mathematics grade 5 book math 05 new.pdf · if you are good in mathematics, then you will enjoy....

National Book Foundationas

Federal Textbook BoardIslamabad

GRADE 5MATHEMATICSMATHEMATICSTextbook of

National Book Foundation as

Federal Textbook BoardIslamabad

Grade

5

Textbook of

MATHEMATICSMATHEMATICS

Content Authors : Dr. Saleem Ullah, Mr. Khalid Mahmood, Mr. Obaid-Ur-RehmanPedagogical Author : Mr. Abdur RashidDesigning / Layout : Hafiz Rafiuddin, Mr. Shahzad AhmadEditor : Mr. Majeed Ur Rehman MalikDesk Officer, Curriculum : Mr. Javaid SaleemManagement of : Mr. Aftab SoomroSupervision of : Prof. Dr. Attash Durrani (T.I), Advisor FTBB (NBF),

Member National Curriculum Council

First Edition : 2014 Qty : 27,000New Developed Edition : Mar. 2015 Qty : 30,0002nd Print : Aug. 2015 Qty : 50003rd Print : Jan. 2016 Qty :Price : Rs. 140/-Code : STE-513ISBN : 978-969-37-0773-7Printer :for Information about other National Book Foundation Publications, visit our Web site http://www.nbf.org.pk or call 92-51-9261125or Email us at: [email protected]

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

Textbook ofMathematics Grade - 5

OUR MOTTO

Standards Outcomes Access Style

PrefacePrefacePrefaceMathematics 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 DirectorNational Book Foundation

ContentsContentsContents

MATHEMATICS

5

Numbers and ArithemticOperation

1

P.6HCF

and LCM

2

P.26

Fractions

3

P.38

Decimals and Percentages

4

P.54

Distance, Time and

Temperature

5

P.80

Unitary Method

6

P.102

Geometry

7

P.116

Perimeter and Area

8

P.138

Information Handling

9

P.145

Answers

P.170

Glossary

P.183

1 Numbers and Arithmetic Operations 4

NUMBERS AND ARITHMETIC OPERATIONS

A number is a mathematical object

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?

If you are good in Mathematics, then you will enjoy.

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.

Look

Reading

1

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.

http://en.wikipedia.org/wiki/Mathematical_object

http://en.wikipedia.org/wiki/Counting

http://en.wikipedia.org/wiki/Measurement

http://en.wikipedia.org/wiki/Mathematics


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

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

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

We read it as “one billion, two hundred thirty six million, six hundred fifty eight

thousand two hundred eleven”.

(ii) 9,024,579,320

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

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

Bilion Millions Thousands Ones


1 2 3 6 6 5 8 2 1 1



9 0 2 4 5 7 9 3 2 0


We read it as “nine billion, twenty four million, five hundred seventy nine thousand

and three hundred twenty”.

(iii) 3,201,506,056

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)

We write the number as 8,290,403,645.

(ii) 1,588,753,042

(iii) 3,900,404,105



3 2 0 1 5 0 6 0 5 6



8 2 9 0 4 0 3 6 4 5


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.

1.2 ADDITION AND SUBTRACTION The basic operations with whole numbers are addition, subtraction, multiplication

and division. These are also called fundamental operations.

Exercise 1.1

Reading


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

Example 3:

Add 5396273 and 3241628.

Solution:

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

Example 4:

Subtract 24905713 from 36728526.

Solution:

3 6 7 2 8 5 2 6

– 2 4 9 0 5 7 1 3

1 1 8 2 2 8 1 3

5 3 9 6 2 7 3

+ 3 2 4 1 6 2 8

8 6 3 7 9 0 1

2 6 1 3 Addend

+ 1 3 7 4 Addend

3 9 8 7 Sum

8 9 2 9 Minuend

– 1 3 7 4 Subtrahend

7 5 5 5 Difference


1. Solve:

(i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii)

(ix) (x)

Exercise 1.2

3 6 2 4 6 6 7

+ 3 1 6 7 4 6

4 3 5 0 6 7 8 9

+ 7 1 3 4 7 6

6 2 3 4 1 9 2

2 9 5 7 4 2 1

+ 4 5 0 5 8 9

3 1 3 4 7 4 6 5

2 1 3 4 1 3 7 5

+ 7 4 8 9 6 2 3 4

5 1 5 6 9 2 8

1 7 3 2 5 9 4

+ 1 0 3 6 0 3

8 2 1 3 9 1 6 7

1 4 3 2 5 7 4

+ 3 2 6 7 0 5 7 0

5 2 1 4 1 9 6

4 9 6 7 3 2 9

+ 1 4 7 8 5 8 2

3 3 7 4 7 0 2 5

1 6 3 6 9 3 7 8

+ 8 8 7 6 2 4 6

6 1 3 4 4 8 2

2 9 5 7 7 2 8

+ 5 5 1 3 5 9 0

2 4 6 8 0 4 6 8

9 8 7 6 5 4 3 2

+ 1 2 3 4 5 6 7 8


2. Solve:

(i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii)

(ix) (x)

8 2 7 5 4 3 8 0

– 1 0 8 1 7 5 9 3

5 4 7 9 2 4 6 5

– 2 7 7 6 3 5 9

2 7 7 6 3 5 0 8

– 7 2 1 9 4 7 9

4 4 7 1 4 7 9 0 7

– 1 3 6 8 6 9 8 5 1

8 1 9 3 5 8 7 5 7

– 9 7 9 2 1 9 2 1

3 1 3 4 7 4 6 5

– 2 1 3 4 1 3 7 5

2 1 8 3 2 8 7 6 7

– 1 9 7 8 2 5 9 3 2

4 2 4 5 8 5 7 6

– 3 0 3 2 6 2 6 1

9 2 7 0 4 7 6 5 9

– 8 7 8 3 2 1 5 8

7 1 9 0 5 8 0 0 7

– 2 9 7 0 2 6 9 2 4


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

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, 49 × 1000 = 49000

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

Reading

2 6 5 Multiplicand

× 3 Multiplier

7 9 5 Product

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….

It is very simple. I can do it.


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

In general we omit the detail of the procedure

followed.

Example 6:


Solution:

7 2 4 3 9 2

× 2 4

2 8 9 7 5 6 8

1 4 4 8 7 8 4 0

1 7 3 8 5 4 0 8

4 1 3 7

× 2 1 3 (2 0 0 + 1 0 + 3)

1 2 4 1 1 4 1 3 7 × 3

4 1 3 7 0 4 1 3 7 × 1 0

8 2 7 4 0 0 4 1 3 7 × 2 0 0

8 8 1 1 8 1

Locate the mistake in the

solution.

Complete the following

question.


Example 7:


Solution:

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

Example 8: Divide 628354 by 293.

Solution:

4 6 2 9 3 4 5

× 5 8 3

1 3 8 8 8 0 3 5

2 3 1 4 6 7 2 5 0

3 7 0 3 4 7 6 0 0

2 6 9 8 9 0 8 1 3 5

2 1 4 4

2 9 3 6 2 8 3 5 4

- 5 8 6

4 2 3

- 2 9 3

1 3 0 5

- 1 1 7 2

1 3 3 4

- 1 1 7 2

1 6 2

3 5 quotient

divisor 12 4 2 9 dividend

- 3 6

6 9

- 6 0

9 remainder

I can solve…


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:

1. Find the product.

(i) 65 × 10 (ii) 483 × 100 (iii) 6213 × 1000

(iv) 570 × 60 (v) 658 × 500 (vi) 234 × 7000

Exercise 1.3

5 3 4 7

2 6 1 3 9 0 2 2

- 1 3 0

9 0

- 7 8

1 2 2

- 1 0 4

1 8 2

- 1 8 2

0

Fill the blank spaces?


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

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.

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?

Reading

1 6 3 7 8 0 0

+ 7 1 3 4 4 0

2 3 5 1 2 4 0


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

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

= 2 4 3 7 5 ÷ 3 7 5

= 65

Thus, 65 shelves are required.

1 3 2 6 5

× 2 9 5

6 6 3 2 5

1 1 9 3 8 5 0

2 6 5 3 0 0 0

3 9 1 3 1 7 5

65

3 7 5 2 4 3 7 5

- 2 2 5 0

1 8 7 5

- 1 8 7 5

0


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 to

pack 18720 bottles of water?

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.

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.

Exercise 1.4

Reading


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.


For example

5 + (4 × 3 – 9)

= 5 + (12 – 9)

= 5 + 3 = 8

Example 14:

Simplify )45(124128

Solution: )45(124128

44

1232

1124128

Example 15:

Simplify 239)210(80

Solution: 239)210(80

7

34

692080

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

1. 8122)52( 2. 11)610()416(3

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. 4)125()238( 12. )47()2525(239

13. 24)25()339( 14. 13226)3311()510(10

Exercise 1.5


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


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 = 3 6 + 3 4

= 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

Reading

Are you good in Math?


L.H.S = 3 (6 4)

= 3 2 = 6

R.H.S = 3 6 3 4

= 18 12 = 6

L.H.S = R.H.S

3 (6 4) = 3 6 3 4

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) = 2 6 + 2 5

ii. 7 (3 + 1) = 7 3 + 7 1

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

Exercise 1.6


4. Verify a × ( b c ) = a × b a × c , if :

a = 9, b = 6 and c = 3

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.

Key Points


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:


a (b + c) = a b + a c and a (b c) = a b a c

1. Encircle the best answer in the following.

i. The number 1,000,000,000 is read as

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

ii. 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 (d) None of these

v. 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

Review Exercise 1


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


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

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