MATHEMATICS 4 TEXTBOOK – 2 Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________

Preface ClassKlap partners with schools, supporting them with learning materials and processes that are all crafted to work together as an interconnected system to drive learning. Our books strive to ensure inclusiveness in terms of gender and diversity in representation, catering to the heterogeneous Indian classroom. ClassKlap presents the Traveller series, designed specifically to meet the requirements of the new curriculum released in November 2016 by the Council for the Indian School Certificate Examinations (CISCE). Guiding principles: The 2016 CISCE curriculum states the following as a few of its guiding principles for Mathematics teaching: D evelop mathematical thinking and problem-solving skills and apply these skills to formulate and solve problems. A cquire the necessary mathematical concepts and skills for everyday life and for continuous learning in Mathematics and related disciplines. R ecognise and use connections among mathematical ideas and between Mathematics and other disciplines. R eason logically, communicate mathematically and learn cooperatively and independently. Each of these principles resonates with the spirit in which the ClassKlap textbooks, workbooks and teacher companion books have been designed. The ClassKlap team of pedagogy experts has carried out an intensive mapping exercise to create a framework based on the CISCE curriculum document. Key features of ClassKlap Traveller series: Theme-based content that holistically addresses all the learning outcomes specified by the CISCE curriculum. T he textbooks and workbooks are structured as per Bloom’s taxonomy to help organise the learning process according to the different levels involved. Student engagement through simple, age-appropriate content with detailed explanation of steps. Learning is supported through visually appealing images, especially for Grades 1 and 2. Increasing difficulty level in sub-questions for every question. Multiplication tables provided as per CISCE requirement. All in all, the Traveller Mathematics books aim to develop problem-solving and reasoning skills in the learners’ everyday lives while becoming adept at mathematical skills as appropriate to the primary level. – The Authors

Textbook Features I Will Learn About I Think Contains the list of concepts to be covered Arouses the student’s in the chapter along with the learning curiosity before objectives introducing the concept I Recall I RUenmdeermsbtearndand Pin-Up-Note Recapitulates the Elucidates the basic Highlights the key points or prerequisite knowledge for elements that form the definitions the concept learnt previously basis of the concept ? Train My Brain I Apply I Explore(H.O.T.S.) C hecks for learning to gauge Connects the concept E ncourages the student to the understanding level of the to real-life situations by extend the concept learnt student providing an opportunity to more complex scenarios to apply what the student has learnt Maths Munchies Connect the Dots Drill Time Aims at improving speed of Aims at integrating Revises the concepts with calculation and problem Mathematical concepts practice questions at the solving with interesting facts, with other subjects end of the chapter tips or tricks A Note to Parent E ngages the parent in the out-of- classroom learning of their child

Contents 7 Playing with Numbers 7.1 Factors and Multiples������������������������������������������������������������������������������1 7.2 H .C.F. and L.C.M.�����������������������������������������������������������������������������������10 8 Time 8.1 C onversion of Units of Time�������������������������������������������������������������������18 9 Fractions 9.1 Introduction to Fractions����������������������������������������������������������������������28 9.2 Like, Unlike and Equivalent Fractions��������������������������������������������������38 10 Money 10.1 Conversion between Rupees and Paise�������������������������������������������49 11 Measurement 11.1 Conversion of Units of Length�������������������������������������������������������������55 11.2 S tandard Units of Mass and Volume��������������������������������������������������62 11.3 A rea and Perimeter of Simple Shapes�����������������������������������������������69 12 Data Handling 12.1 Bar Graphs�������������������������������������������������������������������������������������������78

7Chapter Playing with Numbers I Will Learn About • finding factors and multiples of numbers. • prime and composite numbers. • finding prime factors of numbers. • calculating the H.C.F. and L.C.M. of numbers. 7.1 Factors and Multiples I Think Manoj and Surbhi were given a set of 200 books each to arrange in racks. Manoj plans to arrange the books in 10 racks while Surbhi plans to organise them in 20 racks. Are both the arrangements possible? Will there be any book left to organise? I Recall We have learnt multiplication and division in the previous grade. In the multiplication of two numbers, • the number that is multiplied is called multiplicand. • the number that multiplies the multiplicand is called the multiplier. • the answer obtained after multiplying is called the product. 1

Multiplication Fact ↓ ↓ ↓ Multiplicand Multiplier Product Similarly, in the division of two numbers, • the number that is divided is called the dividend. • the number that divides is called the divisor. • the answer obtained after division is called the quotient. Division Fact ↓ ↓ ↓ Dividend Divisor Quotient I Remember and Understand Factors of a number Raj arranged a bunch of 12 leaves in different ways as given below. 1 row of 12 leaves 2 rows of 6 leaves 2

3 rows of 4 leaves We can represent the arrangement of the leaves as: 1 row × 12 leaves = 12 leaves 2 rows × 6 leaves = 12 leaves 3 rows × 4 leaves = 12 leaves Similarly, 4 rows of 3 leaves = 12 leaves 6 rows of 2 leaves = 12 leaves 12 rows of 1 leaf = 12 leaves The numbers that divide a given number exactly (that is, without leaving any remainders) are called the factors of that number. In other words, the numbers which are multiplied to give a product are called the factors of the product. For example, in the given example, 1, 2, 3, 4, 6 and 12 are the factors of 12. Let us now find the factors of some numbers. Example 1: Find the factors of: a) 16 b) 18 Solution: a) T o find the factors of a given number, express it as a product of two numbers as shown: 16 = 1 × 16 =2 × 8 =4 × 4 Then write each factor only once. Therefore, the factors of 16 are 1, 2, 4, 8 and 16. b) 18 = 1 × 18 =2 × 9 =3 × 6 Therefore, the factors of 18 are 1, 2, 3, 6, 9 and 18. Playing with Numbers 3

Example 2: Find the common factors of 10 and 15. Solution: First write the factors of 10 and 15. 10 = 1 × 10 and 10 = 2 × 5 So, the factors of 10 are 1, 2, 5 and 10. 15 = 1 × 15 and 15 = 3 × 5 So, the factors of 15 are 1, 3, 5 and 15. Therefore, the common factors of 10 and 15 are 1 and 5. We can find the factors of a number by multiplication or by division. Example 3: Find the factors of 30. Solution: Factors of 30 Using multiplication 1 × 30 = 30 2 × 15 = 30 3 × 10 = 30 5 × 6 = 30 Therefore, the factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Using division 30 ÷ 1 = 30 30 ÷ 2 = 15 30 ÷ 3 = 10 30 ÷ 5 = 6 The different quotients and divisors of the given number are its factors. Therefore, the factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Facts on Factors • 1 is the smallest factor of a number. • 1 is a factor of every number. • Every number is a factor of itself. • The greatest factor of a number is the number itself. • Factors of a number are less than or equal to the number itself. 4

• Every number (other than 1) has at least two factors: 1 and the number itself. • The factors of a number are finite. Multiples of a number In the multiplication table of 2, the products obtained The products obtained are 2, 4, 6, 8, 10 and so on. These are called the multiples when a number is of 2. Similarly, multiplied by 1, 2, 3, 4, 5…. a) 3, 6, 9, 12, 15, 18, … are the multiples of 3. are called the multiples b) 5, 10, 15, 20, 25, 30, … are the multiples of 5. of that number. Let us now find the multiples of some numbers. Example 4: Find the first six multiples of: a) 4 b) 6 c) 7 Solution: The first six multiples of a number are the products when the number is multiplied by 1, 2, 3, 4, 5 and 6. a) 4 × 1 = 4, 4 × 2 = 8, 4 × 3 = 12, 4 × 4 = 16, 4 × 5 = 20, 4 × 6 = 24. Therefore, the first six multiples of 4 are 4, 8, 12, 16, 20 and 24. Now, complete these: b) 6 × 1 = 6, 6 × ______ = ______, 6 × ______ = ______ , 6 × ______ = ______ , ... Therefore, the first six multiples of 6 are ____ , ____ , ____ , ____ , ____ and ____. c) 7 × 1 = 7, 7 × ______ = ______, 7 × ______ = ______ , 7 × ______ = ______ , ... Therefore, the first six multiples of 7 are ____, ____ , ____ , ____ , ____ and ____. Example 5: Find the first three common multiples of 10 and 15. Solution: Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90,100,…. Multiples of 15 are 15, 30, 45, 60, 75, 90, 105,…. Therefore, the first three common multiples of 10 and 15 are 30, 60 and 90. Facts on Multiples • Every number is a multiple of itself. • Every number is a multiple of 1. • The smallest multiple of a number is the number itself. • The multiples of a number are greater than or equal to the number itself. • The multiples of a given natural number is unlimited. • The greatest multiple of a natural number cannot be determined. Playing with Numbers 5

Rules for divisibility We can find the numbers that can divide a given number exactly with the help of the rules for divisibility. Divisor Rule Examples 2 The ones digit of the given number must 10, 42, 56, 48, 24 be 0, 2, 4, 6 or 8. 3 The sum of the digits of the given number 36 (3 + 6 = 9) must be divisible by 3. 48 (4 + 8 = 12) The number formed by the last two digits 4 of the given number must be divisible by 1400, 3364, 2500, 7204 4 or both the digits must be zero. 5 The ones digit of the given number must 230, 375, 100, 25 be 0 or 5. 6 The number must be divisible by both 2 36, 480, 1200 and 3. 9 The sum of the digits of the given number 36 (3 + 6 = 9) must be divisible by 9. 144 (1 + 4 + 4 = 9) 10 The ones digit of the given number must 300, 250, 5670 be 0. Let us now apply the divisibility rules to check if a given number is divisible by 2, 3, 4, 5, 6, 9 or 10. Example 6: Which of the numbers 2, 3, 4, 5, 6, 9 and 10 exactly divide 4260? Solution: To check if 2, 3, 4, 5, 6, 9 or 10, divide 4260, apply their divisibility rules. Divisibility by 2: The ones place of 4260 has 0. So, it is divisible by 2. Divisibility by 3: The sum of the digits of 4260 is 4 + 2 + 6 + 0 = 12. 12 is divisible by 3. So, 4260 is divisible by 3. Divisibility by 4: T he number formed by the digits in the last two places of 4260 is 60, which is exactly divisible by 4. So, 4260 is divisible by 4. Divisibility by 5: The ones place of 4260 has 0. So, it is divisible by 5. Divisibility by 6: 4260 is divisible by 2 and 3. So, it is divisible by 6. Divisibility by 9: The sum of the digits of 4260 is 4 + 2 + 6 + 0 = 12, which is not divisible by 9. So, 4260 is not divisible by 9. Divisibility by 10: The ones place of 4260 has 0. So, it is divisible by 10. T herefore, the numbers that exactly divide 4260 are 2, 3, 4, 5, 6 and 10. 6

Example 7: Complete this table by putting a tick for the numbers which are factors of the given number. Put a cross for the rest of the numbers. Number 2 3 Divisible by 9 10 456 23 464 390 Solution: Apply the divisibility rules to check if the given numbers are divisible by the given factors. Tick the box if the number is a factor and cross the box if it is not. Number 2 3 Divisible by 9 10 456 23 464 390 Prime and composite numbers Observe the following table which gives the factors of numbers from 1 to 10. Number Factors Number of Number Factors Number of factors factors 1 1 1 6 1, 2, 3, 6 4 2 1, 2 2 7 1, 7 2 3 1, 3 8 4 1, 2, 4 2 9 1, 2, 4, 8 4 5 1, 5 10 1, 3, 9 3 3 2 1, 2, 5, 10 4 From the given table, we find that - 1) The numbers 2, 3, 5 and 7 have only two factors (1 and themselves). Such numbers are called prime numbers. 2) The numbers 4, 6, 8, 9 and 10 have three or more factors (more than two factors). Such numbers are called composite numbers. 3) The number 1 has only one factor. So, it is neither prime nor composite. Sieve of Eratosthenes Eratosthenes was a Greek mathematician. He invented a method to find prime numbers between any two given numbers. This is called the sieve of Eratosthenes. Steps to find prime numbers between 1 and 100 using the sieve of Eratosthenes: Playing with Numbers 7

Step 1: Write a grid of numbers 1 to 100. Step 2: Cross out 1 as it is neither prime nor composite. Step 3: Circle 2 as it is the first prime number. Then cross out all the multiples of 2. Step 4: Circle 3 as it is the next prime number. Then cross out all the multiples of 3. Step 5: Circle 5 as it is the next prime number. Then cross out all the multiples of 5. Step 6: Circle 7 as it is the next prime number. Then cross out all the multiples of 7. Continue this process till all the numbers between 1 and 100 are either circled or crossed out. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 The circled numbers are prime numbers and the crossed out numbers are composite numbers. There are 25 prime numbers between 1 and 100. These are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. Note: 1) All prime numbers (except 2) are odd. 2) 2 is the only even prime number. 8

? Train My Brain Answer the following: a) Find the first five multiples of 7 and 12. b) Find all the factors of 26 and 28. c) Find the common factors of 15 and 30. I Apply Let us apply the knowledge of factors and multiples in day-to-day life. Example 8: 24 students are to be lined up in rows for an assembly. Can they be lined up equally in 7 rows? Solution: 7 is not a factor of 24. In other words, 24 is not a multiple of 7. Therefore, 24 students cannot be lined up equally in 7 rows. Example 9: Suman has 15 toys. She wants to arrange them in shelves such that each shelf has the same number of toys. How many shelves will she need? Solution: Number of toys = 15 Factors of 15 = 1, 3, 5 and 15 We can divide 15 by the factors to find the number of shelves. So, 15 ÷ 1 = 15 15 ÷ 3 = 5 15 ÷ 5 = 3 15 ÷ 15 = 1 Therefore, Suman will need 1, 3, 5 or 15 shelves to organise the toys equally in each shelf. I Explore (H.O.T.S.) Consider these examples. Example 10: Ravi has two ribbons. One ribbon is 4 cm long and the other is 10 cm long. He wants to cut both the ribbons in such a way that each piece is of the same length. What will be the length of each piece? Playing with Numbers 9

Solution: The length of two ribbons is 4 cm and 10 cm. Factors of 4 = 1, 2, 4 Factors of 10 = 1, 2, 5, 10 Common factors of 4 and 10 are 1 and 2. Therefore, Ravi can cut both the ribbons in pieces of length 1 cm or 2 cm. Example 11: There are 14 students in Class 4A and 21 students in Class 4B. If the students of both the classes need to be arranged in the same number of rows, what is the maximum number of rows possible in each class? Solution: Number of students in Class 4A = 14 Number of students in Class 4B = 21 Factors of 14 = 1, 2, 7, 14 Factors of 21 = 1, 3, 7, 21 Common factors of 14 and 21 are 1 and 7. Therefore, the teacher will need a maximum of 7 rows to arrange these students. 7.2 H.C.F. and L.C.M. I Think Surbhi starts jumping from 3 and continues her steps in multiples of 3 as 3, 6, 9,... Suraj jumps in steps of 2. At which step-number will both of them meet? I Recall Let us revise the concept of factors by finding the common factors of the following pairs of numbers: a) 12, 9 b) 15, 10 c) 16, 12 d) 24, 16 e) 15, 21 I Remember and Understand Prime numbers have only 1 and themselves as their factors. Composite numbers have more than two factors. So, we can express composite numbers as the products of prime numbers or composite numbers. 10

For example, 5 = 1 × 5; 9 = 1 × 9 and 3 × 3; 20 = 1× 20, 2 × 10 and 4 × 5. We can express all composite numbers as the products of their prime factors. Prime Factorisation of Numbers To prime factorise a number, we use factor trees. Let us see a few examples to understand this better. Example 12: Carry out the prime factorisation of 36 by the factor tree method. Solution: To carry out the prime factorisation of a given number, draw a factor tree Step 1: as shown. Step 2: Express the given number as a product of 36 Step 3: two factors. One of these factors is the least 2 × 18 number (other than 1) that can divide it. The second factor may be prime or composite. If the second factor is a composite number, 2 ×2 ×9 express it as a product of two factors. One of these factors is the least number (other than 2 × 2 × 3 × 3 1) that can divide it. The second factor may be prime or composite. Repeat the process till the factors cannot be split further. In other words, repeat the process till the factors do not have any common factor other than 1. Step 4: Then write the given number as the product of all the prime numbers. Therefore, the prime factorisation of 36 is 2 × 2 × 3 × 3. Note: A factor tree must be drawn using a prime number as one of the factors of the number at each step. Example 13: Carry out the prime factorisation of 54 by the factor tree method. Solution: The prime factorisation of 54 using a factor tree is as shown below: 54 2 × 27 2×3×9 2×3×3×3 Therefore, the prime prime factorisation of 54 is 2 × 3 × 3 × 3. Playing with Numbers 11

Highest Common Factor (H.C.F.) and Least Common Multiple (L.C.M.) Finding factors and multiples helps us to find the Highest Common Factor (H.C.F.) and the Least Common Multiple (L.C.M.) of the given numbers. H.C.F.: The highest common factor of two or more numbers is the greatest number that divides the numbers exactly (without leaving a remainder). L.C.M.: The least common multiple of two or more numbers is the smallest number that can be divided by the numbers exactly (without leaving a remainder). Example 14: Find the highest common factor of 12 and 18. Solution: 12 = 1 × 12, 12 = 2 × 6 and 12 = 3 × 4 So, the factors of 12 are 1, 2, 3, 4, 6 and 12. 18 = 1 × 18, 18 = 2 × 9 and 18 = 3 × 6 So, the factors of 18 are 1, 2, 3, 6, 9 and 18. The common factors of 12 and 18 are 1, 2, 3 and 6. Therefore, the highest common factor of 12 and 18 is 6. Example 15: Find the least common multiple of 12 and 18. Solution: The multiples of 12 are 12, 24, 36, 48, 60, 72, …. The multiples of 18 are 18, 36, 54, 72, …. The common multiples of 12 and 18 are 36, 72, ….. Therefore, the least common multiple of 12 and 18 is 36. Let us now complete these tables of H.C.F. and L.C.M. of the given numbers. Example 16: Complete the H.C.F. table given. Some H.C.F. values are given for you. Numbers 10 12 18 30 2 2 3 12 6 15 15 Solution: Numbers 2 3 10 12 18 30 12 15 2222 1333 2 12 6 6 5 3 3 15 12

Example 17: Complete the L.C.M. table below. Some L.C.M. values are given for you. Solution: Numbers 10 12 18 30 2 18 3 12 12 15 30 Numbers 10 12 18 30 2 10 12 18 30 3 30 12 18 30 12 60 12 36 60 15 30 60 90 30 Finding H.C.F. and L.C.M. using the prime factorisation method Let us now find the H.C.F. of two numbers using the prime factorisation method. Consider these examples. Example 18: Find the H.C.F. of 48 and 54 by the prime • The product of two factorisation method. numbers is equal to the product of H.C.F. and Solution: The prime factorisation of 48 is L.C.M. 2 × 2 × 2 × 2 × 3. • The L.C.M. of the given The prime factorisation of 54 is numbers is always a 2 × 3 × 3 × 3. multiple of their H.C.F. Therefore, the H. C. F of 48 and 54 is 2 × 3 = 6. Example 19: Find the L.C.M. of 18 and 24 by the prime factorisation method. Solution: The prime factorisation of 18 is 2 × 3 × 3. The prime factorisation of 24 is 2 × 2 × 2 × 3. Therefore, the L.C.M. of 18 and 24 is 2 × 3 × 2 × 2 × 3 = 72. ? Train My Brain Find the prime factors of: a) 28 b) 20 c) 14 Playing with Numbers 13

I Apply Let us apply the knowledge of H.C.F. and L.C.M. to some real-life situations. Example 20: Nikki organised a party at her house. She went to a grocery store to buy candles and candle stands. The store only had packs that contained 18 candles and 12 stands each. How many packs would she have to buy to ensure that all the candles were placed on a candle stand? Solution: Number of candle stands in a pack = 12 Number of candles in a pack = 18 T o find the number of packs to buy to avoid any additional candle or candle stand, we find the L.C.M. of both the numbers. Prime factors of 12 = 2 × 2 × 3 Prime factors of 18 = 2 × 3 × 3 L.C.M. of 12 and 18 is 2 × 3 × 2 × 3 = 36. To find the number of packs, we divide 36 by 18 and 12. 36 ÷ 12 = 3; 36 ÷ 18 = 2 Therefore, Nikki would need 3 packs of candle stands and 2 packs of candles. Example 21: Rekha baked 20 butterscotch cakes and 24 mango cakes. She wants to pack equal number of cakes in each container. What is the maximum number of containers that Rekha will need? Solution: Number of butterscotch cakes baked = 20 Number of mango cakes baked = 24 T o find the total number of containers that can fit the maximum of cakes of each type, we find the H.C.F. of both the numbers. 20 24 2 × 10 2 × 12 2 × 5 2 × 6 2 × 3 14

Prime factorisation of 20 is 2 × 2 × 5. Prime factorisation of 24 is 2 × 2 × 2 × 3. H.C.F. of 20 and 24 = 2 × 2 = 4 Therefore, Rekha will need 4 containers. I Explore (H.O.T.S.) Let us see some more applications of H.C.F. and L.C.M. in real life. Example 22: R ita has a chocolate shop. She gives a free strawberry chocolate to every 14th customer and a free mango chocolate to every 16th customer. Who would be the first customer to receive both the free chocolates? Solution: Every 14th customer receives a free strawberry chocolate. Every 16th customer receives a free mango chocolate. The first customer who receives both the free chocolates is the L.C.M. of 14 and 16. Multiples of 14 = 14, 28, 42, 56, 70, 84, 98, 112, 126, ... Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, ... The L.C.M. of 14 and 16 is 112. T herefore, the 112th customer will receive both a free strawberry chocolate and a free mango chocolate. Example 23: Three bells rings at intervals of 2, 3 and 4 minutes respectively. Find the time when they will ring together. Solution: Intervals at which the bells ring = 2, 3 and 4 minutes To find the time when they will ring together, we find the L.C.M. of the numbers. Multiples of 2 = 2, 4, 6, 8, 10, 12, 14, ... Multiples of 3 = 3, 6, 9, 12, 15, 18, ... Multiples of 4 = 4, 8, 12, 16, ... The L.C.M. of 2, 3 and 4 = 12 Therefore, the bells will ring together after every 12 minutes. Playing with Numbers 15

Maths Munchies The numbers 60, 72, 84, 90 and 96 are the six 2-digit numbers which have the maximum number of factors. The factors of each one of them are: 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 Connect the Dots Social Studies Fun How do you think traffic controllers control traffic lights correctly? How do they ensure that all the lights do not turn green at the same time? They do it by finding the L.C.M. of the durations of all the traffic lights at a crossing. English Fun State the part of speech of the underlined word in the following sentence: Prime numbers have only two factors. Drill Time c) 32 7.1 Factors and Multiples 1) Find the factors of the following: a) 36 b) 49 d) 40 e) 28 16

2) Find the multiples of the following as given in the brackets: a) 7 (First 8) b) 15 (First 5) c) 100 (First 10) d) 25 (First 4) e) 30 (First 6) 3) Find the highest common factor of the following pairs of numbers: a) 12, 20 b) 15, 27 c) 24, 48 d) 16, 64 e) 30, 45 4) Find the least common multiple of the following pairs of numbers: a) 8, 10 b) 12, 15 c) 16, 20 d) 6, 18 e) 15, 30 7.2 H.C.F. and L.C.M. 5) Prime factorise using the factor tree: a) 18 b) 16 c) 56 d) 48 e) 63 6) Find the following using the prime factorisation method: a) L.C.M. of 32 and 56 b) H.C.F. of 25 and 75 c) H.C.F. of 96 and 108 d) L.C.M. of 45 and 75 A Note to Parent Help your child to collect a few objects such as leaves, pebbles, plastic bottles, flowers and so on from the surroundings. Ask him or her to make a table with their number. Then check for their divisibility by 2, 3, 4, 5, 6, 9 and 10. Playing with Numbers 17

Time8Chapter I Will Learn About • conversion of days, hours and minutes. • expressing the time in a.m. and p.m. • the 24-hour clock with respect to the 12-hour clock. 8.1 Conversion of Units of Time I Think Surbhi was going to school. When she started from home, the time shown by the clock was . Surbhi was easily able to read it as 8 o’clock. When she reached the school, the time shown by the school clock was . Surbhi found it difficult to read the time from the clock. Can you tell what time it is? I Recall There are 24 hours in a day. In a clock, the shorter hand is the hour hand and it completes one turn in 12 hours. The longer hand is the minute hand and it completes one turn in one hour. 18

Let us recall how to write the time from the given clocks. d) a) b) c) I Remember and Understand Observe this clock. The long hand is called the minute hand. The short hand is called the hour hand. It has numbers from 1 to 12 on its face. Between any two consecutive numbers on the clock, there are four small divisions. They divide the space between the two numbers into five equal parts. Each division between two consecutive numbers indicates a minute. Thus, the sixty divisions together make 60 minutes or 1 hour. Example 1: Let us read the time shown by these clocks. One has been done for you. a) b) c) The hour hand has crossed The hour hand has crossed The hour hand has crossed 10. _________________________. _________________________. The minute hand is on the The minute hand is on the The minute hand is on the third division after 2. So, the _________________________. _________________________. minutes is (2 × 5 + 3) = 13 minutes. So, the minutes is So, the minutes is _________________________. _________________________. Therefore, the time shown is 10:13. The time is ________. The time is ________. Time 19

We have learnt to read and write time in the 12-hour clock. Now, let us learn to write time in the 24-hour clock. In the 12-hour clock time: • The hour hand of the clock goes around the clock face (dial) twice in 24 hours. • To identify morning or evening, we write a.m. or p.m. along with the time. In the 24-hour clock time: • The time is expressed as a 4-digit number (in the form hhmm). 12-hour clock time 24-hour clock time Read as in 24-hour clock 6:24 a.m. 06:24 Six twenty-four hours 11:47 a.m. 11:47 3:31 p.m. 15:31 Eleven forty-seven hours 8:35 p.m. 20:35 Fifteen thirty-one hours Twenty thirty-five hours • Here, the first two digits from the left tell us the hours. The next two digits tell us the minutes. • 12 o’clock midnight is written as 00:00. 12 o’clock noon is written as 12:00. • The time before noon is written in a 12-hour clock but without a.m. For example, 5:34 a.m. is written as 05:34. • The time post noon is written by adding 12 to the number of hours. • For example, 5:37 p.m. is written as To convert the time from 24-hour 17:37, 6:19 p.m. is written as 18:19 and clock to 12-hour clock, we subtract so on. 12 from the number of hours and write p.m. after the difference. • When the hours is 12 and the minutes are more than 00, the time is past noon. To convert time from 12-hour clock In such cases, we do not subtract 12 to 24-hour clock for the time after from hours. We write the time as 12:22. 12 noon, we add 12 to the number of hours and omit writing p.m. Do you know? • Railways/Airlines/Armed forces use the 24-hour clock to keep time. • The 24-hour clock is used in digital watches. Example 2: Convert the given 24-hour clock time to 12-hour clock time. a) 14:23 b) 05:51 c) 09:17 d) 22:04 e) 18:00 f) 19:36 g) 23:55 h) 00:33 i) 00:45 j) 03:12 20

Solution: a) (14 – 12):23 = 2:23 p.m. b) 5:51 a.m. c) 9:17 a.m. d) (22 – 12):04 = 10:04 p.m. e) (18 – 12):00 = 6 p.m. f) (19 – 12):36 = 7:36 p.m. g) (23 – 12):55 = 11:55 p.m. h) (00 + 12):33 = 12:33 a.m. i) (00 + 12):45 = 12:45 a.m. j) 3:12 a.m. Example 3: Draw the hands of a clock to show the given time. a) 13:17 b) 6:53 c) 19:12 d) 22:35 Solution: To draw the hands of a clock, first note the minutes. If the minutes are between 1 and 30, draw the hour hand between the given hour and the next. But care should be taken to draw it closer to the given hour. If the minutes are between 30 and 60, draw the hour hand closer to the next hour. Then, draw the minute hand on the number that shows the given minutes. a) b) c) d) We have learnt how to read and show time, exact to minutes and hours. Let us now learn to convert days, hours and minutes. Example 4: Convert: a) 2 days into hours b) 11 hours into minutes c) 360 minutes into hours d) 96 hours into days Solution: Solved Solve these a) To convert days into hours, multiply by 24. 5 days = _______________ hours 2 days = ____________ hours 1 day = 24 hours Therefore, 2 days = 2 × 24 hours = 48 hours Time 21

Solved Solve these 9 hours = ___________ minutes b) T o convert hours into minutes, multiply by 60. 11 hours = ___________ minutes 1 hour = 60 minutes Therefore, 11 hours = 11 × 60 minutes = 660 minutes c) To convert minutes into hours, divide by 180 minutes = ___________ hours 60. 360 minutes = ___________ hours 60 minutes = 1 hour Therefore, 360 minutes = 360 ÷ 60 hours = 6 hours d) To convert hours into days, divide by 24. 120 hours = ___________ days 96 hours = ___________ days 24 hours = 1 day Therefore, 96 hours = 96 ÷ 24 days = 4 days (96 = 4 × 24) ? Train My Brain Draw the hands of a clock to show the time given: a) 6:12 b) 11:43 c) 3:32 I Apply Let us now understand the addition and subtraction of time. While adding time, we add the minutes (smaller units) first and then the hours (larger units). Let us see an example. Example 5: Add: 2 hours 15 minutes and 3 hours 35 minutes 22

Solution: Steps Solved Solve these Step 1: Write both the numbers one below the other. Hours Minutes Hours Minutes Step 2: Add the minutes first and 2 15 1 20 then the hours. +3 35 +3 10 Hours Minutes 2 15 +3 35 Hours Minutes 5 50 2 25 The sum is 5 hours 50 +2 20 minutes. While subtracting, we subtract the minutes first (smaller units) and then the hours (larger units). Example 6: Subtract: 1 hour 35 minutes from 3 hours 40 minutes Solution: Steps Solved Solve these Step 1: Write both the numbers Hours Minutes Hours Minutes one below the other, such that the smaller number is 3 40 4 30 subtracted from the larger. –1 35 –1 20 Step 2: Subtract the minutes Hours Minutes first and then the hours. 3 40 Hours Minutes –1 35 3 55 2 05 –2 40 The difference is 2 hours 5 minutes. The time between two given times is called the length of time or time duration or time interval. It is given by the difference between the end time and start time. Example 7: Find the duration of time from the given start time and end time. a) Start Time = 14:00; End Time = 17:45 b) Start Time = 6:30 p.m.; End Time = 9:45 p.m. Solution: a) Start Time = 14:00; End Time = 17:45 Time 23

Time from 14:00 to 17:00 = (17 – 14) hours = 3 hours Time from 17:00 to 17:45 = 17:45 – 17:00 = 45 minutes Total time duration = 3 hours + 45 minutes = 3 hours 45 minutes. b) Start Time = 6:30 p.m.; End Time = 9:45 p.m. Time from 6:30 p.m. to 7:00 p.m. = 7:00 – 6: 30 = 30 minutes Time from 7:00 p.m. to 9:00 p.m. = 9:00 – 7:00 = 2 hours Time from 9:00 p.m. to 9:45 p.m. = 9:45 – 9:00 = 45 minutes Total time = 30 minutes + 2 hours + 45 minutes = 2 hours 75 minutes. = 3 hours 15 minutes I Explore (H.O.T.S.) Let us see a few real-life examples involving duration of time. Example 8: The clocks given show the start time and end time of a Maths class respectively. How long was the Maths class? Solution: The start time is 9:00 and the end time is 9:45. So, the time between is the length of the Maths class = 9:45 – 9:00 = 45 minutes Therefore, the length of the Maths class was 45 minutes. Example 9: Anil took a flight from Delhi at 10:10 p.m. and reached Hyderabad in 2 hours 5 minutes. At what time did the flight reach Hyderabad? Solution: Start time of the flight = 10:10 p.m. Duration of travel = 2 hours 5 minutes End time = Start time + Duration of travel = 10:10 p.m. + 2 hours 5 minutes = 12:15 a.m. (After 12 midnight, the time is taken as a.m.) 24

Example 10: A movie began at 5:35 p.m. Lucky switched on the TV at 18:23. For how much time did Lucky miss the movie? Solution: Start time of the movie = 5:35 p.m. Time when Lucky switched on the TV = 18:23 = (18 – 12):23 = 6:23 p.m. Time from 5:35 p.m. to 6 p.m. = 25 minutes Time from 6 p.m. to 6:23 p.m. = 23 minutes Total time = 25 minutes + 23 minutes = 48 minutes. Therefore, Lucky missed 48 minutes of the movie. Maths Munchies One hour is divided into sixty minutes and one minute is divide into sixty seconds. The number 60 was probably chosen for its mathematical convenience. It is exactly divisible by many smaller numbers: 2, 3, 4, 5, 6, 10, 12, 15, 20 and 30. Connect the Dots Social Studies Fun The Earth’s rotation about its axis takes 24 hours. Science Fun A sundial is a tool that uses the position of the Sun to measure time, typically involving a shadow cast across a marked surface. Time 25

Drill Time 8.1 Conversion of Units of Time 1) Read the time on the clocks and write them in 12-hour and 24-hour formats. a) b) c) Evening Afternoon Morning d) e) f) Afternoon Evening Night 2) Convert: a) 240 minutes into hours b) 360 hours into days c) 5 hours into minutes d) 10 days into hours 3) Add: a) 3 hours 40 minutes and 1 hour 13 minutes b) 1 hour 26 minutes and 2 hours 22 minutes c) 2 hours 30 minutes and 2 hours 28 minutes 26

4) Subtract: a) 1 hour 20 minutes from 3 hours 55 minutes b) 3 hours 30 minutes from 4 hours 40 minutes c) 2 hours 40 minutes from 6 hours 49 minutes 5) Find the duration of time from the given start time and end time. a) Start Time = 12:00 and End Time = 14:15 b) Start Time = 3:00 p.m. and End Time = 5:00 p.m. c) Start Time = 3:15 p.m. and End Time = 7:20 p.m. d) Start Time = 7:20 a.m. and End Time = 10:41 a.m. e) Start Time = 15:56 and End Time = 17:57 A Note to Parent Encourage your child to make a note of the time when you go out with him or her. Also ask him or her to make a note of the time when you return. Ask your child to calculate the duration for which you went out based on the time he or she noted. Time 27

CHAPTER Fractions9 9Chapter Fractions 1 4 I Will Learn About • fractions as a part of a whole and its representation. • fractions of a collection. • like, unlike, unit and equivalent fractions. • addition and subtraction of like fractions. 9.1 Introduction to Fractions I Think Surbhi and her three friends, Joseph, Salma and Rehan, went on a picnic. Rehan brought only one apple with him. He wanted to share it equally with everyone. What part of the apple does each of them get? I Recall Look at the rectangle given in the next page. We can divide the whole rectangle into many equal parts as shown. 28

1 part: 2 equal parts: 3 equal parts: 4 equal parts: 5 equal parts: and so on. I Remember and Understand Let us understand the concept of parts of a whole. Suppose we want to share an apple with our friends. First, we count their number. Then, we cut the apple into as many equal pieces as the number of persons. Thus, each person gets an equal part of the apple after division. Fraction as a part of a whole A complete or full object is called a whole. Observe the following parts of a chocolate bar: Whole 2 equal parts 3 equal parts 4 equal parts Fractions 29

We can divide a whole into equal parts as shown. Each such division has a different name. To understand this better, let us do an activity. Activity: Halves Take a square piece of paper. Fold it into two equal parts as shown. Each equal part is called a ‘half’. ‘Half’ means 1 out of 2 equal parts. Putting these 2 equal parts together makes the complete piece of paper. We write ‘1 out of 2 equal parts’ as 1 . 2 In 1 , 1 is the number of parts taken and 2 is the total number of equal parts the whole 2 is divided into. Note: 1 and 1 make a whole. 2 2 Thirds In figure (a), observe that the three parts are not equal. We can also divide a whole into three equal parts as shown in figures (b) and (c). 11 1 33 3 Three unequal parts Three equal parts Fig. (a) Fig. (b) Fig (c) The shaded grey part in figure (c) is one out of three equal parts. So, each equal part is called a third or one-third. Two out of three equal parts of figure (c) are not shaded. We call it two-thirds (short form of 2 one-thirds). We write one-third as 1 and two-thirds as 2 . 3 3 Note: 1 , 1 and 1 or 1 and 2 make a whole. 3 3 3 3 3 30

Fourths Similarly, fold a square piece of paper into four equal parts. Each of them is called a fourth or a quarter. In figure (d), the four parts are not equal. In figure (e), each equal part is called a fourth or a quarter and is written as 1 . 4 1 4 1 4 1 4 1 4 Four parts Four equal parts Fig (d) Fig (e) Two out of four equal parts are called two-fourths and three out of four equal parts are called three-fourths, written as 2 and 3 respectively. 44 Note: Each of 1 and 3 1 1 1 and 1 or 1 1 and 2 make a whole. 4 4; 4, 4, 4 4 4, 4 4 The total number of equal parts a whole is divided into is called the denominator. The number of such equal parts taken is called the numerator. A fraction is a part of a whole. Numbers of the form Numerator are For example, 1 , 1, 1 , 2 and so on are fractions. 2 3 4 3 Denominator Let us now see a few examples. called fractions. Example 1: Identify the numerator and denominator in each of the following fractions: a) 1 b) 1 c) 1 2 3 4 Fractions 31

Solution: S. No Fractions Numerator Denominator a) 1 1 2 2 1 1 3 b) 3 1 1 4 c) 4 Example 2: Identify the fraction for the shaded parts in the figures given. a) b) Solution: Steps a) Solved b) Solve this Total number of equal parts Step 1: Count the number of Total number of = _______ equal parts the figure is divided equal parts = 8 into (Denominator). Number of parts shaded = Step 2: Count the number of Number of parts ______ shaded parts (Numerator). shaded = 5 Step 3: Write the fractions. Fraction = Fraction = 5 Numerator 8 Denominator Fraction of a collection Finding a half We can find different fractions of a collection. Suppose there are 10 pens in a box. To find a half of them, we divide them into two equal parts. Each equal part is a half. 32

Each equal part has 5 pens, as 10 ÷ 2 = 5. So, 1 of 10 pens is 5 pens. 2 Finding a third One-third is 1 out of 3 equal parts.In the given figure, there are12 bananas. To find a third, we divide them into three equal parts. Each equal part is a third. Each equal part has 4 bananas, as 12 ÷ 3 = 4. So, 1 of 12 bananas is 4 bananas. 3 1 3 1 3 1 3 Finding a fourth (or a quarter) One-fourth is 1 out of 4 equal parts. In this figure, there are 8 books. To find a fourth, we divide the number of books into 4 equal parts. Fractions 33

1 1 1 1 4 44 4 Each equal part has 2 books, as 8 ÷ 4 = 2. So, 1 of 8 books is 2 books. 4 Let us see a few examples to find the fraction of a collection. Example 3: Write the fraction of the coloured shapes in each of the following. Shapes Fractions a) b) c) Solution: The fractions of the coloured parts of the given shapes are – a) b) Shapes Fractions c) 2 6 3 6 5 8 Example 4: Colour the given shapes to represent the given fractions. Shapes Fractions a) 1 5 b) 2 7 c) 3 4 34

Solution: We can colour the shapes to represent the fractions as – Shapes Fractions a) 1 5 b) 2 7 c) 3 4 ? Train My Brain What fraction of the collection are: a) Chocolate cupcakes b) Strawberry cupcakes c) Blueberry cupcakes I Apply We have learnt to identify the fraction of a whole using the shaded parts. We can learn to shade a figure to represent a given fraction. Let us see some examples. Example 5: Shade a square to represent these fractions: 1 2 3 d) 1 a) 4 b) 3 c) 5 2 Solution: Follow these steps for shading a required shape to represent the given fractions. Fractions 35

Steps Solved Solve these 1 2 Step 1: Identify the 1 23 denominator and 4 35 Denominator the numerator. = Denominator Denominator Denominator Step 2: Draw the =4 == Numerator required shape. = Numerator = 1 Numerator Numerator == Divide it into as many equal parts as the denominator. Step 3: Shade the number of equal parts as the numerator. This shaded part represents the given fraction. Let us now learn to use fractions in real-life situations. Example 6: A set of 48 pens have 13 blue, 15 red and 11 black ink pens. The remaining are green ink pens. What fraction of the pens is green? Solution: Total number of pens = 48 Total number of blue, red and black ink pens = 13 + 15 + 11 = 39 Number of green ink pens = 48 – 39 = 9 Fraction of green ink pens = Number of green ink pens = 9 Total number of pens 48 Example 7: There is a bunch of balloons with three different colours. Write the fraction of balloons of each colour. Solution: Total number of balloons = 15 Number of green balloons = 2 2 Therefore, fraction of green balloons = 15 Number of yellow balloons = 3 36

3 Therefore, fraction of yellow balloons = 15 Number of red balloons = 10 10 Therefore, fraction of red balloons = 15 I Explore (H.O.T.S.) Let us see some examples of real-life situations involving fractions. Example 8: Answer the following questions: a) How many one-sixths are there in a whole? Represent it in a circle. b) H ow many one-fifths are there in a whole? Represent it in a pentagon (a shape with five sides). c) How many halves make a whole? Represent it in a rectangle. Solution: a) There are 6 one-sixths in a whole. b) There are 5 one-fifths in a whole. c) 2 halves make a whole. 11 66 11 55 1 1 11 11 6 6 55 22 1 1 1 Train My Brain 6 6 5 In some real-life situations, we need to find a fraction of some goods such as fruits, vegetables, milk, oil and so on. Let us now see some such examples. Example 9: One kilogram of apples costs ` 16 and one kilogram of papaya costs ` 20. If Rita buys 1 kg of apples and 1 kg of papaya, how much did she spend? 2 4 Solution: Cost of 1 kg apples = ` 16 Cost of 1 kg apples = 1 of ` 16 = ` 16 ÷ 2=` 8 2 2 Fractions 37

(To find a half, we divide by 2.) Cost of 1 kg papaya = ` 20 Cost of 1 kg papaya = 1 of ` 20 =` 20 ÷4 = `5 4 4 (To find a fourth, we divide by 4.) Therefore, the money spent by Rita = ` 8 + ` 5 = ` 13 Example 10: Sujay completed 2 of his Maths homework. If he had to solve 25 5 Solution: problems, how many did he complete? Fraction of homework completed by Sujay = 2 5 Total number of problems to be solved = 25 Number of problems Sujay solved = 2 of 25 = (25 ÷ 5) × 2 = 5 × 2 = 10 5 Therefore, Sujay has solved 10 problems. 9.2 Like, Unlike and Equivalent Fractions I Think Surbhi cuts 3 cakes into 18 equal pieces. Zeenal cuts a cake into 6 equal pieces. Did both of them cut the cakes into equal number of pieces? I Recall Let us recall the concept of fractions by finding the fraction of the parts not shaded in these figures. a) = ____________ b) = ____________ c) = ___________ d) = ____________ 38

I Remember and Understand In the previous concept, we have learnt about fractions. Now let us learn the different types of fractions. Like fractions: Fractions such as 1 , 2 and 3 , that have the same denominator are 88 8 called like fractions. Unlike fractions: Fractions such as 1 , 2 and 3 that have different denominators are 84 7 called unlike fractions. To understand like and unlike fractions, consider the following example. Example 11: Identify the like and unlike fractions from the following. 3 , 3 , 1 , 5 , 6 , 1 , 4 Fractions with numerator Solution: 7 5 7 7 7 4 11 ‘1’ are called unit 3 15 6 fractions. For example, 7 , 7 , 7 and 7 have the same denominator. 1 , 1 , 1 and so on. 234 Therefore, they are like fractions. 3 , 1 and 4 along with 3 or 1 or 5 or 6 5 4 11 7 7 7 7 have different denominators. Therefore, they are unlike fractions. Add and subtract like fractions While adding or subtracting like fractions, add or subtract only their numerators. Write the sum or difference with the same denominator. Let us understand this through some examples. Example 12: Solve: a) 3 + 1 b) 4 + 5 c) 8 – 4 d) 33 – 25 8 8 13 13 9 9 37 37 Solution: 3 1 3 +1 4 a) 8 + 8 = 8 = 8 4 5 4+5 9 b) 13 + 13 = 13 = 13 84 4 c) 9 – 9 = 9 d) 33 – 25 = 33 − 25 = 8 37 37 37 37 Fractions 39

Equivalent fractions Fractions that denote the same part of a whole are called equivalent fractions. Let us now understand what equivalent fractions are. Suppose a bar of chocolate is cut as shown. Ram eats 1 of the chocolate. 5 Then the piece of chocolate that he eats is: 2 Raj eats 10 of the chocolate. Then the piece of chocolate that he eats is: We see that both the pieces of chocolate are of the same size. So, we say that the fractions 1 and 2 are equivalent. We write them as 1 = 2 . 5 10 5 10 Example 13: Shade the regions to show equivalent fractions. a) 12 [ 3 and 6 ] b) 12 [ 4 and 8 ] Solution: 1 2 a) 3 6 40

b) 1 2 4 8 Example 14: Find the figures that represent equivalent fractions. Also, mention the fractions. a) b) c) d) Solution: The fraction represented by trheeprsehsaednetsd24p.aTrht eofshfigaudreedap) aisrt21o.f figure d) The shaded part of figure b) represents 1 . 2 So, the shaded parts of figures a), b) and d) represent equivalent fractions. Methods to find equivalent fractions There are two methods to find equivalent fractions. Let us learn them through a few examples. Example 15: Find two fractions equivalent to the given fractions. 2 12 a) 11 b) 16 Solution: To find fractions equivalent to the given fractions, we either multiply or divide both the numerator and the denominator by the same number. a) 2 11 W e see that 2 and 11 do not have any common factors. So, we 2 cannot divide them to get an equivalent fraction of 11. T herefore, we multiply both the numerator and the denominator by the same number, say 5. 2 2×5 = 10 11= 11× 5 55 Therefore, 10 is a fraction equivalent to 121. 55 Fractions 41

S imilarly, we multiply both the numerator and the denominator by the same number, say 2. 2×2 = 4 11×2 22 Therefore, 4 is a fraction equivalent to 121. 22 b) 12 16 W e see that 12 and 16 have common factors 2 and 4. So, dividing both the numerator and the denominator by 2 and 4, we get fractions equivalent to the given fraction 12 . 16 12 ÷ 2 = 6; 12 ÷ 4 = 3 16 ÷ 2 8 16 ÷ 4 4 Therefore, 6 and 3 are the fractions equivalent to 12 . 8 4 16 ? Train My Brain Identify the like fractions from each of the following. a) 1 , 3 , 2 , 4 b) 151,161, 2 ,181, 67 ,171, 71 c) 1 , 5 , 2 , 3 , 18 , 3, 3, 6 4 6 6 6 7 3 25 25 5 25 4 7 25 I Apply Let us see some word problems involving like, unlike and equivalent fractions. Example 16: Venu paints four-sixths of a cardboard and Raj paints two-thirds of a similar sized cardboard. Who has painted a larger area? Solution: Fraction of the cardboard painted by Venu and Raj are as follows: Venu Raj It is clear that, both Venu and Raj have painted an equal area on each of the cardboards. 42

Example 17: Colour each figure to represent the given fraction and compare them. 3 2 5 5 Solution: 3 Clearly, the part of the figure represented by 5 is greater than that represented by 2 Therefore, 3 is greater than 2. 5. 5 5 Example 18: Arrange 1 , 6 , 2 , 5 and 4 in the ascending and descending orders. 7 7 7 7 7 Solution: Comparing the numerators of the given like fractions, we have 1 < 2 < 4 < 5 < 6. So, 1 < 2 < 4 < 5 < 6 . 7 7 7 7 7 1 2 4, 5 6 Therefore, the required ascending order is 7, 7, 7 7, 7. We know that, the descending order is just the reverse of the ascending order. 65421 Therefore, the required descending order is 7 , 7 , 7 , 7 , 7 . Example 19: The figure shows some parts of a ribbon coloured in blue and yellow. Find the total part of the ribbon coloured blue and yellow. What part of the ribbon is not coloured? Solution: Total number of parts of the ribbon = 9 Part of the ribbon coloured blue = 2 9 3 Part of the ribbon coloured yellow = 9 2 3 2+3 5 Total part of the ribbon coloured = 9 + 9 = 9 = 9 Part of the ribbon that is not coloured is 9− 5 = 9−5 = 4 9 9 9 9 (Note: This is the same as writing the fraction of the ribbon not coloured from the figure. 4 parts of the 9 parts of the ribbon are not coloured.) Fractions 43

I Explore (H.O.T.S.) Consider the following examples. Example 20: Draw four similar rectangles. Divide them into 2, 4, 6 and 8 equal parts. Then colour 1 , 2 , 3 and 5 parts of the rectangles respectively. 2 4 6 8 Compare these coloured parts and write the fractions using >, = or <. Solution: 1 2 2 4 3 6 5 8 From the coloured parts of these rectangles, we can see that all of 5 12 3 them except 8 are of the same size. So, the fractions, 2 , 4 and 6 are equivalent. 123 Therefore, 2 = 4 = 6 . Example 21: Veena ate 5 of a pizza in the morning and 1 in the evening. What part 8 8 of the pizza is remaining? 44

Solution: Part of the pizza eaten by Veena in the morning = 5 8 1 Part of the pizza eaten by Veena in the evening = 8 T o find the remaining part of pizza, add the parts eaten and subtract the sum from the whole. Total part of the pizza eaten = 5 + 1 = 5 +1 = 6 8 8 8 8 6 86 2 Part of the pizza remaining = 1 – 8 = 8 – 8 = 8 Therefore, 2 part of the pizza is remaining. 8 Maths Munchies Egyptians have a different way to represent fractions. To represent 1 as numerator, they use a mouth picture which literally means ‘part’. So, the fraction one-fifth will be shown as given in the image. On the other hand, fractions were only written in words in Ancient Rome. 1 was called unica 6 was called semis 12 12 1 1 was called scripulum 24 was called semunica 144 Connect the Dots Science Fun Around 7 out of 10 parts of air is nitrogen. Oxygen is at the second position. 2 out of 10 parts of air is oxygen. Fractions 45

English Fun Think of at least two words that rhyme with each ‘numerator’ and ‘denominator’. Drill Time 9.1 Introduction to Fractions 1) Find the numerator and the denominator in each of these fractions. 2 b) 1 2 45 a) 5 7 c) 3 d) 9 e) 7 2) Identify the fractions of the shaded parts in these figures. a) b) c) d) e) 3) Find the fraction of the coloured parts in each of these figures. a) b) c) d) e) 4) Find 1 and 1 of the following collection. 2 4 46

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