# NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.2) Exercise 2.2

The Central Board of Secondary Education is one of the most prestigious and premier educational boards available for students in India. A vast number of schools affiliated with the CBSE board are present all over India. CBSE deploys a particular learning scheme for students who are a part of the CBSE board. There is inherent objectivity to all the questions that are asked on the CBSE exams. An objective study pattern includes learning techniques that ensure students learn something from the syllabus. Students must go through the chapter thoroughly and understand it inside out. Students can grasp the fundamental ideas of the chapter and have their basic ideas about the chapter clear as well. Such a pattern is followed because the kinds of questions that frequent the exams are deeply entrenched in the rudimentary ideas of the subject. The questions that appear in the exams are both long-answer type and short-answer type questions, and therefore, unless students have their basic ideas clear they would fail to answer and solve these questions accurately. This observation has been frequently made that students can retain details and information from a chapter without understanding it and then they score well in the exams. This is a practice that is not encouraged at all by CBSE. This is the primary reason CBSE has developed a structure that directly develops and hones the analytical and cognitive abilities of the students. This is required because the questions from the exam can be answered using these qualities. CBSE very vividly makes it clear for the students that they prioritise understanding the fundamental ideas rather than retention of facts. This way students might be able to score well and make progress in their academic careers, but in the long run, students might find it very difficult to carry on with their careers. Teachers who are a part of Extramarks are highly qualified and they have relevant experience in teaching students Mathematics. These teachers are very aware of all the different kinds of struggles that students go through with regard to learning and therefore they have collaborated with Extramarks to provide students with Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English provide detailed explanations of every problem for which they have provided their answers. These teachers have shared all the various skills and tips that they have observed and formulated in their years of being teachers. Teachers have reassured students that there is nothing to be inhibited about in this curriculum. Teachers have ensured that students would not be struggling with hindrances to learning anymore. The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English is one of the most trusted tools compiled by qualified mentors and they implore the students to use the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English as well.

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## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.2) Exercise 2.2

Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English provide solutions to the second chapter of the Class 8 CBSE curriculum. Various new ideas are being covered in this section and they are as follows –

1. Fraction multiplications.
2. Fraction multiplication with a whole number.
3. Multiplication among more than two fractions.
4. Value calculation of a product.

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

Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English can be accessed through the Extramarks website. They are also available on the mobile application of Extramarks. Since Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English can be even accessed through mobile, these NCERT solutions can be accessed conveniently by students who belong to remote areas where they cannot access premiere coaching. The NCERT solutions are available in  PDF format, after students who have successfully accessed Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English and can use these solutions anytime they like. Students in the possession of the NCERT solutions are not without help. Students whenever they are studying are in reach of the best guidance they can get. Since Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English are compiled and reviewed by top educators in the field, students are always under expert guidance. The Exyramarks website also has various other useful and highly efficient resources.

## Chapter 2 – Fractions and Decimals

Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English provide the solutions for the 2nd chapter of the Class 7 CBSE curriculum. One cannot make significant progress and make advancement in the subject without understanding the concepts of Fractions and decimals. The entire milieu of contemporary Mathematics and the entire branch of Arithmetic is based on this chapter. Therefore, teachers have emphasised the importance of this chapter.

### Introduction

Students have learned about fractions and decimals separately in earlier classes. In the earlier classes whenever fractions and decimals were taught, they were taught in the form of rudimentary ideas. In the Class 7 curriculum, advanced ideas from each chapter are taught and then both ideas are merged. The exercise which focuses more on the alignment of both concepts is the most challenging one. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English do great justice in making sure students have understood the complicated topics discussed in these chapters. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English, therefore, clarifies the ideas of this chapter with great adroitness. The Class 7 Maths Chapter 2 Exercise 2.2 Solutions make sure that students are prepared for any kind of unprecedented situation.

### How Well You Have Learned About Fractions

The ideas that are discussed in Chapter 2 of the Class 7 curriculum are Fractions and Decimals. The ideas of this chapter have real-world applications and since it is an arithmetic concept students were just familiar with basic ideas. In this stage and chapter, students are taught about improper fractions and the various properties that it demonstrates. The chapter is replete with various solved and unsolved examples allowing students to completely grasp the chapter. The sums which are to be solved by students are provided in the exercise. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English provide the solutions for these exercises. After students have rigorously used Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English, they could improve their academic performance in this regard.

### Multiplication of Fractions

Multiplication of fractions is one of the most recurrent properties that is consistently used in every corner of education and real life. Students must have this idea very clear. There are a plethora of questions that regularly frequent the paper. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English help students anticipate the kind of problems that might arise in the examinations.

### Multiplication of a Fraction by a Whole Number

This section of the chapter is pretty similar to the previous section. This section divulges the rules and properties that a fraction would inherit if it were multiplied by a whole number. The idea of decimals comes up here as well because fractions and decimals are interchangeable most of the time. There are various properties of a fraction and decimals, and students must know which one of them must be used. Therefore, this section prepares students exactly for that. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English additionally give more scope for students to develop their skills even more. The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2, in Hindi and English are so lucidly written that students would face no hindrances while using them.

### Exercise 2.2 Questions and Answers

Students are recommended to click on the link below to access Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English. Therefore, the Class 7 Maths Chapter 2 Exercise 2.2 Solutions are very crucial for a student’s performance.

### Multiplication of a Fraction by a Fraction

A fraction can be also multiplied with another fraction. There are exact rules for this and students are suggested to use Extramarks solutions for’ NCERT Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English to get a better understanding of the concepts.

### Value of the Products

The ideas that are generally valid with whole numbers often do not directly correspond with fractions. It is important for students to know the value of the fractions and which instances are bigger and smaller. While students solve questions from this section, they are advised to always use Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.2 in Hindi and English to make sure they understand the nuances of the topic. These solutions focus on the Class 7 Maths Chapter 2 Exercise 2.2.

### NCERT Solutions for Class 7

There are solutions available for each subject and all the constituent exercises. Students can access them through Extramarks’ website.

Q.1

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

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

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

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.4

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

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

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.6

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

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.7

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

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.8

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

Ans.

$\begin{array}{l}\text{Total quantity of water}=\text{5 litres}\\ \text{Vidya consumed}\text{\hspace{0.17em}}\frac{2}{5}\text{\hspace{0.17em}}\text{of water and Pratap consumed rest}\\ \text{of the water}.\\ \text{Water consumed by Vidya}=5×\frac{2}{5}\text{\hspace{0.17em}}\text{litre}=2\text{\hspace{0.17em}}\text{litre}\\ \text{Therefore water consumed by Pratap}\\ =\text{Total water}–\text{Water consumed by Vidya}.\\ =\text{5}\text{\hspace{0.17em}}\text{litre-2}\text{\hspace{0.17em}}\text{litre=}\overline{)\text{3}\text{\hspace{0.17em}}\text{litre}}\\ \text{Therefore},\text{fraction of water consumed by Pratap}=\frac{3}{5}\\ \text{Therefore},\left(\text{i}\right)\text{Vidya consumed 2 litre of water}\\ \left(\text{ii}\right)\text{Pratap consumed}\frac{3}{5}\text{of water}.\end{array}$

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