181010032-Alpine-G4-Textbook-Maths-FY

Example 5: Multiply the largest 4-digit number by the smallest Th H T O 1-digit number. 9999 Solution: The largest 4-digit number = 9999 ×1 9999 The smallest 1-digit number = 1 We know that the product obtained when any number is multiplied by 1 is the Example 6: number itself. Solution: Therefore, 9999 × 1 = 9999. Multiply the largest 4-digit number by the largest T Th Th H T O 1-digit number. The largest 4-digit number = 9999 888 The largest 1-digit number = 9 9999 Therefore, 9999 × 9 = 89991. ×9 8 9991 Concept 5.2: Multiply Using Lattice Algorithm Think Jasleen knows how to multiply a 3-digit number by a 2-digit number. But she makes some mistakes. She wants a simple method for multiplication. Do you know any such method? Recall We know multiplication using standard algorithm. Let us recall the standard algorithm of multiplication by solving the following sum. H TO Th H T O Th H T O Th H T O 22 542 142 1243 ×6 ×4 ×8 ×2 Multiplication 47 Alpine_V3_Maths_G4_TB.indb 47 30-01-2018 16:02:34

& Remembering and Understanding There are two ways to multiply numbers: 1) Standard Algorithm 2) Lattice Algorithm Let us now learn to multiply 2-digit and 3-digit numbers using lattice algorithm. The important features of the lattice algorithm: • Setting up the lattice before we begin multiplying • Doing all the multiplications first, followed by addition • There is no carry over in the multiplication phase of the algorithm Let us use the lattice algorithm to multiply: 1) a 2-digit number by a 1-digit number and a 2-digit number 2) a 3-digit number by a 2-digit number Multiply a 2-digit number by a 1-digit number and a 2-digit number Multiplying a 2-digit number by a 1-digit number and a 2-digit number is similar to multiplying a 1-digit number by a 1-digit number. Let us see an example. Example 7: Multiply: a) 29 × 3 b) 43 × 52 Solution: Construct the lattice as shown. Steps Solved Solved Solve these a) 29 × 3 b) 43 × 52 3 2× Step 1: (a) N umber of rows = 4 Number of digits in the 2 multiplier. (b) N umber of columns = Number of digits in multiplicand. 48 30-01-2018 16:02:34 Alpine_V3_Maths_G4_TB.indb 48

Steps Solved × Solved × Solve these a) 29 × 3 b) 43 × 52 Step 2: Write the 5 2× multiplicand along the 29 43 top of the lattice and the 4 multiplier along the right, 3 5 6 one digit for each row or column. Draw diagonals 2 6 1× to divide each box into parts as shown. 2 9× 4 3× 4 4 Step 3: Multiply each digit 2 2015 5 of the multiplicand by 73 5 7× each digit of the multiplier. 2 Write the products in 3 the cells where the 2 9× 4 3 × 1 corresponding rows and 0 2 columns meet. 2 1 5 6 73 0 5 Step 4: If the product is a 2 single digit number, put 0 0 0 in the tens place. 8 6 (2 × 3 = 6) = 06 Step 5: Add the numbers 2 9× 4 3× 6 3× along the diagonals 2 from the right to find 02 73 1 3 the product. Regroup if 20 55 needed. Write the sum 06 3 from left to right. 87 10 0 2 28 6 Therefore, 29 × 3 = 087 3 6 = 87 Therefore, 43 × 52 = 2236. Multiply a 3-digit number by a 2-digit number Multiplying a 3-digit number by a 2-digit number is similar to multiplying a 2-digit number by a 2-digit number. Multiplication 49 Alpine_V3_Maths_G4_TB.indb 49 30-01-2018 16:02:35

Let us see an example. Example 8: Multiply: 168 × 48 Solution: Construct a lattice as shown such that: (a) Number of rows = Number of digits in the multiplier. (b) Number of columns = Number of digits in multiplicand. Steps Solved 8 × Solve these × 168 × 48 17 2 Step 1: Write the multiplicand 16 4 4 along the top of the lattice. Write the multiplier along the right, one 8 2 digit for each row or column. Draw diagonals to divide each box into parts as shown. Step 2: Multiply each digit of the 1 6 8× 2 6 2× multiplicand by each digit of the multiplier. Write the products in 0 2 34 3 the cells where the corresponding 4 4 2 rows and columns meet. 8 8 Step 3: If the product is a single 1 6 8 × 1 7 1× digit number, put 0 in the tens 2 place. 0 3 4 3 4 4 2 1 4 8 0 6 3 4 2× 8 8 4 3 Step 4: Add the numbers along 1 6 8× 2 the diagonals to find the product and write the sum from left to 0 2 3 4 right. 4 4 2 8 0 4 6 0 8 4 28 6 8 01 4 Therefore, 168 × 48 = 8064. 50 30-01-2018 16:02:36 Alpine_V3_Maths_G4_TB.indb 50

Application Let us now see a few real-life examples involving multiplication of 3-digit numbers. Example 9: There are 345 students in each class. Pooja’s 34 5 × school has 12 such classes. How many 00 0 1 students are there in her school? 34 5 1 2 Solution: Number of students in each class = 345 00 0 Number of such classes in Pooja’s school = 12 4 6 8 Total number of students 1 40 = 345 × 12 = 4140 Therefore, there are 4140 students in Pooja’s school. Example 10: 42 people were sitting in a row of a stadium to enjoy a cricket match. How Solution: many people would be there in all if there were 35 such rows? Number of people sitting in one row = 42 42 × Number of rows = 35 10 3 1 26 5 Total number of people in 35 rows = 42 × 35 = 1470 2 1 40 0 Therefore, there are 1470 people in the stadium. Train My B7 rain 0 Higher Order Thinking Skills (H.O.T.S.) We know how to multiply numbers using lattice algorithm. Let us see if we can analyse and solve the following. Example 11: Find the missing numbers. 2 3? × 23___ × 4___= 9954 0 1 n2 8 4 08 2 0 0 1 94 6 ? 4 954 Multiplication 51 Alpine_V3_Maths_G4_TB.indb 51 30-01-2018 16:02:37

Solution: We can see that the box in the top right corner has the number 28. It is the product of 4 and ?. That is, 4 × ? = 28 4 × 7 = 28 So, 7 is the first unknown number Similarly, the box in the bottom left corner has 04. It is the product of 2 and?. That is, 2 × ? = 04 2 × 2 = 04 So, the second unknown number is 2. So, the required numbers are 7 and 2 so that 237 × 42 = 9954. Concept 5.3: Mental Maths Techniques: Multiplication Think Jasleen’s rose garden has rose plants planted in 7 rows. There are 8 plants in each row. Jasleen wanted to find out the total number of rose plants in her garden. How can she find that mentally Recall To learn how to complete multiplication facts by adding partial products mentally, we must memorise tables from 1 to 5 and 10. For example, we know that 6 × 5 = 30. As 6 = 4 + 2, we have (4 + 2) × 5 = (4 × 5) + (2 × 5) = 20 + 10 = 30. & Remembering and Understanding While multiplying two numbers mentally, we split the larger number into two parts. Let us now understand how to complete multiplication facts by adding the products mentally. Example 12: Find the answer by adding the products. 8×9 52 30-01-2018 16:02:37 Alpine_V3_Maths_G4_TB.indb 52

Solution: Steps Solved Solve this , 8×9 7×6 Step 1: Check by how much is the larger number more than 5. The larger number is 9, The larger number is and from 5, we count 6, 7, 8 and 9. So, 9 is 4 more and from 5, we count than 5. and . So, is more than 5. Step 2: Write the number as the 5 + 4 = 9 + = sum of 5 and the other number. Step 3: Multiply the numbers of 5 × 8 = 40 and 5× = the sum by the smaller number. 4 × 8 = 32 and Use memorised tables of 1 to 5 and 10 to solve mentally. × = Step 4: Add both the products 40 + 32 = 72 + = . from step 3 to get the final Therefore, 8 × 9 = 72. Therefore, 7 × 6 = answer. Example 13: Find the answer by adding the products: 14 × 6 Solution: Steps Solved Solve this 14 × 6 12 × 8 Step 1: Check by how much is The larger number is 14, The larger number is ____, the larger number more than 10. and from 10, we count and from 10 we count 11, 12, 13 and 14. So, 14 is and . So is 4 more than 10. more than 10. Step 2: Write the larger number 10 + 4 = 14 + = as the sum of 10 and the other number. Multiplication 53 Alpine_V3_Maths_G4_TB.indb 53 30-01-2018 16:02:38

Steps Solved Solve this 14 × 6 12 × 8 Step 3: Multiply the sum in the previous step by the smaller 10 × 6 = 60 and 10 × = number given, using memorised 4 × 6 = 24 and = tables of 1 to 5 and 10. × Step 4: Add both the products 60 + 24 = 84 + = . from step 3 mentally to get the Therefore, 14 × 6 = 84. Therefore, 12 × 8 = final answer. Application We have learnt some easy ways of completing multiplication facts by adding the products mentally. Let us now see some examples where we apply this concept. Example 14: Rohit works for 8 hours in a day. He works 6 days in a week. For how many hours does he work in a week? Solution: Number of hours Rohit works in a day = 8 Number of days he works in a week = 6 Total number of hours Rohit works in a week = 8 × 6 The larger number is 8, and it is 3 more than 5. As 8 = 5 + 3, 8 × 6 = (5 × 6) + (3 × 6) = 30 +18 = 48. Therefore, Rohit works for 48 hours in a week. Example 15: Jaya’s father bought 7 boxes of mangoes, with 12 mangoes in each box. How many mangoes did Jaya’s father buy in all? Solution: Number of boxes of mangoes Jaya's father bought = 7 Number of mangoes in each box = 12 Total number of mangoes = 12 × 7 The larger number is 12, and it is 2 more than 10. 54 30-01-2018 16:02:38 Alpine_V3_Maths_G4_TB.indb 54

So, 12 = 10 + 2. Hence, 12 × 7 = (10 × 7) + (2 × 7) = 70 + 14 = 84. Therefore, Jaya’s father bought 84 mangoes in all. Higher Order Thinking Skills (H.O.T.S.) Let us now see some more examples where we multiply larger numbers mentally. Example 16: Find the answer by adding the products: 17 × 7 Solution: Steps Solved Solve this , 17 × 7 19 × 9 Step 1: Check by how much is the larger number more The larger number is 17, The larger number is than 10. and from 10 we count 11, 12, 13, 14, 15, 16 and from 10 we count and 17. So, 17 is 7 more than10. ,,,, , , , , and . So, is more than 10. Step 2: Take the number Number from step 1 is 7, Number from step 1 is , from step 1 and check by and from 5, we count 6 how much it is more than 5. and 7. and from 5, we count ,, and . So, is more So, 7 is 2 more than 5. than 5. Step 3: Write the three 10 + 5 + 2 = 17 + + = numbers whose sum is the larger number. 10 × 7 = 70 10 × = 5 × 7 = 35 5× = Step 4: Multiply each 2 × 7 = 14 = number of the sum in × the previous step by the smaller given number. Use memorised tables of 1 to 5 and 10 to solve mentally. Multiplication 55 Alpine_V3_Maths_G4_TB.indb 55 30-01-2018 16:02:38

Steps Solved Solve this 17 × 7 19 × 9 Step 5: Add all the three products from step 4 to get 70 + 35 + 14 = 119 + + = the final answer. Therefore, 17 × 7 = 119. Therefore, 19 × 9 =. Drill Time Concept 5.1: Multiply 3-digit and 4-digit Numbers 1) Multiply a 3-digit number by a 3-digit number. a) 247 × 567 b) 509 × 121 c) 892 × 469 d) 731 × 691 2) Multiply a 4-digit number by a 1-digit number. a) 6741 × 4 b) 3456 × 8 c) 9258 × 9 d) 5555 × 5 3) Word problems a) P ranav makes 253 cotton bags in a day. How many bags will he be able to make in the year 2017? [Hint: 2017 is not a leap year] b) Tanya bought sweaters as Christmas gifts for her 7 cousins. If one sweater costs ` 2734 , then how much money in all did she spend for the gifts? Concept 5.2: Multiply Using Lattice Algorithm 4) Multiply a 2-digit number by a 2-digit number. a) 24 × 32 b) 56 × 15 c) 13 × 39 d) 67 × 51 5) Multiply a 3-digit number by a 2-digit number. a) 158 × 17 b) 451 × 39 c) 651 × 67 d) 721 × 41 6) Word problems a) A movie theatre sold 127 tickets for a movie. Cost of one ticket was ` 85. How much money did the theatre owner earn from that movie? b) T here are 47 students in Class 3. Answer sheets were given to each student for Maths exam. If one answer sheet has 15 pages, then how many total sheets of paper were used for the exam? 56 30-01-2018 16:02:39 Alpine_V3_Maths_G4_TB.indb 56

Concept 5.3: Mental Maths Techniques: Multiplication 7) Multiply the following: a) 9 × 7 b) 9 × 6 c) 11 × 7 d) 14 × 6 e) 13 × 8 8) Word problems a) T here are 14 players in a football team. If 8 teams are participating in the district level football tournament, then how many pairs of boots are needed for them? b) Megha eats 8 chappatis daily. How many chappatis does she eat in a week? Multiplication 57 Alpine_V3_Maths_G4_TB.indb 57 30-01-2018 16:02:39

Chapter Time 6 Let Us Learn About • reading and writing time. • the 12-hour and the 24-hour clock formats. • converting 12-hour clock to 24-hour clock format and vice versa. • the terms ‘duration’, ‘end time’ and ‘start time’. • problems involving estimation of time. Concept 6.1: Duration of Events Think Jasleen was going to school. When she started from home, the time shown by the clock was . Jasleen was easily able to read it as 8 o’clock. When she reached the school, the time shown by the school clock was Jasleen’s found it difficult to read the time from the clock. Can you tell what time it is? Recall There are 24 hours in a day. In a clock, the hour hand shows hours and completes one turn in 12 hours. 58 30-01-2018 16:02:43 Alpine_V3_Maths_G4_TB.indb 58

The minute hand shows minutes and takes one turn in one hour. We have learnt to read time to the nearest hour and minutes when the minute hand is on any one of the numbers on the clock. Let us recall the concept by writing the time for the clocks shown below: Read the time shown by the clocks given: a) b) c) d) & Remembering and Understanding 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 12 and 1, there are four lines. Between 1 and 2, there are four lines. They divide the space between two consecutive numbers into five equal parts. Each division between these consecutive numbers indicates a minute. Thus, these sixty divisions together make 60 minutes or 1 hour. Example 1: Let us read the time shown by these clocks. One is done for you. a) b) c) Time 59 Alpine_V3_Maths_G4_TB.indb 59 30-01-2018 16:02:54

a) T he hour hand has b) The hour hand has c) T he hour hand has crossed 10. crossed ____________. crossed ____________.. 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 The time is ________. The time is ________. is 10:13. We have learnt to read and write time in the 12-hour clock format. Now, let us learn to read time in the 24-hour clock format. In 12-hour clock format: • 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 24-hour clock format: • The time is expressed as a 4-digit number (hhmm) followed by ‘h’ to denote hours. 12-hour clock time 24-hour clock time Read as 4:20 a.m. 0420 h Four twenty hours 11:40 a.m. 1140 h Eleven forty hours 5:30 p.m. 1730 h Seventeen thirty hours 7:35 p.m. 1935 h Nineteen thirty-five hour • Here, the first two d gits from the left tell us the hours and the next two digits tell us the minutes. • We do not write a.m. or p.m. • 12 o’clock at midnight is written as 0000 h. • 12 o’clock in the afternoon is written as 1200 h. The time before noon is written in the 12-hour format but without a.m. For example, 5:30 a.m. is written as 0530 hours. • The time post noon is written by adding 12 to the number of hours. • When the number of hours is more than 12, then the time indicates post noon. For example, 1730 h, 1815 h, 2210 h and so on. • When the hour hand is at 12 and the minutes are more than 00, the time is past noon and we write p.m. along with the number. For example, 1220 h = 12:20 p.m. (Here, we do not subtract 12 from hours.) 60 Alpine_V3_Maths_G4_TB.indb 60 30-01-2018 16:02:54

To convert the time in 24-hour clock to12-hour clock format, we subtract 12 from the number of hours and write p.m. after the difference. To convert time from 12-hour clock into 24-hour clock for the time after 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 record time. • The 24-hour clock is used in digital watches. Example 2: Convert the given time to 12-hour clock format. a) 1320 h b) 0550 h c) 0915 h d) 2105 h e) 1800 h f) 1945 h g) 2355 h h) 0030 h Solution: The 12-hour clock format are given below. a) (13 – 12):20 = 1:20 p.m. b) 5:50 a.m. c) 9:15 a.m. d) (21 – 12): 05 = 9:05 p.m. e) (18 – 12):00 = 6 p.m. f) (19 – 12): 45 = 7:45 p.m. g) (23 – 12):55 = 11:55 p.m. h) (00 + 12):30 = 12:30 a.m. We have learnt how to read and show time, exact to minutes and hours. Let us now consider an example that involves finding the length of time between two given times. Example 3: The clocks given show the start time and end time of a Maths class in a school. 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 Time 61 Alpine_V3_Maths_G4_TB.indb 61 30-01-2018 16:02:57

The time between two given times is called the length of time or time duration or time interval. It is given by the difference of end time and start time. Application Let us see a real-life example involving duration of time. Example 4: Neha went to the airport to see off her uncle. There she saw the departure time for Flight 142 to Hyderabad as 1102 h. What was the time of departure of the flight in the 12-hour clock time? Solution: Time of departure of the flight = 102 h 1102 h is in hhmm form. Since 11 < 12, the given time is a.m. Therefore, the given time in 12-hour clock is 11: 02 a.m. Higher Order Thinking Skills (H.O.T.S.) Let us now see a few more real-life examples involving the duration of time. Example 5: 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. . Duration of travel = 2 hours 5 minutes End time = Start time + Duration = 10:10 p.m. + 2 hours 5 minutes = 12:15 a.m. (After 12 midnight, time is taken as a.m.) Therefore, Anil’s flight reached H derabad at 12:15 a.m. Example 6: A movie began at 5:35 p.m. Lucky switched on the TV at 6:23 p.m. For how much time did Lucky miss the movie? Solution: Start time of the movie = 5:35 p.m. Time at which Lucky switched on the TV = 6:23 p.m. 5:35 pm to 6 p.m. = 25 minutes 6 pm to 6:23 p.m. = 23 minutes 62 30-01-2018 16:02:57 Alpine_V3_Maths_G4_TB.indb 62

The time for which Lucky missed the movie = (25 + 23) = 48 minutes Therefore, Lucky missed the movie for 48 min. Example 7: When Shruti was having her breakfast, the clock showed 7:45. Express the time in the 12-hour and 24-hour clock formats? Solution: The time when Shruti was having her breakfast = 7:45 This time in the 12 hour clock time is 7:45 a.m. In the 24-hour clock time, it is 0745 h. Concept 6.2: Estimate Time Think Jasleen’s father was trying to book flight tickets from Mangalore to Dubai. He asked Jasleen to see the flight timings. He wanted her to find the time it would take for him t reach Dubai. Do you know how to find that Recall The time from midnight 12 to midday 12 is 12 hours. The time from midday 12 to midnight 12 is 12 hours. Observe this timeline. 12 12 Mid Mid night night 12 hours Mid 12 hours Morning day Afternoon / Evening The time after midnight is written with a.m. after it. The time after midday is written with p.m. after it. So, 4 o’clock in the morning is 4 a.m., and 4 o’clock in the evening is 4 p.m. We can show the time in the morning or evening on a clock face. We know how to find the length of the time between two given times. Time 63 Alpine_V3_Maths_G4_TB.indb 63 30-01-2018 16:02:58

Now, let us compare the different units of time. • A minute is a shorter period of time than an hour. • An hour is shorter than a day. A day is shorter than a week. • A week is shorter than a month. • A month is shorter than a year. Express the following in a.m. or p.m. a) 3:30 in the morning b) 11:45 before noon c) 12:15 at midnight d) 5 in the evening & Remembering and Understanding We have learnt how to find the duration of time with the help of start time and end time. Duration = End time – Start time End time = Start time + Duration Start time = End time – Duration Let us understand this through a few examples. Example 8: If an event starts at 1:15 p.m. and it takes 2 hours to get over, then by what time will the event end? Solution: The start time of the event = 1:15 p.m. Duration of the event = 2 hours End time of the event = Start Time + Duration = 1:15 p.m. + 2 hours = 3:15 p.m. Therefore, the end time of the event is 3:15 p.m. Example 9: If a dance class ends at 9:20 a.m. and has taken 1 hour 15 minutes to complete, when did it begin? Solution: The end time of the dance class = 9:20 a.m. Duration of the class = 1 hour 15 minutes Start time of the class = End Time – Duration = 9:20 a.m. – 1 hour 15 minutes = 8:05 a.m. Therefore, the dance class began at 8:05 a.m. 64 Alpine_V3_Maths_G4_TB.indb 64 30-01-2018 16:02:58

Example 10: Ravi’s swimming class is for a duration of 1 h 50 min. If the class begins at 10:15 a.m., at what time does it end? Solution: Duration of Ravi’s swimming class = 1 h 15 min The start time of the class = 10:15 a.m. The end time of the class = 10:15 a.m. + 1 h 15 min = (10 + 1) h + (15 + 15) min = 11 h 30 min Therefore, Ravi’s swimming class ends at 11:30 a.m. Example 11: On the Sports day of a school, the indoor games competition begins at 11:40 a.m. If the competition goes on for 2 hours, at what time will it end? Solution: Start time of indoor games competition = 11:40 a.m. Duration of competition = 2 hours End time = Start time + Duration = 11:40 a.m. + 2 h = 1:40 p.m. Example 12: Our school’s annual day begins at 5:30 p.m. and would end after 5 h 12 min. At what time will it end? Express the end time in the 24-hour clock format. Solution: Start time of our annual day = 5:30 p.m. Duration of the celebration = 5 h 12 min End time = Start time + Duration = 5:30 p.m. + 5 h 12 min = 10:42 p.m. Therefore, the annual day ended at 10:42 p.m. In 24-hour clock time, it is (10 + 12) 42 h = 2242 h. Application Let us see a few real-life examples involving the estimation of time. Example 13: Radha participated in a drawing competition which was scheduled for one hour starting at 9 a.m. If Radha completes her drawing 15 minutes before the end time, at what time does she complete her drawing? Time 65 Alpine_V3_Maths_G4_TB.indb 65 30-01-2018 16:02:58

Solution: Drawing competition was for 1 hour, starting at 9 a.m. So, the competition was scheduled to end at 10 a.m. Radha completed her drawing 15 minutes before the end time. That is, she took (60 – 15) minutes that is 45 minutes for the drawing. 45 minutes from 9 a.m. is 9:45 a.m. Therefore, the time at which Radha completed her drawing was 9:45 a.m. Example 14: Leela goes for the music class at 4:48 p.m. and comes back at 6:45 p.m. How much time does she spend in the class? Solution: Start time of Leela’s music class is 4:48 p.m. End time of Leela’s music class is 6:45 p.m. 4:48 p.m. _1_2__m__in__u_te__s_ 5 p.m. ___1_h_o__u_r__ 6 p.m. 4__5_m__in__u_t_e_s 6:45 p.m. Time spent by Leela in the class = 1 hour + 45 minutes + 12 minutes = 1 hour 57 minutes Therefore, Leela spent 1 hour 57 minutes in the class. Higher Order Thinking Skills (H.O.T.S.) Let us consider another example that involves estimating time. Example 15: On 12th February, Raju saw the calendar and circled 21st March as his father’s birthday. He wanted to buy a gift for his father. How many days are left for him to buy the gift? Solution: Since it is not mentioned as a leap year, we assume the number of days in February to be 28. Days in February = 28 – 11 = 17 Days in March = 21 Total number of days = 17 + 21 = 38 Therefore, there are 38 days from 12th February to 21st March for Raju to buy a gift for his father. 66 30-01-2018 16:02:58 Alpine_V3_Maths_G4_TB.indb 66

Drill Time Concept 6.1: Duration of Events 1) Find the duration of time (in 24-hour clock) from the given start time and end time. a) Start Time = 12:00 and End Time = 02:15 b) Start Time = 15:00 and End Time = 19:00 c) Start Time = 3:15 and End Time = 7:20 d) Start Time = 7:20 and End Time = 10:41 e) Start Time = 5:56 and End Time = 7:57 2) Read the times on the clocks and write them in the 12-hour and 24-hour formats. a) b) Evening Afternoon c) d) Morning Afternoon Time 67 Alpine_V3_Maths_G4_TB.indb 67 30-01-2018 16:03:05

e) f) Evening Night 3) Word problems a) K arthik started his running race at 8:20 a.m. and finished it at 8:45 a.m. For how long did he run? b) S hirish was eating his dinner when it was 10:36 in the clock. What is the time in 12-hour and 24-hour clock formats? Concept 6.2: Estimate Time 4) Word problems a) If Sohail’s magic show begins at 5:56 p.m. and lasts for 2 hours, at what time does his show end? b) Sunny’s karate class lasted for 4 hours. If it ended at 8:20 p.m., when did it begin? 68 30-01-2018 16:03:09 Alpine_V3_Maths_G4_TB.indb 68

Chapter Division 7 Let Us Learn About • dividing 4-digit numbers by 1-digit and 2-digit numbers. • d ividing 3-digit numbers by 2-digit numbers. • properties of division. Concept 7.1: Divide Large Numbers Think Jasleen and seven of her friends want to share 3540 papers equally among themselves. Do you think the papers can be divided, without some being left over? Recall Recall that we can write two multiplication facts for a division fact. For example, a multiplication fact for 45 ÷ 9 = 5 can be written as 9 × 5 = 45 or 5 × 9 = 45. 45 ÷ 9 = 5 ↓ ↓ ↓ Dividend Divisor Quotient The number that is divided is called the dividend. The number that divides is called the divisor. The number of times the divisor divides the dividend is called the quotient. Alpine_V3_Maths_G4_TB.indb 69 69 30-01-2018 16:03:11

Factors Factors Multiplicand × Multiplier = Product Multiplicand × Multiplier = Product 5 × 9 = 45 9 × 5 = 45 ↓ ↓ ↓ ↓ ↓ ↓ Divisor Quotient Dividend Divisor Quotient Dividend The part of the dividend that remains without being divided is called the remainder. Let us solve the following to revise the concept of division. a) 72 ÷ 9 b) 42 ÷ 3 c) 120 ÷ 5 d) 80 ÷ 4 e) 24 ÷ 1 & Remembering and Understanding In Class 3, we have learnt that division and multiplication are reverse operations. Let us now understand the division of large numbers using multiplication. Division of a 4-digit number by a 1-digit number Dividing a 4-digit number by a 1-digit number is similar to that of a 3-digit number by a 1-digit number. Example 1: Solve: 2065 ÷ 5 Solution: Steps Solved Solve these Step 1: Check if the thousands digit of the dividend is greater than the divisor. If it is )5 2065 )7 3748 not, consider the hundreds digit also. 2 is not greater than Dividend = _____ Step 2: Find the largest number in the 5. So, consider 20. Divisor = ______ multiplication table of the divisor that can Quotient = ____ be subtracted from the 2-digit number of 4 Remainder = ___ the dividend. Write the quotient. Write the product of the quotient and divisor below )5 2065 the dividend. -2 0 Step 3: Subtract and write the difference. 5 × 4 = 20 5 × 5 = 25 25 > 20 4 )5 2065 -20 0 70 30-01-2018 16:07:40 Alpine_V3_Maths_G4_TB.indb 70

Steps Solved Solve these Step 4: Check if difference < 0 < 5 (True) divisor is true. )3 2163 4 If it is false, the division is incorrect. Dividend = _____ Step 5: Bring down the tens digit of the )5 2065 Divisor = ______ dividend and write it near the remainder. Quotient = ____ −20↓ Remainder = ___ 06 )5 1555 Step 6: Find the largest number in the 5×1=5 multiplication table of the divisor that can 5 × 2 = 10 Dividend = _____ be subtracted from the 2-digit number in 5 < 6 < 10 Divisor = ______ the previous step. So, 5 is the required Quotient = ____ number. Remainder = ___ Step 7: Write the factor of the required 41 number, other than the divisor, as the quotient. )5 2 0 6 5 Write the product of the divisor and the − 20 ↓ quotient below the 2-digit number. 06 Then subtract them. − 05 01 Step 8: Repeat steps 6 and 7 till all the digits 1 < 5 (True) of the dividend are brought down. 4 13 Check if remainder < divisor is true. 5) 2 0 65 Stop the division. (If this is false, the division is incorrect.) −2 0 ↓ 06 − 05 015 − 015 000 Step 9: Write the quotient and the Quotient = 413 remainder. Remainder = 0 Step 10: Check if (Divisor × Quotient) + 5 × 413 + 0 = 2065 Remainder = Dividend is true. If this is false, 2065 + 0 = 2065 the division is incorrect. 2065 = 2065 (True) Division 71 Alpine_V3_Maths_G4_TB.indb 71 30-01-2018 16:07:47

Division of a 3-digit number by a 2-digit number Let us understand the division of 3-digit numbers by 2-digit numbers, through some examples. Example 2: Divide: 414 ÷ 12 Solution: )Write the dividend and the divisor as Divisor Dividend Steps Solved Solve these Step 1: Guess the quotient by thinking of )12 414 dividing 41 by 12. )14 324 Find the multiplication fact which has 12 × 3 = 36 the number less than or equal to the 12 × 4 = 48 dividend and the divisor. 36 < 41 < 48 So, 36 is the number to be subtracted from 41. Step 2: Write the factor other than the Write 3 in the quotient and Dividend = _____ Divisor = ______ dividend and the divisor as the quotient. 36 below 41, and subtract. Quotient = ____ Then bring down the next number in the dividend. 3 )12 414 −36↓ 054 Remainder = ___ Step 3: Guess the quotient by thinking of 12 × 4 = 48 )16 548 dividing 54 by 12. 12 × 5 = 60 Dividend = _____ Divisor = ______ Find the multiplication fact which has 48 < 54 < 60 Quotient = ____ the number less than or equal to the So, 48 is the number to be Remainder = ___ dividend and divisor. Write the factor subtracted from 54. other than the dividend and the divisor as the quotient. Write 4 in the quotient and 48 below 54, and subtract. 34 )12 414 −36 ↓ 054 − 048 6 Quotient = 34 Remainder = 6 72 Alpine_V3_Maths_G4_TB.indb 72 30-01-2018 16:07:52

Checking for the correctness of division: We can check whether our division is correct or not using a multiplication fact of the division. Step 1: Compare the remainder and the divisor. [Note: The remainder must always be less than the divisor.] Step 2: Check if (Quotient × Divisor) + Remainder = Dividend Let us now check if our division in example 2 is correct. Steps Checked Step 1: Remainder < Divisor Dividend = 414 Step 2: (Quotient × Divisor) + Divisor = 12 Remainder = Dividend Quotient = 34 Remainder = 6 6 < 12 (True) 34 × 12 + 6 = 414 408 + 6 = 414 414 = 414 (True) Note: a) If remainder > divisor, the division is incorrect. b) If (Quotient × Divisor) + Remainder is not equal to Dividend, the division is incorrect. Dividing a 4-digit number by a 2-digit number Dividing a 4-digit number by a 2-digit number is similar to dividing a 3-digit number by a 2-digit number. Let us understand this through the following example. Example 3: Solve: 2340 ÷ 15 Solution: Steps Solved Solve these Step 1: Check if the thousands digit 2 is not greater than 15. So, )12 5088 of the dividend is greater than the consider 23. divisor. If it is not, consider also the hundreds digit too. )15 2340 Division 73 Alpine_V3_Maths_G4_TB.indb 73 30-01-2018 16:07:55

Steps Solved Solve these Step 2: Guess the quotient by 1 Dividend = _____ thinking of dividing 23 by 15. Divisor = ______ )15 2340 Quotient = _____ Remainder = _____ Find the multiplication fact which has −15 )14 4874 a number less than or equal to the 15 × 1 = 15 dividend and the divisor. 15 × 2 = 30 15 < 23 < 30 So, 15 is the required number. Step 3: Write the factor other than Write 1 in the quotient and 15 the dividend and the divisor as the below 23 and subtract. Then quotient. bring down the next number in the dividend. 15) 1 2340 −15↓ 84 Step 4: Guess the quotient by 15 × 5 = 75 thinking of dividing 84 by 15. 15 × 6 = 90 Find the multiplication fact which has 75 < 84 < 90 Dividend = _____ a number less than or equal to the Divisor = ______ So, 75 is the required number Quotient = _____ dividend and the divisor. Remainder = _____ that is to be subtracted from Write the factor other than the dividend and the divisor as the 84. 156 quotient. )15 2340 − 15↓ 84 − 75 9 74 30-01-2018 16:08:08 Alpine_V3_Maths_G4_TB.indb 74

Steps Solved Solve these Step 5: Subtract and write the 15 × 5 = 75 )16 3744 difference. Repeat till all the digits of 15 × 6 = 90 the dividend are brought down. 90 = 90 So, 90 is the required number. 156 15 )2340 − 15 ↓ Dividend = _____ 84 Divisor = ______ Quotient = _____ − 75 90 − 90 00 Quotient = 156 Remainder = 0 Step 6: Check if (Divisor × Quotient) + 15 × 156 + 0 = 2340 Remainder = _____ Remainder = Dividend is true. If this is 2340 + 0 = 2340 false, the division is incorrect. 2340 = 2340 (True) Let us see some properties of division. Properties of division 1) Dividing a number by 1 gives the same number as the quotient. For example: 15 ÷ 1 = 15; 1257 ÷ 1 = 1257; 1 ÷ 1 = 1; 0 ÷ 1 = 0 2) Dividing a number by itself gives the quotient as 1. For example: 15 ÷ 15 = 1; 1257 ÷ 1257 = 1; 1 ÷ 1 = 1 3) Division by zero is not possible and is not defined For example: 10 ÷ 0; 1257 ÷ 0; 1 ÷ 0 are not define Division 75 Alpine_V3_Maths_G4_TB.indb 75 30-01-2018 16:08:14

Application Division of large numbers can be applied in many real-life situations. Consider these examples. Example 4: 4720 apples are to be packed in 8 baskets. If each basket has the 590 same number of apples, how many apples are packed in each Solution: basket? )8 4720 Total number of apples = 4720 − 40↓ 072 Number of baskets = 8 − 072 The number of apples packed in each basket = 4720 ÷ 8 0000 Example 5: Therefore, 590 apples are packed in each basket. − 0000 Solution: 2825 notebooks were distributed equally among 25 students. How many 0b0o0o0ks did each student get? 113 Number of notebooks = 2825 Example 6: )25 2 8 2 5 Solution: Number of students = 25 −25↓ Number of books each student got = 2825 ÷ 25 032 Therefore, each student got 113 notebooks. −025 0 075 8308 people watched a hockey match. If 10 people watched − 0075 from each cabin in the stadium, how many cabins were full? How 000 many people were there in the remaining cabin? 830 Number of people = 8308 Number of people in each cabin = 10 )10 8 3 0 8 −80↓ 30 Number of cabins = 8308 ÷ 10 = 830 − 30 Number of people in the remaining cabin = 8 (Remainder in the 008 division of 8308 by 10). Therefore, 8 people were remaining in the cabin. Higher Order Thinking Skills (H.O.T.S.) Let us see some more examples of situations where we use division of large numbers. 76 30-01-2018 16:08:21 Alpine_V3_Maths_G4_TB.indb 76

Example 7: A school has 530 students in the primary section, 786 students in the middle school and 658 students in the high school section. If equal number of students Solution: are seated in 6 halls, how many students are seated in each hall? Number of students in the primary section = 530 329 Number of students in the middle school section = 786 Number of students in the high school section = 658 6) 1974 Thus, the total number of students in the school − 18 = 530 + 786 + 658 = 1974 017 Example 8: 1974 children are equally seated in 6 halls. − 012 Solution: 54 − 54 00 Therefore, the number of students in each hall = 1974 ÷ 6 = 329 students. Divide the largest 4-digit number by the largest 2-digit number. Write the quotient and the remainder. 101 The largest 4-digit number is 9999. The largest 2-digit number is 99. 99) 9999 The required division is 9999 ÷ 99 − 99 ↓ 009 − 000 99 Quotient = 101; Remainder = 0 − 99 00 Drill Time Concept 7.1: Divide Large Numbers 1) Divide a 4-digit number by a 1-digit number. a) 1347 ÷ 6 b) 4367 ÷ 5 c) 3865 ÷ 4 d) 5550 ÷ 5 2) Divide a 4-digit and 3-digit numbers by a 2-digit number. a) 3195 ÷ 10 b) 612 ÷ 10 c) 2676 ÷ 12 d) 267 ÷ 11 3) Word Problems a) A n amount of ` 1809 is distributed equally among 9 women. How much money did each of them get? b) 10 boxes have 1560 pencils. How many pencils are there in a box? c) A school has 1254 students, who are equally grouped into 14 groups. How many students are there in each group? How many students are remaining? Division 77 Alpine_V3_Maths_G4_TB.indb 77 30-01-2018 16:08:27

Chapter Fractions - I 8 Let Us Learn About • equivalent fractions. • problems related to equivalent fractions. • like and unlike fractions. • adding and subtracting like fractions. Concept 8.1: Equivalent Fractions Think Jasleen cuts 3 apples into 18 equal pieces. Ravi cuts an apple into 6 equal pieces. Did both of them cut the apples into equal pieces? Recall In Class 3, we have learnt that a fraction is a part of a whole. A whole can be a region or a collection. When a whole is divided into two equal parts, each part is called ‘a half’. 11 ‘Half’ means 1 out of 2 equal parts. We write ‘half’ as 1 . 22 2 78 30-01-2018 16:08:34 Alpine_V3_Maths_G4_TB.indb 78

Two halves make a whole. Numerator Numbers of the form Denominator are called fractions. The total number of equal parts into which a whole is divided is called the denominator. The number of such equal parts taken is called the numerator. Similarly, each of the three equal parts of a whole is called a third. We write one-third as 1 and, two-thirds as 2 . 33 Three-thirds or 3 make a whole. 3 Each of four equal parts of a whole is called a fourth or a quarter written as 1 . 4 Two such equal parts are called two-fourths, and three equal parts are called three-fourths, written as 2 and 3 respectively. Four quarters make a whole. 44 2 halves, 3 thirds, 4 fourths, 5 fifths, …, 10 tenths make a whole. So, we write a whole as 2 , 3 , 4 , 5 ,...,10 and so on. 2 3 4 5 10 & Remembering and Understanding Fractions that denote the same part of a whole are called equivalent fractions. Let us now understand what equivalent fractions are. Suppose there is 1 bar of chocolate with Ram and Raj each as shown. chocolate with Ram chocolate with Raj Ram eats 1 of the chocolate. 5 Then the piece of chocolate he gets is Raj eats 2 of the chocolate. 10 Then the piece of chocolate he gets is Fractions - I 79 Alpine_V3_Maths_G4_TB.indb 79 30-01-2018 16:08:37

We see that both the pieces of chocolates 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 1: Shade the regions to show equivalent fractions. a) [ 1 and 2 ] 36 b) [ 1 and 2 ] 48 Solution: a) 1 3 2 6 b) 1 4 2 8 Example 2: Find the figures that represent e uivalent fractions. Also, mention the fractions. a) b) c) d) 80 30-01-2018 16:08:39 Alpine_V3_Maths_G4_TB.indb 80

Solution: The fraction represented by the shaded part of figure a) is 1 . 2 The shaded part of figure b) repr sents 2 . The shaded part of figure d 4 represents 1 . 2 So, the shaded parts of figures a , b) and d) represent equivalent fractions. Application Let us see a few examples of equivalent fractions. Example 3: Shade the second figure to give a fraction equivalent to the first Solution: The fraction denoted in the first figure is 2 . This is half of the given figure. 4 So, to denote a fraction equivalent to the first, shade half of the the second figure as shown Example 4: 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. Fractions - I 81 Alpine_V3_Maths_G4_TB.indb 81 30-01-2018 16:08:40

Higher Order Thinking Skills (H.O.T.S.) We have learnt how to find equivalent fractions using pictures. Let us see a few more examples involving equivalent fractions. Example 5: Find two fractions equivalent to the given fractions. Solution: a) 2 b) 33 11 66 To find fractions equivalent to th 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 cannot divide them to get an equivalent fraction of 2 . 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 Thus, 10 is a fraction equivalent to 2 . 55 11 2 L ikewise, we can multiply by any number of our choice to get more 11 fractions equivalent to it. b) 33 66 W e see that 33 and 66 have common factors 3, 11 and 33. So, dividing both the numerator and the denominator by 3, 11 or 33, we get fractions equivalent to 33 . 66 33 ÷ 3 = 11 , 33 ÷ 11 = 3 or 33 ÷ 33 = 1 66 ÷ 3 22 66 ÷11 6 66 ÷ 33 2 Therefore, 11 , 3 and 1 are the fractions equivalent to 33 . 22 6 2 66 82 30-01-2018 16:08:45 Alpine_V3_Maths_G4_TB.indb 82

Example 6: Draw four similar rectangles. Divide them into 2, 4, 6 and 8 equal parts. Then Solution: colour 1 2 3 and 5 parts of the rectangles respectively. Compare these , , 246 8 coloured parts and write the fractions using >, = or <. 11 22 22 4 33 66 55 88 From the coloured parts of these rectangles, we can see that all of them except 5 are of the same size. So, the fractions, 12 and 3 are , 8 24 6 equivalent. 123 Therefore, = = . 246 Fractions - I 83 Alpine_V3_Maths_G4_TB.indb 83 30-01-2018 16:08:48

Concept 8.2: Identify and Compare Like Fractions Think Jasleen has a circular disc coloured in blue, green, red and white as shown. She wants to know if there is any special name for the fractions shown by different colours on the circular disc. Do you know any special name for such fractions? Recall In Class 3, we have learnt to represent shaded parts of a whole as fractions. Recall the same through the following example. Jasleen’s colourful circular disc is given here. Find the fractions represented by the following colours: a) Red b) Green c) Blue d) White & Remembering and Understanding 12 3 Fractions such as 8 , 8 and 8 , that have the same denominator are called like fractions. Fractions such as 1, 2 and 3 that have different denominators are called unlike fractions. 8 4 7 Fractions with numerator ‘1’ are called unit fraction. such as 1 , 1 , 1 and so on. 234 To understand these fractions, consider the following examples. Example 7: Identify like and unlike fractions from the following fractions. 3 ,3 , 1, 5 , 6, 1, 4 7 5 7 7 7 4 11 84 30-01-2018 16:08:52 Alpine_V3_Maths_G4_TB.indb 84

Solution: 3 , 1 , 5 and 6 have the same denominator. So, they are like fractions. 777 7 3 , 1 and 4 have different denominators. So, they are unlike fractions. Example 8: 54 11 a) Find the fraction of the parts not shaded in these figures b) c) d) Which of them represent like fractions? Solution: a) Number of parts not shaded = 1 Total number of equal parts = 2 Fraction = Number of parts not shaded = 1 Total number of equal parts 2 b) Number of parts not shaded = 3 Total number of equal parts = 4 Fraction = Number of parts not shaded = 3 Total number of equal parts 4 c) Number of parts not shaded = 3 Total number of equal parts = 5 Fraction = Number of parts not shaded = 3 Total number of equal parts 5 d) Number of parts not shaded = 3 Total number of equal parts = 6 Fraction = Number of parts not shaded = 3 = 1 Total number of equal parts 6 2 a) and d) have denominator equal to 2. They represent like fractions. Fractions - I 85 Alpine_V3_Maths_G4_TB.indb 85 30-01-2018 16:08:54

Application We can compare like fractions and tell which is greater or less than the others. To compare like fractions, we compare their numerators. The fraction with the greater numerator is greater. Let us understand this better through some examples. Example 9: Jai ate 1 of the apple and Vijay ate 2 of the apple. Who ate more? Solution: 33 1 Fraction of apple Jai ate = 3 Fraction of apple Vijay ate = 2 3 21 Since, 2 > 1, 3 > 3 Therefore, Vijay ate more. Example 10: The circular disc shown here is divided into equal parts. The parts are painted in different colours. Write the fraction of each colour on the disc. Compare the fractions and tell which colour is used more and which the least. Solution: Total number of equal parts on the disc is 16. Number of parts painted yellow is 3. Fraction = 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 = 6 Total number of equal parts 16 The fraction of the disc that is painted red = Number of parts painted red = 4 Total number of equal parts 16 The fraction of the disc that is painted blue = Number of parts painted blue = 3 Total number of equal parts 16 86 30-01-2018 16:08:57 Alpine_V3_Maths_G4_TB.indb 86

Comparing the numerators of these fractions, we get 3 < 4 < 6. Since, 6 is 16 the greatest and 3 is the least, white is used the most and blue and yellow 16 are the least. Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples using comparison of like fractions. Example 11: Colour each figure to represent he given fraction and compare them. 3 2 5 5 Solution: Clearly, the part of the figure re resented by 3 is greater than that 5 represented by 2 . Hence, 3 is greater than 2 . 55 5 Let us try to arrange some like fractions in ascending and descending orders. Example 12: Arrange 1 , 6 , 2 , 5 and 4 in the ascending and descending orders. 7777 7 Solution: Comparing the numerators of the given likeTrfraacintionMs, y Brain we have 1 < 2 < 4 < 5 < 6. So, 1 < 2 4 5 6 7 7< 7< 7< 7. Therefore, the required ascending order is 1 , 2 , 4 , 5 , 6 . 77777 We know that, the descending order is just the reverse of the ascending order. So, the required descending order is 6 , 5 , 4 , 2 , 1 . 77 7 7 7 Fractions - I 87 Alpine_V3_Maths_G4_TB.indb 87 30-01-2018 16:09:01

Concept 8.3: Add and Subtract Like Fractions Think Jasleen has a cardboard piece, equal parts of which are coloured in different colours. Some of the equal parts are not coloured. She wants to find the part of the cardboard that has been coloured and left uncoloured. How do you think Jasleen can find that Recall Recall that like fractions have the same denominators. To compare them, we compare their numerators. Let us answer the following to recall the concept of like fractions. Compare the following using >, < and =. a) 2 ____ 1 b) 4 ____ 8 c) 3 ____ 5 d) 7 ____ 3 e) 1 ____ 4 33 10 10 77 88 55 & Remembering and Understanding While adding or subtracting like fractions, add or subtract only their numerators. Write the sum or difference on the same denominator. Let us understand addition and subtraction of like fractions through some examples. Example 13: In the given figures, find the fractions represented by the shaded parts usin addition. Then find the fractions represented by the unshaded parts using subtraction. a) b) c) Solution: We can find the fractions represented by the shaded and the unshaded parts with the following steps. 88 30-01-2018 16:09:03 Alpine_V3_Maths_G4_TB.indb 88

Solved Solve these Steps Step 1: Count the total Total number of equal Total number of Total number of number of equal parts. equal parts = ____ parts = 6 equal parts = ___ Step 2: Count the a) Number of parts a) N umber of parts a) N umber of parts number of parts of each coloured pink = 1 coloured yellow coloured violet = colour. = ______ _______ b) Number of parts coloured blue = 2 b) Number of parts b) N umber of parts coloured violet = coloured brown _______ = ______ Step 3: Write the fraction Pink: 1 , Blue: 2 Yellow: ________ Violet: ________ representing the number 66 Violet: ________ Brown: ________ of parts of each colour. Step 4: To add the like The fraction that The fraction that The fraction that fractions in step 3, add represents the their numerators and represents the shaded represents the shaded part of the write the sum on the given figure is same denominator. part of the given shaded part of the ____ + ____=____. figure is given figure i 1 + 2 =1 + 2 = 3 . ____ + ____=____. 66 6 6 Step 5: Write the whole Like fraction Like fraction Like fraction representing the representing the as a like fraction of the representing the whole = 6 . whole = _______. whole = _______. sum in step 4. Then, to 6 So, the fraction So, the subtract the like fractions, that represents the subtract their numerators. So, the fraction unshaded part of fraction that Write the difference on that represents the the given figure is represents the the same denominator. unshaded part of the unshaded part of given figure i ____ − ____=____. the given figure is 6−3 =6−3 = 3. ____ − ____=_____. 66 6 6 Fractions - I 89 Alpine_V3_Maths_G4_TB.indb 89 30-01-2018 16:09:05

Example 14: Add: a) 3 + 1 45 23 57 Solution: 88 b) 13 + 13 c) 100 + 100 c) 48 – 26 3 1 3 +1 4 Example 15: a) 8 + 8 = 8 = 8 125 125 Solution: b) 4 + 5 = 4 + 5 = 9 13 13 13 13 c) 23 + 57 = 23 + 57 = 80 100 100 100 100 Subtract: a) 8 – 4 b) 33 – 25 99 37 37 a) 8 – 4 = 4 99 9 b) 33 – 25 = 33 − 25 = 8 37 37 37 37 48 26 48 − 26 22 c) 125 – 125 = 125 = 125 Application In some real-life situations, we use addition or subtraction of like fractions. Let us see a few examples. Example 16: 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 Part of the ribbon coloured yellow = 3 9 Total part of the ribbon coloured = 2 + 3 = 2 + 3 = 5 99 9 9 90 30-01-2018 16:09:16 Alpine_V3_Maths_G4_TB.indb 90

Part of the ribbon that is not coloured is 9 − 5 = 9 − 5 = 4 99 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 Example 17: Suman ate a quarter of a chocolate bar on one day and another quarter of Solution: the chocolate on the next day. How much chocolate did Suman eat in all? How much chocolate is remaining? Part of the chocolate eaten by Suman on the first day = 1 4 Part of the chocolate eaten by him on the next day = 1 4 Total chocolate eaten by Suman on both the days = 1 + 1 = 1+1 = 2 44 4 4 He ate 2 chocolate in all. Remaining chocolate = 4 − 2 = 4 − 2 = 2 4 44 4 4 Example 18: Manav painted two-tenths of a strip of chart in one hour and four-tenths of it in the next hour. What part of the strip did he paint in two hours? How much is left unpainted? Solution: Part of the strip of chart painted by Manav in one hour = 2 10 1st hour: Part of the strip painted by him in the next hour = 4 10 2nd hour: Part of the strip painted by him in two hours = 2 + 4 = 2 +4 = 6 10 10 10 10 Part of the strip of chart left without painting = 10 – 6 = 4 10 10 10 [From the figure, the total part of the strip painted = 6 and the part of the 10 strip not painted = 4 .] 10 Fractions - I 91 Alpine_V3_Maths_G4_TB.indb 91 30-01-2018 16:09:21

Higher Order Thinking Skills (H.O.T.S.) Let us see some more examples of addition and subtraction of like fractions. Example 19: Veena ate 5 of a pizza in the morning and 1 in the evening. What part of 88 Solution: the pizza is remaining? Part of the pizza eaten by Veena in the morning = 5 8 Part of the pizza eaten by Veena in the evening = 1 8 To find the remaining part of pizz , add the parts eaten and subtract the sum from the whole. Total part of the pizza eaten = 5 + 1 = 5 +1 = 6 88 8 8 Part of the pizza remaining = 1 – 6 = 8 – 6 = 2 8 8 8 8 Drill Time Concept 8.1: Equivalent Fractions 1) Shade the regions to show equivalent fractions. a) 1 2  2 4 and b) 1 2  5 10 and 92 30-01-2018 16:09:26 Alpine_V3_Maths_G4_TB.indb 92

2) Write four equivalent fractions for each of the following fractions. a) 1 b) 4 c) 3 d) 4 2 7 10 11 Concept 8.2: Identify and Compare Like Fractions 3) Identify like and unlike fractions from the following. a) 2, 2, 1, 5, 2 ,7 ,6 ,2 b) 7,4,4,2,4,2,3,6 83286889 95997449 c) 6 , 5 , 5 , 4 , 8 ,7 , 9 , 2 d) 3, 4 , 1, 3, 1, 4 14 14 17 17 17 14 17 14 5 5 5 7 9 11 4) Arrange the following fractions in the ascending order. a) 3 , 1 , 7 , 4 b) 3, 2 ,9 ,5 c) 1, 3 ,4 ,2 d) 1 , 8 , 7 , 9 11 11 11 11 13 13 13 13 7777 14 14 14 14 5) Arrange the following fractions in descending order. d) 1 , 7 , 8 , 3 20 20 20 20 a) 1, 8, 7, 4 b) 3 , 6 , 10, 8 c) 7 , 9 , 2 ,13 9999 17 17 17 17 21 21 21 21 Concept 8.3: Add and Subtract Like Fractions 6) Add: b) 3 + 16 c) 9 + 4 d) 187+ 147 e) 1 + 2 11 11 55 13 13 a) 2 + 5 77 7) Subtract: a) 15 − 7 b) 9 − 5 c) 11 − 3 d) 7 − 4 e) 13 − 12 66 88 40 40 45 45 30 30 8) Word problems a) Leena paints three-sixths of a cardboard and Rani paints half of similar cardboard. Who has painted a smaller area? b) Colour each fig re to represent the given fraction and compare them. 57 88 c) A jit ate 1 of a cake in the morning and 2 of it in the evening. What part of the cake 5 5 is remaining? Fractions - I 93 Alpine_V3_Maths_G4_TB.indb 93 30-01-2018 16:09:34

Chapter Fractions - II 9 Let Us Learn About • finding fractions of a number. • problems based on finding fractions. • proper, improper and mixed fractions. • converting improper to mixed fractions and vice versa. Concept 9.1: Fraction of a Number Think Jasleen’s father told her that he spends two-thirds of his salary per month and saves the rest. Jasleen calculated the amount her father saves from his salary of ` 25,000 per month. How do you think Jasleen could calculate her father’s savings per month? Recall In Class 3, we have learnt how to find the fraction of a collection. To find the fraction of collection, we find the number of each type of object in the total collection. Let us answer these to recall the concept. a) A half of a dozen bananas = _______________ bananas b) A quarter of 16 books = _______________ books c) A third of 9 balloons = _______________ balloons 94 30-01-2018 16:09:41 Alpine_V3_Maths_G4_TB.indb 94

d) A half of 20 apples = _______________ apples e) A quarter of 8 pencils = _______________ pencils & Remembering and Understanding To find the fraction of a number, we multiply the number by the fraction. Let us now learn to find the fraction of a number. Suppose there are 20 shells in a bowl. Vani wants to take 1 of them. So, she divides the shells 5 into 5 (the number in the denominator) equal groups and takes 1 group (the number in the numerator). This gives 5 groups with 4 shells in each group. So, 1 of 20 is 4. 5 Vani’s sister Rani wants to take 3 of the shells. So, she divides 10 the shells into 10 (the number in the denominator) equal groups, and takes 3 groups (the number in the numerator) of them. This gives 2 shells in each group. Hence, Rani takes 6 shells. Therefore, 3 of 20 is 6. 10 We write 1 of 20 as 1 × 20 = 20 = 4. 5 55 Similarly, 3 of 20 = 3 × 20 = 6. 10 10 Example 1: Find the following: 1 a) 2 of a metre (in cm) b) 10 of a kilogram (in g) 5 Solution: a) 2 of a metre = 2 × 1 m = 2 × 100 cm = 2 × 100 cm = 200 cm = 40 cm 5 55 55 b)  1 of a kilogram = 1 × 1 kg = 1 × 1000 g = 1000 g = 100 g 10 10 10 10 Example 2: Find the following: a) 2 of an hour (in minutes) b) 1 of a day (in hours) Solution: 3 4 a) 2 of an hour = 2 ×1h= 2 × 60 min = 2 × 60 = 120 = 40 min 3 3 3 3 3 b) 1 of a day = 1 × 1 day = 1 × 24 h = 1 × 24 h = 24 hrs = 6 h 4 44 4 Fractions - II 95 Alpine_V3_Maths_G4_TB.indb 95 30-01-2018 16:09:49

Application Let us now see some real-life examples in which we find the fraction of a number. Example 3: Ravi has ` 120 with him. He gave two-thirds of it to his sister. How much money is left with Ravi? Solution: Amount Ravi has = ` 120 Amount Ravi gave his sister = 2 of ` 120 = 2 × ` 120 = 2 × ` 40 = ` 80 33 Difference in the amounts = ` 120 – ` 80 = ` 40 Therefore, ` 40 is left with Ravi. Example 4: Reema completed one-tenth of a distance of 2 kilometres. How much distance (in metres) has she covered? Solution: The total distance to be covered by Reema = 2 km We know that 1 km = 1000 m. So, 2 km = 2000 m. The distance covered by Reema = 1 of 2 kilometres = 1 x 2000 m = 200 m Example 5: 10 10 Therefore, Reema has covered 200 metres of the distance. A school auditorium has 2500 chairs. On the annual day, 4 of the auditorium 5 was occupied. How many chairs were occupied? Solution: Total number of chairs in the auditorium = 2500 4 Fraction of chairs occupied = 5 4 4×2500 10000 Number of chairs occupied = × 2500 = = 5 5 5 Therefore, 2000 chairs were occupied in the auditorium. Higher Order Thinking Skills (H.O.T.S.) Let us now see some more examples where we have to find the fraction of a number. Example 6: Venu paints three-sixths of a cardboard and Raj paints a third of it. If the cardboard has an area of 144 sq.cm, what area of the cardboard did each of them paint? 96 30-01-2018 16:09:51 Alpine_V3_Maths_G4_TB.indb 96