## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.3) Exercise 2.3

Free PDF download of NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.3 (EX 2.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 7 Science , Maths solutions and solutions of other subjects.

**Study without Internet (Offline)**

## Download PDF of NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.3) Exercise 2.3

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

Exercise 2.3

1. Find:

(i) $\dfrac{\text{1}}{\text{4}}\,\text{of}$ (a) $\dfrac{\text{1}}{\text{4}}$ (b) $\dfrac{\text{3}}{\text{5}}$ (c) $\dfrac{\text{4}}{\text{3}}$

Ans:

(a) Calculating the value,

$\dfrac{\text{1}}{\text{4}}\,\text{of}\,\dfrac{\text{1}}{\text{4}}\text{=}\dfrac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{4}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 1}}{\text{4 }\!\!\times\!\!\text{ 4}}\text{=}\dfrac{\text{1}}{\text{16}}$

(b) Calculating the value,

$\dfrac{\text{1}}{\text{4}}\,\text{of}\,\dfrac{\text{3}}{\text{5}}\text{=}\dfrac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{4}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 3}}{\text{4 }\!\!\times\!\!\text{ 4}}\text{=}\dfrac{\text{3}}{\text{16}}$

(c) Calculating the value,

$\dfrac{\text{1}}{\text{4}}\,\text{of}\,\dfrac{\text{4}}{\text{3}}\text{=}\dfrac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{4}}{\text{3}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 4}}{\text{4 }\!\!\times\!\!\text{ 3}}\text{=}\dfrac{\text{1}}{\text{3}}$

(ii) \[\dfrac{\text{1}}{\text{7}}\,\text{of}\] (a) \[\dfrac{\text{2}}{\text{9}}\] (b) \[\dfrac{\text{6}}{\text{5}}\] (c) $\dfrac{\text{3}}{\text{10}}$

Ans:

(a) Calculating the value,

$\dfrac{\text{1}}{\text{7}}\,\text{of}\,\dfrac{\text{2}}{\text{9}}\text{=}\dfrac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{2}}{\text{9}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 2}}{\text{7 }\!\!\times\!\!\text{ 9}}\text{=}\dfrac{\text{2}}{\text{63}}$

(b) Calculating the value,

$\dfrac{\text{1}}{\text{7}}\,\text{of}\,\dfrac{\text{2}}{\text{9}}\text{=}\dfrac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{6}}{\text{5}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 6}}{\text{7 }\!\!\times\!\!\text{ 5}}\text{=}\dfrac{\text{6}}{\text{35}}$

(c) Calculating the value,

$\dfrac{\text{1}}{\text{7}}\,\text{of}\,\dfrac{\text{2}}{\text{9}}\text{=}\dfrac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{10}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 3}}{\text{7 }\!\!\times\!\!\text{ 10}}\text{=}\dfrac{3}{70}$

2. Multiply and reduce to lowest form (if possible):

(i) $\dfrac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 2}\dfrac{\text{2}}{\text{3}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{2}{3}\times 2\dfrac{2}{3}=\dfrac{2}{3}\times \dfrac{8}{3}=\dfrac{2\times 8}{3\times 3}=\dfrac{16}{9}=1\dfrac{7}{9}$

(ii) $\dfrac{\text{2}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{7}}{\text{9}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{2}{7}\times \dfrac{7}{9}=\dfrac{2\times 7}{7\times 9}=\dfrac{2}{9}$

(iii) $\dfrac{\text{3}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{6}}{\text{4}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{3}{8}\times \dfrac{6}{4}=\dfrac{3\times 6}{8\times 4}=\dfrac{3\times 3}{8\times 2}=\dfrac{9}{16}$

(iv) $\dfrac{\text{9}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{5}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{9}{5}\times \dfrac{3}{5}=\dfrac{9\times 3}{5\times 5}=\dfrac{27}{25}=1\dfrac{2}{25}$

(v) $\dfrac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{15}}{\text{8}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{1}{3}\times \dfrac{15}{8}=\dfrac{1\times 15}{3\times 8}=\dfrac{1\times 5}{1\times 8}=\dfrac{5}{8}$

(vi) $\dfrac{\text{11}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{10}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{11}{2}\times \dfrac{3}{10}=\dfrac{11\times 3}{2\times 10}=\dfrac{33}{20}=1\dfrac{3}{20}$

(vii) $\dfrac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{12}}{\text{7}}$

Ans:

Multiplying and reducing to lowest form,

$\dfrac{4}{5}\times \dfrac{12}{7}=\dfrac{4\times 12}{5\times 7}=\dfrac{48}{35}=1\dfrac{13}{35}$

3. Multiply the following fractions:

(i) $\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ 5}\dfrac{\text{1}}{\text{4}}$

Ans:

Performing multiplication,

$\dfrac{2}{5}\times 5\dfrac{1}{4}=\dfrac{2}{5}\times \dfrac{21}{4}=\dfrac{2\times 21}{5\times 4}=\dfrac{1\times 21}{5\times 2}=\dfrac{21}{10}=2\dfrac{1}{10}$

(ii) $\text{6}\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{7}}{\text{9}}$

Ans:

Performing multiplication,

$6\dfrac{2}{5}\times \dfrac{7}{9}=\dfrac{32}{5}\times \dfrac{7}{9}=\dfrac{32\times 7}{5\times 9}=\dfrac{224}{45}=4\dfrac{44}{45}$

(iii) $\dfrac{\text{3}}{\text{2}}\text{ }\!\!\times\!\!\text{ 5}\dfrac{\text{1}}{\text{3}}$

Ans:

Performing multiplication,

$\dfrac{3}{2}\times 5\dfrac{1}{3}=\dfrac{3}{2}\times \dfrac{16}{3}=\dfrac{48}{6}=8$

(iv) $\dfrac{\text{5}}{\text{6}}\text{ }\!\!\times\!\!\text{ 2}\dfrac{\text{3}}{\text{7}}$

Ans:

Performing multiplication,

$\dfrac{5}{6}\times 2\dfrac{3}{7}=\dfrac{5}{6}\times \dfrac{17}{7}=\dfrac{85}{42}=2\dfrac{1}{42}$

(v) $\text{3}\dfrac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{4}}{\text{7}}$

Ans:

Performing multiplication,

$3\dfrac{2}{5}\times \dfrac{4}{7}=\dfrac{17}{7}\times \dfrac{4}{7}=\dfrac{68}{35}=1\dfrac{33}{35}$

(vi) $\text{2}\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ 3}$

Ans:

Performing multiplication,

$2\dfrac{3}{5}\times 3=\dfrac{13}{5}\times \dfrac{3}{1}=\dfrac{13\times 3}{5\times 1}=\dfrac{39}{5}=7\dfrac{4}{5}$

(vii) $\text{3}\dfrac{\text{4}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{5}}$

Ans:

Performing multiplication,

$3\dfrac{4}{7}\times \dfrac{3}{5}=\dfrac{25}{7}\times \dfrac{3}{5}=\dfrac{5\times 3}{7\times 1}=\dfrac{15}{7}=2\dfrac{1}{7}$

4. Which is greater:

(i) $\dfrac{\text{2}}{\text{7}}\,\text{of}\,\dfrac{\text{3}}{\text{4}}\,\text{or}\,\dfrac{\text{3}}{\text{5}}\,\text{of}\,\dfrac{\text{5}}{\text{8}}$

Ans:

Calculating the greater term,

$\dfrac{\text{2}}{\text{7}}\,\text{of}\,\dfrac{\text{3}}{\text{4}}\,\text{or}\,\dfrac{\text{3}}{\text{5}}\,\text{of}\,\dfrac{\text{5}}{\text{8}}$

$\Rightarrow \dfrac{\text{2}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{4}}\,\text{or}\,\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{5}}{\text{8}}$

$\Rightarrow \dfrac{\text{3}}{\text{14}}\,\text{or}\,\dfrac{\text{3}}{\text{8}}$

$\Rightarrow \dfrac{3}{14}<\dfrac{3}{8}$

Hence, $\dfrac{\text{3}}{\text{5}}\,\text{of}\,\dfrac{\text{5}}{\text{8}}$ is greater.

(ii) $\dfrac{\text{1}}{\text{2}}\,\text{of}\,\dfrac{\text{6}}{\text{7}}\,\text{or}\,\dfrac{\text{2}}{\text{3}}\,\text{of}\,\dfrac{\text{3}}{\text{7}}$

Ans:

Calculating the greater term,

$\dfrac{\text{1}}{\text{2}}\,\text{of}\,\dfrac{\text{6}}{\text{7}}\,\text{or}\,\dfrac{\text{2}}{\text{3}}\,\text{of}\,\dfrac{\text{3}}{\text{7}}$

$\Rightarrow \dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{6}}{\text{7}}\,\text{or}\,\dfrac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{7}}$

$\Rightarrow \dfrac{\text{6}}{\text{14}}\,\text{or}\,\dfrac{\text{2}}{\text{7}}$

$\Rightarrow \dfrac{\text{6}}{\text{14}}>\dfrac{\text{2}}{\text{7}}$

Hence, $\dfrac{\text{1}}{\text{2}}\,\text{of}\,\dfrac{\text{6}}{\text{7}}$ is greater.

5. Saili plants \[\text{4}\] saplings in a row in her garden. The distance between two adjacent saplings is \[\dfrac{\text{3}}{\text{4}}\] m. Find the distance between the first and the last sapling.

Ans:

Given: Saili plants \[4\] saplings in a row where the distance between two adjacent saplings $=\dfrac{3}{4}$m.

(Image will be uploaded soon)

The number of gaps in saplings \[=\text{ }3\]

Hence,

The distance between the first and the last saplings$\text{=3 }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{4}}\text{=}\dfrac{\text{9}}{\text{4}}\text{m=2}\dfrac{\text{1}}{\text{4}}\text{m}$

Therefore, the distance between the first and the last saplings is $\text{2}\dfrac{\text{1}}{\text{4}}\,\text{m}$

6. Lipika reads a book for $\text{1}\dfrac{\text{3}}{\text{4}}$ hours every day. She reads the entire book in \[\text{6}\] days. How many hours in all were required by her to read the book?

Ans:

Given: Time taken for reading a book by Lipika $=1\dfrac{3}{4}$ hours.

Lipika reads the entire book in $6$ days

Calculating the Total hours taken by Lipika to read the entire book,

$=1\dfrac{3}{4}\times 6=\dfrac{7}{4}\times 6=\dfrac{21}{2}=10\dfrac{1}{2}$ hours.

Hence, it would take $10$ hours to read the book.

7. A car runs $\text{16}$ km using \[\text{1}\] litre of petrol. How much distance will it cover using $\text{2}\dfrac{\text{3}}{\text{4}}$ litres of petrol?

Ans:

Given: A car covers the distance$\text{=16}\,\text{km}$ in $1$ litre of petrol.

Calculating the distance covered by car in $2\dfrac{3}{4}$ litres of petrol,

Distance$\text{=2}\dfrac{\text{3}}{\text{4}}\,\text{of}\,\text{16}\,\text{km=}\dfrac{\text{11}}{\text{4}}\text{ }\!\!\times\!\!\text{ 16=44}\,\text{km}$

Therefore, car will cover a distance of $44$ km in $2\dfrac{3}{4}$ litres of petrol.

8. (a) (i) Provide the number in the box $___$, such that $\dfrac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\text{=}\dfrac{\text{10}}{\text{30}}$

(ii) The simplest form of the number obtained in $___$ is _____.

Ans:

(i) The number inside the box should be $\dfrac{2}{3}\times =\dfrac{10}{30}$

(ii) The simplest form of the number obtained in $\dfrac{\text{5}}{\text{10}}\,\text{is}\,\dfrac{\text{1}}{\text{2}}$

(b) (i) Provide the numberin the box $ ___$ , such that $\dfrac{3}{5}\times =\dfrac{24}{75}$.

(ii) The simplest form of the number obtained in $___$ is______.

Ans:

(i) The number inside the box should be $\dfrac{3}{5}\times =\dfrac{24}{75}$

(ii) The simplest form of the number obtained in $\dfrac{\text{8}}{\text{15}}\,\text{is}\,\dfrac{\text{8}}{\text{15}}$

## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.3

Opting for the NCERT solutions for Ex 2.3 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 2.3 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 2 Exercise 2.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 2 Exercise 2.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 2 Exercise 2.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.