NCERT Solutions Class 7 Mathematics Chapter 2

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

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?

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

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.

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

$\begin{array}{l}\text{Solve}:\\ \text{(i) 2-}\frac{3}{5}\text{(ii) 4+}\frac{7}{8}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(iii)}\frac{3}{5}+\frac{2}{7}\text{(iv)}\frac{9}{11}-\frac{4}{15}\\ \text{(v)}\frac{7}{10}\text{+}\frac{2}{5}\text{+}\frac{3}{2}\text{(vi) 2}\frac{2}{3}+3\frac{1}{2}\text{(vii) 8}\frac{1}{2}-3\frac{5}{8}\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}2-\frac{3}{5}=\frac{10-3}{5}=\overline{)\frac{7}{5}}\\ \left(ii\right)\text{\hspace{0.17em}}4+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=\overline{)4\frac{7}{8}}\\ \left(iii\right)\text{\hspace{0.17em}}\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\overline{)\frac{31}{35}}\\ \left(iv\right)\text{\hspace{0.17em}}\frac{9}{11}-\frac{4}{15}=\frac{135+44}{165}=\frac{179}{165}=\overline{)1\frac{14}{165}}\\ \left(v\right)\text{\hspace{0.17em}}\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}=\frac{26}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{13}{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\frac{3}{5}\\ \left(vi\right)\text{\hspace{0.17em}}2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16+21}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{37}{6}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6\frac{1}{6}\\ \left(vii\right)\text{\hspace{0.17em}}8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{68-29}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{39}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\frac{7}{8}\end{array}$

Q.2

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

Ans.

$\begin{array}{l}\text{First},\text{we find the LCM of 9},\text{3 and 21}.\\ \text{LCM of 9},\text{3 and 21}=\text{63}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2×7}{9×7},\text{\hspace{0.17em}}\frac{2×21}{3×21},\text{\hspace{0.17em}}\frac{8×3}{21×3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{14}{63},\text{\hspace{0.17em}}\frac{42}{63},\frac{24}{63}\\ \text{Since},\text{42}>\text{24}>\text{14}.\\ \text{Therefore},\\ \frac{42}{63}>\text{\hspace{0.17em}}\frac{24}{63}>\frac{14}{63}\\ ⇒\overline{)\frac{2}{3}>\frac{8}{21}>\frac{2}{9}}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{5},\text{\hspace{0.17em}}\frac{3}{7},\text{\hspace{0.17em}}\frac{7}{10}\\ \text{First},\text{find the LCM of 5},\text{7 and 1}0\\ \text{LCM of denominators 5},\text{7 and 1}0\text{is 7}0.\\ \text{So},\text{\hspace{0.17em}}\\ \frac{1}{5},\text{\hspace{0.17em}}\frac{3}{7},\text{\hspace{0.17em}}\frac{7}{10}\\ ⇒\frac{1×14}{5×14},\frac{3×10}{7×10},\frac{7×7}{10×7}\\ ⇒\frac{14}{70},\frac{30}{70},\frac{49}{70}\\ \text{Since},\text{49}>\text{3}0\text{}>\text{14}.\\ \text{Therefore},\text{\hspace{0.17em}}\\ \frac{49}{70}>\text{\hspace{0.17em}}\frac{30}{70}>\text{\hspace{0.17em}}\frac{14}{70}\\ ⇒\overline{)\frac{7}{10}>\frac{3}{7}>\frac{1}{5}}\end{array}$

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?

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

Ans.

$\begin{array}{l}\text{Finding the sum of first row}:\\ Sum\text{\hspace{0.17em}}of\text{\hspace{0.17em}}1st\text{\hspace{0.17em}}row=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}\\ =\frac{4+9+2}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of second row}:\\ \text{Sum of 2nd row}=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of third row}:\\ \text{Sum of 3rd row}=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}\\ =\frac{8+1+6}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of first column}:\\ \text{Sum of 1st column}=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}\\ =\frac{4+3+8}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of second column}:\\ \text{Sum of 2nd column}=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}\\ =\frac{9+5+1}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of third column}:\\ \text{Sum of 3rd column}=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}\\ =\frac{2+7+6}{11}\\ =\frac{15}{11}\\ \text{Now},\text{we must find diagonal sum}.\\ \text{Sum of}diagonal\text{\hspace{0.17em}}from\text{\hspace{0.17em}}left\text{\hspace{0.17em}}\text{​}bottom\text{\hspace{0.17em}}to\text{\hspace{0.17em}}right\text{\hspace{0.17em}}top\\ =\frac{8}{11}+\frac{5}{11}+\frac{2}{11}\\ =\frac{8+5+2}{11}\\ =\frac{15}{11}\\ \text{Sum of}diagonal\text{\hspace{0.17em}}from\text{\hspace{0.17em}}left\text{\hspace{0.17em}}\text{​}bottom\text{\hspace{0.17em}}to\text{\hspace{0.17em}}right\text{\hspace{0.17em}}top\\ =\frac{6}{11}+\frac{5}{11}+\frac{4}{11}\\ =\frac{6+5+4}{11}\\ =\frac{15}{11}\\ \text{Since the sum of the numbers in each row},\text{in each}\\ \text{column along the diagonal is the same},\text{which is equal}\\ \text{to}\text{\hspace{0.17em}}\frac{15}{11}.\\ \text{Therefore},\text{given square is magical}.\end{array}$

Q.4

$\begin{array}{l}\mathrm{A}\mathrm{rectangular}\mathrm{sheet}\mathrm{of}\mathrm{paper}\mathrm{is}\text{\hspace{0.17em}}12\frac{1}{2}\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{long}\mathrm{and}\\ 10\frac{2}{3}\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{wide}.\mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$

Ans.

$\begin{array}{l}\text{We are given}:\\ \text{Length}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{rectangular}\text{\hspace{0.17em}}\text{sheet}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{paper}=12\frac{1}{2}\text{cm}=\frac{25}{2}\text{cm}\\ \text{Width}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{rectangular}\text{\hspace{0.17em}}\text{sheet}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{paper=10}\frac{2}{3}\text{cm}=\frac{32}{3}\text{cm}\\ \text{Since},\text{\hspace{0.17em}}\overline{)\text{Perimeter of a rectangle}=\text{2}×\left(\text{length}+\text{width}\right)}\\ \text{Therefore},\\ \text{Perimeter of the given rectangular sheet of paper}\\ =2×\left(\frac{25}{2}+\frac{32}{3}\right)\text{cm}\\ =2×\left(\frac{75+64}{6}\right)\text{cm}\\ =2×\frac{139}{6}\text{cm}\\ =\frac{139}{3}\text{cm}=\overline{)46\frac{1}{3}\text{cm}}\end{array}$

Q.4

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

Ans.

$\begin{array}{l}\text{Since},\text{the Perimeter of a triangle}=\text{side}+\text{side}+\text{side}\\ \text{Therefore},\text{Perimeter of triangle ABE}=\text{AB}+\text{BE}+\text{AE}\\ =\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}\text{cm}\\ =\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\text{cm}\\ =\frac{50+55+72}{20}\text{cm}\\ =\frac{177}{20}\text{cm}\\ \text{In rectangle BCDE},\text{length BE}=2\frac{2}{4}\text{cm}=\frac{11}{4}\text{cm}\\ \text{Width ED}=\frac{7}{6}\text{cm}\\ \text{Since},\text{\hspace{0.17em}}\overline{)\text{Perimeter of a rectangle}=\text{2}×\left(\text{length}+\text{width}\right)}\\ \text{Therefore},\text{perimeter of rectangle BCDE}=2×\left(BE+ED\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×\left(\frac{11}{4}+\frac{7}{6}\right)\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=2×\left(\frac{33+14}{12}\right)\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=2×\frac{47}{12}\text{cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=\frac{47}{6}\text{cm}\\ \text{Hence},\text{the perimeter of}\text{\hspace{0.17em}}\Delta ABC=\overline{)\frac{177}{20}\text{cm}}\\ \text{and}\\ \text{The perimeter of rectangle BCDE}=\overline{)\frac{47}{6}\text{cm}}\\ \text{Now},\text{comparing both the perimeter to find which is}\\ \text{greater}.\\ \text{Since},\\ \text{The LCM of denominators of both the perimeter}\left(\text{2}0\text{and 6}\right)\text{is}\\ \text{equal to 6}0.\\ \text{Therefore},\\ \frac{177}{20},\text{\hspace{0.17em}}\frac{47}{6}⇒\frac{177×3}{20×3},\frac{47×10}{6×10}⇒\frac{531}{60},\frac{470}{10}\\ \text{Since},\text{531 is greater than 47}0.\\ \text{Therefore},\\ \frac{531}{60}>\frac{470}{60}⇒\frac{177}{20}>\frac{47}{6}\\ \text{Thus},\text{the perimeter of triangle ABE is greater than}\\ \text{the perimeter of rectangle BCDE}.\end{array}$

Q.5

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

Ans.

$\begin{array}{l}\text{We are given}:\\ \text{The}\text{\hspace{0.17em}}\text{width}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{picture=7}\frac{\text{3}}{\text{5}}\text{cm}\\ \text{Required}\text{\hspace{0.17em}}\text{width}\text{\hspace{0.17em}}\text{of picture to be fit in frame=7}\frac{3}{10}\text{cm}\\ \text{Therefore},\text{picture to be trim}=7\frac{3}{5}-7\frac{3}{10}\text{cm}\\ =\frac{38}{5}-\frac{73}{10}\text{cm}\\ =\frac{3}{10}\text{cm}\\ \text{Thus picture should be trimmed by}\text{\hspace{0.17em}}\overline{)\frac{3}{10}\text{cm}}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Ritu}\mathrm{ate}\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}\mathrm{part}\mathrm{of}\mathrm{an}\mathrm{apple}\mathrm{and}\mathrm{the}\mathrm{remaining}\mathrm{apple}\mathrm{was}\\ \mathrm{eaten}\mathrm{by}\mathrm{her}\mathrm{brother}\mathrm{Somu}.\mathrm{How}\mathrm{much}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{apple}\\ \mathrm{did}\mathrm{Somu}\mathrm{eat}?\mathrm{Who}\mathrm{had}\mathrm{the}\mathrm{larger}\mathrm{share}?\mathrm{By}\mathrm{how}\mathrm{much}?\end{array}$

Ans.

$\begin{array}{l}\text{Since},\text{Ritu ate\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}part of an apple and her brother eaten the}\\ \text{rest part of the apple}.\\ \text{So},\text{the part of apple left after eaten by Ritu}=1-\frac{3}{5}\\ =\frac{5-3}{5}\\ =\frac{2}{5}\\ \text{Therefore},\text{Somu ate\hspace{0.17em}}\frac{2}{5}\text{\hspace{0.17em}parts of apple}.\\ \text{Now comparing the part eaten by them\hspace{0.17em}}\frac{3}{5},\text{\hspace{0.17em}​}\frac{2}{5}\\ \text{Since},\text{3}>\text{2}.\\ \text{Therefore}\\ \frac{3}{5}>\frac{2}{5}\\ \text{That means Ritu ate larger part}.\\ \text{The difference in both parts}=\frac{3}{5}-\frac{2}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3-2}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{5}\\ \text{Thus,}\\ \text{Somu ate}\frac{2}{5}\text{\hspace{0.17em}}\text{part of apple}.\\ \text{Ritu ate the larger part of apple}\\ \text{Ritu ate}\text{part more apple than her brother Somu}.\end{array}$

Q.7

$\begin{array}{l}\mathrm{Michael}\mathrm{finished}\mathrm{colouring}\mathrm{a}\mathrm{picture}\mathrm{in}\text{\hspace{0.17em}}\frac{7}{12}\mathrm{hour}.\\ \mathrm{Vaibhav}\mathrm{finished}\mathrm{colouring}\mathrm{the}\mathrm{same}\mathrm{picture}\mathrm{in}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\mathrm{hour}.\\ \mathrm{Who}\mathrm{worked}\mathrm{longer}?\mathrm{By}\mathrm{what}\mathrm{fraction}\mathrm{was}\mathrm{it}\mathrm{longer}?\end{array}$

Ans.

$\begin{array}{l}\text{We are given}:\\ \text{Michael\hspace{0.17em}worked\hspace{0.17em}for\hspace{0.17em}}\frac{\text{7}}{\text{12}}\text{hour}\\ \mathrm{Vaibhav}\text{\hspace{0.17em}}\mathrm{worked}\text{\hspace{0.17em}}\mathrm{for}\text{\hspace{0.17em}}\frac{3}{4}\mathrm{hour}\phantom{\rule{0ex}{0ex}}\text{Therefore},\\ \text{In order to find the longer hour of work we have to compare}\\ \text{both the fractions\hspace{0.17em}}\frac{7}{12},\text{\hspace{0.17em}}\frac{3}{4}\\ \text{The LCM of 12 and 4}=\text{12}\\ \mathrm{So},\frac{7}{12},\frac{3}{4}=\frac{7}{12},\frac{9}{12}\\ \text{Therefore},\text{\hspace{0.17em}}\frac{9}{12}>\frac{7}{12}\\ \text{That means Vaibhav worked for longer hour}.\\ \text{Difference in their working hour\hspace{0.17em}\hspace{0.17em}}=\frac{9}{12}-\frac{7}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9-7}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\overline{)\frac{1}{6}}\\ \text{Thus},\text{Vaibhav worked for}\frac{1}{6}\text{\hspace{0.17em}hour more than Michael}.\end{array}$

Q.8

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{drawing}\left(\mathrm{a}\right)\mathrm{to}\left(\mathrm{d}\right)\mathrm{show}:\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}2×\frac{1}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}2×\frac{1}{2}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}3×\frac{2}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\text{\hspace{0.17em}}3×\frac{1}{4}\text{\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)\hspace{0.17em}}2×\frac{1}{5}=\frac{2}{5}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}2×\frac{1}{2}=1\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}3×\frac{2}{3}=2\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}3×\frac{1}{4}=\frac{3}{4}\\ \text{(a)\hspace{0.17em}Here},\text{there are three pictures and each pictures}\\ \text{show\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}part}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2+2+2}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{6}{3}=2\\ \text{Hence},\text{}\left(\text{iii}\right)\text{denotes picture}\left(\text{a}\right).\\ \text{(b)\hspace{0.17em}Here},\text{there are two pictures}.\text{Each pictures show}\frac{1}{2}\\ \text{part.}\\ \text{So},\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\\ \text{Hence},\text{}\left(\text{ii}\right)\text{denotes picture}\left(\text{b}\right).\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}Here},\text{there are three pictures}.\text{Each pictures show}\frac{1}{4}\text{\hspace{0.17em}}\mathrm{part}.\\ \text{Therefore},\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1+1+1}{4}=\frac{3}{4}\\ \text{Hence},\text{}\left(\text{iv}\right)\text{denotes picture}\left(\text{c}\right).\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}Here},\text{there are two pictures and each pictures denotes}\\ \frac{1}{5}\text{\hspace{0.17em}part.}\\ \text{Therefore},\frac{1}{5}+\frac{1}{5}=\frac{1+1}{5}=\frac{2}{5}\\ \text{Hence},\text{}\left(\text{i}\right)\text{denotes picture}\left(\text{d}\right).\\ \text{Thus}:\\ \overline{)\begin{array}{l}\left(\text{i}\right)\text{denotes picture}\left(\text{d}\right)\\ \left(\text{ii}\right)\text{denotes picture}\left(\text{b}\right)\\ \left(\text{iii}\right)\text{denotes picture}\left(\text{a}\right)\\ \left(\text{iv}\right)\text{denotes picture}\left(\text{c}\right)\end{array}}\end{array}$

Q.9

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

Ans.

$\begin{array}{l}\text{In picture}\left(\text{a}\right):\\ \text{There are two pictures in left hand side},\text{each denotes}\frac{1}{3}\\ \text{and picture in right hand side denotes}\text{\hspace{0.17em}}\frac{2}{3}.\\ \text{Therefore},\text{it denotes}\frac{1}{3}×2=\frac{2}{3}\text{which denotes}\left(\text{ii}\right).\\ \text{In picture}\left(\text{b}\right)\text{:}\\ \text{Here are three pictures in left hand side},\text{each shows}\frac{3}{4}.\\ \text{There are three pictures in right hand side in which each}\\ \text{of first two denotes 1}.\\ \text{Therefore},\frac{3}{4}×3=2\frac{1}{4}\\ \text{Thus picture}\left(\text{b}\right)\text{denotes}\left(\text{iii}\right).\\ \text{In picture}\left(\text{c}\right):\\ \text{There are three pictures in left hand side and each}\\ \text{of them denotes}\frac{1}{5}.\\ \text{Hence},\\ \overline{)\begin{array}{l}\left(\text{i}\right)\text{denotes picture}\left(\text{c}\right)\\ \left(\text{ii}\right)\text{denotes picture}\left(\text{a}\right)\\ \left(\text{iii}\right)\text{denotes picture}\left(\text{b}\right)\end{array}}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{reduce}\mathrm{to}\mathrm{lowest}\mathrm{form}:\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}7×\frac{3}{5}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}4×\frac{1}{3}\left(\mathrm{iii}\right)\text{\hspace{0.17em}}2×\frac{6}{7}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}5×\frac{2}{9}\left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{2}{3}×4\left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{5}{2}×6\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}11×\frac{4}{7}\left(\mathrm{viii}\right)\text{\hspace{0.17em}}20×\frac{4}{5}\left(\mathrm{ix}\right)\text{\hspace{0.17em}}13×\frac{1}{3}\\ \left(\mathrm{x}\right)\text{\hspace{0.17em}}15×\frac{3}{5}\end{array}$

Ans.

$\begin{array}{c}\left(\mathrm{i}\right)\text{\hspace{0.17em}}7×\frac{3}{5}=\frac{7}{1}×\frac{3}{5}=\frac{21}{5}\\ =\overline{)4\frac{1}{5}}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}4×\frac{1}{3}=\frac{4}{1}×\frac{1}{3}=\frac{4}{3}\\ =\overline{)1\frac{1}{3}}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}2×\frac{6}{7}=\frac{2}{1}×\frac{6}{7}=\frac{12}{7}\\ =\overline{)1\frac{5}{7}}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}5×\frac{2}{9}=\frac{5}{1}×\frac{2}{9}=\frac{10}{9}\\ =\overline{)1\frac{1}{9}}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{2}{3}×4=\frac{2}{3}×\frac{4}{1}=\frac{8}{3}\\ =\overline{)2\frac{2}{3}}\\ \left(\mathrm{vi}\right)\text{​\hspace{0.17em}}\frac{5}{2}×6=\frac{5×6}{2}=\frac{30}{2}\\ =\overline{)15}\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}11×\frac{4}{7}=\frac{11}{1}×\frac{4}{7}=\frac{44}{7}\\ =\overline{)6\frac{2}{7}}\\ \left(\mathrm{viii}\right)\text{\hspace{0.17em}}20×\frac{4}{7}=\frac{20}{1}×\frac{4}{5}=\frac{80}{5}\\ =\overline{)16}\\ \left(\mathrm{ix}\right)\text{\hspace{0.17em}}13×\frac{1}{3}=\frac{13}{1}×\frac{1}{3}=\frac{13}{3}\\ =\overline{)4\frac{1}{3}}\\ \left(\mathrm{x}\right)\text{\hspace{0.17em}}15×\frac{3}{5}=\frac{15}{1}×\frac{3}{5}=\frac{45}{5}\\ =\overline{)9}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Shade}:\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{1}{2}\text{}\mathrm{of}\mathrm{the}\mathrm{circles}\mathrm{in}\mathrm{box}\left(\mathrm{a}\right)\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}\mathrm{of}\mathrm{the}\mathrm{triangles}\mathrm{in}\mathrm{box}\left(\mathrm{b}\right)\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}\mathrm{of}\mathrm{the}\mathrm{squares}\mathrm{in}\mathrm{box}\left(\mathrm{c}\right)\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\\ \text{Here},\text{there are 12 circles in box}\left(\text{a}\right).\\ \text{Therefore},\\ \frac{1}{2}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}12=\frac{1}{2}×12=6\\ \text{Hence 6 circles to be coloured in}\left(\text{a}\right).\\ \left(\mathrm{ii}\right)\\ \text{Here},\text{there are 9 triangles in box}\left(\text{b}\right)\text{and\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}of the}\\ \text{triangles are to be coloured}.\\ \text{So},\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}9=\frac{2}{3}×9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{18}{3}=6\\ \text{Thus},\text{6 triangles are needed to be coloured}.\\ \left(\text{iii}\right)\\ \text{Here},\text{there are 15 rectangles in box}\left(\text{c}\right)\text{and}\frac{3}{5}\text{of them}\\ \text{are needed to be coloured}.\\ \text{So},\\ \frac{3}{5}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}​}15=\frac{3}{5}×15\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{45}{5}=9\\ \text{Thus},\text{9 rectangles are needed to be coloured}.\end{array}$

Q.12

$\begin{array}{l}\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{d}:\\ \left(\mathbf{a}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathbf{i}\right)\text{}\mathbf{2}\mathbf{4}\left(\mathbf{i}\mathbf{i}\right)\text{}\mathbf{4}\mathbf{6}\\ \left(\mathbf{b}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathbf{i}\right)\text{}\mathbf{1}\mathbf{8}\left(\mathbf{i}\mathbf{i}\right)\text{}\mathbf{2}\mathbf{7}\\ \left(\mathbf{c}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\text{of}\left(\mathbf{i}\right)\text{}\mathbf{1}\mathbf{6}\text{\hspace{0.17em}}\left(\mathbf{i}\mathbf{i}\right)\text{}\mathbf{3}\mathbf{6}\\ \left(\mathbf{d}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{4}{5}\text{\hspace{0.17em}}\text{of}\left(\mathbf{i}\right)\text{}\mathbf{2}\mathbf{0}\text{\hspace{0.17em}}\left(\mathbf{i}\mathbf{i}\right)\text{}\mathbf{3}\mathbf{5}\end{array}$

Ans.

$\begin{array}{l}\left(a\right)\\ \left(i\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}24=\frac{1}{2}×24=\frac{24}{2}=\overline{)12}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}of46=\frac{1}{2}×46=\frac{46}{2}=\overline{)23}\\ \left(b\right)\\ \left(i\right)\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}18=\frac{2}{3}×18=\frac{36}{3}=\overline{)12}\\ \left(ii\right)\frac{2}{3}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}27=\frac{2}{3}×27=\frac{54}{3}=\overline{)18}\text{\hspace{0.17em}}\\ \left(c\right)\\ \left(i\right)\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}16=\frac{3}{4}×16=\frac{48}{4}=\overline{)12}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}36=\frac{3}{4}×36=\frac{108}{4}=\overline{)27}\\ \left(d\right)\\ \left(i\right)\text{\hspace{0.17em}}\frac{4}{5}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}20=\frac{4}{5}×20=\frac{80}{5}=\overline{)16}\end{array}$

Q.13

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{express}\mathrm{as}\mathrm{a}\mathrm{mixed}\mathrm{fraction}:\\ \left(a\right)\text{\hspace{0.17em}}3×5\frac{1}{5}\left(b\right)\text{​}\text{\hspace{0.17em}}5×6\frac{3}{4}\left(c\right)\text{\hspace{0.17em}}7×2\frac{1}{4}\\ \left(d\right)\text{\hspace{0.17em}}4×6\frac{1}{3}\left(e\right)\text{​}\text{\hspace{0.17em}}3\frac{1}{4}×6\left(f\right)\text{\hspace{0.17em}}3\frac{2}{5}×8\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}3×5\frac{1}{5}=3×\frac{26}{5}=\frac{78}{5}=\overline{)15\frac{3}{5}}\\ \left(\text{b}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×6\frac{3}{4}=5×\frac{27}{4}=\frac{135}{4}=\overline{)33\frac{3}{4}}\\ \left(\text{c}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}7×2\frac{1}{4}=7×\frac{9}{4}=\frac{63}{4}=\overline{)15\frac{3}{4}}\\ \left(\text{d}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}4×6\frac{1}{3}=4×\frac{19}{3}=\frac{76}{3}=\overline{)25\frac{1}{3}}\\ \left(\text{e}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\frac{1}{4}×6=\frac{13}{4}×6=\frac{78}{7}=\overline{)19\frac{1}{2}}\\ \left(\text{f}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\frac{2}{5}×8=\frac{17}{5}×8=\frac{136}{5}=\overline{)27\frac{1}{5}}\end{array}$

Q.14

$\begin{array}{l}Find\\ \left(a\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}\mathrm{of}\left(\mathrm{i}\right)2\frac{3}{4}\text{\hspace{0.17em}}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}4\frac{2}{9}\\ \left(b\right)\text{\hspace{0.17em}}\frac{5}{8}\text{\hspace{0.17em}}\mathrm{of}\left(\mathrm{i}\right)\text{\hspace{0.17em}}3\frac{5}{6}\text{\hspace{0.17em}}\left(\mathrm{ii}\right)\text{​}9\frac{2}{3}\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}2\frac{3}{4}=\frac{1}{2}×\frac{11}{4}=\frac{11}{8}\text{\hspace{0.17em}}=\overline{)1\frac{3}{8}}\\ \left(\text{ii}\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}4\frac{2}{9}=\frac{1}{2}×\frac{38}{9}=\frac{19}{9}=\overline{)2\frac{1}{9}}\\ \left(\text{b}\right)\\ \left(\text{i}\right)\text{\hspace{0.17em}}\frac{5}{8}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}3\frac{5}{6}=\frac{5}{8}×\frac{23}{6}=\frac{115}{48}\text{\hspace{0.17em}}=\overline{)2\frac{19}{48}}\\ \left(\text{ii}\right)\text{\hspace{0.17em}}\frac{5}{8}\text{\hspace{0.17em}}\text{of}\text{​}\text{\hspace{0.17em}}9\frac{2}{3}=\frac{5}{8}×\frac{29}{3}=\frac{145}{24}=\overline{)6\frac{1}{24}}\end{array}$

Q.15

$\begin{array}{l}\mathrm{Vidya}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{went}\mathrm{}\mathrm{for}\mathrm{}\mathrm{a}\mathrm{}\mathrm{picnic}.\mathrm{}\mathrm{Their}\mathrm{}\mathrm{mother}\\ \mathrm{gave}\mathrm{}\mathrm{them}\mathrm{}\mathrm{a}\mathrm{}\mathrm{water}\mathrm{}\mathrm{bag}\mathrm{}\mathrm{that}\mathrm{}\mathrm{contained}\mathrm{}5\mathrm{}\mathrm{liters}\mathrm{}\mathrm{of}\\ \mathrm{water}.\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{consumed}\frac{2}{5}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{water}.\mathrm{}\mathrm{Pratap}\\ \mathrm{consumed}\mathrm{}\mathrm{the}\mathrm{}\mathrm{remaining}\mathrm{}\mathrm{water}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\mathrm{}\mathrm{How}\mathrm{}\mathrm{much}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{drink}?\\ \phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{}\mathrm{What}\mathrm{}\mathrm{fraction}\mathrm{}\mathrm{of}\mathrm{}\mathrm{total}\\ \mathrm{quantity}\mathrm{}\mathrm{of}\mathrm{}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{drink}?\end{array}$

Ans.

$\begin{array}{l}\mathrm{Vidya}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{went}\mathrm{}\mathrm{for}\mathrm{}\mathrm{a}\mathrm{}\mathrm{picnic}.\mathrm{}\mathrm{Their}\mathrm{}\mathrm{mother}\\ \mathrm{gave}\mathrm{}\mathrm{them}\mathrm{}\mathrm{a}\mathrm{}\mathrm{water}\mathrm{}\mathrm{bag}\mathrm{}\mathrm{that}\mathrm{}\mathrm{contained}\mathrm{}5\mathrm{}\mathrm{liters}\mathrm{}\mathrm{of}\\ \mathrm{water}.\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{consumed}\frac{2}{5}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{water}.\mathrm{}\mathrm{Pratap}\\ \mathrm{consumed}\mathrm{}\mathrm{the}\mathrm{}\mathrm{remaining}\mathrm{}\mathrm{water}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\mathrm{}\mathrm{How}\mathrm{}\mathrm{much}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{drink}?\\ \phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{}\mathrm{What}\mathrm{}\mathrm{fraction}\mathrm{}\mathrm{of}\mathrm{}\mathrm{total}\\ \mathrm{quantity}\mathrm{}\mathrm{of}\mathrm{}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{drink}?\end{array}$

Q.16

$\begin{array}{l}Find:\\ \left(i\right)\text{\hspace{0.17em}}\frac{1}{4}of\text{}\left(a\right)\frac{1}{4}\text{}\text{}\left(b\right)\text{\hspace{0.17em}}\frac{3}{5}\text{}\left(c\right)\text{\hspace{0.17em}}\frac{4}{3}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}of\left(a\right)\text{\hspace{0.17em}}\frac{2}{9}\text{}\left(b\right)\text{​}\text{\hspace{0.17em}}\frac{6}{5}\text{\hspace{0.17em}}\text{}\left(c\right)\text{​}\frac{3}{10}\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\\ \left(a\right)\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{of}\frac{1}{4}=\frac{1}{4}×\frac{1}{4}=\overline{)\frac{1}{16}}\\ \left(b\right)\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{5}=\frac{1}{4}×\frac{3}{5}=\overline{)\frac{3}{20}}\\ \left(c\right)\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{4}{3}=\frac{1}{\overline{)4}}×\frac{\overline{)4}}{3}=\overline{)\frac{1}{3}}\\ \left(ii\right)\\ \left(a\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{2}{9}=\frac{1}{7}×\frac{2}{9}=\overline{)\frac{2}{63}}\\ \left(b\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{5}=\frac{1}{7}×\frac{6}{5}=\overline{)\frac{6}{35}}\\ \left(c\right)\text{\hspace{0.17em}}\frac{1}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{10}=\frac{1}{7}×\frac{3}{10}=\overline{)\frac{3}{70}}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{reduce}\mathrm{to}\mathrm{lowest}\mathrm{form}\left(\mathrm{if}\mathrm{possible}\right).\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{2}{3}×2\frac{2}{3}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{2}{7}×\frac{7}{9}\left(\mathrm{iii}\right)\text{\hspace{0.17em}}\frac{3}{8}×\frac{6}{4}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}\frac{9}{5}×\frac{3}{5}\left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{1}{3}×\frac{15}{8}\left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{11}{2}×\frac{3}{10}\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}\frac{4}{5}×\frac{12}{7}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{2}{3}×2\frac{2}{3}=\frac{2}{3}×\frac{8}{3}=\overline{)\frac{16}{9}}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{2}{7}×\frac{7}{9}=\frac{2}{\overline{)7}}×\frac{\overline{)7}}{9}=\overline{)\frac{2}{9}}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\frac{3}{8}×\frac{6}{4}=\frac{3}{\overline{)2}×4}×\frac{\overline{)2}×3}{4}=\overline{)\frac{9}{16}}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}\frac{9}{5}×\frac{3}{5}=\overline{)\frac{27}{25}}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{1}{3}×\frac{15}{8}=\frac{1}{\overline{)3}}×\frac{\overline{)3}×5}{8}=\overline{)\frac{5}{8}}\\ \left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{11}{2}×\frac{3}{10}=\overline{)\frac{33}{20}}\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}\frac{4}{5}×\frac{12}{7}=\overline{)\frac{48}{35}}\end{array}$

Q.18

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

Ans.

$\begin{array}{l}\left(\text{i}\right)\\ \left(\text{a}\right)\frac{2}{5}×5\frac{1}{4}=\frac{2}{5}×\frac{21}{4}=\frac{\overline{)2}}{5}×\frac{21}{\overline{)2}×2}=\overline{)\frac{21}{10}}\\ \left(\text{b}\right)\text{It is an improper fraction}.\\ \left(\text{c}\right)\frac{21}{10}=\overline{)2\frac{1}{10}}\\ \left(\mathrm{ii}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}6\frac{2}{5}×\frac{7}{9}=\frac{32}{5}×\frac{7}{9}=\overline{)\frac{224}{45}}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\text{It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{224}{45}=\overline{)4\frac{44}{45}}\\ \left(\mathrm{iii}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}\frac{3}{2}×5\frac{1}{3}=\frac{3}{2}×\frac{16}{3}=\frac{\overline{)3}}{\overline{)2}}×\frac{\overline{)2}×8}{\overline{)3}}=\frac{8}{1}=\overline{)8}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It\hspace{0.17em}is whole number.}\\ \text{(iv)}\\ \text{(a)\hspace{0.17em}}\frac{5}{6}×2\frac{3}{7}\\ =\frac{5}{6}×\frac{17}{7}\\ =\overline{)\frac{85}{42}}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{85}{42}=\overline{)2\frac{1}{42}}\\ \left(\mathrm{v}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}​}3\frac{2}{5}×\frac{4}{7}\\ =\frac{17}{5}×\frac{4}{7}\\ =\frac{68}{35}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{68}{35}=\overline{)1\frac{33}{35}}\\ \left(\mathrm{vi}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}2\frac{3}{5}×3\\ =\frac{13}{5}×\frac{3}{1}\\ =\frac{39}{5}\\ \left(\text{b}\right)\text{It is an improper fraction}.\\ \left(\text{c}\right)\text{\hspace{0.17em}}\frac{39}{5}=\overline{)7\frac{4}{5}}\\ \left(\mathrm{vii}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}3\frac{4}{7}×\frac{3}{5}\\ =\frac{25}{7}×\frac{3}{5}\\ =\frac{5×\overline{)5}}{7}×\frac{3}{\overline{)5}}\\ =\frac{15}{7}\\ \text{(b)\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\frac{15}{7}=\overline{)2\frac{1}{7}}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Which}\mathrm{is}\mathrm{greater}:\\ \left(i\right)\text{\hspace{0.17em}}\frac{2}{7}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{5}{8}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{6}{7}\text{\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\frac{3}{7}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}\\ \frac{2}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{5}{8}\\ \frac{2}{7}×\frac{3}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\frac{3}{5}×\frac{5}{8}\\ =\frac{\overline{)2}}{7}×\frac{3}{\overline{)2}×2}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{\overline{)5}}×\frac{\overline{)5}}{8}\\ =\frac{3}{14}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{8}\\ \text{Now},\text{the LCM of 14 and 8 is 56}.\text{So},\text{we get}\\ \frac{3}{14}=\frac{3×4}{14×4}=\frac{12}{56}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{3}{8}=\frac{3×7}{8×7}=\frac{21}{56}\\ \text{Since},\text{21}>\text{12},\text{so}\text{\hspace{0.17em}}\frac{3}{8}>\frac{3}{14}\\ \text{Hence},\text{\hspace{0.17em}}\overline{)\frac{3}{5}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{5}{8}>\frac{2}{7}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{4}}\\ \left(\text{ii}\right)\\ \frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{7}\text{\hspace{0.17em}}or\text{​}\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{7}\\ \frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{7}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{7}\\ =\frac{1}{\overline{)2}}×\frac{\overline{)2}×3}{7}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{2}{\overline{)3}}×\frac{\overline{)3}}{7}\\ =\frac{3}{7}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\frac{2}{7}\\ \text{Since},\text{3}>\text{2},\text{so},\frac{3}{7}>\frac{2}{7}\\ \text{Hence},\text{\hspace{0.17em}}\overline{)\frac{1}{2}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{6}{7}>\frac{2}{3}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\frac{3}{7}}\end{array}$

Q.20

$\mathrm{Saili}\text{}\mathrm{plants}\text{}4\text{}\mathrm{saplings},\text{}\mathrm{in}\text{}\mathrm{a}\text{}\mathrm{row},\text{}\mathrm{in}\text{}\mathrm{her}\text{}\mathrm{garden}.\text{}\mathrm{The}\text{}\mathrm{distance}\text{}\mathrm{between}\text{}\mathrm{two}\text{}\mathrm{adjacent}\mathrm{}\mathrm{saplings}\mathrm{}\mathrm{is}\text{}\frac{3}{4}\text{}m.\phantom{\rule{0ex}{0ex}}\text{\hspace{0.17em}}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{distance}\text{}\mathrm{between}\text{}\mathrm{the}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{the}\text{}\mathrm{last}\text{}\mathrm{sapling}.$

Ans.

$\begin{array}{l}\text{Here},\text{the distance between the two saplings is given as}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\text{m}\\ \text{So},\text{the distance between first and the last sapling is}\\ \frac{3}{4}\text{\hspace{0.17em}}\text{m}×\text{3=}\frac{9}{4}\text{m}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\overline{)\text{2}\frac{1}{4}\text{m}}\end{array}$

Q.21

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

Ans.

$\begin{array}{l}\text{Lipika reads}\text{\hspace{0.17em}}1\frac{3}{4}=\frac{7}{4}\text{hours a day}.\\ \text{So},\text{Number of hours required by her to read the book}\\ \text{in 6 days is}\text{​}\frac{7}{4}\text{hour}×6=\frac{7}{2×\overline{)2}}\text{hour}×\left(\overline{)2}×3\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{21}{2}\text{hour}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\overline{)\text{10}\frac{1}{2}\text{\hspace{0.17em}}\text{hour}}\end{array}$

Q.22

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

Ans.

$\begin{array}{c}\text{In 1 litre of petrol},\text{car runs 16 km}.\\ \text{So},\text{in}\text{\hspace{0.17em}}2\frac{3}{4}\text{litres of petrol},\text{distance covered by car is}\text{\hspace{0.17em}}\\ 2\frac{3}{4}×16\text{\hspace{0.17em}}\text{km}\\ =\frac{11}{\overline{)4}}×\left(\overline{)4}×4\right)\text{\hspace{0.17em}}\text{km}\\ =\overline{)\text{44}\text{\hspace{0.17em}}\text{km}}\end{array}$

Q.23

$\begin{array}{l}\left(\mathrm{a}\right)\\ \left(\mathrm{i}\right)\text{​}\mathrm{Provide}\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{the}\mathrm{box}\text{\hspace{0.17em}​}\overline{)}\text{\hspace{0.17em}},\mathrm{such}\mathrm{that}\text{\hspace{0.17em}}\frac{2}{3}\text{}×\overline{)}=\frac{10}{30}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{The}\mathrm{simplest}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{obtained}\mathrm{in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}____\\ \left(\mathrm{b}\right)\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{Provide}\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{the}\mathrm{box}\text{\hspace{0.17em}}\overline{)},\mathrm{such}\mathrm{that}\frac{3}{5}×\overline{)}=\frac{24}{75}\\ \left(\mathrm{ii}\right)\\ \mathrm{The}\mathrm{simplest}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{obtained}\mathrm{in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}____\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\text{\hspace{0.17em}}\frac{\text{2}}{\text{3}}×\overline{)}\text{=}\frac{\text{10}}{\text{30}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{2×5}}{\text{3×10}}=\frac{\text{2}}{\text{3}}\text{×}\frac{\text{5}}{\text{10}}\\ \text{So,}\text{\hspace{0.17em}}\frac{\text{2}}{\text{3}}×\frac{\text{5}}{\text{10}}=\frac{\text{10}}{\text{30}}\\ \text{Therefore, the number in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\frac{\text{5}}{\text{10}}\text{.}\\ \text{(ii)}\\ \text{The simplest form of}\text{\hspace{0.17em}}\frac{\text{5}}{\text{10}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\frac{\text{5}}{\text{10}}=\frac{}{}=\overline{)\frac{\text{1}}{\text{2}}}\\ \text{(b)}\\ \left(\text{i}\right)\text{\hspace{0.17em}}\frac{\text{3}}{\text{5}}×\overline{)}\text{=}\frac{\text{24}}{\text{75}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{3×8}}{\text{5×15}}=\frac{\text{3}}{\text{5}}\text{×}\frac{\text{8}}{\text{15}}\\ \text{So,}\text{\hspace{0.17em}}\frac{\text{3}}{\text{5}}\text{×}\frac{\text{8}}{\text{15}}=\frac{\text{24}}{\text{75}}\\ \text{Therefore, the number in}\text{\hspace{0.17em}}\overline{)}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\frac{\text{8}}{\text{15}}\text{.}\\ \text{(ii)}\\ \text{The simplest form of}\text{\hspace{0.17em}}\frac{\text{8}}{\text{15}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\overline{)\frac{\text{8}}{\text{15}}}\end{array}$

Q.24

$\begin{array}{l}\text{Find:}\\ \text{(i)}\text{\hspace{0.17em}}\text{12÷}\frac{\text{3}}{\text{4}}\text{(ii)}\text{14÷}\frac{\text{5}}{\text{6}}\text{(iii)}\text{\hspace{0.17em}}\text{8÷}\frac{\text{7}}{\text{3}}\\ \text{(iv)}\text{​}\text{\hspace{0.17em}}\text{4÷}\frac{\text{8}}{\text{3}}\text{(v)}\text{\hspace{0.17em}}\text{3÷2}\frac{\text{1}}{\text{3}}\text{(vi)}\text{\hspace{0.17em}}\text{5÷3}\frac{\text{4}}{\text{7}}\end{array}$

Ans.

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

Q.25

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{reciprocal}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{fractions}.\mathrm{Classify}\mathrm{the}\mathrm{reciprocals}\mathrm{as}\mathrm{proper}\\ \mathrm{fractions},\mathrm{improper}\mathrm{fractions}\mathrm{and}\mathrm{whole}\mathrm{numbers}.\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{3}{7}\text{}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{5}{8}\text{\hspace{0.17em}}\left(\mathrm{iii}\right)\text{\hspace{0.17em}}\frac{9}{7}\text{\hspace{0.17em}}\left(\mathrm{iv}\right)\text{​\hspace{0.17em}}\frac{6}{5}\\ \left(\mathrm{v}\right)\text{​\hspace{0.17em}}\frac{12}{7}\left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{1}{8}\text{\hspace{0.17em}}\left(\mathrm{vii}\right)\text{\hspace{0.17em}}\frac{1}{11}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}\frac{3}{7}\\ \text{Reciprocal of\hspace{0.17em}}\frac{3}{7}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\frac{7}{3}\text{\hspace{0.17em}}\text{and it is an improper fraction}.\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{5}{8}\\ \text{Reciprocal of\hspace{0.17em}}\frac{5}{8}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\frac{8}{5}\text{\hspace{0.17em}}\text{and it is an improper fraction}\\ \text{(iii)\hspace{0.17em}}\frac{9}{7}\\ \text{Reciprocal of\hspace{0.17em}}\frac{9}{7}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\frac{7}{9}\text{\hspace{0.17em}}\text{and it is an proper fraction}\\ \left(\mathrm{iv}\right)\text{​\hspace{0.17em}}\frac{6}{5}\\ \text{Reciprocal of\hspace{0.17em}}\frac{6}{5}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\frac{5}{6}\text{\hspace{0.17em}}\text{and it is an proper fraction}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}}\frac{12}{7}\\ \text{Reciprocal of\hspace{0.17em}}\frac{12}{7}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\frac{7}{12}\text{\hspace{0.17em}}\text{and it is an proper fraction}\\ \left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{1}{8}\\ \text{Reciprocal of\hspace{0.17em}}\frac{1}{8}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\frac{8}{1}\text{\hspace{0.17em}}\text{and it is an whole fraction}\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}\frac{1}{11}\\ \text{Reciprocal of\hspace{0.17em}}\frac{1}{11}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}11\text{\hspace{0.17em}}\text{and it is a\hspace{0.17em}whole number.}\end{array}$

Q.26

$\begin{array}{l}Find:\\ \left(i\right)\text{\hspace{0.17em}}\frac{7}{3}÷2\left(ii\right)\text{\hspace{0.17em}}\frac{4}{9}÷5\left(iii\right)\text{\hspace{0.17em}}\frac{6}{13}÷7\\ \left(iv\right)\text{\hspace{0.17em}}4\frac{1}{3}÷3\left(v\right)\text{\hspace{0.17em}}3\frac{1}{2}÷4\left(vi\right)\text{\hspace{0.17em}}4\frac{3}{7}÷7\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\frac{7}{3}÷2=\frac{7}{3}×\frac{1}{2}=\frac{7}{6}=\overline{)1\frac{1}{6}}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{4}{9}÷5=\frac{4}{9}×\frac{1}{5}=\overline{)\frac{4}{45}}\\ \left(iii\right)\text{\hspace{0.17em}}\frac{6}{13}÷7=\frac{6}{13}×\frac{1}{7}=\overline{)\frac{6}{91}}\\ \left(iv\right)\text{\hspace{0.17em}}4\frac{1}{3}÷3=\frac{13}{3}×\frac{1}{3}=\frac{13}{9}=\overline{)1\frac{4}{9}}\\ \left(v\right)\text{\hspace{0.17em}}3\frac{1}{2}÷4=\frac{7}{2}×\frac{1}{4}=\overline{)\frac{7}{8}}\\ \left(vi\right)\text{\hspace{0.17em}}4\frac{3}{7}÷7=\frac{31}{7}×\frac{1}{7}=\overline{)\frac{31}{49}}\end{array}$

Q.27

$\begin{array}{l}\text{Find}\\ \left(i\right)\text{\hspace{0.17em}}\frac{2}{5}÷\frac{1}{2}\text{}\\ \left(ii\right)\text{​}\text{\hspace{0.17em}}\frac{4}{9}÷\frac{2}{3}\text{}\\ \left(iii\right)\text{\hspace{0.17em}}\frac{3}{7}÷\frac{8}{7}\\ \left(iv\right)\text{\hspace{0.17em}}2\frac{1}{3}÷\frac{3}{5}\\ \left(v\right)\text{\hspace{0.17em}}3\frac{1}{2}÷\frac{8}{3}\\ \left(vi\right)\text{\hspace{0.17em}}\frac{2}{5}÷1\frac{1}{2}\\ \left(vii\right)\text{\hspace{0.17em}}3\frac{1}{5}÷1\frac{2}{3}\\ \left(viii\right)\text{\hspace{0.17em}}2\frac{1}{5}÷1\frac{1}{5}\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\frac{2}{5}÷\frac{1}{2}=\frac{2}{5}×\frac{2}{1}=\overline{)\frac{4}{5}}\\ \left(ii\right)\text{\hspace{0.17em}}\frac{4}{9}÷\frac{2}{3}=\frac{4}{9}×\frac{3}{2}=\frac{\left(2×\overline{)2}\right)}{\left(3×\overline{)3}\right)}×\frac{\overline{)3}}{\overline{)2}}=\overline{)\frac{2}{3}}\\ \left(iii\right)\text{\hspace{0.17em}}\frac{3}{7}÷\frac{8}{7}=\frac{3}{\overline{)7}}×\frac{\overline{)7}}{8}=\overline{)\frac{3}{8}}\\ \left(iv\right)\text{\hspace{0.17em}}2\frac{1}{3}÷\frac{3}{5}=\frac{7}{3}×\frac{5}{3}=\frac{35}{9}=\overline{)3\frac{8}{9}}\\ \left(v\right)\text{\hspace{0.17em}}3\frac{1}{2}÷\frac{8}{3}=\frac{7}{2}×\frac{3}{8}=\frac{21}{16}=\overline{)1\frac{5}{16}}\\ \left(vi\right)\text{\hspace{0.17em}}\frac{2}{5}÷1\frac{1}{2}=\frac{2}{5}÷\frac{3}{2}=\frac{2}{5}×\frac{2}{3}=\overline{)\frac{4}{15}}\\ \left(vii\right)\text{\hspace{0.17em}}3\frac{1}{5}÷1\frac{2}{3}=\frac{16}{5}÷\frac{5}{3}=\frac{16}{5}×\frac{3}{5}=\frac{48}{25}=\overline{)1\frac{23}{25}}\\ \left(viii\right)\text{\hspace{0.17em}}2\frac{1}{5}÷1\frac{1}{5}=\frac{11}{5}÷\frac{6}{5}=\frac{11}{\overline{)5}}×\frac{\overline{)5}}{6}=\frac{11}{6}=\overline{)1\frac{5}{6}}\end{array}$

Q.28

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

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{}0.\text{5 or}0.0\text{5}\\ \overline{)0.5>0.05}\\ \left(\text{ii}\right)\text{}0.\text{7 or}0.\text{5}\\ \overline{)0.7>0.5}\\ \left(\text{iii}\right)\text{7 or}0.\text{7}\\ \overline{)7>0.7}\\ \left(\text{iv}\right)\text{1}.\text{37 or 1}.\text{49}\\ \overline{)1.49>1.37}\\ \left(\text{v}\right)\text{2}.0\text{3 or 2}.\text{3}0\\ \overline{)2.30>2.03}\\ \left(vi\right)\text{\hspace{0.17em}}0.8\text{\hspace{0.17em}}\text{or}\text{​}\text{\hspace{0.17em}}0.88\\ \overline{)0.88>0.8}\end{array}$

Q.29

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

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{7 paise}\\ \text{Since},1\text{\hspace{0.17em}}\text{paise}=\text{\hspace{0.17em}}₹\frac{1}{100}\text{\hspace{0.17em}}\\ \text{So},\text{\hspace{0.17em}}\text{7}\text{\hspace{0.17em}}\text{paise}\text{\hspace{0.17em}}\text{=7×}\text{\hspace{0.17em}}₹\frac{\text{1}}{\text{100}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}₹\text{}\frac{\text{7}}{\text{100}}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\overline{)₹\text{0}\text{.07}}\text{\hspace{0.17em}}\\ \left(\text{ii}\right)\text{7rupees 7 paise}\\ \text{Since},\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{paise}=\text{\hspace{0.17em}}₹\frac{1}{100}\text{\hspace{0.17em}}\\ \text{So,}\text{\hspace{0.17em}}\text{7}\text{\hspace{0.17em}}\text{rupees}\text{\hspace{0.17em}}\text{7}\text{\hspace{0.17em}}\text{paise=₹\hspace{0.17em}}\text{7+}\left(\text{7×}\frac{\text{1}}{\text{100}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=₹\hspace{0.17em}}\text{7+₹\hspace{0.17em}}\text{0}\text{.07}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\overline{)\text{₹\hspace{0.17em}}\text{7}\text{.07}}\\ \left(\text{iii}\right)\text{77 rupees 77 paise}\text{.}\\ \text{Since},\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{paise}=\text{\hspace{0.17em}}₹\frac{1}{100}\text{\hspace{0.17em}}\\ \text{So},\text{77 rupees 77 paise}=\text{\hspace{0.17em}}₹\text{\hspace{0.17em}}77+₹\text{\hspace{0.17em}}\left(\frac{77}{100}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹\hspace{0.17em}}77+₹\text{\hspace{0.17em}}0.77\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\overline{)₹\text{\hspace{0.17em}}77.77}\\ \left(\text{iv}\right)\text{5}0\text{paise}\text{.}\\ \text{Since},\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{paise}=\text{\hspace{0.17em}}₹\frac{1}{100}\text{\hspace{0.17em}}\\ \text{So},\text{\hspace{0.17em}}50\text{\hspace{0.17em}}\text{paise}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=₹\text{\hspace{0.17em}}\left(\frac{50}{100}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\overline{)₹\text{\hspace{0.17em}}0.50}\\ \left(\text{v}\right)\text{235 paise}\text{.}\\ \text{Since},\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{paise}=\text{\hspace{0.17em}}₹\frac{1}{100}\text{\hspace{0.17em}}\\ \text{So},235\text{\hspace{0.17em}}\text{paise}=₹\text{\hspace{0.17em}}\left(\frac{235}{100}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\overline{)₹\text{\hspace{0.17em}}2.35}\end{array}$

Q.30

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

Ans.

$\begin{array}{l}\text{(i)\hspace{0.17em}Since,\hspace{0.17em}1 cm =}\frac{1}{100}\text{m}=\frac{1}{1000}\text{km}\\ \text{So,\hspace{0.17em}we get}\\ \text{5 cm=}\frac{5}{100}\text{m}=\overline{)0.05\text{\hspace{0.17em}m}}\\ \text{and}\\ 5\text{\hspace{0.17em}cm\hspace{0.17em}}=\frac{5}{1000}\text{\hspace{0.17em}km}=\overline{)0.005\text{\hspace{0.17em}km}}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Since,\hspace{0.17em}1 mm =}\frac{1}{10}\text{cm}=\frac{1}{1000}\text{m=}\frac{1}{100000}\text{km}\\ \text{So,\hspace{0.17em}we get}\\ \text{35 mm =}\frac{35}{10}\text{cm}=\overline{)3.5\text{\hspace{0.17em}m}}\\ 35\text{\hspace{0.17em}mm\hspace{0.17em}}=\frac{35}{1000}\text{\hspace{0.17em}m}=\overline{)0.035\text{\hspace{0.17em}m}}\\ \text{and}\\ 35\text{​\hspace{0.17em}mm}=\frac{35}{100000}\text{\hspace{0.17em}km}=\overline{)0.00035\text{\hspace{0.17em}km}}\end{array}$

Q.31

$\begin{array}{l}\mathrm{Express}\mathrm{in}\mathrm{kg}:\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}}200\mathrm{g}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}3470\mathrm{g}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}4\mathrm{kg}8\mathrm{g}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}2598\mathrm{mg}\end{array}$

Ans.

$\begin{array}{l}\text{Since, 1 g=}\frac{\text{1}}{\text{1000}}\text{kg}\\ \text{So, 200 g=}\frac{\text{200}}{\text{1000}}\text{\hspace{0.17em}kg=}\frac{\text{2×}\overline{)\text{100}}}{\text{10×}\overline{)\text{100}}}\text{\hspace{0.17em}kg=}\frac{\text{2}}{\text{10}}\text{\hspace{0.17em}kg=}\overline{)\text{0.2\hspace{0.17em}kg}}\\ \text{(ii)\hspace{0.17em}3470\hspace{0.17em}g}\\ \text{3470\hspace{0.17em}g=}\frac{\text{3470}}{\text{1000}}\text{\hspace{0.17em}kg=}\frac{\text{347×}\overline{)\text{10}}}{\text{100×}\overline{)\text{10}}}\text{kg=}\frac{\text{347}}{\text{100}}\text{kg=}\overline{)\text{3.47\hspace{0.17em}kg}}\\ \text{(iii)\hspace{0.17em}4\hspace{0.17em}kg\hspace{0.17em}8\hspace{0.17em}g}\\ \text{4\hspace{0.17em}kg\hspace{0.17em}8\hspace{0.17em}g=4\hspace{0.17em}kg+}\frac{\text{8}}{\text{1000}}\text{kg=4\hspace{0.17em}kg+0.008\hspace{0.17em}kg=}\overline{)\text{4.008\hspace{0.17em}kg}}\\ \text{(iv)\hspace{0.17em}2598\hspace{0.17em}mg}\\ \text{Since,\hspace{0.17em}1\hspace{0.17em}mg\hspace{0.17em}​=}\frac{\text{1}}{\text{1000000}}\text{kg}\\ \text{So,\hspace{0.17em}2598\hspace{0.17em}mg=}\frac{\text{2598}}{\text{1000000}}\text{\hspace{0.17em}kg\hspace{0.17em}=}\overline{)\text{0.002598\hspace{0.17em}kg}}\end{array}$

Q.32

$\begin{array}{l}\mathrm{Write}\mathrm{the}\mathrm{following}\mathrm{decimal}\mathrm{numbers}\mathrm{in}\mathrm{the}\mathrm{expanded}\mathrm{form}:\\ \begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20.03\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}2.03\\ \left(\mathrm{iii}\right)200.03\\ \left(\mathrm{iv}\right)2.034\end{array}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{20}\text{.03=}\text{\hspace{0.17em}}\overline{)\text{2×10+0×1+0×}\frac{\text{1}}{\text{10}}\text{+3×}\frac{\text{1}}{\text{100}}}\\ \text{(ii)}\text{\hspace{0.17em}}\text{2}\text{.02=}\text{\hspace{0.17em}}\overline{)\text{2×1+0×}\frac{\text{1}}{\text{10}}\text{+3×}\frac{\text{1}}{\text{100}}}\\ \text{(iii)}\text{\hspace{0.17em}}\text{200}\text{.03=}\text{\hspace{0.17em}}\overline{)\text{2×100+0×10+0×1+0×}\frac{\text{1}}{\text{10}}\text{+3×}\frac{\text{1}}{\text{100}}}\\ \text{(iv)}\text{\hspace{0.17em}}\text{2}\text{.034=}\text{\hspace{0.17em}}\overline{)\text{2×1+0×}\frac{\text{1}}{\text{10}}\text{+3×}\frac{\text{1}}{\text{100}}\text{+4×}\frac{\text{1}}{\text{1000}}}\end{array}$

Q.33

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

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}2.56\\ \text{The place value of 2 in 2}.\text{56}=\text{}\overline{)\text{2 Ones}}\\ \left(\text{ii}\right)\text{21}.\text{37}\\ \text{The place value of 2 in 21}.\text{37}=\text{}\overline{)\text{2 Tens}}\\ \left(\text{iii}\right)\text{1}0.\text{25}\\ \text{The place value of 2 in 1}0.\text{25}=\text{}\overline{)\text{2 Tenths}}\\ \left(\text{iv}\right)\text{9}.\text{42}\\ \text{The place value of 2 in 9}.\text{42}=\text{}\overline{)\text{2 Hundredths}}\\ \left(\text{v}\right)\text{63}.\text{352}\\ \text{The place value of 2 in 63}.\text{352}=\text{}\overline{)\text{2 Thousandths}}\end{array}$

Q.34

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

Ans.

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

Q.35

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

Ans.

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

Q.36

$\mathrm{How}\mathrm{much}\mathrm{less}\mathrm{is}28\mathrm{km}\mathrm{than}42.6\mathrm{km}?$

Ans.

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

Q.37

$\begin{array}{l}Find:\\ \left(i\right)0.2×6\\ \left(ii\right)8×4.6\\ \left(iii\right)2.71×5\\ \left(iv\right)20.1×4\\ \left(v\right)0.05×7\\ \left(vi\right)211.02×4\\ \left(vii\right)2×0.86\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{}0.\text{2}×\text{6 =}\overline{)1.2}\\ \left(\text{ii}\right)\text{8}×\text{4}.\text{6}\text{\hspace{0.17em}}=\overline{)36.8}\\ \left(\text{iii}\right)\text{2}.\text{71}×\text{5}=\overline{)13.55}\\ \left(\text{iv}\right)\text{2}0.\text{1}×\text{4}\text{\hspace{0.17em}}\text{=}\overline{)\text{80}\text{.4}}\\ \left(\text{v}\right)\text{}0.0\text{5}×\text{7}\text{\hspace{0.17em}}\text{=}\overline{)\text{0}\text{.35}}\\ \left(\text{vi}\right)\text{211}.0\text{2}×\text{4}\text{\hspace{0.17em}}=\overline{)844.08}\\ \left(\text{vii}\right)\text{2}×\text{}0.\text{86}\text{\hspace{0.17em}}\text{=}\overline{)\text{1}\text{.72}}\end{array}$

Q.38

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

Ans.

$\begin{array}{l}\text{The area of a rectangle is given by length}×\text{breadth.}\\ \text{So, the area of given rectangle would be 5.7 cm}×3\text{\hspace{0.17em}cm}\\ \text{=}\overline{){\text{17.1 cm}}^{2}}\end{array}$

Q.39

$\begin{array}{l}Find:\\ \left(i\right)1.3×10\text{}\left(ii\right)36.8×10\text{}\left(iii\right)153.7×10\\ \left(iv\right)168.07×10\text{}\left(v\right)31.1×100\text{}\left(vi\right)156.1×100\\ \left(vii\right)3.62×100\left(viii\right)43.07×100\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(ix\right)0.5×10\\ \left(x\right)0.08×10\text{\hspace{0.17em}}\left(xi\right)0.9×100\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(xii\right)0.03×1000\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{1}.\text{3}×\text{1}0=\overline{)13}\\ \left(\text{ii}\right)\text{36}.\text{8}×\text{1}0=\overline{)368}\\ \left(\text{iii}\right)\text{153}.\text{7}×\text{1}0=\overline{)1537}\\ \left(\text{iv}\right)\text{168}.0\text{7}×\text{1}0=\overline{)1680.7}\\ \left(\text{v}\right)\text{31}.\text{1}×\text{1}00=\overline{)3110}\\ \left(\text{vi}\right)\text{156}.\text{1}×\text{1}00=\overline{)15610}\\ \left(\text{vii}\right)\text{3}.\text{62}×\text{1}00=\overline{)362}\\ \left(\text{viii}\right)\text{43}.0\text{7}×\text{1}00=\overline{)4307}\\ \left(\text{ix}\right)\text{}0.\text{5}×\text{1}0=\overline{)5}\\ \left(\text{x}\right)\text{}0.0\text{8}×\text{1}0=\overline{)0.8}\\ \left(\text{xi}\right)\text{}0.\text{9}×\text{1}00=\overline{)90}\\ \left(\text{xii}\right)\text{}0.0\text{3}×\text{1}000=\overline{)30}\end{array}$

Q.40

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

Ans.

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

Q.41

$\begin{array}{l}Find:\\ \left(i\right)2.5×0.3\text{}\text{}\left(ii\right)0.1×51.7\text{}\left(iii\right)0.2×316.8\\ \left(iv\right)1.3×3.1\text{}\text{}\left(v\right)0.5×0.05\text{}\left(vi\right)11.2×0.15\\ \left(vii\right)1.07×0.02\left(viii\right)10.05×1.05\text{}\left(ix\right)101.01×0.01\\ \left(x\right)100.01×1.1\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{2}.\text{5}×\text{}0.\text{3 =}\overline{)\text{0}\text{.75}}\\ \left(\text{ii}\right)\text{}0.\text{1}×\text{51}.\text{7}=\overline{)5.17}\\ \left(\text{iii}\right)\text{}0.\text{2}×\text{316}.\text{8}=\overline{)63.36}\\ \left(\text{iv}\right)\text{1}.\text{3}×\text{3}.\text{1=}\overline{)\text{4}\text{.03}}\\ \left(\text{v}\right)\text{}0.\text{5}×\text{}0.0\text{5}=\overline{)0.025}\\ \left(\text{vi}\right)\text{11}.\text{2}×\text{}0.\text{15=}\overline{)\text{1}\text{.68}}\\ \left(\text{vii}\right)\text{1}.0\text{7}×\text{}0.0\text{2=}\overline{)\text{0}\text{.0214}}\\ \left(\text{viii}\right)\text{1}0.0\text{5}×\text{1}.0\text{5=}\overline{)\text{10}\text{.5525}}\\ \left(\text{ix}\right)\text{1}0\text{1}.0\text{1}×\text{}0.0\text{1=}\overline{)\text{1}\text{.0101}}\\ \left(\text{x}\right)\text{1}00.0\text{1}×\text{1}.\text{1=}\overline{)\text{110}\text{.011}}\end{array}$

Q.42

$\begin{array}{l}Find:\\ \left(i\right)0.4÷2\text{}\left(ii\right)0.35÷5\text{}\left(iii\right)2.48÷4\text{}\left(iv\right)65.4÷6\\ \left(v\right)651.2÷4\text{}\left(vi\right)14.49÷7\left(vii\right)3.96÷4\text{}\left(viii\right)0.80÷5\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{}0.\text{4}÷\text{2}=\overline{)0.2}\\ \left(\text{ii}\right)\text{}0.\text{35}÷\text{5}=\overline{)0.07}\\ \left(\text{iii}\right)\text{2}.\text{48}÷\text{4=}\overline{)\text{0}\text{.62}}\\ \left(\text{iv}\right)\text{65}.\text{4}÷\text{6=}\overline{)\text{10}\text{.9}}\\ \left(\text{v}\right)\text{651}.\text{2}÷\text{4=}\overline{)\text{162}\text{.8}}\\ \left(\text{vi}\right)\text{14}.\text{49}÷\text{7}=\overline{)2.07}\\ \left(\text{vii}\right)\text{3}.\text{96}÷\text{4=}\overline{)\text{0}\text{.99}}\\ \left(\text{viii}\right)0.\text{8}0÷\text{5=}\overline{)\text{0}\text{.16}}\end{array}$

Q.43

$\begin{array}{l}Find:\\ \left(i\right)4.8÷10\left(ii\right)52.5÷10\left(iii\right)0.7÷10\text{}\left(iv\right)33.1÷10\\ \left(v\right)272.23÷10\left(vi\right)0.56÷10\left(vii\right)3.97÷10\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{4}.\text{8}÷\text{1}0=\overline{)0.48}\\ \left(\text{ii}\right)\text{52}.\text{5}÷\text{1}0=\overline{)5.25}\\ \left(\text{iii}\right)\text{}0.\text{7}÷\text{1}0=\overline{)0.07}\\ \left(\text{iv}\right)\text{33}.\text{1}÷\text{1}0=\overline{)3.31}\\ \left(\text{v}\right)\text{272}.\text{23}÷\text{1}0=\overline{)27.223}\\ \left(\text{vi}\right)\text{}0.\text{56}÷\text{1}0=\overline{)0.056}\\ \left(\text{vii}\right)\text{3}.\text{97}÷\text{1}0=\overline{)0.397}\end{array}$

Q.44

$\begin{array}{l}Find:\\ \left(i\right)2.7÷100\text{}\left(ii\right)0.3÷100\text{}\text{}\left(iii\right)0.78÷100\\ \left(iv\right)432.6÷100\text{}\left(v\right)23.6÷100\text{}\text{}\left(vi\right)\text{\hspace{0.17em}}98.53÷100\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{2}.\text{7}÷\text{10}0=\overline{)0.027}\\ \left(\text{ii}\right)\text{}0.3÷\text{10}0=\overline{)0.003}\\ \left(\text{iii}\right)\text{}0.\text{78}÷\text{1}00=\overline{)0.0078}\\ \left(\text{iv}\right)\text{}432.6÷\text{10}0=\overline{)4.326}\\ \left(\text{v}\right)\text{}23.6÷\text{1}00=\overline{)0.236}\\ \left(vi\right)\text{\hspace{0.17em}}98.53÷\text{10}0=\overline{)0.9853}\end{array}$

Q.45

$\begin{array}{l}Find:\\ \left(i\right)7.9÷1000\text{}\left(ii\right)26.3÷1000\text{}\text{}\left(iii\right)38.53÷1000\\ \left(iv\right)128.9÷1000\text{}\left(v\right)0.5÷1000\text{}\text{}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{}7.9÷\text{10}00=\overline{)0.0079}\\ \left(\text{ii}\right)\text{}26.3÷\text{100}0=\overline{)0.0263}\\ \left(\text{iii}\right)\text{}38.\text{53}÷\text{1}000=\overline{)0.03853}\\ \left(\text{iv}\right)\text{}128.9÷\text{100}0=\overline{)0.1289}\\ \left(\text{v}\right)\text{}0.5÷\text{10}00=\overline{)0.0005}\end{array}$

Q.46

$\begin{array}{l}Find:\\ \left(i\right)7÷3.5\text{}\left(ii\right)36÷0.2\text{}\left(iii\right)3.25÷0.5\text{}\left(iv\right)30.94÷0.7\\ \left(v\right)0.5÷0.25\text{}\text{}\text{}\left(vi\right)7.75÷0.25\left(vii\right)76.5÷0.15\\ \left(viii\right)37.8÷14\text{}\text{}\text{}\left(ix\right)2.73÷1.3\end{array}$

Ans.

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

Q.47

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

Ans.

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