51704171_BGM_Traveller G4 Mathematics Textbook FY_Text

= 23 × 1000 g + 240 g (As 1 kg = 1000 g) = 23240 g Quantity of flour bought by Venu = 11 kg 750 g + 12 kg 400 g = 24 kg 150 g = 24 × 1000 g + 150 g (As 1 kg = 1000 g) = 24150 g = 24150 g > 23240 g Therefore, Venu bought more quantity of flour than Kiran. The difference in their weights is (24150 – 23240) g = 910 g. Example 16: Suresh bought a few apples, grapes and a watermelon. The total weight of the fruits in his bag is 3 kg 750 g. The weight of the apples is 1 kg 100 g and that of the grapes is 1 kg 150 g. What is the weight of the watermelon? Solution: Suresh had 3 kinds of fruits, apples, grapes and a watermelon in his bag. Weight of apples = 1 kg 100 g kg g Weight of grapes = 1 kg 150 g 1 100 Total weight of apples and grapes +1 150 = 1 kg 100 g + 1 kg 150 g 2 250 Therefore, the weight of apples and grapes together is 2 kg 250 g. Weight of watermelon = Total weight of the fruits − Total weight of apples and grapes kg g The total weight of all the fruits in the bag = 3 kg 750 g 3 750 Weight of apples and grapes together = 2 kg 250 g − 2 250 Weight of the watermelon = 3 kg 750 g – 2 kg 250 g 1 500 Therefore, the weight of the watermelon is 1 kg 500 g. Example 17: Chandu, the milkman, has only 5 ℓ and 3 ℓ measures. How will he sell 4 ℓ of milk to Gita? (Hint: Find the difference between 5 ℓ and 3 ℓ.) Solution: Chandu first pours milk in 5 ℓ measure. He then transfers some of it into the 3 ℓ measure till its full. Then the quantity of milk left in the 5 ℓ measure is 2 ℓ. This 2 ℓ milk can be transferred into Gita’s vessel. He repeats the same process once more. Thus, in this way, he sells 4 ℓ of milk to Gita. Measurement 147

Example 18: A container has some juice. A glass has a capacity of 200 mℓ. How many glasses of juice must be poured to have 2 ℓ of juice? Solution: Capacity of the glass = 200 mℓ Quantity of juice needed = 2 ℓ = 2 × 1000 mℓ = 2000 mℓ 2000 = 200 × 10 Therefore, 10 glasses of juice must be poured to make 2 ℓ. 11.3 Area and Perimeter of Simple Shapes I Think Surbhi has to paste a chart paper on a cardboard and then decorate its borders with a colourful tape. She knows the quantity of paper and the length of the tape that she needs. How do you think she could find that? I Recall Recall that figures or shapes such as triangle, square and rectangle are called flat shapes. These are also known as 2D shapes. All these shapes have different features. Write down the features of the shapes given. Shape Features AB D C X Y WZ 148

Shape Features T SU •O I Remember and Understand We have learnt about the simple closed figures The perimeter is the length made of straight line segments. of the outline of a shape. To find the perimeter of a Let us now learn about their perimeters. rectangle or square, we add the lengths of all its The sum of the lengths of the sides of a closed figure four sides. is called its perimeter. It is denoted by ‘P’. Example 19: Find the perimeter of each of the given shapes. a) J 4 cm K b) D 3 cm E c) X 2 cm 2 cm 3 cm 3 cm 5 cm M 4 cm Solution: 4 cm L G 3 cm F a) Perimeter of rectangle = JK + KL + LM + MJ Z 3 cm y = 4 cm + 2 cm + 4 cm + 2 cm = 12 cm b) Perimeter of square = DE + EF + FG + GD = 3 cm + 3 cm + 3 cm + 3 cm = 12 cm c) Perimeter of triangle = XY + YZ + ZX = 5 cm + 3 cm + 4 cm = 12 cm Let us now learn a common method to find the perimeters of a few 2D shapes. Perimeter of a rectangle Consider the given rectangle ABCD. Measurement 149

The two opposite sides AB & CD are called lengths. They are represented by ‘ℓ’ units. The two opposite sides BC & DA are called breadths. They are represented by ‘b’ units. Perimeter of a rectangle = Sum of all its sides ℓ AB = AB + BC + CD + DA = length (ℓ) + breadth (b) b b + length (ℓ) + breadth (b) D C Therefore, perimeter (P) = 2 × ℓ + 2 × b units. ℓ Example 20: Find the perimeter of the given rectangle. M 5 cm N 3 cm 3 cm P 5 cm O Perimeter of rectangle = 2 × ℓ + 2 × b Solution: = 2 × 5 + 2 × 3 cm = 10 + 6 cm = 16 cm Therefore, the perimeter of the rectangle MNOP is 16 cm. Perimeter of a square Consider the given square ABCD. s B A s Perimeter of a square = Sum of all its sides s C = AB + BC + CD + DA = 4 × side units Ds Therefore, perimeter (P) of a square is 4 × s units. Example 21: Find the perimeter of the given square. Q 2 cm R 2 cm 2 cm T 2 cm S Solution: Perimeter of a square = 4 × s = 4 × 2 cm = 8 cm Therefore, the perimeter of the square QRST is 8 cm. 150

? Train My Brain Find the perimeter of the following: V a) S 4 cm T b) N 3 cm O c) 3 cm 4 cm 4 cm 4 cm 4 cm 2 cm 2 cm V 4 cm U Q P X 3 cm W I Apply Let us now see how to find the area of a given shape. Area specifies the region of the total number of squares of unit side it covers. So, the units of area is square units, written in short as sq. units. The lengths of the sides of a 2D figure are usually in cm or m. So, the units of its area are sq. cm or cm2 or sq. m or m2 respectively. Area of an object is the amount of surface or region covered by it. It is denoted by ‘A’. Area of a rectangle: Observe the given rectangle. The area of 1 square = 1 cm × 1 cm = 1 sq. cm The region covered by the rectangle = the total region covered by 15 squares of size 1 sq. cm each = 15 × 1 sq. cm 1 cm = 15 cm2 or 15 sq. cm 1 cm The rectangle can be considered to have a length of 5 cm and breadth of 3 cm. Thus, area of a rectangle = length × breadth =  × b sq. units = 5 cm × 3 cm = 15 sq. cm Area of a square: Observe the given square. The area of 1 square = 1 cm × 1 cm = 1 sq. cm The surface region covered by the square = The total region covered by 9 squares of size 1 sq. cm each = 9 × 1 sq. cm Measurement 151

= 9 cm2 or 9 sq. cm 1 cm = 3 cm × 3 cm 1 cm The square can be considered to have a side of 3 cm. Thus, area of a square = 9 sq. cm = 3 cm × 3 cm = side × side = s × s sq. units Example 22: Find the areas of the given figures: a) b) 9 cm 4 cm 11 cm 9 cm Solution: a) Area of a rectangle =  × b b) Area of a square = s × s = 11 cm × 4 cm = 9 cm × 9 cm = 44 sq. cm = 81 sq. cm The areas of the other shapes can be found out by adding the number of unit squares they cover. Example 23: Find the areas of each of given figures if each square is of area 1 sq. cm. a) b) c) Solution: a) This figure has 6 full and 6 half squares (3 squares). Therefore, its area = (6 + 3) sq. cm = 9 sq. cm. b) This figure has 6 full and 4 half squares (2 squares). Therefore, its area = (6 + 2) sq. cm = 8 sq. cm. c) This figure has 14 full and 2 half squares (1 square). Therefore, its area = (14 + 1) sq. cm = 15 sq. cm. 152

I Explore (H.O.T.S.) Let us see some more examples of perimeter and area. Example 24: Find the length of a rectangle with breadth 6 cm and perimeter 32 cm. Solution: Breadth of a rectangle = 6 cm Its perimeter = 32 cm Perimeter of a rectangle (P) = 2 × length + 2 × breadth 32 = 2 × length + 2 × 6 32 = 2 × length + 12 32 = 20 + 12 = (2 × 10) + 12 Therefore, the length of the rectangle is 10 cm. Example 25: Find the length of the side and area of a square whose perimeter is 68 m. Solution: Perimeter of the square = 4 × side = 68 m Side of the square = 4 × 17 m = 68 m Side of the square = 17 m Area of the square = s × s = 17 m × 17 m = 289 sq. m. Therefore the side of the square is 17 m and its area measures 289 sq. m. Maths Munchies The blood in our body also has a unit of measurement called ‘pint’ or ‘unit’. An adult body contains 8 to 10 pints of blood. 1 pint is equal to 473 mℓ. Therefore, our body has 3784 mℓ to 4730 mℓ of blood. Connect the Dots Social Studies Fun India measures about 3,200 kilometres from north to south. The length from west to east is about 2,900 kilometres. Measurement 153

Science Fun Dwarf Willow is one of the smallest woody plants in the world. It grows to only 1 to 6 cm in height. It has round and shiny green leaves, 1 to 2 cm in length and breadth. Drill Time 11.1 Conversion of Units of Length 1) Word problem Roopa’s house and the places close to it are shown on the map. 2 km 2 km Hospital Playground 4 km 2 km 1 km 250 m Market Post Office 5 km 500 m Roopa’s house School 2 km 450 m 4 km 6 km 3 km 300 m Airport Study the map and answer these questions. a) The shortest route from Roopa’s house to the market is through __________ and is __________ km. b) The shortest route from Roopa’s house to the airport is _________ km. c) Which is the shortest route from the post office to the market? 154

d) Roopa went to the post office from her school through the shortest route. What is the distance she travelled? 2) Convert into centimetres. d) 5 m 20 cm e) 8 m 36 cm a) 3 m b) 9 m c) 2 m 45 cm 3) Convert into metres. d) 6 km 112 m e) 1 km 100 m a) 4 km b) 15 km c) 5 km 555 m 4) Solve the following: b) 31 m + 18 m 59 cm a) 24 m 13 cm + 13 m 45 cm c) 10 km 100 m + 20 km 200 m d) 88 km 100 m − 10 km 800 m e) 26 m 14 cm – 20 m 10 cm 11.2 Standard Units of Mass and Volume 5) Convert into grams. a) 14 kg b) 29 kg c) 14 kg 300 g d) 75 kg 226 g e) 10 kg 112 g 6) Solve the following: a) 28 kg 421 g + 30 kg 232 g b) 42 kg 876 g + 31 kg 111 g c) 44 kg 444 g – 22 kg 222 g d) 43 g 230 mg – 11 g 100 mg 7) Word problem Maya bought a few vegetables with the given quantities: Brinjal – 2 kg 250 g; Onion – 1 kg 750 g; Potato – 1 kg 250 g Find the total weight of vegetables in her bag. 8) Convert into millilitres. a) 13 ℓ b) 28 ℓ c) 13 ℓ 400 mℓ d) 64 ℓ 206 mℓ e) 14 ℓ 142 mℓ 9) Solve the following: a) 28 ℓ 421 mℓ + 40 ℓ 262 mℓ b) 41 ℓ 836 mℓ + 41 ℓ 113 mℓ c) 30 ℓ 320 mℓ + 20 ℓ 300 mℓ d) 33 ℓ 530 mℓ – 11 ℓ 300 mℓ e) 66 ℓ 666 mℓ – 44 ℓ 444 mℓ Measurement 155

10) Word problem A arthi has a jug with some buttermilk. She uses glasses which can hold 150 mℓ. How many glasses must she fill so that she has 3 ℓ of buttermilk? 11.3 Area and Perimeter of Simple Shapes 11) Find the perimeter of the given figures.2 cm c) a) b) 3 cm 4 cm 8 cm 8 cm d) 3 cm 4 cm 13 cm 3 cm 12) Find the area of the given figures. c) a) b) 4 cm 8 cm 4 cm 13 cm 8 cm 15 cm A Note to Parent Ask your child to weigh different things present at home such as a pencil, a flower vase or some utensils. This will help him or her to understand the usage of mg, g and kg for different objects. 156

CHAPTER55 13 12Chapter Data Handling 50 45 40 35 30 I Will Learn About 25 20 15 10 5 0 • reading bar graphs. • drawing bar graphs based on the given data. 12.1 Bar Graphs I Think Surbhi attended a fruit festival conducted for a week in her school. She was asked to give a report on the sale of different fruits per day in the form of a graph. Surbhi knew how to represent the data as a pictograph. She knew that it would take a lot of effort to draw a pictograph. She wanted to represent it in an easier and simpler way. How do you think Surbhi would have given the report? I Recall We know that: • the information collected for a specific purpose is called data. • the information given as numbers is called the numerical data. • the information shown in the form of pictures is called a pictograph. 157

Let us recall the pictographs through the following example. The favourite sports of Class 4 students are given. Read the pictograph and answer the questions. Key: 1 = 6 students Favourite game of Class 4 students Volleyball Cricket Basketball Kabaddi Football a) The most favourite game of Class 4 students is _____________ . b) The least favourite game of Class 4 students is _____________ . c) The number of students who like to play basketball is___________ . d) The number of students who like to play football is _____________ . e) The number of students who like to play kabaddi is _____________ . I Remember and Understand While drawing pictographs, we choose a relevant picture to represent the given data. If the data is large, it is difficult and time consuming to draw a pictograph. 158

An easier way of representing data is the bar graph. In a bar graph, we draw rectangular bars of the same width. We draw bar graphs on a graph paper and give them The bars in a bar graph suitable titles. can be drawn either horizontally or vertically. First, we decide the scale of the graph. The markings on Bar graphs are useful in the axes of the graphs is the scale. We can adjust the comparing data. scale to draw the bar graph within the given range. Let us understand how to read and interpret bar graphs. Example 1: The marks scored by Kamala in a monthly test are represented using a bar graph as given. Understand the graph and answer the questions that follow. Scale: X-axis: 1 cm = 1 subject; Y-axis: 1 cm = 5 marks Kamala’s Performance in a Monthly Test Marks English Maths Science Social Music Hindi Studies Subjects a) What is the title of the graph? b) In which subject did Kamala perform the best? Data Handling 159

c) In which subject does Kamala need to improve? d) What are Kamala’s total marks? Solution: a) The title of the graph is “Kamala’s Performance in a Monthly Test”. b) T he height of the bar representing Maths is maximum. It means that, Kamala performed the best in Maths. c) T he height of the bar representing Social Studies is the minimum. So, Kamala needs to improve in Social Studies. d) Kamala’s total marks are 35 + 47 + 42 + 28 + 32 + 40 = 224 Example 2: Information about a primary school is represented in the form of a bar graph as shown. Observe the graph carefully and answer the questions that follow. Scale: X-axis: 1 cm = 1 class; Y-axis: 1 cm = 5 students Strength of Primary School l oohcS Class strength Class a) What is the total strength of all the 5 classes? b) Which class has the least strength? 160

c) Which class has the greatest strength? d) What is the title of the graph? Solution: a) Total strength is 42 + 36 + 38 + 43 + 45 = 204 b) Class 2 c) Class 5 d) Strength of a Primary School ? Train My Brain Fill in the blanks. a) A bar graph is used to represent ___________________. b) _________________ bars are used in a bar graph. c) Bar graphs are drawn on __________. I Apply We have learnt how to read and interpret bar graphs. Now, let us learn to draw a bar graph using a graph paper. Steps to draw a bar graph on a graph paper: Step 1: Draw a horizontal line and vertical line, called the axes. They meet at a point called the origin. Step 2: Take a suitable scale such as 1 cm = 5 units. Step 3: On the X-axis, show the items of the data. On the Y-axis show the values of the items. Step 4: Draw bars of equal width on the X-axis. The heights of the rectangles represent the values of the data which are given on the Y-axis. Step 5: Give a relevant title to the bar graph. Let us understand this through an example. Example 3: The following pictograph shows the number of scooters manufactured by a factory in a week. Complete the pictograph. Then draw a bar graph for the same data. Data Handling 161

Key: 1 = 5 scooters Number of Weekday Scooters manufactured in a week scooters Monday Tuesday Wednesday Thursday Friday Saturday Solution: Total Step 1: Let us follow these steps to draw a bar graph. Step 2: Count the number of pictures in the pictograph. Complete the table by writing the product of the number of pictures and the number of scooters per key. Take a graph paper and draw the X and Y axes meeting each other at one corner as shown. 162

Step 3: Choose a suitable scale. Since the maximum number of scooters is 30 and the minimum is 10, we can take the scale as 1 cm = 5 scooters. Mark weekdays on the X-axis as 1 cm = 1 weekday. Mark the number of scooters manufactured on the Y-axis from 0 to 35. Number of scooters manufactured Mon Tues Wed Thurs Fri Sat Weekdays Data Handling 163

Step 4: On the X-axis, mark 30, 15, 20, 25, 20 and 10 against the Y-axis as shown. We can plot these points two points apart. Number of scooters manufactured Step 5: Monday Tuesday Wednesday Thusday Friday Saturday Weekdays Draw vertical rectangular bars from these points for each weekday on the X-axis. Give a suitable title to the graph. Weekly Manufacturing of Scooters Number of scooters manufactured Monday Tuesday Wednesday Thusday Friday Saturday Weekdays 164

We can draw the same graph using horizontal bars by interchanging the values on X and Y axes. Weekly Manufacturing of Scooters Weekdays Number of scooters manufactured Example 4: The number of roses sold during a month in Roopa’s shop is given in the table Week Number of roses sold 1st week 148 2nd week 165 Solution: 3rd week 130 4th week 172 Represent the data in a bar graph. Scale: X-axis: 1 cm = 1 week;Y-axis: 1 cm = 20 roses Data Handling 165

Sale of Roses Roses sold Weeks I Explore (H.O.T.S.) Consider a few real-life examples where we represent data using a bar graph. Example 5: In 2010, the heights of Ramu, Somu, RcamdrheaspTaernacdtiiSvnweleMy.thAyaftweBrertraweoinnoyeteadrs, as 130 cm, 125 cm, 115 cm and 120 their heights were again noted as 140 cm, 132 cm, 124 cm and 128 cm respectively. Draw a bar graph to represent the data and answer the questions that follow. a) Who was the tallest among the friends in 2010? b) Who was the shortest among them during 2012? c) How much taller was Ramu than Somu in 2010? d) Whose height has increased the maximum in 2 years? e) A rrange the children’s heights in 2010 in ascending order and their heights in 2012 in descending order. 166

Solution: Name Height in 2010 Height in 2012 Ramu 130 cm 140 cm Somu 125 cm 132 cm Radha 115 cm 124 cm Swetha 120 cm 128 cm Scale: On X-axis: 2 cm = 1 student On Y-axis: 1 cm = 20 cm Comparison of Heights Height (in cm) Names of children a) As the bar for Ramu’s height in 2010 is the highest, Ramu is the tallest among the children. b) R adha is the shortest among them in 2012. (Shortest bar in 2012). c) Ramu is 5 cm (130 – 125) taller than Somu. Data Handling 167

d) Increase in the heights of the children in the two years: Ramu: (140 – 130) cm = 10 cm S omu: (132 – 125) cm = 7 cm Radha: (124 – 115) cm = 9 cm Swetha: (128 – 120) cm = 8 cm 7 cm < 8 cm < 9 cm < 10 cm Therefore, Ramu’s height increased the maximum in 2 years. e) Heights of the children in 2010: 130 cm, 125 cm, 115 cm, 120 cm Ascending order: 115 cm, 120 cm, 125 cm, 130 cm Heights of the children in 2012: 140 cm, 132 cm, 124 cm, 128 cm Descending order: 140 cm, 132 cm, 128 cm, 124 cm Example 6: The weights of four children are noted in 2014 and 2016 as given. Draw a bar graph and answer the questions that follow. Name Weight in 2014 Weight in 2016 Ram 30 kg 34 kg Shyam 34 kg 32 kg Reema 28 kg 31 kg Seema 29 kg 31 kg a) Who weighed the most in 2014 and 2016? b) Whose weight has decreased in 2016 from 2014? c) Name the two children who were of the same weight in 2016. d) Whose weight in 2014 is the same as that of another child in 2016? e) Write the weights of the children in 2014 in descending order and their Solution: weights in 2016 in ascending order. Scale: O n X-axis: 2 cm = 1 student; Y-axis: 1 cm = 5 kg 168

Comparison of Weights Comparison of Weights Weights (in kg) Ram Shyam Reema Seema Names of children a) Shyam was the heaviest in 2014 and Ram was the heaviest in 2016. b) Shyam’s weight decreased in 2016 from 2014. c) Reema and Seema are of the same weight in 2016. d) Shyam’s weight in 2014 is equal to Ram’s weight in 2016. e) Weights in 2014: 30 kg, 34 kg, 28 kg, 29 kg Descending order: 34 kg, 30 kg, 29 kg, 28 kg Weights in 2016: 34 kg, 32 kg, 31 kg and 31 kg Ascending order: 31 kg, 31 kg, 32 kg, 34 kg Maths Munchies We can show the bar graphs as histograms when the data keeps changing with respect to the time. The bar graphs are usually separated while the histograms do not have space between the bars and are adjacent to each other. Sometimes, you can have bar graphs without any space. However, a histogram cannot have space between the bars. Data Handling 169

Connect the Dots Social Studies Fun A Climate graph is usually shown as a bar graph. It is a combination of temperature and rainfall. The months are shown on the X-axis while the amount of rainfall and temperatures are shown on the Y-axis. English Fun Make a list of your favourite authors. Count the number of books that you know of each author. Use this data to draw a bar graph. (Some names of authors for reference: J K Rowling, Ruskin Bond, C S Lewis, Charles Dickens, R K Narayan and so on.) Drill Time 12.1 Bar Graphs 1) The score of students in an essay writing competition are given in the table. Draw a bar graph. Subject Marks scored Piyush 65 Suman 72 Vaishnavi 82 Pooja 93 2) The table shows the marks secured by Rajeev in Test 1 and Test 2. Subject Marks in Test 1 Marks in Test 2 Hindi 65 68 English 78 80 Mathematics 60 85 Science 88 80 Social Studies 54 65 170

Compare his performance in the two tests by drawing a bar graph and answer the questions that follow. a) Find Rajeev’s total marks in Test 1 and Test 2 separately. b) In which of the two tests did he perform well with respect to Mathematics? c) In which subject(s) has he improved from Test 1 to Test 2? d) In which of the two tests has Rajeev got less marks? 3) The approximate monthly attendance of Grade 4 is shown in the pictograph given. Draw a bar graph and answer the questions that follow. Month Attendance June July August September October November Key: 1 = 10 students Data Handling 171

a) In which month is the attendance maximum? b) In which month is the attendance minimum? c) In which months is the attendance less than 45? A Note to Parent From newspapers or magazines, find out the bar graphs and explain what they are about. Explain the terms mentioned in the bar graph first, to give the background and then form basic questions from the same. You may choose articles of common interest like cars, bikes, movies, travel, hobbies and so on. 172