NCERT Solutions Class 7 Mathematics Chapter 2

Fractions and decimals are the focus of NCERT Class 7 Mathematics Chapter 2. We all understand the difference between a fraction and a decimal. We even use fractions and decimals in our daily lives, whether we realise it or not. However , we might wonder how it helps the students to learn these things?

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All the concepts like fixed fractions, proper fractions, and improper fractions with their addition and subtraction, equivalent fractions, fraction comparisons, fraction ordering, and fraction representation on a number line are studied under fractions. In previous classes, we learned fractions and decimals, as well as how to add and subtract them. We will be learning how to solve questions related to multiplication and division of fractions and decimals in NCERT Class 7 Chapter 2 and going through our NCERT Solutions by Extramarks will assist you to learn these topics in a systematic and organised manner to excel in academics.

NCERT Solutions for Class 7 Mathematics Chapter 2 Extramarks have detailed answers to every question in the NCERT textbook. They will aid in scoring better marks in exams. These solutions are prepared by subject-matter experts, thus you can count on them as they are reliable and genuine resources.

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

Access NCERT Solutions for Class 7 Mathematics Chapter 2 – Fractions and Decimals

NCERT Solutions for Class 7 Chapter 2 Mathematics

Chapter 2 – Fractions and Decimals Exercises
Exercise 2.1	Questions & Solutions
Exercise 2.2	Questions & Solutions
Exercise 2.3	Questions & Solutions
Exercise 2.4	Questions & Solutions
Exercise 2.5	Questions & Solutions
Exercise 2.6	Questions & Solutions
Exercise 2.7	Questions & Solutions

Students can refer to the NCERT Class 7 Mathematics Chapter 2 solutions here at Extramarks. All the questions are solved in such a way so that students can understand them well.

2.1 Introduction

Students have already learned about fraction and its types in previous classes. They will learn about multiplication and division of fractions in Class 7 Chapter 2. The concept of fractions focuses on the ratios and proportions, their distribution etc. Students will learn all these Mathematical concepts in this chapter.

2.2 Recollect

How well have you learnt about fractions? While solving questions of NCERT Class 7 Mathematics Chapter 2, students will recall what they learned in their previous classes.

2.3 Multiplication of Fractions

Students will learn how to multiply two fractions in this section. For example, if they have values like x and y, they can say xy is the product of x and y. If the values are a/b, how will we multiply? In this scenario, there are two methods – by using a whole number and by using a fraction. .

2.3.1 Multiplication of Fractions Using the Whole Number

This section explains the parts of a fraction. The lower part of a fraction is a whole number (except zero) and the upper part is an integer. The lower part of a fraction is called the denominator, while the upper part is called the numerator. If the fractions are the same, we can multiply them with a whole number. Let’s say we have a/b as an example. Then we can multiply by 3×a/b since 3 is a whole number. A fraction can also be multiplied with improper or mixed fractions. However, before multiplying, students must reduce them to their simplest forms.

2.3.1 Multiplication of Fractions Using the Fraction

This section deals with how to multiply two fractions that are dissimilar to each other. To multiply them, students need to use this formula:

(Product of the numerators) / (Product of the denominators)

However, it must be noted that when two proper fractions are multiplied, the resultant product is less than the two fractions. When two improper fractions are multiplied, the result is greater than the two fractions.

2.4 Division of Fractions

It is possible to divide a whole number by a fraction and a fraction by a whole number.

Similarly, when mixed fractions are divided with a whole number, mixed fractions should be changed to improper fractions. It makes division easy. Students also learn how to divide a fraction with another fraction by changing one of the fractions to its reciprocal form.

2.5 Recall of Decimals

In this section, with the help of a quick recap, students will be forced to recall their decimal knowledge. Height, distance, weight, measuring values, interest rates, shares, and fractions can all be expressed with decimals. We can multiply with 10,100, etc. to change the place value of a point.

2.6 Multiplication of Decimals

The purpose of this section is for students to practise multiplication of decimals. Even though multiplication is simple, students may have doubts. To do so, count the number of values after a decimal point in both numbers and keep the point before that number of places in the result. Another variation of decimal multiplication is changing the decimal point’s place value by multiplying it by 10 multiples.

2.7 Division of Decimals

Students will learn how to divide decimals and its variations.
If a whole number is divided with 10 multiples, we derive decimals.
If decimals are divided, we get whole numbers
Students also learn the division of decimals with a whole number. In this case, the place value of the decimal point will not change in the result as well.

Key Features of NCERT Solutions for Class 7 Mathematics Chapter 2

Fractions and decimals, as well as their concepts, are covered in NCERT Class 7 Mathematics Chapter 2 to help students improve their fundamental knowledge and analytical skills. Some benefits of NCERT Solutions for Class 7 Mathematics Chapter 2 provided by Extramarks are:

It incorporates answers to textbook questions for assisting students in solving textbook questions easily.
The solutions are prepared by subject-matter experts and experienced faculty, so they are error-free and reliable.
They provide exam-style answers so that students can learn how to solve each question properly in the exam and practise religiously to come out with flying colours.

NCERT Solutions for Class 7 Mathematics

The NCERT Solutions Class 7 Mathematics Chapter 2 has been carefully curated by subject experts to ensure that the entire Mathematics syllabus is solved and has enough practice exercises. Students can easily plan their exam preparation and revision with the help of these competent resources to achieve the best-desired exam results.

You will also find exercise solutions for Fractions and Decimals Class 7 here, which will allow you to search for answers to all questions which are not so easy and may require time and effort to solve.

Mathematics can be a rewarding subject if you know how to solve the given problems. Using the NCERT Class 7 Solutions, you can perform confidently in your exams. The Mathematics textbook for Class 7 has 15 chapters, each with problems and questions at the end of the chapter. The chapter-by-chapter NCERT Solutions for Class 7 Mathematics provided here can be used to answer questions from the NCERT textbook.

NCERT Solutions Class 7 Maths Chapter-wise List

Chapter 1 – Integers

Chapter 2 – Fractions and Decimals

Chapter 3 – Data Handling

Chapter 4 – Simple Equations

Chapter 5 – Lines and Angles

Chapter 6 – The Triangle and Its Properties

Chapter 7 – Congruence of Triangles

Chapter 8 – Comparing Quantities

Chapter 9 – Rational Numbers

Chapter 10 – Practical Geometry

Chapter 11 – Perimeter and Area

Chapter 12 – Algebraic Expressions

Chapter 13 – Exponents and Powers

Chapter 14 – Symmetry

Chapter 15 – Visualising Solid Shapes

NCERT Solutions for Class 7

Do you need guidance with your Class 7 homework or textbook questions? You’ve come to the right place! NCERT Solutions Class 7 will not only assist you with your homework but will also provide accurate and reliable answers to all questions in the NCERT textbook.

Science, Mathematics, Social Science, Hindi, and English are the five major subjects in NCERT Class 7. NCERT Class 7 Solutions covers answers to all subject related questions of the NCERT textbook.

Q.1

$\begin{array}{l} Solve : \\ (i) 2- \frac{3}{5} (ii) 4+ \frac{7}{8} (iii) \frac{3}{5} + \frac{2}{7} (iv) \frac{9}{11} - \frac{4}{15} \\ (v) \frac{7}{10} + \frac{2}{5} + \frac{3}{2} (vi) 2 \frac{2}{3} + 3 \frac{1}{2} (vii) 8 \frac{1}{2} - 3 \frac{5}{8} \end{array}$

Ans.

$\begin{array}{l} (i) 2 - \frac{3}{5} = \frac{10 - 3}{5} = \frac{7}{5} \\ (i i) 4 + \frac{7}{8} = \frac{32 + 7}{8} = \frac{39}{8} = 4 \frac{7}{8} \\ (i i i) \frac{3}{5} + \frac{2}{7} = \frac{21 + 10}{35} = \frac{31}{35} \\ (i v) \frac{9}{11} - \frac{4}{15} = \frac{135 + 44}{165} = \frac{179}{165} = 1 \frac{14}{165} \\ (v) \frac{7}{10} + \frac{2}{5} + \frac{3}{2} = \frac{7 + 4 + 15}{10} = \frac{26}{10} \\ = \frac{13}{5} \\ = 2 \frac{3}{5} \\ (v i) 2 \frac{2}{3} + 3 \frac{1}{2} = \frac{8}{3} + \frac{7}{2} = \frac{16 + 21}{6} \\ = \frac{37}{6} \\ = 6 \frac{1}{6} \\ (v i i) 8 \frac{1}{2} - 3 \frac{5}{8} = \frac{17}{2} - \frac{29}{8} \\ = \frac{68 - 29}{8} \\ = \frac{39}{8} \\ = 4 \frac{7}{8} \end{array}$

Q.2

$\begin{array}{l} Arrange the following in descending order . \\ (i) \frac{2}{9}, \frac{2}{3}, \frac{8}{21} (ii) \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \end{array}$

Ans.

$\begin{array}{l} First, we find the LCM of 9, 3 and 21 . \\ LCM of 9, 3 and 21 = 63 \\ = \frac{2 \times 7}{9 \times 7}, \frac{2 \times 21}{3 \times 21}, \frac{8 \times 3}{21 \times 3} \\ = \frac{14}{63}, \frac{42}{63}, \frac{24}{63} \\ Since, 42 > 24 > 14 . \\ Therefore, \\ \frac{42}{63} > \frac{24}{63} > \frac{14}{63} \\ \Rightarrow \frac{2}{3} > \frac{8}{21} > \frac{2}{9} \\ (i i) \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \\ First, find the LCM of 5, 7 and 1 0 \\ LCM of denominators 5, 7 and 1 0 is 7 0. \\ So, \\ \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \\ \Rightarrow \frac{1 \times 14}{5 \times 14}, \frac{3 \times 10}{7 \times 10}, \frac{7 \times 7}{10 \times 7} \\ \Rightarrow \frac{14}{70}, \frac{30}{70}, \frac{49}{70} \\ Since, 49 > 30 > 14 . \\ Therefore, \\ \frac{49}{70} > \frac{30}{70} > \frac{14}{70} \\ \Rightarrow \frac{7}{10} > \frac{3}{7} > \frac{1}{5} \end{array}$

Q.3 In a “magic square”, the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?

$\begin{matrix} \frac{4}{11} & \frac{9}{11} & \frac{2}{11} \\ \frac{3}{11} & \frac{5}{11} & \frac{7}{11} \\ \frac{8}{11} & \frac{1}{11} & \frac{6}{11} \end{matrix}$

Ans.

$\begin{array}{l} Finding the sum of first row : \\ S u m o f 1 s t r o w = \frac{4}{11} + \frac{9}{11} + \frac{2}{11} \\ = \frac{4 + 9 + 2}{11} \\ = \frac{15}{11} \\ Finding the sum of second row : \\ Sum of 2nd row = \frac{3}{11} + \frac{5}{11} + \frac{7}{11} \\ = \frac{15}{11} \\ Finding the sum of third row : \\ Sum of 3rd row = \frac{8}{11} + \frac{1}{11} + \frac{6}{11} \\ = \frac{8 + 1 + 6}{11} \\ = \frac{15}{11} \\ Finding the sum of first column : \\ Sum of 1st column = \frac{4}{11} + \frac{3}{11} + \frac{8}{11} \\ = \frac{4 + 3 + 8}{11} \\ = \frac{15}{11} \\ Finding the sum of second column : \\ Sum of 2nd column = \frac{9}{11} + \frac{5}{11} + \frac{1}{11} \\ = \frac{9 + 5 + 1}{11} \\ = \frac{15}{11} \\ Finding the sum of third column : \\ Sum of 3rd column = \frac{2}{11} + \frac{7}{11} + \frac{6}{11} \\ = \frac{2 + 7 + 6}{11} \\ = \frac{15}{11} \\ Now, we must find diagonal sum . \\ Sum of d i a g o n a l f r o m l e f t b o t t o m t o r i g h t t o p \\ = \frac{8}{11} + \frac{5}{11} + \frac{2}{11} \\ = \frac{8 + 5 + 2}{11} \\ = \frac{15}{11} \\ Sum of d i a g o n a l f r o m l e f t b o t t o m t o r i g h t t o p \\ = \frac{6}{11} + \frac{5}{11} + \frac{4}{11} \\ = \frac{6 + 5 + 4}{11} \\ = \frac{15}{11} \\ Since the sum of the numbers in each row, in each \\ column along the diagonal is the same, which is equal \\ to \frac{15}{11} . \\ Therefore, given square is magical . \end{array}$

Q.4

$\begin{array}{l} A rectangular sheet of paper is 12 \frac{1}{2} cm long and \\ 10 \frac{2}{3} cm wide . Find its perimeter . \end{array}$

Ans.

$\begin{array}{l} We are given : \\ Length of rectangular sheet of paper = 12 \frac{1}{2} cm = \frac{25}{2} cm \\ Width of rectangular sheet of paper=10 \frac{2}{3} cm = \frac{32}{3} cm \\ Since, Perimeter of a rectangle = 2 \times (length + width) \\ Therefore, \\ Perimeter of the given rectangular sheet of paper \\ = 2 \times (\frac{25}{2} + \frac{32}{3}) cm \\ = 2 \times (\frac{75 + 64}{6}) cm \\ = 2 \times \frac{139}{6} cm \\ = \frac{139}{3} cm = 46 \frac{1}{3} cm \end{array}$

Q.4

$\begin{array}{l} Find the perimeters of (i) ΔABE (ii) the rectangle \\ BCDE in this figure . Whose perimeter is greater ? \end{array}$

Ans.

$\begin{array}{l} Since, the Perimeter of a triangle = side + side + side \\ Therefore, Perimeter of triangle ABE = AB + BE + AE \\ = \frac{5}{2} + 2 \frac{3}{4} + 3 \frac{3}{5} cm \\ = \frac{5}{2} + \frac{11}{4} + \frac{18}{5} cm \\ = \frac{50 + 55 + 72}{20} cm \\ = \frac{177}{20} cm \\ In rectangle BCDE, length BE = 2 \frac{2}{4} cm = \frac{11}{4} cm \\ Width ED = \frac{7}{6} cm \\ Since, Perimeter of a rectangle = 2 \times (length + width) \\ Therefore, perimeter of rectangle BCDE = 2 \times (B E + E D) \\ = 2 \times (\frac{11}{4} + \frac{7}{6}) cm \\ = 2 \times (\frac{33 + 14}{12}) cm \\ = 2 \times \frac{47}{12} cm \\ = \frac{47}{6} cm \\ Hence, the perimeter of Δ A B C = \frac{177}{20} cm \\ and \\ The perimeter of rectangle BCDE = \frac{47}{6} cm \\ Now, comparing both the perimeter to find which is \\ greater . \\ Since, \\ The LCM of denominators of both the perimeter (20 and 6) is \\ equal to 6 0. \\ Therefore, \\ \frac{177}{20}, \frac{47}{6} \Rightarrow \frac{177 \times 3}{20 \times 3}, \frac{47 \times 10}{6 \times 10} \Rightarrow \frac{531}{60}, \frac{470}{10} \\ Since, 531 is greater than 47 0. \\ Therefore, \\ \frac{531}{60} > \frac{470}{60} \Rightarrow \frac{177}{20} > \frac{47}{6} \\ Thus, the perimeter of triangle ABE is greater than \\ the perimeter of rectangle BCDE . \end{array}$

Q.5

$\begin{array}{l} Sali wants to put a picture in a frame . The picture is 7 \frac{3}{5} cm wide . \\ To fit in the frame the picture cannot be more than 7 \frac{3}{10} cm wide . \\ How much should the picture be trimmed ? \end{array}$

Ans.

$\begin{array}{l} We are given : \\ The width of the picture=7 \frac{3}{5} cm \\ Required width of picture to be fit in frame=7 \frac{3}{10} cm \\ Therefore, picture to be trim = 7 \frac{3}{5} - 7 \frac{3}{10} cm \\ = \frac{38}{5} - \frac{73}{10} cm \\ = \frac{3}{10} cm \\ Thus picture should be trimmed by \frac{3}{10} cm \end{array}$

Q.6

$\begin{array}{l} Ritu ate \frac{3}{5} part of an apple and the remaining apple was \\ eaten by her brother Somu . How much part of the apple \\ did Somu eat ? Who had the larger share ? By how much ? \end{array}$

Ans.

$\begin{array}{l} Since, Ritu ate \frac{3}{5} part of an apple and her brother eaten the \\ rest part of the apple . \\ So, the part of apple left after eaten by Ritu = 1 - \frac{3}{5} \\ = \frac{5 - 3}{5} \\ = \frac{2}{5} \\ Therefore, Somu ate \frac{2}{5} parts of apple . \\ Now comparing the part eaten by them \frac{3}{5}, \frac{2}{5} \\ Since, 3 > 2 . \\ Therefore \\ \frac{3}{5} > \frac{2}{5} \\ That means Ritu ate larger part . \\ The difference in both parts = \frac{3}{5} - \frac{2}{5} \\ = \frac{3 - 2}{5} \\ = \frac{1}{5} \\ Thus, \\ Somu ate \frac{2}{5} part of apple . \\ Ritu ate the larger part of apple \\ Ritu ate part more apple than her brother Somu . \end{array}$

Q.7

$\begin{array}{l} Michael finished colouring a picture in \frac{7}{12} hour . \\ Vaibhav finished colouring the same picture in \frac{3}{4} hour . \\ Who worked longer ? By what fraction was it longer ? \end{array}$

Ans.

$\begin{array}{l} We are given : \\ Michael worked for \frac{7}{12} hour \\ Vaibhav worked for \frac{3}{4} hour Therefore, \\ In order to find the longer hour of work we have to compare \\ both the fractions \frac{7}{12}, \frac{3}{4} \\ The LCM of 12 and 4 = 12 \\ So, \frac{7}{12}, \frac{3}{4} = \frac{7}{12}, \frac{9}{12} \\ Therefore, \frac{9}{12} > \frac{7}{12} \\ That means Vaibhav worked for longer hour . \\ Difference in their working hour = \frac{9}{12} - \frac{7}{12} \\ = \frac{9 - 7}{12} \\ = \frac{2}{12} \\ = \frac{1}{6} \\ Thus, Vaibhav worked for \frac{1}{6} hour more than Michael . \end{array}$

Q.8

$\begin{array}{l} Which of the following drawing (a) to (d) show : \\ (i) 2 \times \frac{1}{5} (ii) 2 \times \frac{1}{2} \\ (iii) 3 \times \frac{2}{3} (iv) 3 \times \frac{1}{4} \end{array}$

Ans.

$\begin{array}{l} (i) 2 \times \frac{1}{5} = \frac{2}{5} \\ (ii) 2 \times \frac{1}{2} = 1 \\ (iii) 3 \times \frac{2}{3} = 2 \\ (iv) 3 \times \frac{1}{4} = \frac{3}{4} \\ (a) Here, there are three pictures and each pictures \\ show \frac{2}{3} part = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} \\ = \frac{2 + 2 + 2}{3} \\ = \frac{6}{3} = 2 \\ Hence, (iii) denotes picture (a) . \\ (b) Here, there are two pictures . Each pictures show \frac{1}{2} \\ part. \\ So, \frac{1}{2} + \frac{1}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1 \\ Hence, (ii) denotes picture (b) . \\ (c) Here, there are three pictures . Each pictures show \frac{1}{4} part . \\ Therefore, \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{1 + 1 + 1}{4} = \frac{3}{4} \\ Hence, (iv) denotes picture (c) . \\ (d) Here, there are two pictures and each pictures denotes \\ \frac{1}{5} part. \\ Therefore, \frac{1}{5} + \frac{1}{5} = \frac{1 + 1}{5} = \frac{2}{5} \\ Hence, (i) denotes picture (d) . \\ Thus : \\ \begin{array}{l} (i) denotes picture (d) \\ (ii) denotes picture (b) \\ (iii) denotes picture (a) \\ (iv) denotes picture (c) \end{array} \end{array}$

Q.9

$\begin{array}{l} Some pictures (a) to (c) are given below. \\ Tell which of them show \\ (i) 3 \times \frac{1}{5} = \frac{3}{5} (ii) 2 \times \frac{1}{3} = \frac{2}{3} (iii) 3 \times \frac{3}{4} = 2 \frac{1}{4} \end{array}$

Ans.

$\begin{array}{l} In picture (a) : \\ There are two pictures in left hand side, each denotes \frac{1}{3} \\ and picture in right hand side denotes \frac{2}{3} . \\ Therefore, it denotes \frac{1}{3} \times 2 = \frac{2}{3} which denotes (ii) . \\ In picture (b) : \\ Here are three pictures in left hand side, each shows \frac{3}{4} . \\ There are three pictures in right hand side in which each \\ of first two denotes 1 . \\ Therefore, \frac{3}{4} \times 3 = 2 \frac{1}{4} \\ Thus picture (b) denotes (iii) . \\ In picture (c) : \\ There are three pictures in left hand side and each \\ of them denotes \frac{1}{5} . \\ Hence, \\ \begin{array}{l} (i) denotes picture (c) \\ (ii) denotes picture (a) \\ (iii) denotes picture (b) \end{array} \end{array}$

Q.10

$\begin{array}{l} Multiply and reduce to lowest form : \\ (i) 7 \times \frac{3}{5} (ii) 4 \times \frac{1}{3} (iii) 2 \times \frac{6}{7} \\ (iv) 5 \times \frac{2}{9} (v) \frac{2}{3} \times 4 (vi) \frac{5}{2} \times 6 \\ (vii) 11 \times \frac{4}{7} (viii) 20 \times \frac{4}{5} (ix) 13 \times \frac{1}{3} \\ (x) 15 \times \frac{3}{5} \end{array}$

Ans.

$\begin{matrix} (i) 7 \times \frac{3}{5} = \frac{7}{1} \times \frac{3}{5} = \frac{21}{5} \\ = 4 \frac{1}{5} \\ (ii) 4 \times \frac{1}{3} = \frac{4}{1} \times \frac{1}{3} = \frac{4}{3} \\ = 1 \frac{1}{3} \\ (iii) 2 \times \frac{6}{7} = \frac{2}{1} \times \frac{6}{7} = \frac{12}{7} \\ = 1 \frac{5}{7} \\ (iv) 5 \times \frac{2}{9} = \frac{5}{1} \times \frac{2}{9} = \frac{10}{9} \\ = 1 \frac{1}{9} \\ (v) \frac{2}{3} \times 4 = \frac{2}{3} \times \frac{4}{1} = \frac{8}{3} \\ = 2 \frac{2}{3} \\ (vi) \frac{5}{2} \times 6 = \frac{5 \times 6}{2} = \frac{30}{2} \\ = 15 \\ (vii) 11 \times \frac{4}{7} = \frac{11}{1} \times \frac{4}{7} = \frac{44}{7} \\ = 6 \frac{2}{7} \\ (viii) 20 \times \frac{4}{7} = \frac{20}{1} \times \frac{4}{5} = \frac{80}{5} \\ = 16 \\ (ix) 13 \times \frac{1}{3} = \frac{13}{1} \times \frac{1}{3} = \frac{13}{3} \\ = 4 \frac{1}{3} \\ (x) 15 \times \frac{3}{5} = \frac{15}{1} \times \frac{3}{5} = \frac{45}{5} \\ = 9 \end{matrix}$

Q.11

$\begin{array}{l} Shade : \\ (i) \frac{1}{2} of the circles in box (a) \\ (ii) \frac{2}{3} of the triangles in box (b) \\ (iii) \frac{3}{5} of the squares in box (c) \end{array}$

Ans.

$\begin{array}{l} (i) \\ Here, there are 12 circles in box (a) . \\ Therefore, \\ \frac{1}{2} of 12 = \frac{1}{2} \times 12 = 6 \\ Hence 6 circles to be coloured in (a) . \\ (ii) \\ Here, there are 9 triangles in box (b) and \frac{2}{3} of the \\ triangles are to be coloured . \\ So, \frac{2}{3} of 9 = \frac{2}{3} \times 9 \\ = \frac{18}{3} = 6 \\ Thus, 6 triangles are needed to be coloured . \\ (iii) \\ Here, there are 15 rectangles in box (c) and \frac{3}{5} of them \\ are needed to be coloured . \\ So, \\ \frac{3}{5} of 15 = \frac{3}{5} \times 15 \\ = \frac{45}{5} = 9 \\ Thus, 9 rectangles are needed to be coloured . \end{array}$

Q.12

$\begin{array}{l} F i n d : \\ (a) \frac{1}{2} of (i) 24 (i i) 46 \\ (b) \frac{2}{3} of (i) 18 (i i) 27 \\ (c) \frac{3}{4} of (i) 16 (i i) 36 \\ (d) \frac{4}{5} of (i) 20 (i i) 35 \end{array}$

Ans.

$\begin{array}{l} (a) \\ (i) \frac{1}{2} o f 24 = \frac{1}{2} \times 24 = \frac{24}{2} = 12 \\ (i i) \frac{1}{2} o f 46 = \frac{1}{2} \times 46 = \frac{46}{2} = 23 \\ (b) \\ (i) \frac{2}{3} o f 18 = \frac{2}{3} \times 18 = \frac{36}{3} = 12 \\ (i i) \frac{2}{3} o f 27 = \frac{2}{3} \times 27 = \frac{54}{3} = 18 \\ (c) \\ (i) \frac{3}{4} o f 16 = \frac{3}{4} \times 16 = \frac{48}{4} = 12 \\ (i i) \frac{3}{4} o f 36 = \frac{3}{4} \times 36 = \frac{108}{4} = 27 \\ (d) \\ (i) \frac{4}{5} o f 20 = \frac{4}{5} \times 20 = \frac{80}{5} = 16 \end{array}$

Q.13

$\begin{array}{l} Multiply and express as a mixed fraction : \\ (a) 3 \times 5 \frac{1}{5} (b) 5 \times 6 \frac{3}{4} (c) 7 \times 2 \frac{1}{4} \\ (d) 4 \times 6 \frac{1}{3} (e) 3 \frac{1}{4} \times 6 (f) 3 \frac{2}{5} \times 8 \end{array}$

Ans.

$\begin{array}{l} (a) 3 \times 5 \frac{1}{5} = 3 \times \frac{26}{5} = \frac{78}{5} = 15 \frac{3}{5} \\ (b) 5 \times 6 \frac{3}{4} = 5 \times \frac{27}{4} = \frac{135}{4} = 33 \frac{3}{4} \\ (c) 7 \times 2 \frac{1}{4} = 7 \times \frac{9}{4} = \frac{63}{4} = 15 \frac{3}{4} \\ (d) 4 \times 6 \frac{1}{3} = 4 \times \frac{19}{3} = \frac{76}{3} = 25 \frac{1}{3} \\ (e) 3 \frac{1}{4} \times 6 = \frac{13}{4} \times 6 = \frac{78}{7} = 19 \frac{1}{2} \\ (f) 3 \frac{2}{5} \times 8 = \frac{17}{5} \times 8 = \frac{136}{5} = 27 \frac{1}{5} \end{array}$

Q.14

$\begin{array}{l} F i n d \\ (a) \frac{1}{2} of (i) 2 \frac{3}{4} (ii) 4 \frac{2}{9} \\ (b) \frac{5}{8} of (i) 3 \frac{5}{6} (ii) 9 \frac{2}{3} \end{array}$

Ans.

$\begin{array}{l} (a) \\ (i) \frac{1}{2} of 2 \frac{3}{4} = \frac{1}{2} \times \frac{11}{4} = \frac{11}{8} = 1 \frac{3}{8} \\ (ii) \frac{1}{2} of 4 \frac{2}{9} = \frac{1}{2} \times \frac{38}{9} = \frac{19}{9} = 2 \frac{1}{9} \\ (b) \\ (i) \frac{5}{8} of 3 \frac{5}{6} = \frac{5}{8} \times \frac{23}{6} = \frac{115}{48} = 2 \frac{19}{48} \\ (ii) \frac{5}{8} of 9 \frac{2}{3} = \frac{5}{8} \times \frac{29}{3} = \frac{145}{24} = 6 \frac{1}{24} \end{array}$

Q.15

$\begin{array}{l} Vidya and Pratab went for a picnic . Their mother \\ gave them a water bag that contained 5 liters of \\ water . Vidya consumed \frac{2}{5} of the water . Pratap \\ consumed the remaining water . (i) How much water did Vidya drink ? \\ (ii) What fraction of total \\ quantity of water did Pratab drink ? \end{array}$

Ans.

Q.16

$\begin{array}{l} F i n d : \\ (i) \frac{1}{4} o f (a) \frac{1}{4} (b) \frac{3}{5} (c) \frac{4}{3} \\ (i i) \frac{1}{7} o f (a) \frac{2}{9} (b) \frac{6}{5} (c) \frac{3}{10} \end{array}$

Ans.

$\begin{array}{l} (i) \\ (a) \frac{1}{4} of \frac{1}{4} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \\ (b) \frac{1}{4} of \frac{3}{5} = \frac{1}{4} \times \frac{3}{5} = \frac{3}{20} \\ (c) \frac{1}{4} of \frac{4}{3} = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3} \\ (i i) \\ (a) \frac{1}{7} of \frac{2}{9} = \frac{1}{7} \times \frac{2}{9} = \frac{2}{63} \\ (b) \frac{1}{7} of \frac{6}{5} = \frac{1}{7} \times \frac{6}{5} = \frac{6}{35} \\ (c) \frac{1}{7} of \frac{3}{10} = \frac{1}{7} \times \frac{3}{10} = \frac{3}{70} \end{array}$

Q.17

$\begin{array}{l} Multiply and reduce to lowest form (if possible) . \\ (i) \frac{2}{3} \times 2 \frac{2}{3} (ii) \frac{2}{7} \times \frac{7}{9} (iii) \frac{3}{8} \times \frac{6}{4} \\ (iv) \frac{9}{5} \times \frac{3}{5} (v) \frac{1}{3} \times \frac{15}{8} (vi) \frac{11}{2} \times \frac{3}{10} \\ (vii) \frac{4}{5} \times \frac{12}{7} \end{array}$

Ans.

$\begin{array}{l} (i) \frac{2}{3} \times 2 \frac{2}{3} = \frac{2}{3} \times \frac{8}{3} = \frac{16}{9} \\ (ii) \frac{2}{7} \times \frac{7}{9} = \frac{2}{7} \times \frac{7}{9} = \frac{2}{9} \\ (iii) \frac{3}{8} \times \frac{6}{4} = \frac{3}{2 \times 4} \times \frac{2 \times 3}{4} = \frac{9}{16} \\ (iv) \frac{9}{5} \times \frac{3}{5} = \frac{27}{25} \\ (v) \frac{1}{3} \times \frac{15}{8} = \frac{1}{3} \times \frac{3 \times 5}{8} = \frac{5}{8} \\ (vi) \frac{11}{2} \times \frac{3}{10} = \frac{33}{20} \\ (vii) \frac{4}{5} \times \frac{12}{7} = \frac{48}{35} \end{array}$

Q.18

$\begin{array}{l} For the fractions given below : \\ (a) Multiply and reduce the product to lowest \\ form (if possible) \\ (b) Tell whether the fraction obtained is proper \\ or improper . \\ (c) If the fraction obtained is improper then \\ convert it into a mixed fraction . \\ (i) \frac{2}{5} \times 5 \frac{1}{4} \\ (ii) 6 \frac{2}{5} \times \frac{7}{9} \\ (iii) \frac{3}{2} \times 5 \frac{1}{3} \\ (iv) \frac{5}{6} \times 2 \frac{3}{7} \\ (v) 3 \frac{2}{5} \times \frac{4}{7} \\ (vi) 2 \frac{3}{5} \times 3 \\ (vii) 3 \frac{4}{7} \times \frac{3}{5} \end{array}$

Ans.

$\begin{array}{l} (i) \\ (a) \frac{2}{5} \times 5 \frac{1}{4} = \frac{2}{5} \times \frac{21}{4} = \frac{2}{5} \times \frac{21}{2 \times 2} = \frac{21}{10} \\ (b) It is an improper fraction . \\ (c) \frac{21}{10} = 2 \frac{1}{10} \\ (ii) \\ (a) 6 \frac{2}{5} \times \frac{7}{9} = \frac{32}{5} \times \frac{7}{9} = \frac{224}{45} \\ (b) It is an improper fraction . \\ (c) \frac{224}{45} = 4 \frac{44}{45} \\ (iii) \\ (a) \frac{3}{2} \times 5 \frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = \frac{3}{2} \times \frac{2 \times 8}{3} = \frac{8}{1} = 8 \\ (b) It is whole number. \\ (iv) \\ (a) \frac{5}{6} \times 2 \frac{3}{7} \\ = \frac{5}{6} \times \frac{17}{7} \\ = \frac{85}{42} \\ (b) It is an improper fraction . \\ (c) \frac{85}{42} = 2 \frac{1}{42} \\ (v) \\ (a) 3 \frac{2}{5} \times \frac{4}{7} \\ = \frac{17}{5} \times \frac{4}{7} \\ = \frac{68}{35} \\ (b) It is an improper fraction . \\ (c) \frac{68}{35} = 1 \frac{33}{35} \\ (vi) \\ (a) 2 \frac{3}{5} \times 3 \\ = \frac{13}{5} \times \frac{3}{1} \\ = \frac{39}{5} \\ (b) It is an improper fraction . \\ (c) \frac{39}{5} = 7 \frac{4}{5} \\ (vii) \\ (a) 3 \frac{4}{7} \times \frac{3}{5} \\ = \frac{25}{7} \times \frac{3}{5} \\ = \frac{5 \times 5}{7} \times \frac{3}{5} \\ = \frac{15}{7} \\ (b) It is an improper fraction . \\ (c) \frac{15}{7} = 2 \frac{1}{7} \end{array}$

Q.19

$\begin{array}{l} Which is greater : \\ (i) \frac{2}{7} o f \frac{3}{4} or \frac{3}{5} o f \frac{5}{8} \\ (i i) \frac{1}{2} o f \frac{6}{7} or \frac{2}{3} o f \frac{3}{7} \end{array}$

Ans.

$\begin{array}{l} (i) \\ \frac{2}{7} of \frac{3}{4} or \frac{3}{5} of \frac{5}{8} \\ \frac{2}{7} \times \frac{3}{4} o r \frac{3}{5} \times \frac{5}{8} \\ = \frac{2}{7} \times \frac{3}{2 \times 2} or \frac{3}{5} \times \frac{5}{8} \\ = \frac{3}{14} or \frac{3}{8} \\ Now, the LCM of 14 and 8 is 56 . So, we get \\ \frac{3}{14} = \frac{3 \times 4}{14 \times 4} = \frac{12}{56} or \frac{3}{8} = \frac{3 \times 7}{8 \times 7} = \frac{21}{56} \\ Since, 21 > 12, so \frac{3}{8} > \frac{3}{14} \\ Hence, \frac{3}{5} of \frac{5}{8} > \frac{2}{7} of \frac{3}{4} \\ (ii) \\ \frac{1}{2} of \frac{6}{7} o r \frac{2}{3} of \frac{3}{7} \\ \frac{1}{2} of \frac{6}{7} o r \frac{2}{3} of \frac{3}{7} \\ = \frac{1}{2} \times \frac{2 \times 3}{7} or \frac{2}{3} \times \frac{3}{7} \\ = \frac{3}{7} or \frac{2}{7} \\ Since, 3 > 2, so, \frac{3}{7} > \frac{2}{7} \\ Hence, \frac{1}{2} of \frac{6}{7} > \frac{2}{3} of \frac{3}{7} \end{array}$

Q.20

$Saili plants 4 saplings, in a row, in her garden . The distance between two adjacent saplings is \frac{3}{4} m . Find the distance between the first and the last sapling .$

Ans.

$\begin{array}{l} Here, the distance between the two saplings is given as \frac{3}{4} m \\ So, the distance between first and the last sapling is \\ \frac{3}{4} m \times 3= \frac{9}{4} m \\ = 2 \frac{1}{4} m \end{array}$

Q.21

$\begin{array}{l} Lipika reads a book for 1 \frac{3}{4} hours every day . She \\ reads the entire book in 6 days . How many hours \\ in all were required by her to read the book ? \end{array}$

Ans.

$\begin{array}{l} Lipika reads 1 \frac{3}{4} = \frac{7}{4} hours a day . \\ So, Number of hours required by her to read the book \\ in 6 days is \frac{7}{4} hour \times 6 = \frac{7}{2 \times 2} hour \times (2 \times 3) \\ = \frac{21}{2} hour \\ = 10 \frac{1}{2} hour \end{array}$

Q.22

$\begin{array}{l} A car runs 16 km using 1 litre of petrol . How \\ much distance will it cover using 2 \frac{3}{4} litres of petrol ? \end{array}$

Ans.

$\begin{matrix} In 1 litre of petrol, car runs 16 km . \\ So, in 2 \frac{3}{4} litres of petrol, distance covered by car is \\ 2 \frac{3}{4} \times 16 km \\ = \frac{11}{4} \times (4 \times 4) km \\ = 44 km \end{matrix}$

Q.23

$\begin{array}{l} (a) \\ (i) Provide the number in the box , such that \frac{2}{3} \times = \frac{10}{30} \\ (ii) The simplest form of the number obtained in is____ \\ (b) \\ (i) Provide the number in the box, such that \frac{3}{5} \times = \frac{24}{75} \\ (ii) \\ The simplest form of the number obtained in is____ \end{array}$

Ans.

$\begin{array}{l} (a) \\ (i) \frac{2}{3} \times = \frac{10}{30} \\ = \frac{2×5}{3×10} = \frac{2}{3} × \frac{5}{10} \\ So, \frac{2}{3} \times \frac{5}{10} = \frac{10}{30} \\ Therefore, the number in is \frac{5}{10} . \\ (ii) \\ The simplest form of \frac{5}{10} is \frac{5}{10} = \frac{}{} = \frac{1}{2} \\ (b) \\ (i) \frac{3}{5} \times = \frac{24}{75} \\ = \frac{3×8}{5×15} = \frac{3}{5} × \frac{8}{15} \\ So, \frac{3}{5} × \frac{8}{15} = \frac{24}{75} \\ Therefore, the number in is \frac{8}{15} . \\ (ii) \\ The simplest form of \frac{8}{15} is \frac{8}{15} \end{array}$

Q.24

$\begin{array}{l} Find: \\ (i) 12÷ \frac{3}{4} (ii) 14÷ \frac{5}{6} (iii) 8÷ \frac{7}{3} \\ (iv) 4÷ \frac{8}{3} (v) 3÷2 \frac{1}{3} (vi) 5÷3 \frac{4}{7} \end{array}$

Ans.

$\begin{array}{l} (i) 12÷ \frac{3}{4} =12× \frac{4}{3} = \frac{48}{3} = 16 \\ (ii) 14÷ \frac{5}{6} =14× \frac{6}{5} = \frac{84}{5} = 16 \frac{4}{5} \\ (iii) 8÷ \frac{7}{3} =8× \frac{3}{7} = \frac{24}{7} = 3 \frac{3}{7} \\ (iv) 4÷ \frac{8}{3} =4× \frac{3}{8} \frac{3}{} = \frac{3}{2} = 1 \frac{1}{2} \\ (v) 3÷2 \frac{1}{3} =3÷ \frac{7}{3} =3× \frac{3}{7} = \frac{9}{7} = 1 \frac{2}{7} \\ (vi) 5÷3 \frac{4}{7} =5÷ \frac{25}{7} =5× \frac{7}{25} \frac{7}{} = \frac{7}{5} = 1 \frac{2}{5} \end{array}$

Q.25

$\begin{array}{l} Find the reciprocal of each of the following \\ fractions . Classify the reciprocals as proper \\ fractions, improper fractions and whole numbers . \\ (i) \frac{3}{7} (ii) \frac{5}{8} (iii) \frac{9}{7} (iv) \frac{6}{5} \\ (v) \frac{12}{7} (vi) \frac{1}{8} (vii) \frac{1}{11} \end{array}$

Ans.

$\begin{array}{l} (i) \frac{3}{7} \\ Reciprocal of \frac{3}{7} is \frac{7}{3} and it is an improper fraction . \\ (ii) \frac{5}{8} \\ Reciprocal of \frac{5}{8} is \frac{8}{5} and it is an improper fraction \\ (iii) \frac{9}{7} \\ Reciprocal of \frac{9}{7} is \frac{7}{9} and it is an proper fraction \\ (iv) \frac{6}{5} \\ Reciprocal of \frac{6}{5} is \frac{5}{6} and it is an proper fraction \\ (v) \frac{12}{7} \\ Reciprocal of \frac{12}{7} is \frac{7}{12} and it is an proper fraction \\ (vi) \frac{1}{8} \\ Reciprocal of \frac{1}{8} is \frac{8}{1} and it is an whole fraction \\ (vii) \frac{1}{11} \\ Reciprocal of \frac{1}{11} is 11 and it is a whole number. \end{array}$

Q.26

$\begin{array}{l} F i n d : \\ (i) \frac{7}{3} \div 2 (i i) \frac{4}{9} \div 5 (i i i) \frac{6}{13} \div 7 \\ (i v) 4 \frac{1}{3} \div 3 (v) 3 \frac{1}{2} \div 4 (v i) 4 \frac{3}{7} \div 7 \end{array}$

Ans.

$\begin{array}{l} (i) \frac{7}{3} \div 2 = \frac{7}{3} \times \frac{1}{2} = \frac{7}{6} = 1 \frac{1}{6} \\ (i i) \frac{4}{9} \div 5 = \frac{4}{9} \times \frac{1}{5} = \frac{4}{45} \\ (i i i) \frac{6}{13} \div 7 = \frac{6}{13} \times \frac{1}{7} = \frac{6}{91} \\ (i v) 4 \frac{1}{3} \div 3 = \frac{13}{3} \times \frac{1}{3} = \frac{13}{9} = 1 \frac{4}{9} \\ (v) 3 \frac{1}{2} \div 4 = \frac{7}{2} \times \frac{1}{4} = \frac{7}{8} \\ (v i) 4 \frac{3}{7} \div 7 = \frac{31}{7} \times \frac{1}{7} = \frac{31}{49} \end{array}$

Q.27

$\begin{array}{l} Find \\ (i) \frac{2}{5} \div \frac{1}{2} \\ (i i) \frac{4}{9} \div \frac{2}{3} \\ (i i i) \frac{3}{7} \div \frac{8}{7} \\ (i v) 2 \frac{1}{3} \div \frac{3}{5} \\ (v) 3 \frac{1}{2} \div \frac{8}{3} \\ (v i) \frac{2}{5} \div 1 \frac{1}{2} \\ (v i i) 3 \frac{1}{5} \div 1 \frac{2}{3} \\ (v i i i) 2 \frac{1}{5} \div 1 \frac{1}{5} \end{array}$

Ans.

$\begin{array}{l} (i) \frac{2}{5} \div \frac{1}{2} = \frac{2}{5} \times \frac{2}{1} = \frac{4}{5} \\ (i i) \frac{4}{9} \div \frac{2}{3} = \frac{4}{9} \times \frac{3}{2} = \frac{(2 \times 2)}{(3 \times 3)} \times \frac{3}{2} = \frac{2}{3} \\ (i i i) \frac{3}{7} \div \frac{8}{7} = \frac{3}{7} \times \frac{7}{8} = \frac{3}{8} \\ (i v) 2 \frac{1}{3} \div \frac{3}{5} = \frac{7}{3} \times \frac{5}{3} = \frac{35}{9} = 3 \frac{8}{9} \\ (v) 3 \frac{1}{2} \div \frac{8}{3} = \frac{7}{2} \times \frac{3}{8} = \frac{21}{16} = 1 \frac{5}{16} \\ (v i) \frac{2}{5} \div 1 \frac{1}{2} = \frac{2}{5} \div \frac{3}{2} = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15} \\ (v i i) 3 \frac{1}{5} \div 1 \frac{2}{3} = \frac{16}{5} \div \frac{5}{3} = \frac{16}{5} \times \frac{3}{5} = \frac{48}{25} = 1 \frac{23}{25} \\ (v i i i) 2 \frac{1}{5} \div 1 \frac{1}{5} = \frac{11}{5} \div \frac{6}{5} = \frac{11}{5} \times \frac{5}{6} = \frac{11}{6} = 1 \frac{5}{6} \end{array}$

Q.28

$\begin{array}{l} Which is greater ? \\ (i) 0.5 or 0.05 \\ (ii) 0.7 or 0.5 \\ (iii) 7 or 0.7 \\ (iv) 1.37 or 1.49 \\ (v) 2.03 or 2.30 \\ (vi) 0.8 or 0.88 \end{array}$

Ans.

$\begin{array}{l} (i) 0. 5 or 0.0 5 \\ 0.5 > 0.05 \\ (ii) 0. 7 or 0. 5 \\ 0.7 > 0.5 \\ (iii) 7 or 0. 7 \\ 7 > 0.7 \\ (iv) 1 . 37 or 1 . 49 \\ 1.49 > 1.37 \\ (v) 2 .0 3 or 2 . 30 \\ 2.30 > 2.03 \\ (v i) 0.8 or 0.88 \\ 0.88 > 0.8 \end{array}$

Q.29

$\begin{array}{l} Express as rupees using decimals : \\ (i) 7 paise \\ (ii) 7 rupees 7 paise \\ (iii) 77 rupees 77 paise \\ (iv) 50 paise \\ (v) 235 paise \end{array}$

Ans.

$\begin{array}{l} (i) 7 paise \\ Since, 1 paise = ₹ \frac{1}{100} \\ So, 7 paise =7× ₹ \frac{1}{100} \\ = ₹ \frac{7}{100} \\ = ₹ 0 .07 \\ (ii) 7rupees 7 paise \\ Since, 1 paise = ₹ \frac{1}{100} \\ So, 7 rupees 7 paise=₹ 7+ (7× \frac{1}{100}) \\ =₹ 7+₹ 0 .07 \\ = ₹ 7 .07 \\ (iii) 77 rupees 77 paise . \\ Since, 1 paise = ₹ \frac{1}{100} \\ So, 77 rupees 77 paise = ₹ 77 + ₹ (\frac{77}{100}) \\ = ₹ 77 + ₹ 0.77 \\ = ₹ 77.77 \\ (iv) 50 paise . \\ Since, 1 paise = ₹ \frac{1}{100} \\ So, 50 paise = ₹ (\frac{50}{100}) \\ = ₹ 0.50 \\ (v) 235 paise . \\ Since, 1 paise = ₹ \frac{1}{100} \\ So, 235 paise = ₹ (\frac{235}{100}) \\ = ₹ 2.35 \end{array}$

Q.30

$\begin{array}{l} (i) Express 5 cm in metre and kilometre \\ (ii) Express 35 mm in cm, m and km . \end{array}$

Ans.

$\begin{array}{l} (i) Since, 1 cm = \frac{1}{100} m = \frac{1}{1000} km \\ So, we get \\ 5 cm= \frac{5}{100} m = 0 .05 m \\ and \\ 5 cm = \frac{5}{1000} km = 0 .005 km \\ (ii) Since, 1 mm = \frac{1}{10} cm = \frac{1}{1000} m= \frac{1}{100000} km \\ So, we get \\ 35 mm = \frac{35}{10} cm = 3 .5 m \\ 35 mm = \frac{35}{1000} m = 0 .035 m \\ and \\ 35 mm = \frac{35}{100000} km = 0 .00035 km \end{array}$

Q.31

$\begin{array}{l} Express in kg : \\ (i) 200 g \\ (ii) 3470 g \\ (iii) 4 kg 8 g \\ (iv) 2598 mg \end{array}$

Ans.

$\begin{array}{l} Since, 1 g= \frac{1}{1000} kg \\ So, 200 g= \frac{200}{1000} kg= \frac{2× 100}{10× 100} kg= \frac{2}{10} kg= 0.2 kg \\ (ii) 3470 g \\ 3470 g= \frac{3470}{1000} kg= \frac{347× 10}{100× 10} kg= \frac{347}{100} kg= 3.47 kg \\ (iii) 4 kg 8 g \\ 4 kg 8 g=4 kg+ \frac{8}{1000} kg=4 kg+0.008 kg= 4.008 kg \\ (iv) 2598 mg \\ Since, 1 mg = \frac{1}{1000000} kg \\ So, 2598 mg= \frac{2598}{1000000} kg = 0.002598 kg \end{array}$

Q.32

$\begin{array}{l} Write the following decimal numbers in the expanded form : \\ \begin{array}{l} (i) 20.03 \\ (ii) 2.03 \\ (iii) 200.03 \\ (iv) 2.034 \end{array} \end{array}$

Ans.

$\begin{array}{l} (i) 20 .03= 2×10+0×1+0× \frac{1}{10} +3× \frac{1}{100} \\ (ii) 2 .02= 2×1+0× \frac{1}{10} +3× \frac{1}{100} \\ (iii) 200 .03= 2×100+0×10+0×1+0× \frac{1}{10} +3× \frac{1}{100} \\ (iv) 2 .034= 2×1+0× \frac{1}{10} +3× \frac{1}{100} +4× \frac{1}{1000} \end{array}$

Q.33

$\begin{array}{l} Write the place value of 2 in the following decimal numbers : \\ (i) 2.56 (ii) 21.37 (iii) 10.25 (iv) 9.42 (v) 63.352 \end{array}$

Ans.

$\begin{array}{l} (i) 2 .56 \\ The place value of 2 in 2 . 56 = 2 Ones \\ (ii) 21.37 \\ The place value of 2 in 21 . 37 = 2 Tens \\ (iii) 10.25 \\ The place value of 2 in 1 0.25 = 2 Tenths \\ (iv) 9.42 \\ The place value of 2 in 9 . 42 = 2 Hundredths \\ (v) 63.352 \\ The place value of 2 in 63 . 352 = 2 Thousandths \end{array}$

Q.34

$\begin{array}{l} Dinesh went from place A to place B and from there to \\ place C . A is 7.5 km from B and B is 12.7 km from C . \\ A yub went from place A to place D and from there to \\ place C . D is 9.3 km from A and C is 11.8 km from D . \\ Who travelled more and by how much ? \end{array}$

Ans.

$\begin{array}{l} We are given that: \\ Dinesh went from A to B and from B to C . \\ Ayub went from A to D and from D to C . \\ So, the distance from A to B =7 .5 km \\ Distance from B to C=12 .7 km \\ Distance from C to D=11 .8 km \\ Distance from A to D=9 .3 km \\ Distance covered by Dinesh=7 .5+12 .7 km=20 .2 km \\ Distance covered by Ayub = 9 .3+11 .8 km=21 .1 km \\ Hence, Ayun travelled 21 .1 km and 0 .9 km more than Dinesh . \end{array}$

Q.35

$\begin{array}{l} Shyama bought 5 kg 300 g apples and 3 kg 250 g \\ mangoes . Sarala bought 4 kg 800 g oranges and \\ 4 kg 150 g bananas . Who bought more fruit ? \end{array}$

Ans.

$\begin{array}{l} Apples bought by Shyama \\ =5 kg 300 g \\ =5 kg+ \frac{300}{1000} kg \\ =5 kg+ 0.3 kg \\ =5.3 kg \\ Mangoes bought by Shyama \\ = 3 kg 250 g \\ = 3 kg+ \frac{250}{1000} kg \\ = 3 kg+ 0.25 kg \\ =3.25 kg \\ Thus, total fruits bought by Shyama \\ = 5.3 kg+3.25 kg \\ = 8.55 kg \\ Oranges bought by Sarla \\ = 4 kg 800 g \\ = 4 kg+ \frac{800}{1000} kg \\ = 4 kg+ 0.8 kg \\ = 4.8 kg \end{array}$ $\begin{array}{l} Bananas bought by Sarla \\ = 4 kg 150 g \\ = 4 kg+ \frac{150}{1000} kg \\ = 4 kg+ 0 .15 kg \\ = 4 .15 kg \\ Thus, total fruits bought by Sarla \\ = 4 .8 kg+4 .15 kg \\ = 8 .95 kg \\ Thus, Sarla bought more fruits . \end{array}$

Q.36

$How much less is 28 km than 42.6 km ?$

Ans.

$\begin{array}{l} Since, \\ 42 . 6 km - 28 km = 14 . 6 km \\ Hence, 28 km is 14 . 6 km less than 42 . 6 km. \end{array}$

Q.37

$\begin{array}{l} F i n d : \\ (i) 0.2 \times 6 \\ (i i) 8 \times 4.6 \\ (i i i) 2.71 \times 5 \\ (i v) 20.1 \times 4 \\ (v) 0.05 \times 7 \\ (v i) 211.02 \times 4 \\ (v i i) 2 \times 0.86 \end{array}$

Ans.

$\begin{array}{l} (i) 0. 2 \times 6 = 1.2 \\ (ii) 8 \times 4.6 = 36.8 \\ (iii) 2.71 \times 5 = 13.55 \\ (iv) 2 0. 1 \times 4 = 80 .4 \\ (v) 0.0 5 \times 7 = 0 .35 \\ (vi) 211 .0 2 \times 4 = 844.08 \\ (vii) 2 \times 0. 86 = 1 .72 \end{array}$

Q.38

$\begin{array}{l} Find the are a of rectangle whose length is 5.7 cm \\ and breadth is 3 cm . \end{array}$

Ans.

$\begin{array}{l} The area of a rectangle is given by length \times breadth. \\ So, the area of given rectangle would be 5.7 cm \times 3 cm \\ = {17.1 cm}^{2} \end{array}$

Q.39

$\begin{array}{l} F i n d : \\ (i) 1.3 \times 10 (i i) 36.8 \times 10 (i i i) 153.7 \times 10 \\ (i v) 168.07 \times 10 (v) 31.1 \times 100 (v i) 156.1 \times 100 \\ (v i i) 3.62 \times 100 (v i i i) 43.07 \times 100 (i x) 0.5 \times 10 \\ (x) 0.08 \times 10 (x i) 0.9 \times 100 (x i i) 0.03 \times 1000 \end{array}$

Ans.

$\begin{array}{l} (i) 1.3 \times 10 = 13 \\ (ii) 36.8 \times 10 = 368 \\ (iii) 153.7 \times 10 = 1537 \\ (iv) 168 .0 7 \times 10 = 1680.7 \\ (v) 31.1 \times 100 = 3110 \\ (vi) 156.1 \times 100 = 15610 \\ (vii) 3.62 \times 100 = 362 \\ (viii) 43 .0 7 \times 100 = 4307 \\ (ix) 0. 5 \times 10 = 5 \\ (x) 0.0 8 \times 10 = 0.8 \\ (xi) 0. 9 \times 100 = 90 \\ (xii) 0.0 3 \times 1000 = 30 \end{array}$

Q.40

$\begin{array}{l} A two - wheeler covers a distance of 55.3 km in one \\ litre of petrol . How much distance will it cover in 10 litres \\ of petrol ? \end{array}$

Ans.

$\begin{array}{l} Distance covered by a two-wheeler in 1 litres \\ = 55.3 km \\ So, distance covered by a two-wheeler in 10 litres \\ =55.3 km \times 10 \\ = 553 km \end{array}$

Q.41

$\begin{array}{l} F i n d : \\ (i) 2.5 \times 0.3 (i i) 0.1 \times 51.7 (i i i) 0.2 \times 316.8 \\ (i v) 1.3 \times 3.1 (v) 0.5 \times 0.05 (v i) 11.2 \times 0.15 \\ (v i i) 1.07 \times 0.02 (v i i i) 10.05 \times 1.05 (i x) 101.01 \times 0.01 \\ (x) 100.01 \times 1.1 \end{array}$

Ans.

$\begin{array}{l} (i) 2.5 \times 0. 3 = 0 .75 \\ (ii) 0. 1 \times 51.7 = 5.17 \\ (iii) 0. 2 \times 316.8 = 63.36 \\ (iv) 1.3 \times 3 . 1= 4 .03 \\ (v) 0. 5 \times 0.0 5 = 0.025 \\ (vi) 11.2 \times 0. 15= 1 .68 \\ (vii) 1 .0 7 \times 0.0 2= 0 .0214 \\ (viii) 1 0.0 5 \times 1 .0 5= 10 .5525 \\ (ix) 101 .0 1 \times 0.0 1= 1 .0101 \\ (x) 1 00.0 1 \times 1 . 1= 110 .011 \end{array}$

Q.42

$\begin{array}{l} F i n d : \\ (i) 0.4 \div 2 (i i) 0.35 \div 5 (i i i) 2.48 \div 4 (i v) 65.4 \div 6 \\ (v) 651.2 \div 4 (v i) 14.49 \div 7 (v i i) 3.96 \div 4 (v i i i) 0.80 \div 5 \end{array}$

Ans.

$\begin{array}{l} (i) 0. 4 \div 2 = 0.2 \\ (ii) 0. 35 \div 5 = 0.07 \\ (iii) 2.48 \div 4= 0 .62 \\ (iv) 65.4 \div 6= 10 .9 \\ (v) 651.2 \div 4= 162 .8 \\ (vi) 14.49 \div 7 = 2.07 \\ (vii) 3.96 \div 4= 0 .99 \\ (viii) 0. 80 \div 5= 0 .16 \end{array}$

Q.43

$\begin{array}{l} F i n d : \\ (i) 4.8 \div 10 (i i) 52.5 \div 10 (i i i) 0.7 \div 10 (i v) 33.1 \div 10 \\ (v) 272.23 \div 10 (v i) 0.56 \div 10 (v i i) 3.97 \div 10 \end{array}$

Ans.

$\begin{array}{l} (i) 4.8 \div 10 = 0.48 \\ (ii) 52.5 \div 10 = 5.25 \\ (iii) 0. 7 \div 10 = 0.07 \\ (iv) 33.1 \div 10 = 3.31 \\ (v) 272.23 \div 10 = 27.223 \\ (vi) 0. 56 \div 10 = 0.056 \\ (vii) 3.97 \div 10 = 0.397 \end{array}$

Q.44

$\begin{array}{l} F i n d : \\ (i) 2.7 \div 100 (i i) 0.3 \div 100 (i i i) 0.78 \div 100 \\ (i v) 432.6 \div 100 (v) 23.6 \div 10 0 (v i) 98.53 \div 100 \end{array}$

Ans.

$\begin{array}{l} (i) 2.7 \div 100 = 0.027 \\ (ii) 0.3 \div 100 = 0.003 \\ (iii) 0. 78 \div 100 = 0.0078 \\ (iv) 432.6 \div 100 = 4.326 \\ (v) 23.6 \div 10 0 = 0.236 \\ (v i) 98.53 \div 100 = 0.9853 \end{array}$

Q.45

$\begin{array}{l} F i n d : \\ (i) 7.9 \div 1000 (i i) 26.3 \div 1000 (i i i) 38.53 \div 1000 \\ (i v) 128.9 \div 1000 (v) 0.5 \div 1000 \end{array}$

Ans.

$\begin{array}{l} (i) 7.9 \div 1000 = 0.0079 \\ (ii) 26.3 \div 1000 = 0.0263 \\ (iii) 38. 53 \div 1000 = 0.03853 \\ (iv) 128.9 \div 1000 = 0.1289 \\ (v) 0.5 \div 1000 = 0.0005 \end{array}$

Q.46

$\begin{array}{l} F i n d : \\ (i) 7 \div 3.5 (i i) 36 \div 0.2 (i i i) 3.25 \div 0.5 (i v) 30.94 \div 0.7 \\ (v) 0.5 \div 0.25 (v i) 7.75 \div 0.25 (v i i) 76.5 \div 0.15 \\ (v i i i) 37.8 \div 14 (i x) 2.73 \div 1.3 \end{array}$

Ans.

$\begin{array}{l} (i) 7 \div 3 .5= 2 \\ (ii) 36 \div 0 .2= 180 \\ (iii) 3.25 \div 0. 5= 6 .5 \\ (iv) 30.94 \div 0 .7= 44 .2 \\ (v) 0.5 \div 0 .25=2 \\ (vi) 7.75 \div 0 .25= 31 \\ (vii) 76.5 \div 0 .15= 510 \\ (viii) 37.8 \div 14= 2 .7 \\ (ix) 2 .73 \div 1.3 = 2.1 \end{array}$

Q.47

$\begin{array}{l} A vehicle covers a distance of 43.2 km in 2.4 litres of \\ petrol . How much distance will it cover in one litre of petrol ? \end{array}$

Ans.

$\begin{array}{l} Distance coverd by a vehicle in 2.4 litres \\ = 43.2 km \\ So, the distance vehicle will cover in 1 litre of petrol \\ = \frac{43 .2}{2 .4} km \\ = 18 km \end{array}$

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FAQs (Frequently Asked Questions)

1. What are the main points covered in Chapter 2 of NCERT Solutions for Class 7 Mathematics?

The main topics covered in Fractions and Decimals Class 7 are- addition and subtraction of decimals and fractions, multiplication of decimals and fractions, multiplication of fractions with whole numbers, multiplication of fractions with fractions, multiplication of decimal numbers by 10, 100, and 1000, division of fractions and decimals, division of whole numbers by fractions, division of a fraction by a whole number, reciprocal of fraction, division of a fraction by another fraction, and division of decimals by 10, 100, 1000. The chapter is well-structured, covering all the important sub-topics to help students prepare for their final exams.

2. Why should I refer to NCERT Solutions for Class 7 Mathematics Fractions and Decimals Chapter 2?

The benefits of NCERT Solutions for Class 7 Mathematics Chapter 2 provided by Extramarks are:

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NCERT Solutions Class 7 Mathematics Chapter 2

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

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NCERT Solutions for Class 7 Chapter 2 Mathematics

2.1 Introduction

2.2 Recollect

2.3 Multiplication of Fractions

2.3.1 Multiplication of Fractions Using the Whole Number

2.3.1 Multiplication of Fractions Using the Fraction

2.4 Division of Fractions

2.5 Recall of Decimals

2.6 Multiplication of Decimals

2.7 Division of Decimals

Key Features of NCERT Solutions for Class 7 Mathematics Chapter 2

NCERT Solutions for Class 7 Mathematics

NCERT Solutions for Class 7

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NCERT Solutions Class 7 Mathematics Chapter 2

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

Access NCERT Solutions for Class 7 Mathematics Chapter 2 – Fractions and Decimals

NCERT Solutions for Class 7 Chapter 2 Mathematics

2.1 Introduction

2.2 Recollect

2.3 Multiplication of Fractions

2.3.1 Multiplication of Fractions Using the Whole Number

2.3.1 Multiplication of Fractions Using the Fraction

2.4 Division of Fractions

2.5 Recall of Decimals

2.6 Multiplication of Decimals

2.7 Division of Decimals

Key Features of NCERT Solutions for Class 7 Mathematics Chapter 2

NCERT Solutions for Class 7 Mathematics

NCERT Solutions for Class 7

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1. What are the main points covered in Chapter 2 of NCERT Solutions for Class 7 Mathematics?

2. Why should I refer to NCERT Solutions for Class 7 Mathematics Fractions and Decimals Chapter 2?

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