MATHEMATICS TEXTBOOK – PART 2 3 Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________
Preface IMAX partners with schools, supporting them with learning materials and processes that are all crafted to work together as an interconnected system to drive learning. IMAX Program presents the latest version of this series – updated and revised after considering the perceptive feedback and comments shared by our experienced reviewers and users. This series endeavours to be faithful to the spirit of the prescribed board curriculum. Our books strive to ensure inclusiveness in terms of gender and diversity in representation, catering to the heterogeneous Indian classroom. The books are split into two parts to manage the bag weight. The larger aim of the curriculum regarding Mathematics teaching is to develop the abilities of a student to think and reason mathematically, pursue assumptions to their logical conclusion and handle abstraction. The Mathematics textbooks and workbooks offer the following features: Structured as per Bloom’s taxonomy to help organise the learning process according to the different levels involved Student engagement through simple, age-appropriate language S upported learning through visually appealing images, especially for grades 1 and 2 Increasing rigour in sub-questions for every question in order to scaffold learning for students W ord problems based on real-life scenarios, which help students to relate Mathematics to their everyday experiences Mental Maths to inculcate level-appropriate mental calculation skills S tepwise breakdown of solutions to provide an easier premise for learning of problem-solving skills Overall, the IMAX Mathematics textbooks, workbooks and teacher companion books aim to enhance logical reasoning and critical thinking skills that are at the heart of Mathematics teaching and learning. – The Authors
Textbook Features Let Us Learn About Think Contains the list of learning objectives Introduces the concept and to be covered in the chapter arouses curiosity among students Recall Discusses the prerequisite knowledge for the concept from the previous academic year/chapter/ concept/term Remembering and Understanding Explains the elements in detail that form the Application basis of the concept Ensures that students are engaged in learning throughout Connects the concept to real-life situations by enabling students to apply what has been learnt through the practice questions Higher Order Thinking Skills (H.O.T.S.) Encourages students to extend the concept learnt to advanced scenarios Drill Time Additional practice questions at the end of every chapter
Contents 3Class 7 Time 1 6 7.1 Read a Calendar 7.2 Read Time Correct to the Hour 13 19 8 Division 29 8.1 Division as Equal Grouping 35 8.2 Divide 2-digit and 3-digit Numbers by 1-digit Numbers 41 9 Fractions 44 48 9.1 Fraction as a Part of a Whole 50 9.2 Fraction of a Collection 56 10 Money 62 67 10.1 Convert Rupees to Paise 10.2 Add and Subtract Money with Conversion 73 10.3 Multiply and Divide Money 10.4 Rate Charts and Bills 11 Measurements 11.1 Conversion of Standard Units of Length 11.2 Conversion of Standard Units of Weight 11.3 Conversion of Standard Units of Volume 12 Data Handling 12.1 Record Data Using Tally Marks
Chapter Time 7 Let Us Learn About • identifying a day and a date on a calendar • reading the time correctly to the hour. Concept 7.1: Read a Calendar Think Farida and her friends are playing a game using a calendar. They split into two groups. Each group says a date or a day of a particular month. The other group answers with the corresponding day or date of another month. Can you also play such a game? Recall Let us recall the days in a week and the months in a year. There are 7 days in a week. They are: 1) Sunday 2) Monday 3) Tuesday 4) Wednesday 5) Thursday 6) Friday 7) Saturday 1
There are 12 months in a year. They are: 1) January 2) February 3) March 4) April 7) July 8) August 5) May 6) June 11) November 12) December 9) September 10) October Recall While reading a calendar we can find the day of a given date. We can also find dates that fall on a particular day of the month. The calendar that we use is called the Gregorian calendar. Let us do an activity to understand this concept better. Activity: 1) List out the birthdays of your parents, grandparents, brothers and sisters. 2) Arrange them in a table as they appear in a calendar month-wise. 3) Note the days on which the birthdays appear. Stick this on your writing table. This will remind you to wish your family members “HAPPY BIRTHDAY” on their birthdays. Your tables could be similar to the one given below. Birthdays of my family members Birthday Member of the family Day 08-January Brother Sunday 10-March Mother Friday MINE Friday 16-June Father Thursday 03-August Wednesday 04-October Grand father Tuesday 12-December Grand mother Example 1: From the calendar for the year 2019, write the days of the following events. a) Independence Day - ____________ b) Republic Day - _____________ c) Christmas - ____________ d) Teacher’s Day - _____________ 2
e) Children’s Day - _____________ Solution: a) Independence Day - Thursday b) Republic Day - Saturday c) Christmas - Wednesday d) Teacher’s Day - Thursday e) Children’s Day - Thursday Example 2: Observe the given calendar and answer the questions that follow. a) How many days are there in this month? JANUARY 2019 b) How many Sundays are there in this SUN MON TUE WED THU FRI SAT month? 12345 6 7 8 9 10 11 12 c) Which day appears 5 times? 13 14 15 16 17 18 19 d) On which day is the Republic day? 20 21 22 23 24 25 26 27 28 29 30 31 e) On which date is the second Saturday? Solution: a) There are 31 days in this month. b) There are four Sundays in this month. c) T uesday, Wednesday and Thursday appear five times. d) The Republic day is on a Saturday. e) The second Saturday is on 12th. Application We use the calendar on a daily basis. Events like planning holidays, conducting sports and examinations in school are a few examples. October 2019 Example 3: Renu wants to plan her holiday in October SUN MON TUE WED THU FRI SAT from Friday to Wednesday to New Delhi. 12345 On the calendar, mark the days when 6 7 8 9 10 11 12 Renu can plan her holiday. 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Time 3
Solution: Renu’s trip will start on a Friday and end on a Wednesday. Fridays in this month: 4, 11, 18, 25 October 2019 Wednesdays in this month: 2, 9, 16, 23, 30 SUN MON TUE WED THU FRI SAT 1 2 345 Renu’s trip could be planned for 4th to 9th, 11th to 16th, 18th to 23rd or 25th to 30th 6 7 8 9 10 11 12 as marked on the calendar. 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Example 4: Use the January 2019 calendar shown to answer the question. Rupali is a clerk in a bank. She has January 2019 Solution: holidays on Sundays and on the first and the third Saturdays of the month. She also SUN MON TUE WED THU FRI SAT has holidays on the New Year’s Day and 12345 Republic Day. How many holidays does she have in the month of January? 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Republic day is on 26th January. 20 21 22 23 24 25 26 27 28 29 30 31 New Year day is on 1st January. The first and the third Saturday falls on 5th and 19th January respectively. Sundays fall on 6th, 13th, 20th and 27thJanuary. Rupali has holidays on 1st, 5th, 6th, 13th, 19th, 20th, 26th and 27th January. Therefore, she has 8 holidays in January. Higher Order Thinking Skills (H.O.T.S.) Observe the calendar for February of different years. February 2012 February 2013 February 2014 SUN MON TUE WED THU FRI SAT SUN MON TUE WED THU FRI SAT SUN MON TUE WED THU FRI SAT 1234 12 1 5 6 7 8 9 10 11 3456789 12 13 14 15 16 17 18 10 11 12 13 14 15 16 2345678 19 20 21 22 23 24 25 17 18 19 20 21 22 23 26 27 28 29 24 25 26 27 28 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 4
February 2015 February 2016 February 2017 SUN MON TUE WED THU FRI SAT SUN MON TUE WED THU FRI SAT SUN MON TUE WED THU FRI SAT 123 456 7 12 345 6 1234 5 6 7 8 9 10 11 8 9 10 11 12 13 14 7 8 9 10 11 12 13 12 13 14 15 16 17 18 19 20 21 22 23 24 25 15 16 17 18 19 20 21 14 15 16 17 18 19 20 26 27 28 22 23 24 25 26 27 28 21 22 23 24 25 26 27 28 29 February 2018 February 2019 February 2020 SUN MON TUE WED THU FRI SAT SUN MON TUE WED THU FRI SAT SUN MON TUE WED THU FRI SAT 123 12 1 4 5 6 7 8 9 10 3456789 11 12 13 14 15 16 17 10 11 12 13 14 15 16 2345678 18 19 20 21 22 23 24 17 18 19 20 21 22 23 25 26 27 28 24 25 26 27 28 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 We observe that February has 29 days in the years 2012, 2016 and 2020. In the other years, February has 28 days. Every four years, an extra day is added to the month of February. This is due to the revolution of the Earth around the Sun. The Earth takes 365 days and 6 hours to go around the Sun. An ordinary year is taken as 365 days only. 6 hours put together four times make an extra day for every four years. This is added on to get the leap year. So, there are 365 + 1 = 366 days in a leap year. Example 5: Find the leap years in the following years. 2020, 2021, 2022, 2024, 2025 Solution: In a leap year, the number formed by the last two digits is an exact multiple of 4. In 2020, the number formed by the last two digits is 20, which is a multiple of 4. In 2021, the number formed by the last two digits is 21, which is not a multiple of 4. In 2022, 22 is not a multiple of 4. In 2024, 24 is a multiple of 4. In 2025, 25 is not a multiple of 4. Thus, 2020 and 2024 are the leap years. Time 5
Example 6: How many days were there from Christmas 2010 to Christmas 2011? Solution: 2011 was not a leap year. So, the number of days from Christmas 2010 to Christmas 2011 was 365. Concept 7.2: Read Time Correct to the Hour Think Farida’s teacher taught her to read time. She now knows the units of time. Farida reads time when her father moves the hands of a clock to different numbers. Can you also read time from a clock? Recall We learnt that the long hand on the clock shows minutes and the short hand shows hours. In some clocks, we see another hand, thinner than the hour and the minute hands. This is the seconds hand. Let us recall reading time from a clock. a) 7 o’clock is _______________ hours more than 4 o’clock. b) The _______________ hand takes one hour to go around the clock. c) The _______________ hand is the shortest hand on the clock. d) The time is _______________ when both the hour hand and the minute hand are on 12. e) 2 hours before 10 o’clock is _______________. & Remembering and Understanding We see numbers 1 to 12 on the clock. These numbers are for counting hours. There are 60 parts or small lines between these numbers. They stand for minutes. The minute hand takes 1 hour to go around the clock face once. 1 hour is equal to 60 minutes. The minute hand takes 5 minutes to go from one number to the next number on the clock face. 6
We multiply the number to which the minute hand points by 5 to get the minutes. For example, the minute hand in the figure is at 6. So, it denotes 6 × 5 = 30 minutes past the hour (here, after 3). Therefore, the time is read as 3:30. The hour hand takes one hour to move from one number to the other. Let us now read the time shown by these clocks. Fig. (a) Fig. (b) Fig. (c) Fig. (d) In figure (a), the minute hand is at 9. The hour hand is in between 5 and 6 . The number of minutes is 9 × 5 = 45. Thus, the time shown is 5:45. In figure (b), the minute hand is at 6. The number of minutes is 6 × 5 = 30. The hour hand is between 7 and 8. Therefore, the time shown is 7:30. In figure (c), the minute hand is at 3. The number of minutes is 3 × 5 = 15. The hour hand has just passed 9. Therefore, the time shown is 9:15. In figure (d), the minute hand is at 4. So, the number of minutes is 4 × 5 = 20. The hour hand has just passed 2. Therefore, the time shown is 2:20. Example 7: On which number is the minute hand if the time is as given? a) 35 minutes b) 15 minutes c) 40 minutes d) 30 minutes Solution: To find minutes when the minute hand is at a number, we multiply by 5. So, to get the number from the given minutes, we must divide it by 5. a) 35 ÷ 5 = 7. So, the minute hand is at 7. b) 15 ÷ 5 = 3. So, the minute hand is at 3. c) 40 ÷ 5 = 8. So, the minute hand is at 8. d) 30 ÷ 5 = 6. So, the minute hand is at 6. Time 7
Quarter past, half past and quarter to the hour We know that, ‘quarter’ means 1 . 4 In Fig (a), the minute hand of the clock has travelled a quarter of an hour. So, we call it quarter past the hour. The time shown is 2:15 or 15 minutes past 2 or quarter past 2. Fig. (a) Fig. (b) ‘Half’ means 1 Fig. (c) 2 In Fig. (b), the minute hand has travelled half the clock after an hour. So, we call it half past the hour. The time shown is 2:30 or 30 minutes past 2 or half past 2. In Fig. (c), the minute hand has to travel a quarter of the clock before it completes one hour. We call it quarter to the hour. The time shown is 7:45 or 45 minutes past 7 or quarter to 8. Example 8: Read the time in each of the given clocks and write it in two different ways. Solved Solve these Fig. (a) Fig. (b) Fig. (c) Fig. (d) The hour hand is The hour hand is The hour hand is The hour hand is between 3 and 4. between _____ and between _____ and between _____ and _____. The minutes _____. The minutes _____. The minutes So, the minutes are are after ____hours. are after ____hours. are after ____hours. after 3 hours. The The minute hand The minute hand The minute hand minute hand is at is at _____. So, is at _____. So, is at _____. So, 6. So, the time is 30 the time is _____ the time is _____ the time is _____ minutes after 3. We minutes after _____. minutes after _____. minutes after _____. write it as 3:30 or We write it as _____ We write it as _____ We write it as _____ half past 3. or _____. or _____. or _____. 8
Application We have learnt how to read the time. Now let us draw hands on the clocks when the time is given. Example 9: Draw the hands of a clock to show the given time. a) 1:15 b) 6:15 c) 7:30 d) 9:45 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) Example 10: Draw the hands of a clock to show the given time. a) Quarter to 7 b) Half past 4 Solution: a) b) Time 9
Higher Order Thinking Skills (H.O.T.S.) We have learnt to read and show time, exact to minutes and hours. Let us now learn to find the length of time between two given times. Example 11: The clocks given show the start time and the end time of the Maths class. How long was the class? Solution: The start time is 10:00 and the end time is 10:45. The difference in the given times = 10:45 – 10:00 = 45 minutes Therefore, the length of the Maths class was 45 minutes. Example 12: Sanjay spends an hour between 4:30 and 5:30 for different activities. The start time for each activity is as shown. Train My Brain playing drinking milk homework TV on TV off Read the clocks and answer the following questions. a) When did Sanjay begin drinking milk? b) For how long did he play? c) For how long did he watch TV? d) When did he switch off the TV? Solution: From the given figures, a) Sanjay began drinking milk at 4:45. 10
b) S anjay began playing at 4:30 and ended at 4:45. So, he played for a quarter hour (15 minutes) as 4:45 – 4:30 = 15 minutes. c) The time for which he watched TV was 5:30 – 5:20 = 10 minutes. d) Sanjay switched off the TV at 5:30. The time between two given times is called the length of time. It is also called time duration or time interval. It is given by the difference of end time and start time. Drill Time Concept 7.1: Read a Calendar 1) Observe the calendar and answer the following questions. a) How many weekends and weekdays are 2020 JANUARY there in the month shown in the calendar? SUN MON TUES WED THU FRI SAT (Consider Saturday and Sunday as weekend 1234 days.) 5 6 7 8 9 10 11 b) Write the day and date before two days of 12 13 14 15 16 17 18 19 20 21 22 23 24 25 the fourth Saturday of January. 26 27 28 29 30 31 c) On which day does the month end? 2) Word Problems a) Raju bought a new dress on 1st September. 2019 SEPTEMBER He bought another new dress 10 days after SUN MON TUES WED THU FRI SAT first day of the same month. On which date 1 2 34567 did he buy the other dress? 8 9 10 11 12 13 14 b) Shane’s birthday was on 2nd September. 15 16 17 18 19 20 21 What is the date, if he celebrated it on the 22 23 24 25 26 27 28 same day of the third week. 29 30 c) Arif solved problems from one chapter of his book on 9th of September. He solved problems from the next chapter 5 days later. On which day did he solve problems from the next chapter? Concept 7.2: Read Time Correct to the Hour 3) Draw the hands of the clock to show the given time. a) Half past 2 b) 4:15 c) Quarter to 12 d) 4:25 e) 6:20 Time 11
4) What is the time shown on each of these clocks? 5) Word problems a) O n which number is the minute hand if the time is as given? A) 25 minutes B) 45 minutes C) 20 minutes D) 50 minutes b) The start time of Ram’s activities are shown in these figures. wake up brush have bath Wear uniform study breakfast From the figures, answer the following questions. A) When did Ram wake up? B) How much time did Ram spend for wearing his school uniform? C) When did Ram start studying? D) At what time did Ram had his breakfast? 12
Chapter Division 8 Let Us Learn About • equal grouping and sharing. • repeated subtraction and division facts. • dividing 2-digit number by 1-digit number. • checking the correctness of division. Concept 8.1: Division as Equal Grouping Think Farida and Piyush got a chocolate bar with 14 pieces for Christmas. Piyush divided it and gave Farida 6 pieces. Do you think Farida got an equal share? How can we find out? Recall In the previous chapter, we have learnt multiplication. Multiplication is finding the total number of objects that have been grouped equally. Let us use this to distribute objects equally in groups. Consider 12 bars of chocolate. The different ways in which they can be distributed are as follows. 13
Distributing in 1 group: 1 × 12 = 12 Distributing in 2 groups: 2 × 6 = 12 Distributing in 3 groups: 3 × 4 = 12 Distributing in 4 groups: 4 × 3 = 12 Distributing in 6 groups: 6 × 2 = 12 Distributing in 12 groups: 12 × 1 = 12 Distributing a given number of objects into equal groups is called division. We can understand division better by using equal sharing and equal grouping. & Remembering and Understanding Equal sharing means having equal number of objects or things in a group. We use division to find the number of things in a group and the number of groups. 14
Suppose 9 balloons are to be shared 1st round: 1 balloon is taken by each equally among 3 friends. Let us use friend. repeated subtraction to distribute the balloons. 9 – 3 = 6. So, 6 balloons remain. 2nd round: From the remaining 6 balloons, 3rd round: From the remaining 3 balloons, 1 more balloon is taken by each friend. 1 more balloon is taken by each friend. Now, each friend has 2 balloons. Now, each of them has 3 balloons. 6 – 3 = 3. So, 3 balloons remain. 3 – 3 = 0. So, 0 balloons remain. Each friend gets 3 balloons. We can write it as 9 divided by 3 equals 3. 9 divided by 3 equals 3 is written as ↓ ↓ ↓ Total Number of Number of number of objects in each groups objects group Quotient Dividend Divisor Division 15
In a division, the number that is divided is called the dividend. The number that divides is called the divisor. The answer in division is called the quotient. The number (part of the dividend) that remains is called the remainder. The symbol for ‘is divided by’ is ÷. 9 ÷ 3 = 3 is called a division fact. In this, 9 is the dividend, 3 is the divisor and 3 is the quotient. Note: Representing the dividend, divisor and quotient using the symbols ÷ and = is called a division fact. We use multiplication tables to find the quotient in a division. We find the factor which when multiplied by the divisor gives the dividend. Let us understand this through a few examples. Example 1: 18 pens are to be shared equally by 3 children. How many pens does each of them get? Solution: Total number of pens = 18 Number of children = 3 Number of pens each child gets = 18 ÷ 3 = 6 (since 6 × 3 = 18) Therefore, each child gets 6 pens. Example 2: 10 flowers are put in some vases. If each vase has 2 flowers, how many vases are used? Solution: Number of flowers = 10 Number of flowers in each vase = 2 Number of vases used = 10 ÷ 2 = 5 (since 5 × 2 = 10) Therefore, 5 vases are used to put 10 flowers. We get two division facts from a multiplication fact. The divisor and the quotient are the factors of the dividend. 16
Observe the following table: Dividend ÷ Divisor = Quotient Multiplicand × Multiplier = Product 6 × 3 = 18 18 ÷ 6 = 3 ↓↓ ↓ ↓↓ ↓ Divisor Quotient Dividend Product Factor Factor (Multiplicand) (Multiplier) From the multiplication fact 6 × 3 = 18, we can write two division facts: a) 18 ÷ 3 = 6 and b) 18 ÷ 6 = 3 Multiplication and division are reverse operations. Let us now understand this through an activity. We can show a multiplication fact on the number line. For example, 5 × 3 = 15 means 5 times 3 is 15. To show 5 times 3 on the number line, we take steps of 3 for 5 times. We go forward from 0 to 15. Similarly, we can show the division fact 15 ÷ 3 = 5 on the number line. To show 15 divided by 3 on the number line, we take steps of 3 for 5 times. We go backward from 15 to 0 as shown. Division 17
Application Equal sharing and equal grouping are used in some real-life situations. Consider the following situations. Example 3: 25 buttons are to be stitched on 5 shirts. If each shirt has the same number of buttons, how many buttons are there on each shirt? Solution: Total number of buttons = 25 Number of shirts = 5 The division fact for 25 buttons distributed among 5 shirts = 25 ÷ 5 = 5 Therefore, each shirt has 5 buttons on it. Example 4: 24 marbles are to be divided among 4 friends. How many marbles will each friend get? Solution: Total number of marbles = 24 Number of friends = 4 Number of marbles each friend will get = 24 ÷ 4 = 6 Therefore, each friend will get 6 marbles. Higher Order Thinking Skills (H.O.T.S.) Division is used in many situations in our day-to-day lives. Let us see some examples. Example 5: Aman spends 14 hours a week for tennis practice. He spends 21 hours a week for doing homework and 48 hours a week at school. How much time does he spend in a day for these activities? (Hint: 1 week = 7 days. The school works for 6 days a week.) Solution: Time spent for tennis practice per day = 14 hours ÷ 7 = 2 hours Time spent for doing homework per day = 21 hours ÷ 7 = 3 hours Time spent at school per day = 48 hours ÷ 6 = 8 hours (School works for 6 days a week) Thus, the total time spent by Aman in a day for all the activities = (2 + 3 + 8) hours = 13 hours (except Sunday) 18
Example 6: Deepa shares 15 lollipops among her 5 friends. Instead, if she shares the lollipops among only 3 of them, how many more lollipops does each of them get? Solution: Number of lollipops = 15 If Deepa shares the lollipops among her five friends, the number of lollipops each of them would get = 15 ÷ 5 = 3 If Deepa shares the lollipops among only three of them, the number of lollipops each of them gets = 15 ÷ 3 = 5 Difference in the number of lollipops = 5 – 3 = 2 Therefore, her friends would get 2 more lollipops. Concept 8.2: Divide 2-digit and 3-digit Numbers by 1-digit Numbers Think Farida has 732 stickers. She wants to distribute them equally among her three friends. How will she distribute? Recall In the previous section, we have learnt that division is related to multiplication. For every division fact, we can write two multiplication facts. For example, the two multiplication facts of 35 ÷ 7 = 5 are: a) 7 × 5 = 35 and b) 5 × 7 = 35. Let us answer these to recall the concept of division. a) The number which divides a given number is called _________________. b) The answer we get when we divide a number by another is called ______________________. c) The division facts for the multiplication fact 2 × 4 = 8 are ________________ and __________________. Division 19
& Remembering and Understanding We can make equal shares or groups and divide with the help of vertical arrangement. A number divided by the same number is always 1. Let us see some examples. D ividing a 2-digit number by a 1-digit number (1-digit quotient) Example 7: Solve: 45 ÷ 5 Solution: Follow these steps to divide a 2-digit number by a 1-digit number. Steps Solved Solve these Step 1: Write the dividend and 5)45 Dividend = _____ Divisor = ______ )divisor as shown: Divisor Dividend Quotient = ____ Remainder = _____ Step 2: Find the multiplication fact 45 = 5 × 9 8) 56 which has the dividend and divisor. - Step 3: Write the other factor as the 9 quotient. Write the product of the factors below the dividend. 5)45 − 45 Step 4: Subtract the product 9 4) 36 Dividend = _____ from the dividend and write the Divisor = ______ difference below the product. 5)45 - Quotient = ____ This difference is called the Remainder = _____ remainder. − 45 00 45 = Dividend 5 = Divisor 9 = Quotient 0 = Remainder Note: If the remainder is zero, the divisor is said to divide the dividend exactly. Checking for correctness of division: The multiplication fact of the division is used to check its correctness. Step 1: Compare the remainder and divisor. The remainder must always be less than the divisor. 20
Step 2: Check if (Quotient × Divisor) + Remainder = Dividend Let us now check if our division in example 7 is correct or not. Step 1: Remainder < Divisor 0 < 5 (True) Step 2: Quotient × Divisor 9×5 Step 3: (Quotient × Divisor) + Remainder = Dividend 45 + 0 = 45 = Dividend Note: The division is incorrect if: a) Remainder > or = divisor b) (Quotient × Divisor) + Remainder ≠ Dividend 2-digit quotient In the examples we have seen so far, the quotients are 1-digit numbers. In some divisions, the quotients may be 2-digit numbers. Let us see some examples. Example 8: Solve: 57 ÷ 3 Solution: Follow these steps to divide a 2-digit number by a 1-digit number. Steps Solved Solve these Step 1: Check if the tens digit of the dividend is greater than the divisor. 5>3 5) 60 Step 2: Divide the tens and write the quotient. 1 − Write the product of quotient and divisor, below the tens digit of the dividend. 3)57 − Step 3: Subtract and write the difference −3 Step 4: Check if difference < divisor is true. 1 Dividend = _____ Divisor = ______ 3)57 Quotient = ____ Remainder = ___ −3 2 2 < 3 (True) Division 21
Steps Solved Solve these Step 5: Bring down the ones digit of the 1 3) 42 dividend and write it beside the remainder. 3)57 − − − 3↓ 27 Step 6: Find the largest number in the 3 × 8 = 24 1 multiplication table of the divisor that can be subtracted from the 2-digit number in )3 × 9 = 27 3 57 the previous step. 3 × 10 = 30 24 < 27 < 30. − 3↓ So, 27 is the 27 required number. Step 7: Write the factor of required number, 19 Dividend = _____ other than the divisor, as the quotient. Write Divisor = ______ the product of the divisor and the quotient 3)57 Quotient = ____ below the 2-digit number. Subtract and Remainder = ___ write the difference. − 3↓ 27 Step 8: Check if remainder < divisor is true. Stop the division. − 27 00 0 < 3 (True) (If this is false, the division is incorrect.) RQeumotaieinndte=Trr1=a90in My Brain Step 9: Write the quotient and the remainder. Step 10: Check if (Divisor × Quotient) + 3 × 19 + 0 = 57 Remainder = Dividend is true. 57 + 0 = 57 57 = 57 (True) (If this is false, the division is incorrect.) Divide 3-digit numbers by 1-digit numbers (2-digit quotient) Dividing a 3-digit number by a 1-digit number is similar to dividing a 2-digit number by a 1-digit number. Let us understand this through a few examples. Example 9: Solve: a) 265 ÷ 5 Solution: Follow these steps to divide a 3-digit number by a 1-digit number. 22
Steps Solved Solve these Step 1: Check if the hundreds digit of 4) 244 the dividend is greater than the divisor. 5)265 − If it is not, consider the tens digit too. 2 is not greater than 5. So, consider 26. Step 2: Find the largest number that 5 − can be subtracted from the 2-digit number of the dividend. Write the 5)265 Dividend = _____ quotient. Divisor = ______ − 25 Quotient = ____ Remainder = ___ Write the product of the quotient and 5 × 4 = 20 the divisor below the dividend. 5 × 5 = 25 9) 378 5 × 6 = 30 Step 3: Subtract and write the − difference. 25 < 26 − 5 5)265 − 25 1 Step 4: Check if difference < divisor 1 < 5 (True) is true. (If it is false, the division is incorrect.) Step 5: Bring down the ones digit 5 of the dividend. Write it beside the remainder. 5)265 − 25↓ 15 Step 6: Find the largest number in the 5 multiplication table of the divisor that can be subtracted from the 2-digit 5)265 number in the previous step. − 25↓ 15 5 × 2 = 10 5 × 3 = 15 5 × 4 = 20 15 is the required number. Division 23
Steps Solved Solve these Step 7: Write the factor of required 53 number, other than the divisor, as Dividend = _____ quotient. Write the product of divisor 5)265 Divisor = ______ and quotient below the 2-digit Quotient = ____ number. Then, subtract them. − 25↓ Remainder = ___ 15 Step 8: Check if remainder < divisor is true. Stop the division. (If this is false, − 15 the division is incorrect.) 00 0 < 5 (True) Step 9: Write the quotient and Quotient = 53 remainder. Remainder = 0 Step 10: Check if (Divisor × Quotient) + 5 × 53 + 0 = 265 Remainder = Dividend is true. (If this is 265 + 0 = 265 false, the division is incorrect.) 265 = 265 (True) 3-digit quotient Example 10: Solve: 784 by 7 Solution: Follow these steps to divide a 3-digit number by a 1-digit number. Steps Solved Solve these Step 1: Check if the hundreds digit of the dividend is greater than or equal to the 7)784 8) 984 divisor. Step 2: Divide the hundreds and write the 7=7 − quotient in the hundreds place. 1 − Write the product of the quotient and the divisor under the hundreds place of the 7)784 − dividend. −7 24
Steps Solved Solve these Step 3: Subtract and write the difference. Dividend = _____ 1 Divisor = ______ Quotient = ____ 7)784 Remainder = ___ −7 5) 965 0 − Step 4: Check if difference < divisor is true. 0 < 7 (True) Step 5: Bring down the next digit of the 1 − dividend. Check if it is greater than or − equal to the divisor. 7)7 84 Dividend = _____ − 7↓ Divisor = ______ 08 Quotient = ____ 8>7 Remainder = ___ Step 6: Find the largest number in the 11 multiplication table of the divisor that can be subtracted from the 2-digit number in 7)784 the previous step. − 7↓ Write the factor other than the divisor as 08 quotient. −7 Write the product of the quotient and the divisor below it. 7×1=7<8 The required number is 7. Step 7: Subtract and write the difference. 11 Bring down the next digit (ones digit) of the dividend. 7)784 Check if the dividend is greater than or − 7↓ equal to the divisor. 08 −7 14 14 > 7 Division 25
Steps Solved Solve these Step 8: Find the largest number in the multiplication table of the divisor that can 112 2) 246 be subtracted from the 2-digit number in the previous step. 7)784 − Write the factor other than the divisor as − the quotient. − 7↓ − Write the product of the quotient and the 08 divisor below it. Dividend = _____ −7 Divisor = ______ Step 9: Subtract and write the difference. 14 Quotient = ____ Check if it is less than the divisor. Stop the Remainder = ___ division. − 14 Step 10: Write the quotient and the 7 × 2 = 14 remainder. The required Step 11: Check if (Divisor × Quotient) + number is 14. Remainder = Dividend is true. (If it is false, the division is incorrect.) 112 7)784 − 7↓ 08 −7 14 − 14 00 Quotient = 112 Remainder = 0 7 × 112 + 0 = 784 784 + 0 = 784 784 = 784 (True) Application Division of 2-digit numbers and 3-digit numbers is used in many real-life situations. Let us consider a few examples. Example 11: A school has 634 students, who are equally grouped into 4 houses. How many students are there in a house? Are there any students who are not grouped into a house? 26
Solution: Number of students = 634 158 Number of houses = 4 Number of students in a house = 634 ÷ 4 4)634 − 4↓ 23 Number of students in each house = 158 − 20 The remainder in the division is 2. 34 Therefore, 2 students are not grouped into any house. − 32 02 Example 12: A football game had 99 spectators. If each row has only 9 seats, how Solution: many rows would the spectators occupy? 11 Number of spectators = 99 9) 99 Number of seats in each row = 9 − 9↓ Number of rows occupied by the spectators = 99 ÷ 9 = 11 09 Therefore, 11 rows were occupied by the spectators. −9 0 Higher Order Thinking Skills (H.O.T.S.) In all the division sums we have seen so far, we did not have a 0 (zero) in dividend or quotient. When a dividend has a zero, we place a 0 in the quotient in the corresponding place. Then, get the next digit of the dividend down and continue the division. Let us now understand division of numbers that have a 0 (zero) in dividend or quotient, through these examples. Example 13: Solve: 505 ÷ 5 Solution: Follow these steps for division of numbers having 0 in dividend. Solved Solve this 101 4) 804 5)505 − − 5↓ − 00 − − 00 05 − 05 00 Division 27
Drill Time Concept 8.1: Division as Equal Grouping 1) Divide the number in equal groups. a) 16 in 4 equal groups b) 18 in 9 equal groups c) 20 in 5 equal groups d) 32 in 8 equal groups e) 10 in 2 equal groups 2) Word Problems a) 26 students are to be divided into 2 groups. How many students will be there in each group? b) 1 4 pencils must be distributed among 7 children. How many pencils will each student receive? Concept 8.2: Divide 2-digit and 3-digit Numbers by 1-digit Numbers 3) Divide 2-digit numbers by 1-digit numbers (1-digit quotient). a) 12 ÷ 2 b) 24 ÷ 6 c) 36 ÷ 6 d) 40 ÷ 8 e) 10 ÷ 5 4) Divide 2-digit numbers by 1-digit numbers (2-digit quotient). a) 12 ÷ 1 b) 99 ÷ 3 c) 48 ÷ 2 d) 65 ÷ 5 e) 52 ÷ 4 5) Divide 3-digit numbers by 1-digit numbers (2-digit quotient). a) 123 ÷ 3 b) 102 ÷ 2 c) 497 ÷ 7 d) 111 ÷ 3 e) 256 ÷ 4 6) Divide 3-digit numbers by 1-digit numbers (3-digit quotient). a) 456 ÷ 2 b) 112 ÷ 1 c) 306 ÷ 3 d) 448 ÷ 4 e) 555 ÷ 5 7) Word Problems a) 260 chocolates have to be equally distributed among 4 students. How many chocolates will each student receive? b) There are 24 people in a bus. Each row in the bus can seat 2 people. How many rows in the bus are occupied? 28
Chapter Fractions 9 Let Us Learn About • fractions as a part of a whole and their representation. • identify parts of fractions. • fractions of a collection. • applying the knowledge of fractions in real life. Concept 9.1: Fraction as a Part of a Whole Think Farida and her three friends, Joseph, Salma and Rehan, went on a picnic. Farida had only one apple with him. He wanted to share it equally with everyone. What part of the apple does each of them get? Recall Look at the rectangle shown below. We can divide the whole rectangle into many equal parts. Consider the following: 29
1 part: 2 equal parts: 3 equal parts: 4 equal parts: 5 equal parts: and so on. Let us understand the concept of parts of a whole through an activity. & Remembering and Understanding Suppose we want to share an apple with our friends. First, we count our friends with whom we want to share the apple. Then, we cut it into as many equal pieces as the number of persons. Thus, each person gets an equal part of the apple after division. Parts 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 We can divide a whole into equal parts as shown above. Each such division has a different name. To understand this better, let us do an activity. 30
Activity: Halves Take a square piece of paper. Fold it into two equal parts as shown. Each of the equal parts 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. Fold a rectangular piece of paper as shown in figures (b) and (c). 11 1 33 3 three parts three equal parts Fig. (c) Fig. (a) Fig. (b) Each equal part is called a third or one-third. The shaded part in figure (c) is one out of three equal parts. So, we call it a 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 makes a whole. 3 3 3 3 3 Fractions 31
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 Four parts 4 Fig. (d) 1 4 1 4 1 4 Four equal parts 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 and 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. Representing the parts of a whole as Numerator is called a fraction. Numbers of Denominator the form Numerator are called fractions. Thus, a fraction is a part of a whole. Denominator 32
For example, 1 , 1, 1, 2 and so on are fractions. 2 3 4 3 Let us now see a few examples. Example 1: Identify the fraction for the shaded parts in the figures below. a) b) Solution: Steps Solved Solve this a) b) Step 1: Count the number of equal parts, the figure is divided into Total number of Total number of equal (Denominator). parts = _______ equal parts = 8 Number of parts shaded Step 2: Count the number of Number of parts = ______ shaded parts (Numerator). shaded = 5 Step 3: Write the fraction Fraction = 5 Numerator . 8 Denominator Fraction = Example 2: The circular disc shown in the figure is divided into equal parts. What fraction of the disc is painted yellow? Write the fraction of the disc that is painted white. Solution: Total number of equal parts of the disc is 16. The fraction of the disc that is painted yellow = Number of parts painted yellow = 3 Total number of equal parts 16 The fraction of the disc that is painted white = Number of parts painted white = 7 Total number of equal parts 16 Fractions 33
Application 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 3: Shade a square to represent these fractions: 1 2 3 d) 1 a) 4 b) 3 c) 5 2 Solution: We can represent the fractions as: Steps Solved 2 Solve these 1 1 3 3 2 Step 1: Identify the Denominator 5 Denominator denominator and the 4 = = numerator. Denominator Numerator Denominator Numerator =4 = = Step 2: Draw the Numerator = 1 = required shape. Divide it into as many Numerator equal parts as the denominator. = Step 3: Shade the number of equal parts as the numerator. This shaded part represents the given fraction. Example 4: Colour the shapes to represent the given fractions. Fractions 1 2 1 4 5 2 Shapes 34
Solution: We can represent the fractions as: Fractions 1 2 1 4 5 2 Shapes Higher Order Thinking Skills (H.O.T.S.) Let us see an example of a real-life situation involving fractions. Example 5: A square shaped garden has coconut trees in a quarter of its land. It has mango trees in two quarters and neem trees in another quarter. Draw a figure of the garden and represent its parts. Solution: Fraction of the garden covered by Coconut coconut trees = Quarter = 1 trees 4 Neem trees Fraction of the garden covered by Mango 1 trees mango trees = 2 Quarters = 2 Fraction of the garden covered by neem trees = Quarter = 1 4 So, the square garden is as shown in the figure. Concept 9.2: Fraction of a Collection Think Farida has a bunch of roses. Some of them are red, some white and some yellow. Farida wants to find the fraction of roses of each colour. How can she find that? Fractions 35
Recall We know that a complete or a full object is called a whole. We also know that we can divide a whole into equal number of parts. Let us answer these to revise the concept. Divide these into equal number of groups as given in the brackets. Draw circles around them. a) [3 groups] b) [2 groups] c) [5 groups] & Remembering and Understanding To find the part or the fraction of a collection, find the number of each type of object out of the total 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. Each equal part has 5 pens, as 10 ÷ 2 = 5. So, 1 of 10 is 5. 2 Finding a third One-third is 1 out of 3 equal parts. In the given figure, there are 12 bananas. 36
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 is 4. 3 11 1 33 3 Finding a fourth (or a quarter) One-fourth is 1 out of 4 equal parts. In the figure, there are 8 books. To find a fourth, divide the number of books into 4 equal parts. 1 1 1 1 444 4 Each equal part has 2 books, as 8 ÷ 4 = 2. So, 1 of 8 is 2. 4 Let us see an example to find the fraction of a collection. Example 6: Find the fraction of the coloured parts of the shapes. Shapes Fractions Fractions 37
Solution: The fractions of the coloured parts of the shapes are – Shapes Fractions 2 6 3 6 5 8 Application We can apply the knowledge of fractions in many real-life situations. Let us see a few examples. Example 7: A basket has 64 flowers. Half of them are roses, a quarter of them are marigolds and a quarter of them are lotus. How many roses, marigolds and lotus are there in the basket? Solution: Total number of flowers = 64 Half of the flowers are roses. The number of roses = 1 of 64 = 64 ÷ 2 = 32 2 A quarter of the flowers are marigolds. Train My Brain 1 The number of marigolds = 4 of 64 = 64 ÷ 4 = 16 A quarter of the flowers are lotus. 1 The number of lotus = 4 of 64 = 64 ÷ 4 = 16 Therefore, there are 32 roses, 16 marigolds and 16 lotus in the basket. Example 8: 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 is 15 . 38
Number of yellow balloons = 3 3 Therefore, fraction of yellow balloons is 15 . Number of red balloons = 10 10 Therefore, fraction of red balloons = 15 Higher Order Thinking Skills (H.O.T.S.) 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 one such example. 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 2 4 money did she spend? Solution: Cost of 1 kg apples = ` 16 Cost of 1 kkg apples = 1 of ` 16 = ` 16 ÷ 2= `8 2 2 (To find a half, we divide by 2) Cost of 1 kg papaya = ` 20 Cost of 1 kkg papaya = 1 ooff ` 20 = ` 20 ÷ 4 = ` 5 4 4 (To find a fourth, we divide by 4) Therefore, the money spent by Rita = ` 8 + ` 5 = ` 13 Therefore, Sujay has solved 10 problems. Drill Time Concept 9.1: Fraction as a Part of a Whole 1) Find the numerator and the denominator in each of these fractions. 2 b) 1 2 a) 5 7 c) 3 45 d) 9 e) 7 Fractions 39
2) Identify the fractions of the shaded parts. a) b) c) d) e) Concept 9.2: Fraction of a Collection 3) Find fraction of coloured parts. a) b) c) d) e) 4) Find 1 and 1 of the following collection. 2 4 5) Word Problems a) A circular disc is divided into 12 equal parts. Venu shaded 1 of the disc 4 pink and 1 of the disc green. How many parts of the disc are shaded? How 3 many parts are not shaded? them are unruled and 1 of them are four-ruled. b) John has 24 notebooks. 1 of 6 2 How many books are (a) unruled and (b) four-ruled? 40
Chapter Money 10 Let Us Learn About • converting rupees to paise and vice-versa. • adding and subtracting money. • multiplying and dividing money. • making rate charts and bills. Concept 10.1: Convert Rupees to Paise Think Farida has ` 38 in her piggy bank. She wants to know how many paise she has. Do you know? Recall We have learnt to identify different coins and currency notes. We have also learnt that 100 paise make a rupee. Let us learn more about money. 1 rupee = 100 paise 100 p = 1 rupee Let us revise the concept about money. a) Identify the value of the given coin. [ ] (A) ` 1 (B) ` 2 (C) ` 5 (D) ` 10 41
b) The ` 500 note among the following is: [] (A) (B) (C) (D) c) The combination that has the greatest value is: [] (A) (B) (C) (D) & Remembering and Understanding Let us understand the conversion of rupees to paise through an activity. Activity: The students must use their play money (having all play notes and coins). As the teacher writes the rupees on the board, each student picks the exact number of paise in it. There can be many combinations for the same amount of rupees. For example, 1 rupee is 100 paise. So, the students may take two 50 paise coins. 42
Let us understand the conversion through some examples. Example 1: Convert the given rupees into paise: a) ` 2 b) ` 5 c) ` 9 Solution: We know that 1 rupee = 100 paise a) ` 2 = 2 × 100 paise = 200 paise b) ` 5 = 5 × 100 paise = 500 paise c) ` 9 = 9 × 100 paise = 900 paise Similarly, we can convert paise into rupees. Converting paise into rupees is the reverse process of converting rupees into paise. Example 2: Convert 360 paise to rupees. Solution: We can convert paise to rupees as: Steps Solved Solve this 360 paise 380 paise Step 1: Write the given 360 paise paise as hundreds of paise. = 300 paise + 60 paise Step 2: Rearrange 300 paise as a product of 100 300 paise paise. = (3 × 100) paise + 60 paise Step 3: Write in rupees. ` 3 + 60 paise = 3 rupees 60 paise Application Let us see some real-life examples involving the conversion of rupees into paise and paise to rupees. Example 3: Anil has ` 10 with him. How many paise does he have? Solution: 1 rupee = 100 paise So, 10 rupees = 10 × 100 paise = 1000 paise Therefore, Anil has 1000 paise with him. Example 4: Raj has 670 paise. How many rupees does he have? Solution: Amount with Raj = 670 paise Money 43
= 600 paise + 70 paise = (6 × 100) paise + 70 paise = ` 6 + 70 paise = 6 rupees 70 paise Therefore, Raj has 6 rupees 70 paise. Higher Order Thinking Skills (H.O.T.S.) Observe this example where conversion of rupees to paise and that of paise to rupees are mostly useful. Example 5: Vani has ` 4, Gita has ` 5 and Ravi has 470 paise. Who has the least amount of money? Solution: Amount Vani has = ` 4 Amount Gita has = ` 5 Amount Ravi has = 470 paise To compare money, all the amounts must be in the same unit. So, let us first convert the amounts from rupees to paise. ` 4 = (4 × 100) = 400 paise ` 5 = (5 × 100) = 500 paise Now, arranging the money in ascending order, we get 400 < 470 < 500. Therefore, Vani has the least amount of money. Concept 10.2: Add and Subtract Money with Conversion Think Farida’s father bought a toy car for ` 56 and a toy bus for ` 43. How much did he spend altogether? How much change does he get if he gives ` 100 to the shopkeeper? 44
Recall Recall that two or more numbers are added by writing them one below the other. This method of addition is called the column method. We know that rupees and paise are separated using a dot or a point. In the column method, we write money in such a way that the dots or points are placed exactly one below the other. The rupees are placed under rupees and the paise under paise. Let us recall a few concepts about money through these questions. a) 50 paise + 50 paise = ________________ b) ` 50 – ` 10 = _______________ c) ` 20 + ` 5 + 50 paise = ______________ d) ` 20 + ` 10 = _______________ e) ` 50 – ` 20 = _______________ & Remembering and Understanding While adding and subtracting money, we write numbers one below the other and add or subtract as needed. Let us understand this through some examples. Example 6: Add: ` 14.65 and ` 23.80 Solution: We can add two amounts as: Steps Solved Solve these Step 1: Write the given numbers `p `p with the points exactly one below 1 4. 6 5 the other, as shown. + 2 3. 8 0 4 1. 5 0 + 4 5. 7 5 Money 45
Steps Solved Solve these `p Step 2: First add the paise. `p Regroup the sum if needed. Write 1 3 8. 4 5 the sum under paise. Place the 1 4. 6 5 + 3 5. 6 0 dot just below the dot. + 2 3. 8 0 . 45 Step 3: Add the rupees. Add `p `p the carry forward (if any) from 1 2 3. 6 5 the previous step. Write the sum 1 4. 6 5 + 1 4. 5 2 under rupees. + 2 3. 8 0 3 8. 4 5 Solve these `p Step 4: Write the sum of the given ` 14.65 + ` 23. 80 amounts. Paise is always written in = ` 38.45 8 0. 7 5 two digits after the point. − 4 1. 5 0 Example 7: Write in columns and subtract ` 56.50 from ` 73.50. Solution: We can subtract the amounts as: Steps Solved Step 1: Write the given numbers with the ` p dots exactly one below the other, as 7 3. 50 shown. − 5 6. 50 Step 2: First subtract the paise. Regroup `p `p if needed. Write the difference under 7 3. 5 0 paise. Place the dot just below the dot. − 5 6. 5 0 6 0. 7 5 Step 3: Subtract the rupees. Write the − 3 2. 5 0 difference under rupees. 00 Step 4: Write the difference of the given `p amounts. 6 13 7 3. 5 0 − 5 6. 5 0 1 7. 0 0 ` 73. 50 – ` 56. 50 = ` 17.00 46
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