MATHEMATICS TEXTBOOK – PART 2 4 Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________
Preface ClassKlap partners with schools, supporting them with learning materials and processes that are all crafted to work together as an interconnected system to drive learning. ClassKlap 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: S tructured as per Bloom’s taxonomy to help organise the learning process according to the different levels involved Student engagement through simple, age-appropriate language Supported 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 Stepwise breakdown of solutions to provide an easier premise for learning of problem-solving skills Overall, the ClassKlap 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 4Class 8 Fractions - I 8.1 Equivalent Fractions..................................................................................................... 1 8.2 Identify and Compare Like Fractions......................................................................... 7 8.3 Add and Subtract Like Fractions............................................................................... 11 9 Fractions - II 9.1 Fraction of a Number................................................................................................. 18 9.2 Conversions of Fractions............................................................................................ 21 10 Decimals 10.1 Conversion Involving Fractions ................................................................................ 28 11 Money 11.1 Conversion of Rupees and Paise ............................................................................. 39 11.2 Add and Subtract Money with Conversion.............................................................. 43 11.3 Multiply and Divide Money ....................................................................................... 45 12 Measurements 12.1 Multiply and Divide Lengths, Weights and Capacities........................................... 51 13 Data Handling 13.1 Bar Graphs................................................................................................................... 58
Chapter Fractions - I 8 Let Us Learn About • e quivalent fractions. • p roblems related to equivalent fractions. • like and unlike fractions. • a dding 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 22 ‘Half’ means 1 out of 2 equal parts. We write ‘half’ as 1 . 2 1
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 2
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 2] [ and 48 Solution: a) 1 3 2 6 b) 1 4 2 8 Example 2: Find the figures that represent equivalent fractions. Also, mention the fractions. a) b) c) d) Fractions - I 3
Solution: The fraction represented by the shaded part of figure a) is 1 . 2 The shaded part of figure b) represents 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. 4
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. a) 2 b) 33 11 66 Solution: To find fractions equivalent to the given fractions, we either multiply or divide both the numerator and the denominator by the same number. a) 2 11 W e see that 2 and 11 do not have any common factors. So, we 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 . 2 66 22 6 Fractions - I 5
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, 1 , 2 and 3 are 8 24 6 equivalent. Therefore, 1 = 2 = 3 . 246 6
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 , and , that have the same denominator are called like fractions. 88 8 Fractions such as 1 , 2 and 3 that have different denominators are called unlike fractions. 84 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 Fractions - I 7
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 a) 11 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. 8
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 Fraction of apple Jai ate = 1 3 Fraction of apple Vijay ate = 2 3 21 Since, 2 > 1, > 33 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 Fractions - I 9
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 the given fraction and compare them. 3 2 5 5 Solution: Clearly, the part of the figure represented 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. 777 7 7 Solution: Comparing the numerators of the given likeTrfraacintionMs, y Brain we have 1 < 2 < 4 < 5 < 6. So, 1 < 2 < 4 < 5 < 6 . 77777 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 10
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 using 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. Fractions - I 11
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) Number 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 is 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 fraction So, the subtract the like fractions, that represents the that represents the subtract their numerators. unshaded part of the unshaded part of fraction that Write the difference on given figure is the given figure is represents the the same denominator. unshaded part of ____ − ____=____. the given figure is 6−3 =6−3 = 3. 66 6 6 ____ − ____=_____. 12
Example 14: Add: a) 3 + 1 45 23 57 88 b) + c) + Solution: a) 3 + 1 = 3 + 1 = 4 13 13 100 100 88 8 8 c) 48 – 26 b) 4 + 5 = 4 + 5 = 9 125 125 13 13 13 13 c) 23 + 57 = 23 + 57 = 80 100 100 100 100 Example 15: Subtract: a) 8 – 4 b) 33 – 25 99 37 37 Solution: 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 Fractions - I 13
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 14
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 pizza, 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 6 86 2 Part of the pizza remaining = 1 – = – = 8 88 8 Drill Time Concept 8.1: Equivalent Fractions 1) Shade the regions to show equivalent fractions. a) 1 and 2 2 4 Fractions - I 15
b) 1 and 2 5 10 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 9 5 9 9 7 4 4 9 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 11 11 11 11 13 13 13 13 c) 1 , 3 , 4 , 2 d) 1 , 8 , 7 , 9 7777 14 14 14 14 5) Arrange the following fractions in descending order. a) 1 , 8 , 7 , 4 b) 3 , 6 ,10 , 8 9999 17 17 17 17 c) 7 , 9 , 2 ,13 d) 1 , 7 , 8 , 3 21 21 21 21 20 20 20 20 Concept 8.3: Add and Subtract Like Fractions 6) Add: a) 2 + 5 b) 3 + 16 c) 9 + 4 d) 8 + 4 e) 1+2 77 11 11 55 13 13 17 17 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 16
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 figure to represent the given fraction and compare them. 57 88 c) Ajit ate 1 of a cake in the morning and 2 of it in the evening. What part of the cake 55 is remaining? Fractions - I 17
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 a 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 18
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 5 shells 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. Vani’s 5 sister Rani wants to take 3 of the shells. So, she divides the shells into 10 (the number in the 10 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 × 1 h = 2 × 60 min = 2 × 60 = 120 = 40 min 3 3 3 33 b) 1 of a day = 1 × 1 day = 1 × 24 h = 1 × 24 h = 24 hrs = 6 h 4 44 4 Fractions - II 19
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 55 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? 20
Solution: Fraction of the cardboard painted by Venu = 3 6 Fraction of the cardboard painted by Raj = 1 3 Area of the cardboard = 144 sq. cm 3 Example 7: Area of the cardboard painted by Venu = × 144 sq.cm 6 Solution: =3 × 144 sq.cm = 432 sq.cm = 72 sq.cm 66 Area of the cardboard painted by Raj = 1 × 144 sq.cm 3 =1× 144 sq.cm = 144 sq.cm = 48 sq.cm 33 Therefore, Venu painted 72 sq.cm of the cardboard and Raj painted 48 sq.cm of the cardboard. Find if 2 of 154 and 4 of 49 are equal to each other or one of them is greater 11 7 than the other. To find if the fractions of the numbers are equal, we first find their values and compare them. 2 of 154 = 2 × 154 = 2 × 154 = 308 = 28 11 11 11 11 4 of 49 = 4 × 49 = 4 × 49 = 196 = 28 77 77 24 Therefore, of 154 = of 49. 11 7 Concept 9.2: Conversions of Fractions Think Jasleen knew about fractions in which the numerators were less than their denominators. She wondered if there could be some fractions in which the numerators are greater than their denominators. Do you know of such fractions? Fractions - II 21
Recall In the previous chapter, we have learnt about addition and subtraction of like fractions. Recall that the sum of two like fractions is a like fraction. Let us answer these to recall the concept. a) 2 + 1 = ______ 41 5 5 b) 7 + 7 = ______ c) 1 + 5 = ______ d) 3 + 1 = _______ 11 11 22 e) 1 + 3 = ______ f) 2 + 1 = _______ 88 99 & Remembering and Understanding Consider 1 + 5 = 6 . Here, the sum of two like fractions is a like fraction with its numerator less 888 than its denominator. Such fractions are called proper fractions. Sometimes ,it is possible that we get the sum with its numerator greater than the denominator. For example, 7 + 5 = 12 . Here, the sum of two like fractions is a like fraction with its 88 8 numerator greater than its denominator. Such fractions are called improper fractions. Note: In some cases, the sum of the numerators of the like fractions may be equal to the denominator. Then, the fraction is said to be an improper fraction. For example, 3 + 4 = 7 , 3 + 5 = 8 and so on. 7 7 78 8 8 Fractions such as 7 , 8 and so on can also be written as a whole, that is 1. 78 12 8 + 4 . This has a whole We can write 8 as the sum of like fractions as 88 8 and a proper 8 = 14 . Such fractions are called mixed fractions. fraction 4 . That is, 12 =1+ 4 8 8 8 8 A mixed fraction is also called a mixed number. For example, in the mixed fraction 12 3 , 12 is the whole and 3 is the proper fraction. 88 22
In short, we can say that, Proper fractions – Fractions having the numerators less than the denominators. Improper fractions – Fractions having the numerators greater than the denominators. Mixed fractions – Fractions having whole numbers and proper fractions. Example 8: List out proper fractions, improper fractions and mixed fractions from the following: Solution: 13 ,15 7 , 11 , 37 , 9 , 65 13 , 143 , 75 3 ,107 27 , 72 , 68 2 , 29 , 50 23 , 69 , 53 18 9 34 6 14 17 98 4 49 59 5 32 35 32 30 From the given fractions, Proper fractions: 13 , 11 , 9 , 29 18 34 14 32 Improper fractions: 37 , 143 , 72 , 69 , 53 6 98 59 32 30 Mixed fractions: 15 7 , 65 13 , 75 3 , 107 27 , 68 2 , 50 23 9 17 4 49 5 35 We usually write fractions as proper or mixed fractions. So, we need to learn to convert improper fractions to mixed fractions and mixed fractions to improper fractions. Conversion of improper fractions to mixed fractions Let us understand the conversion of improper fractions to mixed fractions by solving a few examples. Example 9: Convert 37 to its mixed fraction form. Solution: 6 To convert improper fractions into mixed fractions, follow these steps. Solved Solve these Steps 37 143 72 69 53 6 98 59 32 30 Step 1: Divide the numerator 6 by the denominator. )6 37 − 36 1 Fractions - II 23
Solved Solve these 37 Steps 6 143 72 69 53 30 98 59 32 Step 2: Write the quotient as The mixed the whole. The remainder is fraction form of the numerator of the proper fraction and the divisor is its 37 is 6 1 . denominator. This gives the 66 required mixed fraction. Conversion of mixed fractions to improper fractions Let us understand the conversion of mixed fractions into improper fractions by solving a few examples. Example 10: Convert 15 7 into its improper fraction. 9 Solution: To convert mixed fractions into improper fractions, follow these steps. Solved Solve these 75 3 Steps 15 7 65 13 4 107 27 9 17 49 Step 1: Multiply the whole by the 15 × 9 = 135 denominator. Step 2: Add the numerator of the proper 135 + 7 = 142 fraction to the product in Step 1. Step 3: Write the sum as the denominator The improper of the proper fraction. fraction form of This given the required improper fraction. 15 7 is 142 9 9. Application Let us now see a few real-life examples involving conversions of fractions. Example 11: Rohan wants to arrange 60 books in his shelf. If only 13 books can be put in a rack, how many racks will be filled by the books? Give your answer as a mixed fraction and as an improper fraction. 24
Solution: Number of books Rohan wants to arrange = 60 Number of books that can be arranged on each rack = 13 Number of racks that are filled = 60 ÷ 13 = 4 8 13 Example 12: Improper fraction equivalent to 4 8 = 60 Solution: 13 13 On a science fair day, a group of students prepared 12 1 litres of orange 2 juice. Express the number of litres of orange juice as an improper fraction. Number of litres of orange juice made = 12 1 2 Improper fraction equivalent to 12 1 = 12 × 2 +1 = 25 2 2 2 Higher Order Thinking Skills (H.O.T.S.) Conversion of fractions is done when we need to add and subtract fractions. In the previous chapter, we have already learnt addition and subtraction of like (proper) fractions. Let us see a few examples that involve addition and subtraction of improper and mixed fractions. Example 13: Add: a) 42 + 35 b) 50 23 + 16 Solution: 25 25 35 35 a) 42 + 35 25 25 T o add the given like improper fractions, we add their numerators and write the sum on the same denominator. Therefore, 42 + 35 = 42 + 35 = 77 25 25 25 25 W e usually write fractions as proper or mixed fractions. So, we convert the sum into a mixed fraction by dividing the numerator by the denominator. 77 = 3 2 (77 ÷ 25 gives the quotient as 3 and remainder as 2.) 25 25 Therefore, the sum of the given fractions is 3 2 . 25 Fractions - II 25
b) 50 23 + 16 35 35 T o add the given fractions, we have to convert the mixed fraction into improper fraction. So, 50 23 = (50×35)+23 =11777530 + 23 =1773 35 35 35 35 35 T hen add their numerators and write the sum as the numerator. Therefore, 50 23 + 16 = 1773 + 16 = 1773 + 16 = 1789 . 35 35 35 35 35 35 Convert the improper fraction into a mixed fraction. 1789 = 51 4 (1789 ÷ 35 gives the quotient as 51 and the remainder as 4.) 35 35 Therefore, the sum of the given fractions is 51 4 . 35 Example 14: Subtract: a) 342 - 135 b) 34 17 - 37 Solution: 25 25 42 42 a) 342 - 135 25 25 To subtract the given improper fractions, we subtract their numerators. We then write the difference as the numerator. Therefore, 342 - 135 = 342 −135 = 207 25 25 25 25 A s we usually write fractions as proper or mixed fractions, we convert the difference into a mixed fraction. 207 = 8 7 (207 ÷ 25 the quotient as 8 and the remainder as 7.) 25 25 Therefore, the difference of the given fractions is 8 7 . 25 b) 34 17 - 37 42 42 T o subtract the given fractions, we first convert the mixed fraction into an improper fraction. So, 34 17 = 34 × 42 + 17 = 1445 42 42 42 T hen subtract their numerators and write the difference as the numerator. 26
Therefore, 34 17 − 37 = 1445 − 37 = 1445 − 37 = 1408 42 42 42 42 42 42 Again convert the improper fraction into a mixed fraction. 11440485 = 33 22 (1408 ÷ 42 the quotient as 33 and the remainder as 22.) 4422 42 Therefore, the difference of the given fractions is 33 22 . 42 Drill Time Concept 9.1: Fraction of a Number 1) Find the following: a) 1 of 20 b) 3 of 24 c) 3 of 20 d) 4 of 12 e) 2 of 18 2 4563 Concept 9.2: Conversions of Fractions 2) Convert the following improper fractions to mixed fractions. a) 35 b) 121 c) 93 d) 100 e) 115 4 12 12 26 20 3) Convert the following mixed fractions to improper fractions. a) 15 6 b) 23 2 c) 40 4 d) 125 9 Traei)n40M3 y Brain 8 3 5 10 5 4) Word Problems a) A t Sudhir’s birthday party, there are 19 sandwiches to be shared equally among 13 children. What part of the sandwiches will each child get? Give your answer as a mixed fraction. b) I bought 2 1 litres of paint but used only 3 litres. How much paint is left with me? 22 Give your answer as an improper fraction. Fractions - II 27
Chapter Decimals 10 Let Us Learn About • the term ‘decimal’ and its parts. • u nderstanding decimal system. • expanding decimal numbers with place value charts. • converting fractions to decimals and vice versa. Concept 10.1: Conversion Involving Fractions Think Jasleen and her friends participated in the long jump event in their Jasleen – 4.1m Ravi – 2.85m games period. Her sports teacher noted the distance they jumped on a Rajiv – 3.05 m piece of paper as shown here. Amit – 2.50m Jasleen wondered why the numbers had a point between them as in the case of writing money. Do you know what the point means? Recall Recall that in Class 3 we have learnt to measure the lengths, weights and volumes of objects. For example, a pencil is 12.5 cm long. 12. 5 cm 28
A crayon is 5.4 cm long. 5.4 cm The weight of your mathematics textbook is 0.905 kg. A milk packet has 0.250 of milk, and so on. In all these values, we see numbers with a point between them. Have you read price tags on some items when you go shopping? ` 300.75 ` 439.08 They also have numbers with a point between them. Let us learn why a point is used in such numbers. & Remembering and Understanding We know how to write fractions. In this figure, 3 portion is coloured and 7 portion is not coloured. 10 10 3 or 0.3 and the We can write the coloured portion of the figure as 10 portion that is not coloured as 7 or 0.7. 10 Numbers such as 0.3, 0.7, 3.0, 3.1, 4.7, 58.2 and so on are called decimal numbers or simply decimals. Tenths: The figure below is divided into ten equal parts. 1 111 1 1 1 1 1 1 10 10 10 10 10 10 10 10 10 10 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 Each column is of the same size. Then, each of the ten equal parts is 1 . It is read as one-tenth. Fractional form of each equal part is 1 . 10 10 Decimal form of each equal part is 0.1. Decimals 29
We read numbers like 0.1, 0.2, 0.3 … as ‘zero point one’, ‘zero point two’, ‘zero point three’ and so on. Zero is written to indicate the place of the whole number. A decimal number has two parts. 48 . 35 Whole or integral part Decimal part (= or > 0) (< 1) Decimal Point Note: T he numbers in the decimal part are read as separate digits. Recall the place value chart of numbers. 100 × 10 10 × 10 1 × 10 1 Thousands Hundreds Tens Ones 6 2 5 5 3 2 2 6 5 2 We know that in this chart, as we move from right to left, the value of the digit increases 10 1 times. Also, as we move from left to right, the value of a digit becomes times. The place 10 value of the digit becomes one-tenth, read as a tenth. Its value is 0.1 read as ‘zero point one’. 2 is read as ‘two-tenths’, 7 is read as ‘seven–tenths’ and so on. 10 10 We can extend the place value chart to the right as follows: 1 × 1000 1 × 100 1 × 10 1 . 1 10 Thousands Hundreds Tens Ones Decimal Tenths 7 . 2 1 2 4 . 3 30 4 3 . 6 1 5 . 7 The number 3015.7 is read as three thousand and fifteen point seven. Similarly, the other numbers are read as follows: 30
Seven point two; twenty-four point three and one hundred and forty-three point six. The point placed in between the number is called the decimal point. The system of writing numbers using a decimal point is called the decimal system. [Note: ‘Deci’ means 10.] Hundredths: Study this place value chart. Thousands Hundreds Tens Ones Decimal Tenths Hundredths 1 × 10 1 point 1 × 1000 1 × 100 2 1 1 2 8 6 . 10 100 3 . 9 When the number moves right from the tenths place, we get a new place, which is 1 of the tenths place. It is called the ‘hundredths’ place written as 1 and read 10 100 as one-hundredths. Its value is 0.01, read as ‘zero point zero one’. 2 is read as two-hundredths, 5 is read as five-hundredths and so on. 100 100 So, the number in the place value chart is read as ‘two thousand eight hundred and sixty-two point three nine’. Expansion of decimal numbers Using the place value chart, we can expand decimal numbers. Let us see a few examples. Example 1: Expand these decimals. a) 1430.8 b) 359.65 c) 90045.75 d) 654.08 Solution: To expand the given decimal numbers, first write them in the place value chart as shown. S. no Ten Thousands Hundreds Tens Ones Decimal Tenths Hundredths thousands 1 point a) 4 3 0 . 8 b) 9 0 3 5 9 65 c) 0 4 5 . 75 d) 6 5 4 08 . . Decimals 31
Expansions: 1 a) 1430.8 = 1 × 1000 + 4 × 100 + 3 × 10 + 0 × 1 + 8 × 10 Example 2: b) 359.65 = 3 × 100 + 5 × 10 + 9 × 1 + 6 × 1 + 5 × 1 10 100 c) 90045.75 = 9 × 10000 + 0 × 1000 + 0 × 100 + 4 × 10 + 5 × 1 + 7 × 1 + 5 × 1 10 100 1 1 Solution: d) 654.08 = 6 × 100 + 5 × 10 + 4 × 1 + 0 × + 8 × 10 100 Write these as decimals. a) 7 × 1000 + 2 × 100 + 6 × 10 + 3 × 1 + 9 × 1 + 3 × 1 10 100 b) 3 × 10000 + 0 × 1000 + 1 × 100 + 9 × 10 + 6 × 1 + 4 × 1 + 5 × 1 10 100 c) 2 × 1000 + 2 × 100 + 2 × 10 + 2 × 1 + 2 × 1 + 2 × 1 10 100 d) 5 × 100 + 0 × 10 + 0 × 1 + 0 × 1 + 5 × 1 10 100 First write the numbers in the place value chart as shown. S. no Ten Thousands Hundreds Tens Ones Decimal Tenths Hundredths thousands point a) 7 2 63 . 93 b) 3 0 1 96 . 45 c) 2 2 22 . 22 d) 5 00 . 05 Standard forms of the given decimals are: a) 7263.93 b) 30196.45 c) 2222.22 d) 500.05 Conversion of fractions to decimals Fractions can be written as decimals. Consider an example. Example 3: Express these fractions as decimals. a) 18 2 b) 43 5 c) 26 1 d) 4 9 10 10 10 10 Solution: To write the given fractions as decimals, follow these steps. Step 1: Write the integral part as it is. Step 2: Place a point to its right. 32
Step 3: Write the numerator of the proper fraction part. a) 18 2 = 18.2 b) 43 5 = 43.5 10 10 c) 26 1 = 26.1 d) 4 9 = 4.9 10 10 Example 4: Express these fractions as decimals. a) 25 b) 17 2 c) 43 d) 5 92 100 100 100 100 Solution: a) 25 = 25 hundredths = 0.25 100 b) 17 2 = 17 and 2 hundredths = 17.02 100 c) 43 = 43 hundredths = 0.43 100 d) 5 92 = 5 and 92 hundredths = 5.92 100 Shortcut method: Fractions having 10 or 100 as their denominators, can be expressed in their decimal form by following the steps given below. Step 1: Write the numerator. Step 2: Then count the number of zeros in the denominator. Step 3: Place the decimal point after the same number of digits from the right as the number of zeros. For example, the decimal form of 232 = 2.32 100 Note: F or the decimal equivalent of a proper fraction, place a 0 as the integral part of the decimal number. Conversion of decimals to fractions To convert a decimal into a fraction, follow these steps. Step 1: Write the number without the decimal. Step 2: Count the number of decimal places (that is, the number of places to the right of the decimal number). Step 3: Write the denominator with 1 followed by as many zeros as the decimal point. Decimals 33
Example 5: Write these decimals as fractions. a) 2.3 b) 13.07 c) 105.43 d) 0.52 Solution: a) 2.3 = 23 b) 13.07 = 1307 10 100 c) 105.43 = 10543 d) 0.52 = 52 Alternate method: 100 100 A decimal having an integral part can be written as a mixed fraction. So, 2.3 = 2 and 3 tenths = 2 3 10 13.07 = 13 and 7 hundredths = 13 7 100 105.43 = 105 and 43 hundredths = 105 43 100 Application Let us see a few real-life examples of decimals. Example 6: The amount of money with Sneha and her friends are given in the table. Sneha ` 432.50 Anjali ` 233.20 Rohan ` 515.60 Jay ` 670.80 Write the amounts in words. Solution: To write the decimals in words, the integral part is read as usual. The decimal part is read as digits. Amount In words ` 432.50 four hundred and thirty-two rupees and fifty paise ` 233.20 two hundred and thirty-three rupees and twenty paise ` 515.60 five hundred and fifteen rupees and sixty paise ` 670.80 six hundred and seventy rupees and eighty paise 34
Example 7: The weights of some children in grams are given in the table below: Name Weight in grams Solution: Rahul 23456 Anil 34340 Anjali 28930 Soham 25670 Convert these weights into kilograms. We know that 1 kg = 1000 g. To convert grams to kilograms, we divide it by 1000. So, the weights in kilograms are as follows. Name Weight in grams Weight in kilograms Rahul 23456 23456 1000 = 23.456 Anil 34340 34340 = 34.340 Anjali 28930 1000 28930 = 28.930 1000 Soham 25670 25670 = 25.670 1000 Example 8: Complete this table. S. No Fraction Read as Decimal Read as a) 7 7 tenths 0.7 Zero point seven b) 10 c) 47 100 35 10 d) 0.34 e) 12 and 65 hundredths Decimals 35
Solution: S. No. Fraction Read as Decimal Read as a) 7 tenths 0.7 Zero point seven b) 7 47 hundredths 0.47 Zero point four seven c) 10 3 and 5 tenths 3.5 Three point five d) 47 34 hundredths 0.34 Zero point three four e) 100 12 and 65 hundredths Twelve point six five 35 12.65 10 34 100 12 65 100 Example 9: Ajay and Vijay represented the coloured part of the figure given as follows: Ajay: 3 Vijay: 0.03 10 Solution: Whose representation is correct? The number of shaded parts as a fraction is 3 or 3 tenths. 10 As a decimal it is 0.3 and not 0.03. So, Ajay’s representation is correct. Higher Order Thinking Skills (H.O.T.S.) Observe the following: 2 tenth=s =2 0.2 10 5 tent=hs =5 0.5 10 8 tent=hs =8 0.8 10 10 tent=hs 1=0 1 10 36
57 hundredths = 57 =0.57 100 hundredths = 100 = 1 100 100 Example 10: Write the decimals that represent the shaded part. a) b) c) d) Solution: a) The fully shaded part represents a whole. So, the decimal that represents the given figure is 1.3. Decimals 37
b) The required decimal is 0.6. c) 10 + 43 = 143 = 114.433 d) 100 + 100 + 29 = 22.2299 10 100 100 100 100 100 100 100 Example 11: Observe the pattern in these decimals and write the next three numbers in each. a) 0.12, 0.13, 0.14, _________, _________, _________ b) 2.00, 2.10, 2.20, _________, _________, _________ c) 8.5, 9.5, 10.5, _________, _________, _________ d) 23.31, 23.41, 23.51, _________, _________, _________ Solution: a) 0.12, 0.13, 0.14, 0.15, 0.16, 0.17 (increases by 1 hundredths) b) 2.00, 2.10, 2.20, 2.30, 2.40, 2.50 (increases by 1 tenths) c) 8.5, 9.5, 10.5, 11.5, 12.5, 13.5 (increases by ones) d) 23.31, 23.41, 23.51, 23.61, 23.71, 23.81 (increases by 1 tenths) Drill Time Concept 10.1: Conversion involving Fractions 1) Convert the following into fractions: a) 2.56 b) 14.02 c) 105.89 d) 52.60 e) 8.01 2) Convert the following into decimals: a) 2 b) 23 23 d) 73 e) 834 c) 1000 10 100 10 100 3) Write the following decimals in words: a) 73.5 b) 413.45 c) 0.73 d) 13.45 e) 1.87 4) Word problem The measures of some objects are given in the table. Height of a flag pole 9.50 m Side of a dining table 1.20 m Distance between the two cities 325.75 km Height of a plant 127.80 cm Write these lengths in words. 38
Chapter Money 11 Let Us Learn About • c onverting rupees to paise and vice versa. • problems involving conversion of money. • a dding and subtracting money with column method. • m ultiplying and dividing money. Concept 11.1: Conversion of Rupees and Paise Think Jasleen had some play money in the form of notes and coins. While playing, her friend gave her ` 10. Jasleen has to give paise for the amount her friend gave her. How many paise should Jasleen give her friend? Recall We have already learnt to identify currency and coins, conversion of rupees to paise and also that 1 ` = 100 p. Let us answer these to revise the concept of conversion of money. a) ` 62 = __________ paise b) 500 paise = ` __________ c) ` 28 = __________ paise d) 900 paise = ` __________ e) ` 76 = __________ paise f) 200 paise = ` __________ 39
& Remembering and Understanding We already know that to change rupees into paise we multiply the rupees by 100. For example, as ` 1 = 100 paise, ` 3 = 3 × 100 paise = 300 paise To convert an amount into paise we multiply the rupees given in the amount by 100 and add the product to the number of paise. To convert paise to rupee just add a decimal point two digits from the right. Let us see a few examples involving conversion between rupee and paise. Example 1: Convert ` 132.28 into paise. Solution: ` 132.28 = ` 132 + 28 p = ` 132 × 100 p + 28 p = 13200 p + 28 p = 13228 p Note: An easy way to convert rupees into paise is to remove the symbol (` and p)and the dot (.) between the rupees and the paise and write the number together. So, ` 132.28 = 13228 p. An amount of more than 100 paise, can be expressed in rupees and paise. To convert paise into rupees and paise, divide the number by 100. Write the quotient as rupees and remainder as paise. Example 2: Convert 24365 paise into rupees and paise. Solution: 24365 p = 24300 p + 65 p = ` 243 + 65 p = ` 243.65 Note: An easy way to convert ‘paise’ into ‘rupees’ and paise is to just put a dot (.) after two digits (ones and tens places) from the right and express it as `. TTh Th H T O So, 24365 = ` 243.65 2 4 36 5 a) Convert ` 477.95 to paise. Solve these c) Convert 44390 paise into b) Convert ` 892.95 into rupees. paise. _________________________ _________________________ _________________________ 40
Application Now let us solve some examples involving conversion of money. Example 3: Sheeba has ` 223.57. How many paise does she have in all? Solution: Amount with Sheeba = ` 223.57 We know that, ` 1 = 100 paise. ` 223.57 = ` 223 + 57 p = 223 × 100 p + 57 p = (22300 + 57) p = 22357 p Hence, Sheeba has 22357 paise. Example 4: Anish has 2435 p and Beena has ` 23.75. Who has more money? Solution: Amount with Anish = 2435 p Amount with Beena = ` 23.75 To compare the money they have, both the amounts must be in the same units. So, we convert rupees to paise. ` 23.75 = ` 23 × 100 p + 75 p (Since ` 1 = 100 p.) = (2300 + 75) p = 2375 p Clearly, 2435 > 2375. Therefore, Anish has more money. Example 5: Ram has ` 374.50 and Chandu has ` 365.75 in their kiddy banks. Who has less amount and by how much? Solution: Amount with Ram = ` 374.50 Amount with Chandu = ` 365.75 Comparing the rupee part of the amounts, we get 365 < 374. So, ` 365.75 < ` 374.50. Therefore, Chandu has less money. The difference in their amounts = ` 374.50 – ` 365.75 = ` 8.75 Therefore, Chandu has ` 8.75 less than Ram. Money 41
Higher Order Thinking Skills (H.O.T.S.) Let us see a few more examples of conversion of money. Example 6: Complete the following by writing the number of different coins that can be used to pay ` 10 using different coins. 50 paise coins ` 10 1-rupee coins 2 - rupee coins 2-rupee coins and 1-rupee coins 5-rupee coins Solution: 20 50 paise coins ` 10 10 1-rupee coins 5 2-rupee coins 3 2-rupee coins and 4 1-rupee coins 2 5-rupee coins Example 7: Solution: Write two different ways in which you can pay ` 50. Combination 1: ` 50 = ` 20 + ` 20 + ` 10 Combination 2: ` 50 = ` 10 + ` 10 + ` 10 + ` 10 + ` 10 42
Concept 11.2: Add and Subtract Money with Conversion Think Jasleen went shopping with her elder sister. She bought some groceries for ` 110.50, vegetables for ` 105.50 and stationery for ` 40. They had ` 300. Do you know how much money was left with them after shopping? Recall Recollect that we can add or subtract money just as we add or subtract numbers. 1) To find the total amount, we write one amount below the other. We see to it that the decimal points are exactly one below the other. We then add the amounts just as we add numbers. 2) To find the difference in amounts, we write the smaller amount below the bigger one. We see to it that the decimal points are exactly one below the other. We then subtract the smaller amount from the bigger one. Answer the following to revise the concept of addition and subtraction of money. a) ` 22.10 – ` 11.10 = ___________ b) ` 15.30 + ` 31.45 = ___________ c) ` 82.45 – ` 42.30 = __________ d) ` 15.30 – ` 5.20 = __________ e) ` 32 + ` 7.20 = ___________ & Remembering and Understanding To add or subtract a given amount of money, we follow the steps given below. Step 1: Express the given amounts in figures as decimal numbers. Step 2: Arrange the given amounts in a column. Place the decimal points exactly below one another. Step 3: Add or subtract the amounts as usual. Step 4: In the sum or difference so obtained, put the decimal point exactly below the other decimal points. Money 43
Let us see some examples. Example 8: Add: a) ` 547.38 + ` 130.83 b) ` 239.74 + ` 355.54 Solution: a) `p b) ` p 11 11 74 54 5 4 7.38 2 3 9. 28 + 3 5 5. + 1 3 0.83 ` 5 9 5. ` 6 7 8.21 Example 9: Subtract: a) ` 53354 − ` 24765 b) ` 866.95 − ` 492.58 Solution: a) ` b) ` p 12 12 14 4⁄ 2⁄ 2⁄ 4⁄ 1⁄4 7⁄ 1⁄6 8⁄ 1⁄5 8 6 6.9 5 53354 − 4 9 2.5 8 −24765 ` 28 5 8 9 ` 3 7 4.3 7 Application Let us now see a few real-life situations where addition and subtraction of money are used. Example 10: Anita saved ` 213.60, ` 105.30 and ` 305.45 in three months from her pocket money. How much did she save in all? Solution: Amount saved in the 1st month = ` 213.60 Amount saved in the 2nd month = + ` 105.30 Amount saved in the 3rd month = + ` 305.45 Therefore, the total amount saved in 3 months = ` 624.35 Example 11: Mrs. Gupta had ` 5000 with her. She spent ` 3520.50 for buying different food items. How much money is left with her? Solution: Amount with Mrs. Gupta = ` 5000.00 Amount spent on food items = – ` 3520.50 Therefore, the amount left with Mrs. Gupta = ` 1479.50 44
Higher Order Thinking Skills (H.O.T.S.) Let us solve a few more real-life examples involving addition and subtraction of money. Example 12: Tanya had ` 525 and her friend Arpan had ` 330. They bought a gift for their brother’s birthday costing ` 495.75. How much amount is left with Tanya and Arpan so that they can continue their shopping? Solution: Amount Tanya had = ` 525 Amount Arpan had = ` 330 Total amount = ` 525 + ` 330 = ` 855 `p Total amount = 855 . 00 The amount spent for the gift = – 495 . 75 359 . 25 Therefore, ` 359.25 is left with Tanya and Arpan. Example 13: The cost of three items are ` 125, ` 150 and ` 175. Suresh has only notes of Solution: ` 100. If he buys the three items, how many notes must he give the shopkeeper? Does he get any change? If yes, how much change does he get? Total cost of the three items = ` 125 + ` 150 + ` 175 = ` 450 The denomination of money Suresh has = ` 100 The nearest hundred, greater than the cost of the three items is ` 500. So, the number of notes that Suresh has to give the shopkeeper is 5. ` 450 < ` 500. So, Suresh gets change from the shopkeeper. The change he gets = ` 500 − ` 450 = ` 50 Concept 11.3: Multiply and Divide Money Think Jasleen knows the cost of one dairy milk chocolate and the cost of five biscuit packets. She could quickly find the cost of 10 dairy milk chocolates and 1 biscuit packet. Can you do such quick calculations? Money 45
Recall Remember that we use multiplication to find cost of many items from the cost of one. Similarly, we use division to find the cost of one item from the cost of many. Multiplying or dividing an amount by a number is similar to the usual multiplication and division of numbers. Answer the following to revise the multiplication and division of numbers. a) 2356 × 10 = __________ b) 72 × 3 = ____________ c) 200 ÷ 4 = ___________ d) 549 ÷ 3 = ___________ e) 621 × 2 = ___________ & Remembering and Understanding Let us understand how to multiply or divide the given amounts of money. When 1 or more items are of the same price, multiply the amount by the number of items to get the total amount. To find out the price of one item, divide the total amount by the number of items. Multiplying money To multiply an amount of money by a number, we follow these steps. Step 1: Write the amount in figures without the decimal point. Step 2: Multiply it by the given number, as we multiply any two numbers. Step 3: In the product, place the decimal point ( if the amount is a decimal number) after the second digit from the right. Example 14: Multiply: a) ` 14105 by 7 b) ` 312. 97 by 34 c) ` 506. 75 by 125 46
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