 17385408

Complete the following question.

13

1 Numbers and Arithmetic Operations

Example 7: Multiply 4629345 by 583. Solution:

4629345 ×

583

13888035 231467250 370347600 2698908135

1.3.4 Division Remember that division is the shortest way to repeat subtraction. The number to be divided is known as dividend. The number which divides is known as divisor. The number which tells how many times the divisor is contained in the dividend is known as quotient. Sometimes the dividend does not contain an exact number of times of the divisor, and then the number leftover is called remainder. For example 35 divisor

12

quotient

429

dividend

-36 69 -60 Example 8: Divide 628354 by 293.9 Solution:

2144 293 628354 -586 423 -293 1305 -1172 1334 -1172 162

remainder

I can solve…

14

1 Numbers and Arithmetic Operations

Step 1: How many times does 293 divides 628 293 × 2 = 586 586 is less than 628. Step 2: Take down 3 Hence 293 × 1 = 293 is less than 423. Step 3: Take down 5 293 × 4 = 1172 Step 4: Take down 4 293 × 4 = 1172 Note: Generally we don’t write these steps while dividing the numbers. Example 9: Divide 139022 by 26. Solution:

Fill the blank spaces?

5347

26

139022 -130 90 -78 122 -104 182 - 182

0

Exercise 1.3 1. Find the product. (i)

65 × 10

(iv) 570 × 60

(ii) 483 × 100

(iii) 6213 × 1000

(v) 658 × 500

(vi) 234 × 7000

15

1 Numbers and Arithmetic Operations

2. Find the product. (i) 10265 × 297

(ii) 548345 × 92

(iii) 62139 × 583

(iv) 57002 × 260

(v) 236458 × 253

(vi) 84670 × 372

(vi) 827601 × 47

(vii) 321752 × 78

3. Divide and find the quotient and the remainder. (i) 7725 ÷ 25

(ii) 152725 ÷ 149

(iii) 681736 ÷ 254

(iv) 13632 ÷ 24

(v) 1855032 ÷ 296

(vi) 478234 ÷ 234

Reading

1.3.5 Real Life Problems involving Mixed Operation of Addition, Subtraction, Multiplication and Division We have learnt how to perform the four fundamental operations namely, addition, subtraction, multiplication and division. Let us solve some real life problems using these operations. Example 10: A man purchased a plot of for Rs.1637800 to build a house. He spent Rs.713440 on the construction of the house. How much money did he spend in all? Solution: The total money will be the sum of amount he spent for the plot and construction of the house. 1637800 + 713440 2351240 Thus the total money he spent is Rs.2351240. Example 11: A soap factory produces 13265 soap cakes in a day. If there are 70 holidays in a year. How many soap cakes will the factory produce in the working days of a year?

16

1 Numbers and Arithmetic Operations

Solution: Number of working days in a year = 365 –70 = 295 Number of soap cakes produced in a day

= 13265

Number of soap cakes produced in 295 working days

= 13265 × 295 = 3913175

13265

×

295 66325

1193850 2653000 3913175 Example 12: There are 24375 books of same size to be arranged in shelves. If 375 books can be arranged in a shelf, how many shelves are required to set them arrange? Solution: Number of shelves needed to accommodate 24375 books =24375÷375 = 65 65 375 24375

-2250 1875 -1875 0 Thus, 65 shelves are required.

1 Numbers and Arithmetic Operations

17

Exercise 1.4 1. In an election, the winning candidate got 73846 votes. His rival got 24573 votes, 125 votes were rejected. Find the total number of votes polled. 2. In an examination, 3718351 students appeared. Only 3587602 students passed the examination. How many students were failed? 3. A factory manufactured 62786 bicycles in May, 73471 bicycles in June and 78471 in July. Find the total number of bicycles produced in three months? 4. A builder constructs 360 houses in a colony. If each house costs Rs.945060, find the total cost of 360 houres. 5. A carton can hold 30 bottles of water. How many cartons are required pack 18720 bottles of water?

to

6. A school has 850 students on its roll. If the annual fee per student is Rs.9750, what will be the total fee collected annually by the school? 7. What number must be multiplied by 384 to get 28032? 8. Asim wrote 300508 instead of writing 358. How much is 300508 greater than 358? 9. A man bought a house for Rs.7654900 and sold it for Rs.8000000. Find his profit. 10. There are 24 sections of different classes in a school. If each section consists of 35 students, find the total number of students.

Reading

1.4 USE OF BODMAS RULE We know how to add, subtract, multiply and divide using numbers. Sometimes, we need to perform two or more operations to simplify expressions like 2 + 3 –1, 8 – 2 × 3 –1 and 6 + 8 ÷ 2 + 3.

18

1 Numbers and Arithmetic Operations

To simplify expressions, we follow the following order of operations. First

Division

---------- D

Second

Multiplication

---------- M

Third

Addition

---------- A

Fourth

Subtraction

---------- S

The order in which different operations are carried out in simplifying expressions is called DMAS rule. Example 13: Simplify 5 × 3 – 4 × 2 using DMAS rule. Solution: 5 × 3 – 4 × 2 = 15 – 8 =7 Brackets have their own significance in solving questions. Brackets tell us which part of the sum is to be done first. For example to simplify 9 × (4 – 2), we solve inside the brackets first, that is 9 × (4 – 2) = 9 × 2 = 18 Brackets help us to solve equations involving two or more operations by telling us which part of the question is to do first. Such kinds of expressions are solved through BODMAS rule. We solve questions involving brackets in the following way. i. When brackets occur in numerical expressions, we simplify the expressions within the brackets first. ii. After removing brackets we simplify the expressions according to DMAS rule.

1 Numbers and Arithmetic Operations

19

For example 5 + (4 × 3 – 9) = 5 + (12 – 9) =5+3=8 Example 14: Simplify 128 4 12 (5 4) Solution:

128 4 12 (5 4) 128 4 12 1 32 12 44

Example 15: Simplify 80 (10 2) 9 3 2 Solution:

80 (10 2) 9 3 2 80 20 9 6 43 7

Exercise 1.5 Simplify each of the following numerical expressions, using BODMAS rule. 1.

(2 5) 2 12 8

2.

3 (16 4) (10 6) 11

3.

3 (12 3) 4 8 2

4.

30 6 80 4 (10 2 3)

5.

24 (8 4 5 2)

6.

12 4 6 2 10

7.

56 7 9 (6 2 3)

8.

200 56 (190 83) 50

9.

80 3 30 (7 3) 5

10.

73 3 (84 7) 27

11.

(8 3 2) (5 2 1) 4

12.

9 3 2 (25 5 2) (7 4)

13.

(9 3 3) (5 2) 4 2

14.

10 (10 5) (11 3 3) 26 2 13

20

1 Numbers and Arithmetic Operations

Reading

1.4.1 Verification of Distributive Laws of Numbers 1.

Distributive Law of Multiplication over Addition For any three numbers a, b and c, we have a (b + c) = a b + a c

2.

Distributive Law of Multiplication over Subtraction For any three numbers a, b and c, we have a (b c) = a b a c

Example 16: By taking a = 3, b = 6 and c = 4, prove the distributive law of multiplication over (i)

Addition

(ii)

Subtraction

Solution: (i).

Distribution law of multiplication over addition. i.e.

3 (6 + 4) = 3 6 + 3 4

L.H.S

= 3 (6 + 4) = 3 10 = 30

R.H.S

=

36+34

=

18 + 12 = 30

L.H.S = R.H.S 3 (6 + 4) = 3 6 + 3 4 (ii). Distributive law of multiplication over subtraction. i.e.

3 (6 4) = 3 6 3 4

Are you good in Math?

1 Numbers and Arithmetic Operations

L.H.S

= 3 (6 4) = 32 = 6

R.H.S

=

3634

=

18 12 =

6

L.H.S = R.H.S 3 (6 4) = 3 6 3 4

Exercise 1.6 1. Tick (√) the correct one in the following. i. (6 + 4) = 2 6 + 2 4 ii. 8 (3 + 1) = (8 3) + 1 iii. 7 (5 3) = 7 5 7 3 iv. 10 (6 4) = 10 6 10 4 v. 11 + (6 9) = 11 + 6 11 + 9 vi. (5 + 3) = 4 5 4 3 vii. Distributive law of multiplication over subtraction is true for all numbers. 2. Prove that: i. 2 (6 + 5)

=

26+25

ii. 7 (3 + 1)

=

73+71

iii. 12 (5 3) = 12 5 12 3 iv. 10 (9 4) = 10 9 10 4 v. 15 (5 + 0) = 15 5 + 15 0 3. Verify a × ( b + c ) = a × b + a × c , if : a = 20, b = 10 and c = 5

21

22

1 Numbers and Arithmetic Operations

4. Verify a × ( b c ) = a × b a × c , if : a = 9, b = 6 and c = 3

Key Points i.

100,000,000 is a 9-digits number. We read it as one hundred million.

ii.

The greatest 9-digits number is 999,999,999. The number next to 999,999,999 is 1,000,000,000. It is 10-digits number and is read as “one billion”.

iii. The four basic operations with whole numbers are addition, subtraction, multiplication and division. These operations are also called fundamental operations. iv. For addition and subtraction we generally write the numbers in columns according to their place values. v.

We start addition and subtraction with the ones, tens hundreds and so on. The numbers to be added are called addends and the result is called the sum.

vi. The numbers to be subtracted is known as subtrahend and the number from which it is being subtracted is known as minuend. vii. Multiplication is the process of repeated addition. viii. The number, by which any number is multiplied, is known as the multiplier. The number which is to be multiplied is known as multiplicand. The result of multiplication is known as the product of the multiplier and the multiplicand. ix. To multiply a number by 10, we write zero to the right of the number and similarly to multiply a number by 100 and 1000, we write two and three zeros to the right of the number respectively. x.

Division is the process to do repeated subtraction.

xi. The number to be divided is known as dividend. The number by which it is to be divided is known as the divisor. The number which tells how many times the divisor is contained in the dividend is known as the quotient.

23

1 Numbers and Arithmetic Operations

xii. Sometimes the dividend does not contain an exact number of times of the divisor, and then the number leftover is called the remainder. xiii. Distributive law of multiplication over addition and subtraction: For any three numbers a, b and c, we have a (b + c) = a b + a c and a (b c) = a b a c

Review Exercise 1 1. Encircle the best answer in the following. i.

The number 1,000,000,000 is read as (a) One million

ii.

(b) One billion (c) 10 billion (d) 100 billion

We start addition and subtraction with (a) One

(b) Two

(c) Hundreds (d) Thousands

iii. The number to be subtracted is known as (a) Minuend

(b) Difference (c) Addend

(d) Subtrahend

iv. Multiplication is the process of (a) Repeated subtraction (b) Repeated addition (c) Repeated division v.

(d) None of these

To multiply a number by 100, we write (a) One zero to the right

(b) Two zeros to the right

(c) One zero to the left

(d) Two zeros to the left

vi. The number which tells how many times the divisor is contained in the dividend is known as (a) Quotient

(b) Dividend

(c) Remainder (d) Addend

vii. The product of 213 × 100 is (a) 2.13

(b) 21.3

(c) 2130

(d) 21300

24

1 Numbers and Arithmetic Operations

viii. 2500 ÷ 100 is equal to (a)

2.5

(b)

25

(c)

250

(d) 250000

ix. To simplify the expression, we follow the following order of operation. a. Multiplication, addition, subtraction, division b. Multiplication, division, subtraction, addition c. Division, Multiplication, addition, subtraction d. Division, addition, subtraction, Multiplication x.

After removing the brackets we simplify the expression according to (a) MDAS rule

(b) DMAS rule

(c) MADS rule

(d) DASM rule

2. Solve the problems Written inside the open mouth.

Solve the questions in 10 minutes otherwise I will bite you. i.

1234567890 + 987654321

ii. 987654321–123456789

iii. 123456 × 333 iv. 3075 ÷ 25 v. 88888÷88 vi. 2 + (3 × 7) – (3 + 8) + 4 – 3 vii. 2222222 + 3333333 + 5555555

1 Numbers and Arithmetic Operations

3. Prove that: i. 12 (6 + 3) = 12 6 + 12 3 ii. 4 (6 4) = 4 6 4 4 4. Verify a × ( b + c ) = a × b + a × c , if : a = 15, b = 6 and c = 2 5. Verify a × ( b c ) = a × b a × c , if : a = 7, b = 9 and c = 1

25

Textbook of
MATHEMATICS GRADE
National Book Foundation as
Federal Textbook Board Islamabad
5
Textbook of
MATHEMATICS Grade
5
National Book Fo...

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