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NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals

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• Chapter 2 Fractions And Decimals

NCERT Class 7 Chapter 2 Maths Fractions and Decimals - FREE PDF Download

Here are the NCERT solutions for fractions and decimals Class 7. In order to improve their exam scores, students can practice answering various kinds of questions linked to this chapter either online or by downloading these files. In previous lessons, students have learned how to add and subtract decimals as well as fractions. In this lesson, students will learn how to multiply and divide decimals, as well as fractions. These Class 7 Maths Chapter 2 Solutions, which include fractions and decimals, have been created by experts in the field to help students prepare for their exams.

Glance on Maths Chapter 2 Class 7 - Fractions and Decimals

Basic calculations like adding, subtracting, comparing, and ranking fractions are covered in this chapter.

We work on word problems, such as applying concepts like fractions to real-life scenarios.

We cover higher-order thinking questions that test your ability to evaluate and resolve issues with decimals and fractions.

Focus on concepts like adding, subtracting, multiplying, and dividing fractions, and confidently convert between fractions and decimals.

This article contains chapter notes, important questions, and exercise links for Chapter 2 Fractions and Decimals, which you can download  as PDFs.

There are five exercises (31 fully solved questions) in class 7th maths chapter 2 Fractions and Decimal.

Access Exercise Wise NCERT Solutions for Chapter 2 Maths Class 7

Exercises Under NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

Exercise 2.1: multiplication of fractions.

Multiplying Fractions by Whole Numbers : Understanding how to multiply fractions by whole numbers.

Multiplying Two Fractions : Learning the method to multiply two fractions.

Multiplying Mixed Numbers : Converting mixed numbers to improper fractions and then multiplying them.

Exercise 2.2: Division of Fractions

Reciprocal of a Fraction : Introduction to the concept of reciprocals.

Dividing Fractions by Whole Numbers : Method to divide fractions by whole numbers using reciprocals.

Dividing a Fraction by Another Fraction : Understanding the steps to divide one fraction by another fraction.

Word Problems : Applying division of fractions to solve real-life problems.

Exercise 2.3: Multiplication of Decimal Numbers

This section in Maths Class 7 Chapter 2 covers the rules and methods for multiplying decimal numbers. It includes step-by-step instructions for aligning the decimal points and performing multiplication, followed by placing the decimal point in the product correctly. Practical examples and problems help reinforce the concept and provide practice in multiplying decimals in various contexts.

Exercise 2.4: Division of Decimal Numbers

This section focuses on dividing decimal numbers, explaining how to handle the decimal point in both the dividend and the divisor. It includes methods for converting the divisor into a whole number by multiplying both the dividend and the divisor by a power of ten. The section in Class 7 Fractions and Decimals provides numerous examples and exercises to practice the division of decimals, ensuring a clear understanding of the process.

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

Exercise - 2.1.

1. Which of the drawings $(a)\,to\,(d)$ show:

(i). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{5}}$

Ans: corresponds to $\text{(d)}$  

Because,  $2\times \frac{1}{5}=\frac{1}{5}+\frac{1}{5}$                                                                

(ii). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{2}}$

Ans: corresponds to $\text{(b)}$

Because, $2\times \frac{1}{2}=\frac{1}{2}+\frac{1}{2}$

(iii). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{3}}$

Ans:   corresponds to $\text{(a)}$   

Because, $3\times \frac{2}{3}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}$ 

(iv). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{4}}$

Ans: Corresponds to $\text{(c)}$

Because, $3\times \frac{1}{4}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}$ 

2. Some pictures $\left( \text{a} \right)\,\text{to}\,\left( \text{c} \right)$ are given below. Tell which of them show: 

(i). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{5}}\text{=}\frac{\text{3}}{\text{5}}$

Ans: Corresponds to $\text{(c)}$                    

Because, $3\times \frac{1}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$  

(ii). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}$ 

Ans: Corresponds to $\text{(a)}$                   

Because, $2\times \frac{1}{3}=\frac{1}{3}+\frac{1}{3}$ 

(iii). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}}\text{=2}\frac{\text{1}}{\text{4}}$

Ans: Corresponds to $\text{(b)}$                  

Because, $3\times \frac{3}{4}=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$      

3. Multiply and reduce to lowest form and convert into a mixed fraction:

(i). $\text{7 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{5}}$        

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,   

$7\times \frac{3}{5}=\frac{7\times 3}{5}=\frac{21}{5}=4\frac{1}{5}$

(ii). $\text{4 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{3}}$ 

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$4\times \frac{1}{3}=\frac{4\times 1}{3}=\frac{4}{3}=1\frac{1}{3}$ 

(iii). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{7}}$    

$2\times \frac{6}{7}=\frac{2\times 6}{7}=\frac{12}{7}=1\frac{5}{7}$       

(iv). $\text{5 }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{9}}$ 

$5\times \frac{2}{9}=\frac{5\times 2}{9}=\frac{10}{9}=1\frac{1}{9}$

(v). $\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 4}$     

$\frac{2}{3}\times 4=\frac{2\times 4}{3}=\frac{8}{3}=2\frac{2}{3}$   

(vi)  $\frac{\text{5}}{\text{2}}\text{ }\!\!\times\!\!\text{ 6}$    

$\frac{5}{2}\times 6=5\times 3=15$

(vii)   $\text{11 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{7}}$    

$11\times \frac{4}{7}=\frac{11\times 4}{7}=\frac{44}{7}=6\frac{2}{7}$ 

(viii)  $\text{20 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{5}}$ 

$20\times \frac{4}{5}=4\times 4=16$

(ix)  $\text{13 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{3}}$      

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,    

$13\times \frac{1}{3}=\frac{13\times 1}{3}=\frac{13}{3}=4\frac{1}{3}$ 

(x)  $\text{15 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{5}}$ 

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,   

$15\times \frac{3}{5}=3\times 3=9$                

4. Shade: 

(i). $\frac{\text{1}}{\text{2}}$ of the circles in box

Ans: Half of the circles in the box are,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{12}\,\text{circles=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 12=6}\,\text{circles}$

(iii). $\frac{\text{2}}{\text{3}}$ of the triangles in box 

Ans: Two-third of the triangles in the box are, $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{9}\,\text{triangles=}\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 9=2 }\!\!\times\!\!\text{ 3=6}\,\text{triangles}$

(iv). $\frac{\text{3}}{\text{5}}$ of the squares inbox

Ans: Three-fifth of the squares in the box are,

$\frac{\text{3}}{\text{5}}\,\text{of}\,\text{15}\,\text{squares=}\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ 15=3 }\!\!\times\!\!\text{ 3=9}\,\text{squares}$

(a).$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{(i)}\,\text{24}\,\text{(ii)}\,\text{46}$

(i) Calculating the value,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{24=12}$

(ii) Calculating the value,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{46=23}$ 

(b). $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{(i)}\,\text{18}\,\text{(ii)}\,\text{27}$ 

(i) Calculating the value, $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{18=}\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 18=2 }\!\!\times\!\!\text{ 6=12}$

(ii) Calculating the value, $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{27=}\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 27=2 }\!\!\times\!\!\text{ 9=18}$ 

(c)  $\frac{\text{3}}{\text{4}}\,\text{of}\,\text{(i)}\,\text{16}\,\text{(ii)}\,\text{36}$          

$\frac{\text{3}}{\text{4}}\,\text{of}\,\text{16=}\frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ 16=3 }\!\!\times\!\!\text{ 4=12}$

$\frac{\text{3}}{\text{4}}\,\text{of}\,36\text{=}\frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ 36=3 }\!\!\times\!\!\text{ 9=27}$  

(d)  $\frac{\text{4}}{\text{5}}\,\text{of}\,\text{(i)}\,\text{20}\,\text{(ii)}\,\text{35}$ 

$\frac{\text{4}}{\text{5}}\,\text{of}\,20\text{=}\frac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ 20=4 }\!\!\times\!\!\text{ 4=16}$

$\frac{\text{4}}{\text{5}}\,\text{of}\,35\text{=}\frac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ 35=4 }\!\!\times\!\!\text{ 7=28}$ 

6.   Multiply and express as a mixed fraction:

(a)  $\text{3 }\!\!\times\!\!\text{ 5}\frac{\text{1}}{\text{5}}$       

Ans: Multiplying and expressing the term as mixed fraction,

$3\times 5\frac{1}{5}=3\times \frac{26}{5}=\frac{3\times 26}{5}=\frac{78}{5}=15\frac{3}{5}$

(b) $\text{5 }\!\!\times\!\!\text{ 6}\frac{\text{3}}{\text{4}}$          

$5\times 6\frac{3}{4}=5\times \frac{27}{4}=\frac{5\times 27}{4}=\frac{135}{4}=33\frac{3}{4}$ 

(c) $\text{7 }\!\!\times\!\!\text{ 2}\frac{\text{1}}{\text{4}}$

$7\times 2\frac{1}{4}=7\times \frac{9}{4}=\frac{7\times 9}{4}=\frac{63}{4}=15\frac{3}{4}$        

(d)  $\text{4 }\!\!\times\!\!\text{ 6}\frac{\text{1}}{\text{3}}$ 

$4\times 6\frac{1}{3}=4\times \frac{19}{3}=\frac{4\times 19}{3}=\frac{76}{3}=25\frac{1}{3}$ 

(e)  $\text{3}\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ 6}$ 

$3\frac{1}{4}\times 6=\frac{13}{4}\times 6=\frac{13\times 3}{2}=\frac{39}{2}=19\frac{1}{2}$  

(f)  $\text{3}\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ 8}$ 

$3\frac{2}{5}\times 8=\frac{17}{5}\times 8=\frac{17\times 8}{5}=\frac{136}{5}=27\frac{1}{5}$

(a)  $\frac{\text{1}}{\text{2}}\,\text{of}\,\text{(i)}\,\text{2}\frac{\text{3}}{\text{4}}\,\text{(ii)}\,\text{4}\frac{\text{2}}{\text{9}}$             

(i) Calculating the value, \[\frac{\text{1}}{\text{2}}\,\text{of}\,\text{2}\frac{\text{3}}{\text{4}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 2}\frac{\text{3}}{\text{4}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{11}}{\text{4}}\text{=}\frac{\text{11}}{\text{8}}\text{=1}\frac{\text{3}}{\text{8}}\] 

(ii) Calculating the value, \[\frac{\text{1}}{\text{2}}\,\text{of}\,\text{4}\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 4}\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{38}}{\text{9}}\text{=}\frac{\text{19}}{\text{9}}\text{=2}\frac{\text{1}}{\text{9}}\] 

(b)  $\frac{\text{5}}{\text{8}}\,\text{of}\,\text{(i)}\,\text{3}\frac{\text{5}}{\text{6}}\,\text{(ii)}\,\text{9}\frac{\text{2}}{\text{3}}$ 

(i) Calculating the value, \[\frac{\text{5}}{\text{8}}\,\text{of}\,\text{3}\frac{\text{5}}{\text{6}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ 3}\frac{\text{5}}{\text{6}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{23}}{\text{6}}\text{=}\frac{\text{115}}{\text{48}}\text{=2}\frac{\text{19}}{\text{48}}\] 

(ii) Calculating the value, \[\frac{\text{5}}{\text{8}}\,\text{of}\,\text{9}\frac{\text{2}}{\text{3}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ 9}\frac{\text{2}}{\text{3}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{29}}{\text{3}}\text{=}\frac{\text{145}}{\text{24}}\text{=6}\frac{\text{1}}{\text{24}}\] 

8.  Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained $\text{5}$ liters of water. Vidya consumed $\frac{\text{2}}{\text{5}}$ of the water. Pratap consumed the remaining water.

(i). How much water did Vidya drink?

Ans:   Water consumed by Vidya is, $\text{=}\frac{\text{2}}{\text{5}}\,\text{of}\,\text{5}\,\text{litres=}\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ 5=2}\,\text{litres}$ 

Hence, Vidya drank $2$ litres of water from the bottle.

(ii). What fraction of the total quantity of water did Pratap drink?

Ans: Water consumed by Pratap \[\text{= }\left( \text{1-}\frac{\text{2}}{\text{5}} \right)\text{  }\]part of bottle

Pratap consumed $\frac{\text{3}}{\text{5}}\,\text{of}\,\text{5}\,\text{litres}\,\text{water=}\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ 5=3}\,\text{lites}$ 

Hence, Pratap drank $\frac{3}{5}$ part of the total quantity of water present in the bottle.

Exercise - 2.2

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

(a) Calculating the value,

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

(b) Calculating the value,

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

(c) Calculating the value,

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

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

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

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

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

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

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

Ans: Multiplying and reducing to lowest form,  

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

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

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

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

Ans: Multiplying and reducing to lowest form, 

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

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

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

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

Ans: Multiplying and reducing to lowest form,

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

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

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

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

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

3. Multiply the following fractions:

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

Ans: Performing multiplication,

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

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

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

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

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

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

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

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

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

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

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

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

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

4. Which is greater:

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

Ans: Calculating the greater term,

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

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

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

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

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

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

Calculating the greater term,

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

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

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

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

5. Saili plants \[\text{4}\] saplings in a row in her garden. The distance  between

two adjacent saplings is \[\frac{\text{3}}{\text{4}}\] 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 $=\frac{3}{4}$m.

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

Hence, 

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

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

6. Lipika reads a book for $\text{1}\frac{\text{3}}{\text{4}}$ 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 $=1\frac{3}{4}$ hours.

Lipika reads the entire book in $6$ days

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

$=1\frac{3}{4}\times 6=\frac{7}{4}\times 6=\frac{21}{2}=10\frac{1}{2}$ 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   $\text{2}\frac{\text{3}}{\text{4}}$ litres of 

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

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

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

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

(i) Provide the number in the box , such that 

$\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\text{=}\frac{\text{10}}{\text{30}}$  

Ans: The number inside the box should be $\frac{2}{3}\times =\frac{10}{30}$ 

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

Ans: The simplest form of the number obtained in 

$\frac{\text{5}}{\text{10}}\,\text{is}\,\frac{\text{1}}{\text{2}}$

(i) Provide the number in the box $$ , such that $\frac{3}{5}\times =\frac{24}{75}$ .

Ans: The number inside the box should be $\frac{3}{5}\times =\frac{24}{75}$ 

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

$\frac{\text{8}}{\text{15}}\,\text{is}\,\frac{\text{8}}{\text{15}}$

Exercise - 2.3

(i) $\text{12 }\!\!\div\!\!\text{ }\frac{\text{3}}{\text{4}}$ 

Ans: Calculating the value,

$12\div \frac{3}{4}=12\times \frac{4}{3}=16$

(ii) $\text{14 }\!\!\div\!\!\text{ }\frac{\text{5}}{\text{6}}$ 

$14\div \frac{5}{6}=14\times \frac{6}{5}=\frac{84}{5}=16\frac{4}{5}$ 

(iii) $\text{8 }\!\!\div\!\!\text{ }\frac{\text{7}}{\text{3}}$ 

Ans:  Calculating the value,

$8\div \frac{7}{3}=8\times \frac{3}{7}=\frac{24}{7}=3\frac{3}{7}$

(iv) $\text{4 }\!\!\div\!\!\text{ }\frac{\text{8}}{\text{3}}$ 

$4\div \frac{8}{3}=4\times \frac{3}{8}=\frac{3}{2}=1\frac{1}{2}$

(v) $\text{3 }\!\!\div\!\!\text{ 2}\frac{\text{1}}{\text{3}}$ 

$3\div 2\frac{1}{3}=3\div \frac{7}{3}=3\times \frac{3}{7}=\frac{9}{7}=1\frac{2}{7}$           

(vi) \[\text{5 }\!\!\div\!\!\text{ 3}\frac{\text{4}}{\text{7}}\] 

$5\div 3\frac{4}{7}=5\div \frac{25}{7}=5\times \frac{7}{25}=\frac{7}{5}=1\frac{2}{5}$

2. Find the reciprocal of each of the following fractions. Classify the 

reciprocals as proper fraction, improper fractions and whole numbers.

(i) $\frac{\text{3}}{\text{7}}$

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{3}}{\text{7}}\text{=}\frac{\text{7}}{\text{3}}\to \text{Improper}\,\text{fraction}$  

(ii) $\frac{\text{5}}{\text{8}}$

Reciprocal of$\frac{\text{5}}{\text{8}}\text{=}\frac{\text{8}}{\text{5}}\to \text{Improper}\,\text{fraction}$

(iii) $\frac{\text{9}}{\text{7}}$

Reciprocal of $\frac{\text{9}}{\text{7}}\text{=}\frac{\text{7}}{\text{9}}\to \text{Proper}\,\text{fraction}$

(iv) $\frac{\text{6}}{\text{5}}$ 

Reciprocal of $\frac{\text{6}}{\text{5}}\text{=}\frac{\text{5}}{\text{6}}\to \text{Proper}\,\text{fraction}$

(v) $\frac{\text{12}}{\text{7}}$ 

Reciprocal of $\frac{\text{12}}{\text{7}}\text{=}\frac{\text{7}}{\text{12}}\to \text{Proper}\,\text{fraction}$  

(vi) $\frac{\text{1}}{\text{8}}$

Reciprocal of $\frac{\text{9}}{\text{7}}\text{=8}\to \text{Whole number}$

(vii) $\frac{\text{1}}{\text{11}}$ 

Reciprocal of $\frac{\text{1}}{\text{11}}\text{=11}\to \text{Whole number}$

(i) $\frac{\text{7}}{\text{3}}\text{ }\!\!\div\!\!\text{ 2}$ 

$\frac{7}{3}\div 2=\frac{7}{3}\times \frac{1}{2}=\frac{7\times 1}{3\times 2}=\frac{7}{6}=1\frac{1}{6}$

(ii) $\frac{\text{4}}{\text{9}}\text{ }\!\!\div\!\!\text{ 5}$

$\frac{4}{9}\div 5=\frac{4}{9}\times \frac{1}{5}=\frac{4\times 1}{9\times 5}=\frac{4}{45}$ 

(iii) $\frac{\text{6}}{\text{13}}\text{ }\!\!\div\!\!\text{ 7}$ 

$\frac{6}{13}\div 7=\frac{6}{13}\times \frac{1}{7}=\frac{6\times 1}{13\times 7}=\frac{6}{91}$ 

(iv) $\text{4}\frac{\text{1}}{\text{3}}\text{ }\!\!\div\!\!\text{ 3}$

Ans:  Calculating the value,

$4\frac{1}{3}\div 3=\frac{13}{3}\div 3=\frac{13}{3}\times \frac{1}{3}=\frac{13}{9}=1\frac{4}{9}$

(v) $\text{3}\frac{\text{1}}{\text{2}}\text{ }\!\!\div\!\!\text{ 4}$ 

$3\frac{1}{2}\div 4=\frac{7}{2}\div 4=\frac{7}{2}\times \frac{1}{4}=\frac{7}{8}$

(vi) $\text{4}\frac{\text{3}}{\text{7}}\text{ }\!\!\div\!\!\text{ 7}$ 

$4\frac{3}{7}\div 7=\frac{31}{7}\div 7=\frac{31}{7}\times \frac{1}{7}=\frac{31}{49}$

(i) $\frac{\text{2}}{\text{5}}\text{ }\!\!\div\!\!\text{ }\frac{\text{1}}{\text{2}}$

$\frac{2}{5}\div \frac{1}{2}=\frac{2}{5}\times \frac{2}{1}=\frac{2\times 2}{5\times 1}=\frac{4}{5}$ 

(ii) $\frac{\text{4}}{\text{9}}\text{ }\!\!\div\!\!\text{ }\frac{\text{2}}{\text{3}}$ 

$\frac{4}{9}\div \frac{2}{3}=\frac{4}{9}\times \frac{3}{2}=\frac{2}{3}$

(iii) $\frac{\text{3}}{\text{7}}\text{ }\!\!\div\!\!\text{ }\frac{\text{8}}{\text{7}}$

$\frac{3}{7}\div \frac{8}{7}=\frac{3}{7}\times \frac{7}{8}=\frac{3}{8}$

(iv) $\text{2}\frac{\text{1}}{\text{3}}\text{ }\!\!\div\!\!\text{ }\frac{\text{3}}{\text{5}}$ 

$2\frac{1}{3}\div \frac{3}{5}=\frac{7}{3}\div \frac{3}{5}=\frac{7}{3}\times \frac{5}{3}=\frac{35}{9}=3\frac{8}{9}$

(v) $\text{3}\frac{\text{1}}{\text{2}}\text{ }\!\!\div\!\!\text{ }\frac{\text{8}}{\text{3}}$ 

$3\frac{1}{2}\div \frac{8}{3}=\frac{7}{2}\div \frac{3}{8}=\frac{7}{2}\times \frac{3}{8}=\frac{7\times 3}{2\times 8}=\frac{21}{16}=1\frac{5}{16}$

(vi) $\frac{\text{2}}{\text{5}}\text{ }\!\!\div\!\!\text{ 1}\frac{\text{1}}{\text{2}}$

$2\frac{1}{3}\div \frac{3}{5}=\frac{2}{5}\div 1\frac{1}{2}=\frac{2}{5}\div \frac{3}{2}=\frac{2}{5}\times \frac{2}{3}=\frac{2\times 2}{5\times 3}=\frac{4}{15}$

(vii) $\text{3}\frac{\text{1}}{\text{5}}\text{ }\!\!\div\!\!\text{ 1}\frac{\text{2}}{\text{3}}$

Ans:   Calculating the value,

$3\frac{1}{5}\div 1\frac{2}{3}=\frac{16}{5}\div \frac{5}{3}=\frac{16}{5}\times \frac{3}{5}=\frac{16\times 3}{5\times 5}=\frac{48}{25}=1\frac{23}{25}$

(viii) $\text{2}\frac{\text{1}}{\text{5}}\text{ }\!\!\div\!\!\text{ 1}\frac{\text{1}}{\text{5}}$ 

$2\frac{1}{5}\div 1\frac{1}{5}=\frac{11}{5}\div \frac{6}{5}=\frac{11}{5}\times \frac{5}{6}=\frac{11}{6}=1\frac{5}{6}$

Exercise 2.4

(i) $\text{0}\text{.2 }\!\!\times\!\!\text{ 6}$ 

\[0.2\times 6=1.2\]

(ii) $\text{8 }\!\!\times\!\!\text{ 4}\text{.6}$

\[8\times 4.6=36.8\]

(iii) $\text{2}\text{.71 }\!\!\times\!\!\text{ 5}$ 

\[2.71\times 5=13.55\]

(iv) $\text{20}\text{.1 }\!\!\times\!\!\text{ 4}$ 

\[20.1\times 4=80.4\]

(v) $\text{0}\text{.05 }\!\!\times\!\!\text{ 7}$ 

\[0.05\times 7=0.35\]

(vi) $\text{211}\text{.02 }\!\!\times\!\!\text{ 4}$ 

\[211.02\times 4=844.08\]

(vii) $\text{2 }\!\!\times\!\!\text{ 0}\text{.86}$ 

\[2\times 0.86=1.72\]

2. Find the area of rectangle whose length is \[\text{5}\text{.7 cm}\] and 

breadth is \[\text{3 cm}\text{.}\]

Ans: Given: The \[\text{Length of rectangle = 5}\text{.7 cm and Breadth of 

rectangle = 3 cm}\] 

Applying the area of rectangle formula,

\[\text{Area of rectangle = Length x Breadth}\] 

\[\text{= 5}\text{.7 x 3 = 17}\text{.1 c}{{\text{m}}^{2}}\]  

Hence, the area of rectangle is $\text{17}\text{.1}\,\text{c}{{\text{m}}^{\text{2}}}$.

(i) \[\text{1}\text{.3 }\!\!\times\!\!\text{ 10}\] 

$1.3\times 10=13.0$

(ii) \[\text{36}\text{.8 }\!\!\times\!\!\text{ 10}\] 

$36.8\times 10=368.0$

(iii) \[\text{153}\text{.7 }\!\!\times\!\!\text{ 10}\] 

$153.7\times 10=1537.0$

(iv) \[\text{168}\text{.07 }\!\!\times\!\!\text{ 10}\]

$168.07\times 10=1680.7$

(v) \[\text{31}\text{.1 }\!\!\times\!\!\text{ 100}\] 

$31.1\times 100=3110.0$

(vi) \[\text{156}\text{.1 }\!\!\times\!\!\text{ 100}\] 

$156.1\times 100=15610.0$

(vii) \[\text{3}\text{.62 }\!\!\times\!\!\text{ 100}\] 

$3.62\times 100=362.0$

(viii) \[\text{43}\text{.07 }\!\!\times\!\!\text{ 100}\] 

$43.07\times 100=4307.0$

(ix) \[\text{0}\text{.5 }\!\!\times\!\!\text{ 10}\] 

$0.5\times 10=5.0$ 

(x) \[\text{0}\text{.08 }\!\!\times\!\!\text{ 10}\] 

$0.08\times 10=0.80$ 

(xi) \[\text{0}\text{.9 }\!\!\times\!\!\text{ 100}\]

$0.9\times 100=90.0$

(xii) \[\text{0}\text{.03 }\!\!\times\!\!\text{ 1000}\] 

$0.03\times 1000=30.0$ 

4. A two-wheeler covers a distance of\[\text{ }\!\!~\!\!\text{ 55}\text{.3 

km}\] 

in one litre of petrol. How much distance will it cover in \[\text{10 litres}\] of 

Ans: Given: In one litre a two-wheeler covers a distance\[\text{ = 55}\text{.3 

Since distance covered in one litre by a two-wheeler\[\text{ = 55}\text{.3 km}\]

\[\therefore \,\,\text{In 10 litrs, a two- wheeler covers a distance = 55}\text{.3 x 10 = 553}\text{.0 km}\] 

Hence, $553$ km distance will be covered by two-wheeler in $10$ litres of petrol.

5.  Find:

(i) $\text{2}\text{.5 }\!\!\times\!\!\text{ 0}\text{.3}$

\[\text{2}\text{.5 x 0}\text{.3 = 0}\text{.75}\] 

(ii) $\text{0}\text{.1 }\!\!\times\!\!\text{ 51}\text{.7}$ 

\[\text{0}\text{.1 x 51}\text{.7 = 5}\text{.17}\]

(iii) $\text{0}\text{.2 }\!\!\times\!\!\text{ 316}\text{.8}$ 

\[\text{0}\text{.2 x 316}\text{.8 = 63}\text{.36}\]

(iv) $\text{1}\text{.3 }\!\!\times\!\!\text{ 1}\text{.3}$ 

\[\text{1}\text{.3 x 3}\text{.1 = 4}\text{.03}\]

(v) $\text{0}\text{.5 }\!\!\times\!\!\text{ 0}\text{.05}$ 

\[\text{0}\text{.5 x 0}\text{.05 = 0}\text{.025}\]

(vi) $\text{11}\text{.2 }\!\!\times\!\!\text{ 0}\text{.15}$ 

\[\text{11}\text{.2 x 0}\text{.15 = 1}\text{.680 }\]

(vii) $\text{1}\text{.07 }\!\!\times\!\!\text{ 0}\text{.02}$ 

\[\text{1}\text{.07 x 0}\text{.02 = 0}\text{.0214}\]

(viii) $\text{10}\text{.05 }\!\!\times\!\!\text{ 1}\text{.05}$

\[\text{10}\text{.05 x 1}\text{.05 = 10}\text{.5525}\]

(ix) $\text{101}\text{.01 }\!\!\times\!\!\text{ 0}\text{.01}$ 

\[\text{101}\text{.01 x 0}\text{.01 = 1}\text{.0101}\]

(x) $\text{100}\text{.01 }\!\!\times\!\!\text{ 1}\text{.1}$

\[\text{100}\text{.01 x 1}\text{.1 = 110}\text{.11 }\]

Exercise 2.5

(i) \[\text{0}\text{.4  }\!\!\div\!\!\text{  2}\] 

\[0.4\div 2=\frac{4}{10}\times \frac{1}{2}=\frac{2}{10}=0.2\]

(ii) \[\text{0}\text{.35  }\!\!\div\!\!\text{  5}\] 

\[0.35\div 5=\frac{35}{100}\times \frac{1}{5}=\frac{7}{100}=0.07\]

(iii) \[\text{2}\text{.48  }\!\!\div\!\!\text{  4}\] 

\[2.48\div 4=\frac{248}{100}\times \frac{1}{4}=\frac{62}{100}=0.62\]

(iv) \[\text{65}\text{.4  }\!\!\div\!\!\text{  6}\] 

\[65.4\div 6=\frac{654}{10}\times \frac{1}{6}=\frac{109}{10}=10.9\]

(v) \[\text{651}\text{.2  }\!\!\div\!\!\text{  4}\] 

\[651.2\div 4=\frac{6512}{10}\times \frac{1}{4}=\frac{1628}{10}=162.8\]

(vi) \[\text{14}\text{.49  }\!\!\div\!\!\text{  7 }\] 

\[14.49\div 7=\frac{1449}{100}\times \frac{1}{7}=\frac{207}{100}=2.07\]

(vii) \[\text{3}\text{.96  }\!\!\div\!\!\text{  4}\]

\[3.96\div 4=\frac{396}{100}\times \frac{1}{4}=\frac{99}{100}=0.99\]

(viii) \[\text{0}\text{.80  }\!\!\div\!\!\text{  5}\] 

\[0.80\div 5=\frac{80}{100}\times \frac{1}{5}=\frac{16}{100}=0.16\]

(i) \[\text{4}\text{.8  }\!\!\div\!\!\text{  10}\] 

Ans: Performing the given calculation,

$4.8\div 10=\frac{4.8}{10}=0.48$

(ii) \[\text{52}\text{.5  }\!\!\div\!\!\text{  10}\] 

$52.5\div 10=\frac{52.5}{10}=5.25$

(iii) \[\text{0}\text{.7  }\!\!\div\!\!\text{  10}\] 

$0.7\div 10=\frac{0.7}{10}=0.07$

(iv) \[\text{33}\text{.1  }\!\!\div\!\!\text{  10}\]

$33.1\div 10=\frac{33.1}{10}=3.31$

(v) \[\text{272}\text{.23  }\!\!\div\!\!\text{  10}\] 

$272.23\div 10=\frac{272.23}{10}=27.223$

(vi) \[\text{0}\text{.56  }\!\!\div\!\!\text{  10 }\] 

$0.56\div 10=\frac{0.56}{10}=0.056$

(vii) \[\text{3}\text{.97  }\!\!\div\!\!\text{  10}\] 

$3.97\div 10=\frac{3.97}{10}=0.397$

(i) \[\text{2}\text{.7  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,

$2.7\div 100=\frac{27}{10}\times \frac{1}{100}=\frac{27}{1000}=0.027$ 

(ii) \[\text{0}\text{.3  }\!\!\div\!\!\text{  100 }\]

$0.3\div 100=\frac{3}{10}\times \frac{1}{100}=\frac{3}{1000}=0.003$

(iii) \[\text{0}\text{.78  }\!\!\div\!\!\text{  100}\] 

$0.78\div 100=\frac{78}{10}\times \frac{1}{100}=\frac{78}{1000}=0.0078$

(iv) \[\text{432}\text{.6  }\!\!\div\!\!\text{  100}\] 

$432.6\div 100=\frac{4326}{10}\times \frac{1}{100}=\frac{4326}{1000}=4.326$

(v) \[\text{23}\text{.6  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,$23.6\div 100=\frac{236}{10}\times \frac{1}{100}=\frac{236}{1000}=0.236$

(vi) \[\text{98}\text{.53  }\!\!\div\!\!\text{  100}\] 

$98.53\div 100=\frac{9853}{10}\times \frac{1}{100}=\frac{9853}{1000}=0.9853$

(i) \[\text{7}\text{.9  }\!\!\div\!\!\text{  1000}\] 

$7.9\div 1000=\frac{79}{10}\times \frac{1}{1000}=\frac{79}{10000}=0.0079$ 

(ii) \[\text{26}\text{.3  }\!\!\div\!\!\text{  1000}\]

$26.3\div 1000=\frac{263}{10}\times \frac{1}{1000}=\frac{263}{10000}=0.0263$

(iii) \[\text{38}\text{.53  }\!\!\div\!\!\text{  1000}\] 

$38.53\div 1000=\frac{3853}{10}\times \frac{1}{1000}=\frac{3853}{10000}=0.03853$

(iv) \[\text{128}\text{.9  }\!\!\div\!\!\text{  1000}\] 

$128.9\div 1000=\frac{1289}{10}\times \frac{1}{1000}=\frac{1289}{10000}=0.1289$

(v) \[\text{0}\text{.5  }\!\!\div\!\!\text{  1000}\] 

$0.5\div 1000=\frac{5}{10}\times \frac{1}{1000}=\frac{5}{10000}=0.0005$

(i) \[\text{7  }\!\!\div\!\!\text{  3}\text{.5}\] 

$7\div 3.5=7\div \frac{35}{10}=7\times \frac{10}{35}=\frac{10}{5}=2$

(ii) \[\text{36  }\!\!\div\!\!\text{  0}\text{.2 }\]

$36\div 0.2=36\div \frac{2}{10}=36\times \frac{10}{2}=18\times 10=180$

(iii) \[\text{3}\text{.25  }\!\!\div\!\!\text{  0}\text{.5}\]  

Ans: Converting the terms in fraction form and calculating the value,$3.25\div 0.5=\frac{325}{100}\div \frac{5}{10}=\frac{325}{100}\times \frac{10}{5}=\frac{65}{10}=6.5$

(iv) \[\text{30}\text{.94  }\!\!\div\!\!\text{  0}\text{.7}\]

$30.94\div 0.7=\frac{3094}{100}\div \frac{7}{10}=\frac{3094}{100}\times \frac{10}{7}=\frac{442}{10}=44.2$

(v) \[\text{0}\text{.5  }\!\!\div\!\!\text{  0}\text{.25 }\] 

Ans: Converting the terms in fraction form and calculating the value,$0.5\div 0.25=\frac{5}{10}\div \frac{25}{100}=\frac{5}{10}\times \frac{100}{25}=\frac{10}{5}=2$

(vi) \[\text{7}\text{.75  }\!\!\div\!\!\text{  0}\text{.25}\] 

$7.75\div 0.25=\frac{775}{100}\div \frac{25}{100}=\frac{775}{100}\times \frac{100}{25}=31$

(vii) \[\text{76}\text{.5  }\!\!\div\!\!\text{  0}\text{.15}\] 

$76.5\div 0.15=\frac{765}{100}\div \frac{15}{100}=\frac{765}{10}\times \frac{100}{15}=51\times 10=510$

(viii) \[\text{37}\text{.8  }\!\!\div\!\!\text{  1}\text{.4}\]

$37.8\div 1.4=\frac{378}{10}\div \frac{14}{10}=\frac{378}{10}\times \frac{10}{14}=27$

(ix) \[\text{2}\text{.73  }\!\!\div\!\!\text{  1}\text{.3 }\] 

$2.73\div 1.3=\frac{273}{100}\div \frac{13}{10}=\frac{273}{100}\times \frac{10}{13}=\frac{21}{10}=2.1$

6. A vehicle covers a distance of \[\text{43}\text{.2 km}\] in

\[\text{2}\text{.4}\]litres of petrol. How much distance will it cover in one 

litre 

Ans: Given: \[\,\,\,\text{In 2}\text{.4 litres of petrol, distance covered by the vehicle = 43}\text{.2 km}\]

Since,\[\,\,\,\text{In 2}\text{.4 litres of petrol, distance covered by the vehicle = 43}\text{.2 km}\]

\[\therefore \,\,\text{In 1 litre of petrol, distance covered by the vehicle = 43}\text{.2  }\!\!\div\!\!\text{  2}\text{.4}\] 

Performing the required calculations,

$=\frac{432}{10}\div \frac{24}{10}=\frac{432}{10}\times \frac{24}{10}$ 

$\text{=18}\,\text{km}$ 

Hence, the vehicle can cover \[\text{18 km}\] distance in one litre of petrol.

NCERT Solutions for Class 7 Chapter 2 Maths PDF Download

2.1 introduction.

In NCERT Solutions Class 7 Chapter 2 Maths, students will learn about fractions and decimals. In junior classes, students have learned about what is a fraction and its types: proper, improper, mixed fractions, etc. Now, in class 7, we are going to learn about multiplication and division of fractions. The concept of fractions mainly focuses on the ratios and proportions, how to distribute etc. At the same time, decimals are the accurate values obtained after the division.

2.2 Recollect

In NCERT Solutions Class 7 Maths Chapter 2, students need to think again on the topics they have learned so far in the previous classes. These include representation of fractions on the number line, ordering of fractions, addition and subtraction of fractions, decimals and their additions, how to keep a point, etc. These are reminded in the first two exercises.

2.3 Multiplication of Fractions

In this section, students can understand how to multiply two fractions. If students have values like a and b, they can say ab is the product of a and b. If the values are like p/q, a/b then, how can we multiply? To multiply these fractions, it has two different methods. One is by using a  whole number and the other is by using a portion.

2.3.1 Multiplication of Fractions Using the Whole Number

Here, let us see what Fraction tells us? It explains that a down part is a whole number (except zero) and the upper part is the integer. In a fraction, the down part is known as the denominator whereas the upper part is the numerator. We use a whole number to multiply fractions if they are the same. For instance, let's say we have p/q. Then we can multiply with the whole number as 3*p/q. It is also applicable for improper or mixed fractions. But students need to make them into simpler forms before multiplying.

2.3.1 Multiplication of Fractions Using the Fraction

In this section, students can learn how to multiply two fractions when they are dissimilar. Students use a fraction to multiply them. The formula for multiplying two fractions is,(product of numerators)/(product of denominators).

The resultant product is less than the two fractions if we multiply two proper fractions. On the other hand, the result is greater than the two fractions if we multiply two improper fractions.

2.4 Division of Fractions

Let's discuss the division of fractions. Students can divide a fraction by a whole number and a whole number by a fraction. Here is a particular case to keep in mind. If two fractions for which numerator and denominator are in reverse order, then they are called reciprocals to each other.  Their product is always 1.

In the same way, while dividing mixed fractions with a whole number, students need to change the mixed fraction into improper fractions. Then it is easy to divide and solve. Next, we have to learn to divide a fraction with another fraction by changing one of the fractions into its reciprocal form.

The three concepts are explained differently in the NCERT Solutions for Class 7 Maths Chapter 2 PDF book available on Vedantu for students to go through if necessary.

2.5 Recalling Decimals

Decimals are the proper forms to represent the results obtained from multiplication and division. Placing the point in between numbers plays a vital role. One can express the heights, distances, weights,  measuring values, interest rates, shares, and fractions, also using decimals. To change the place value of the point, we can multiply by 10,100,...... Let's have a glance at the addition and subtraction of decimals.

Overview of Deleted Syllabus for CBSE Class 7 Maths Chapter 2 Fractions and Decimals

Class 7 Maths Chapter 2: Exercises Breakdown

NCERT Solutions for Chapter 2 fraction and decimals Class 7 offers practice problems and clear explanations to help master these concepts. Students should focus on fraction addition, subtraction, multiplication, and division for both like and unlike denominators. They should also be able to convert between fractions and decimals easily and understand relative fraction sizes. By carefully working through NCERT Class 7 Maths Chapter 2 Exercises, students can improve their knowledge and prepare for exams.

Other Study Material for CBSE Class 7 Maths Chapter 2

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals

1. What are the main topics and subtopics covered in chapter 2 of Class 7 Maths?

Topics and sub-topics discussed in Chapter 2 Fractions and Decimals of NCERT Solutions Class 7 Maths are: Addition and Subtraction of Fractions, Multiplication of Fraction, Multiplication of a Fraction by a Whole Number, Multiplication of a Fraction by a Fraction, Division of Fraction, Division of Whole Number by a Fraction, Reciprocal of Fraction, Division of a fraction by a Whole Number, Division of Fraction by Another Fraction, Multiplication of Decimal Numbers, Multiplication of Decimal Numbers by 10, 100 and 1000, Division of Decimal Numbers, Division of Decimals by 10, 100 and 1000, Division of a Decimal Number by a Whole Number and Division of a Decimal Number by Another Decimal Number.

2. How many questions are there in exercises of chapter 2 of Class 7 Maths?

There are a total of seven exercises given in the second chapter of Class 7 Maths. Exercise 2.1 has 8 questions, exercise 2.2 has 8 questions, exercise 2.3 has 8 questions, exercise 2.4 has 4 questions, exercise 2.5 has 9 questions, exercise 2.6 has 5 questions, and exercise 2.7 has 6 questions.

3. How do I compare two fractions class 7 maths ch 2?

By examining the numerators, fractions with the same denominator can be compared. It is a larger fraction with a larger numerator. You can convert fractions with different denominators to equivalent fractions with the same denominator so that you can compare them.

4. What are the different types of fractions?

There are three main types of fractions: proper fractions, improper fractions, and mixed numbers.

Proper fractions: These are fractions whose numerator is smaller than the denominator. For example, 1/2 and 3/5 are proper fractions.

Improper fractions: These are fractions whose numerator is larger than or equal to the denominator. For example, 5/3 and 8/2 are improper fractions.

Mixed numbers: These are numbers that are made up of a whole number and a fraction. For example, 2 1/2 is a mixed number.

5. What are the different types of decimals?

There are two main types of decimals: terminating decimals and non-terminating decimals.

Terminating decimals: These are decimals that end after a finite number of digits. For example, 0.5 and 1.23 are terminating decimals.

Non-terminating decimals: These are decimals that do not end after a finite number of digits. For example, 1/3 and 1/2 are non-terminating decimals.

6. What are the uses of fractions and decimals?

Fractions and decimals are used in a variety of ways in mathematics and everyday life.

In mathematics, fractions and decimals are used to represent parts of a whole, to perform arithmetic operations, and to solve equations.

In everyday life, fractions and decimals are used to represent quantities such as money, time, and measurements.

7. What's important in class 7th maths chapter 2 ?

This chapter covers various aspects of fractions and decimals in class 7th maths chapter 2 . Here's what to focus on:

Understanding different types of fractions (proper, improper, mixed)

Converting between fractions and decimals

Comparing and ordering fractions

Adding and subtracting fractions and decimals

Representing fractions on the number line

8. What are like terms in addition and subtraction of fractions in class 7 chapter 2 maths?

Like terms in fractions, they have the same denominator. When adding or subtracting fractions, make sure they have the same denominator before performing the operation.

9. How do I represent fractions on the number line in fraction and decimals class7?

Divide the space between two whole numbers on the number line according to the denominator of the fraction. Each small interval represents 1/denominator. Shade the required number of intervals based on the numerator.

NCERT Solutions for Class 7 Maths

Ncert solutions for class 7.

• NCERT Exemplar
• NCERT Exemplar Class 7
• Class 7 Maths
• Class 7 Maths Chapter 2

NCERT Exemplar Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT Exemplar Solutions for Class 7 Maths Chapter 2 Fractions and Decimals is the best study material for those students who have difficulties in solving problems. These solutions can help students clear their doubts quickly and also understand the topic effectively. Our expert faculty formulated these solutions to assist them with their exam preparation to attain good marks in the annual exam. Students who wish to score good marks in Maths subject should practise NCERT Exemplar Solutions for Class 7 Maths.

A fraction simply tells us how many parts of a whole we have. You can represent a fraction by the slash that is written in between two numbers. The top number is called the numerator, and the bottom number is called the denominator. A decimal is defined as a fraction whose denominator is a power of ten and whose numerator is expressed by figures placed to the right of a decimal point. NCERT Exemplar Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals PDF are available here. Now, let us have a look at the topics discussed in this chapter.

• Addition, Subtraction, Division and Multiplication of Fractions
• Multiplication of a Fraction by a Whole Number
• Division of Whole Number by a Fraction
• Reciprocal of Fraction
• Division of a Fraction by a Whole Number
• Multiplication and Division of Decimal Numbers

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Access Answers to Maths NCERT Exemplar Solutions for Class 7 Chapter 2 Fractions and Decimals

Exercise Page: 38

In questions 1 to 20, out of four options, only one is correct. Write the correct answer.

(a) 26/25 (b) 52/25 (c) 2/5 (d) 6

First, we have to convert the mixed fraction into an improper fraction

Then, (2/5) × (26/5)

2. 3¾ ÷ ¾ is equal to:

(a) 3 (b) 4 c) 5 (d) 45/16

First, we have to convert the mixed fraction into improper fraction 3¾ = 15/4

Then, 15/4 ÷ ¾

= (15/4)/ (¾)

= (15/4) × (4/3)

= (15 × 4)/ (4 ×3)

= (5 × 1)/ (1 ×1)

3. A ribbon of length 5¼ m is cut into small pieces, each of length ¾m. The number of pieces will be:

(a) 5 (b) 6 (c) 7 (d) 8

First, we have to convert the mixed fraction into improper fraction 5¼ = 21/4

Then, 21/4 ÷ ¾

= (21/4)/ (¾)

= (21/4) × (4/3)

= (21 × 4)/ (4 × 3)

= (7 × 1)/ (1 × 1)

4. The ascending arrangement of (2/3), (6/7), (13/21) is:

(a) 6/7, 2/3, 13/21 (b) 13/21, 2/3, 6/7

(c) 6/7, 13/21, 2/3 (d) 2/3, 6/7, 13/21

(b) 13/21, 2/3, 6/7

LCM of 21, 3, 7 = 21

Now, let us change each of the given fractions into an equivalent fraction having 21 as the denominator.

(13/21) < (14/21) < (18/21)

(13/21) < (2/3) < (6/7)

Hence, the given fractions in ascending order are (13/21), (2/3), (6/7)

5. Reciprocal of the fraction 2/3 is:

(a) 2 (b) 3 (c) 2/3 (d) 3/2

The reciprocal of a non-zero fraction is obtained by interchanging its numerator and denominator.

6. The product of 11/13 and 4 is:

8. Pictorial representation of 3 × 2/3 is:

In the above figure, three circles are divided into 3 equal parts.

Out of 3 equal parts 2 equal parts are hatched.

9. 1/5 ÷ 4/5 equal to:

(a) 4/5 (b) 1/5 (c) 5/4 (d) ¼

= 1/5 ÷ 4/5

= (1/5)/ (4/5)

= (1/5) × (5/4)

10. The product of 0.03 × 0.9 is:

(a) 2.7 (b) 0.27 (c) 0.027 (d) 0.0027

0.03 × 0.9 can be written as = (3/100) × (9/10)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

11. (5/7) ÷ 6 is equal to:

(a) 30/7 (b) 5/42 (c) 30/42 (d) 6/7

= 5/7 ÷ 6/1

= (5/7)/ (6/1)

= (5/7) × (1/6)

= 31/6 ÷ 9/2

= (31/6)/ (9/2)

= (31/6) × (2/9)

= (31/3) × (1/9)

13. Which of the following represents 1/3 of 1/6?

(a) (1/3) + (1/6) (b) (1/3) – (1/6)

(c) (1/3) × (1/6) (d) (1/3) ÷ (1/6)

(c) (1/3) × (1/6)

14. 3/7 of 2/5 is equal to

(a) 5/12 (b) 5/35 (c) 1/35 (d) 6/35

3/7 of 2/5 is equal to = (3/7) × (2/5)

(a) less than 30 cups

(b) between 30 cups and 40 cups

(c) between 40 cups and 50 cups

(d) above 50 cups

From the question, it is given that,

One packet of biscuits requires 2½ cups of flour = 5/2

Total ingredients for one packet of biscuits = (5/2) + (5/3)

= (15 + 10)/6

Then, the total quantity of both ingredients used in 10 such packets of biscuits = 10 × (25/6)

= 5 × (25/3)

16. The product of 7 and 6¾ is

(a) 42¼ (b) 47¼ (c) 42¾ (d) 47¾

First, we have to convert the mixed fraction into improper fraction 6¾ = 27/4

= 7 × (27/4)

17. On dividing 7 by 2/5, the result is

(a) 14/2 (b) 35/4 (c) 14/5 (d) 35/2

= 7 × (5/2)

(a) 8/15 (b) 40/3 (c) 40/5 (d) 8/3

= (8/3) ÷ 5

= (8/3)/(5/1)

= (8/3) × (1/5)

19. 4/5 of 5 kg apples were used on Monday. The next day 1/3 of what was left was used. Weight (in kg) of apples left now is

(a) 2/7 (b) 1/14 (c) 2/3 (d) 4/21

4/5 of 5 kg apples were used on Monday = (4/5) × 5

The next day 1/3 of what was left was used = (1/3) × 1

So, the Weight (in kg) of apples left now is = 1 – (1/3)

= (3 – 1)/3

= 2/3 kg of apples

20. The picture

(a) ¼ ÷ 3 (b) 3 × ¼ (c) ¾ × 3 (d) 3 ÷ ¼

From the given picture, ¼ + ¼ + ¼ = ¾

In Questions 21 to 44, fill in the blanks to make the statements true.

21. Rani ate 2/7 part of a cake while her brother Ravi ate 4/5 of the remaining. Part of the cake left is

Rani ate 2/7 part of a cake while her brother Ravi ate 4/5 of the remaining. Part of the cake left is 1/7.

Now, let us assume total part of cake be 1.

Then, from the question, given that Rani ate 2/7 part of a cake = 1 – (2/7)

= (7 – 2)/7

So, Ravi ate 4/5 of the remaining cake = (4/5) × (5/7)

Therefore, the part of the cake left = (5/7) – (4/7)

22. The reciprocal of 3/7 is

The reciprocal of 3/7 is 7/3

23. 2/3 of 27 is

2/3 of 27 is 18.

24. 4/5 of 45 is

4/5 of 45 is 36.

Then, 4 × 19/3

Then, ½ × (30/7)

27. 1/9 of 6/5 is

1/9 of 6/5 is 2/15.

= (1/9) × (6/5)

= (1/3) × (2/5)

Then, (17/7) × (7/9)

29. (4/5) ÷ 4 is equal to

(4/5) ÷ 4 is equal to 1/5.

= (4/5) ÷ 4

= (4/5) × (¼)

= (1/5) × (1/1)

30. 2/5 of 25 is

2/5 of 25 is 10

31. (1/5) ÷ (5/6) = (1/5) (6/5)

(1/5) ÷ (5/6) = (1/5) × (6/5)

While dividing one fraction by another fraction, we multiply the first fraction by the reciprocal of the other.

32. 3.2 × 10 = _______

3.2 × 10 = 32

To multiply a decimal number by 10, we move the decimal point in the number to the right by as many places as many zeros (0) are at the right of one.

33. 25.4 × 1000 = _______

25.4 × 1000 = 25400

To multiply a decimal number by 1000, we move the decimal point in the number to the right by as many places as many zeros (0) are at the right of one.

34. 93.5 × 100 = _______

93.5 × 100 = 9350

To multiply a decimal number by 100, we move the decimal point in the number to the right by as many places as many zeros (0) are at the right of one.

35. 4.7 ÷ 10 = ______

4.7 ÷ 10 = 0.47

To divide a decimal number by 10, shift the decimal point in the decimal number to the left by as many places as there are zeros over 1, to get the quotient.

36. 4.7 ÷ 100 = _____

4.7 ÷ 100 = 0.047

To divide a decimal number by 100, shift the decimal point in the decimal number to the left by as many places as there are zeros over 1, to get the quotient.

37. 4.7 ÷ 1000 = ______

4.7 ÷ 1000 = 0.0047

To divide a decimal number by 1000, shift the decimal point in the decimal number to the left by as many places as there are zeros over 1, to get the quotient.

38. The product of two proper fractions is _______ than each of the fractions that are multiplied.

The product of two proper fractions is less than each of the fractions that are multiplied.

Consider the two proper fractions, 4/5 and 2/4

= (4/5) × (2/4)

Then, 0.4 is multiplied to the proper fraction less than each of the fractions that are multiplied = (4/5) × 0.4

39. While dividing a fraction by another fraction, we _________ the first fraction by the _______ of the other fraction.

While dividing a fraction by another fraction, we multiply the first fraction by the reciprocal of the other fraction.

= (1/5) ÷ (5/6)

= (1/5) × (6/5)

40. 8.4 ÷ = 2.1

8.4 ÷ 4 = 2.1

Let us assume the missing fraction be x,

8.4 ÷ x = 2.1

8.4/x = 2.1

By cross multiplication, we get,

x = 8.4/2.1

41. 52.7 ÷ _______ = 0.527

52.7 ÷ 100 = 0.527

52.7 ÷ x = 0.527

52.7/x = 0.527

x = 52.7/0.527

42. 0.5 _____ 0.7 = 0.35

0.5 × 0.7 = 0.35

While multiplying two decimal numbers, first multiply them as whole numbers. Count the number of digits to the right of the decimal point in both the decimal numbers. Add the number of digits counted. Put the decimal point in the product by counting the number of digits equal to the sum obtained from its rightmost place.

= 0.5 × 0.7

43. 2 (5/3) = 10/3

2 × (5/3) = 10/3

44. 2.001 ÷ 0.003 = __________

2.001 ÷ 0.003 = 667

= 2.001/0.003

In each of the Questions 45 to 54, state whether the statement is True or False.

45. The reciprocal of a proper fraction is a proper fraction.

Consider the proper fraction 5/8.

Then, reciprocal of 5/8 = 8/5.

Therefore the obtained fraction is improper fraction.

46. The reciprocal of an improper fraction is an improper fraction.

Consider the improper fraction 5/3.

Then, the reciprocal of 5/3 = 3/5.

Therefore the obtained fraction is a proper fraction.

47. Product of two fractions = (Product of their denominators)/ (Product of their numerators)

Product of two fractions = (Product of their numerators)/ (Product of their denominators)

48. The product of two improper fractions is less than both the fractions.

The product of two improper fractions is more than each of the fractions that are multiplied.

Consider the two improper fractions, 5/4 and 4/2

= (5/4) × (4/2)

Then, 2.5 is multiplied to the improper fraction more than each of the fractions that are multiplied = (5/4) × 2.5

And (4/2) × 2.5 = 5

49. A reciprocal of a fraction is obtained by inverting it upside down.

Reciprocal of 8/9 = 9/8

50. To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.

2.5 × 1000 = 2500

51. To divide a decimal number by 100, we move the decimal point in the number to the left by two places.

Example: 3.4/100 = 0.034

52. 1 is the only number which is its own reciprocal.

We know that, if the denominator is not given, then we have to assume 1 always.

So, the reciprocal of 1/1 = 1/1

53. 2/3 of 8 is same as (2/3) ÷ 8

2/3 of 8 = (2/3) × 8

54. The reciprocal of 4/7 is 4/7

The reciprocal of 4/7 is 7/4.

55. If 5 is added to both the numerator and the denominator of the fraction 5/9, will the value of the fraction be changed? If so, will the value increase or decrease?

If 5 is added to both the numerator and denominator of the fraction 5/9 = 10/14

But, 5/9 ≠ 10/14

Yes, the value of the fraction is changed, and also, the value is increased.

56. What happens to the value of a fraction if the denominator of the fraction is decreased while the numerator is kept unchanged?

The value of a fraction is increased when the denominator of the fraction is decreased while the numerator is kept unchanged.

Example: ¼ = 0.25

57. Which letter comes 2/5 of the way among A and J?

There are 10 letters from A to J

So, 2/5 of 10 = 2/5 × 10

The 4 th letter from A to J is D.

Therefore, D comes 2/5 of the way among A and J.

58. If 2/3 of a number is 10, then what is 1.75 times of that number?

2/3 of a number is 10.

Let us assume the number be ‘P’.

2/3 of P = 10

2/3 × P = 10

P = 10 × 3/2

So, the number is 15

Again it is given in the question that, 1.75 times of that number = ?

= 1.75 of 15

= 1.75 × 15

59. In a class of 40 students, 1/5 of the total number of students like to eat rice only, 2/5 of the total number of students like to eat chapati only, and the remaining students like to eat both. What fraction of the total number of students like to eat both?

Number of students in a class = 40 students

1/5 of the total number of students like to eat rice only = 1/5 × 40

= 8 students

2/5 of the total number of students like to eat chapati only = 2/5 × 40

= 16 students

Number of students like to eat both = 40 – (8 + 16)

Fraction of the total number of students like to eat both = 16/40 = 2/5

60. Renu completed 2/3 part of her homework in 2 hours. How much part of her homework had she completed in 1¼ hours?

Renu completed 2/3 part of her homework in 2 hours.

Let us assume the total part of the homework be ‘P’.

2/3 of P = 2

2/3 × P = 2

P = 2 × 3/2

P = 3 homework

So, part of her homework she had completed in 1¼ hours i.e. 5/4 hours

Let us assume part of the homework she had completed in 5/4 hours be = Q

Q of 3 = 5/4

Q × 3 = 5/4

Q = (5/4) × (1/3)

Therefore, Renu completed 5/12 part of her homework in 5/4 hours.

61. Reemu read (1/5) th pages of a book. If she reads further 40 pages, she would have read (7/10) th pages of the book. How many pages are left to be read?

From the question it is given that,

Reemu read (1/5) th pages of a book.

Let us assume the total number of pages in the book be ‘P’.

Then, number of pages read by Reemu = (1/5) of P

= (1/5) × P

And also, it is given in the question, If she reads further 40 pages, she would have read (7/10) th pages of the book. = (7/10) × P

((1/5) × P) + 40 = (7/10) × P

(P + 200)/5 = (7P/10)

By cross mutliplication we get,

2P + 400 = 7P

Then, pages read by Reemu = Total pages – Pages read

P – (7P/10) = (3P/10)

(3/10) × 80

Therefore, 24 pages are left to be read.

Let us assume the missing number be P.

(3/7) × P = 15/98

By cross multiplication we get,

P = (15/98) × (7/3)

P = (5/14) × (1/1)

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Chapter 2 Class 7 Fractions and Decimals

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Updated from 2023-24 NCERT Book.

Get solutions of all questions of Chapter 2 Class 7 Fractions & Decimals free at teachoo. All NCERT exercise questions and examples have been solved with detailed explanation of each solution. Concepts have also been explained in the concept wise.

In this chapter, we will study

• What is a fraction
• What is proper , improper and mixed fraction
• What are equivalent fractions
• Comparing fractions
• Adding and Subtracting Fractions
• Then, we will learn how to Multiply Fractions and Mixed Fractions
• And how to divide Fractions
• And do some statement questions on multiplication and division of fractions
• What are Decimal Numbers
• Place value of Decimals
• Comparing Decimal Numbers
• Converting g → kg, mm → cm, mm → m, mm → km, cm → m, cm → km
• Addition and Subtraction of Decimal Numbers
• We will learn how to Multiply Decimal Numbers
• and How to divide Decimals
• And do some statement questions

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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals and Class 7 Maths Chapter 2 Try These Solutions in Hindi and English Medium revised and updated for academic year 2024-25. According to new syllabus and latest NCERT book, there are only five exercises in chapter 2 of 7th mathematics for 2024-25 exams.

7th Maths Chapter 2 Solutions in English Medium

• Class 7 Maths Chapter 2 Try These
• Class 7 Maths Exercise 2.1 in English
• Class 7 Maths Exercise 2.2 in English
• Class 7 Maths Exercise 2.3 in English
• Class 7 Maths Exercise 2.4 in English
• Class 7 Maths Exercise 2.5 in English

7th Maths Chapter 2 Solutions in Hindi Medium

• Class 7 Maths Exercise 2.1 in Hindi
• Class 7 Maths Exercise 2.2 in Hindi
• Class 7 Maths Exercise 2.3 in Hindi
• Class 7 Maths Exercise 2.4 in Hindi
• Class 7 Maths Exercise 2.5 in Hindi
• Class 7 Maths Chapter 2 NCERT Book
• Class 7th Maths Solutions Page
• Class 7 all Subjects Solutions

Learning fractions and decimals in Class 7 (Chapter 2) can be made more effective by following try these solutions. Ensure you have a clear understanding of basic concepts like numerator, denominator, proper fractions, improper fractions, mixed numbers, and decimals.

Class 7 Maths Chapter 2 Fractions and decimals can be abstract concepts. Utilize visual aids like fraction bars, number lines, and pie charts to help you visualize the relationships between different fractions and decimals. Start by reading the chapter in your textbook. Pay attention to explanations, examples, and solved problems to understand the concepts.

Consistent practice is key. Allocate a specific time each day to practice fractions and decimals. Solve the example problems given in the textbook. These problems are designed to guide you through the application of concepts. Begin with simpler problems and gradually move on to more complex ones. Practice converting between fractions and decimals. This skill is essential and helps you understand the relationship between the two. Learn to compare and order fractions and decimals. Work on problems that involve arranging them from least to greatest or vice versa.

7th Maths Exercise 2.1, Exercise 2.2, Exercise 2.3, Exercise 2.4 and Exercise 2.5 in English Medium and Hindi Medium for 2024-25. Practice addition, subtraction, multiplication, and division of fractions and decimals. Understand the rules and processes for each operation. Download Prashnavali 2.1, Prashnavali 2.2, Prashnavali 2.3, Prashnavali 2.4 and Prashnavali 2.5 in Hindi Medium free in PDF file format. You may use these solutions offline after downloading or study online without download. NCERT (https://ncert.nic.in/) Solutions 2024-25 for all other subjects are also available to download in updated form according to new CBSE NCERT Books . Class 7 Maths Chapter 2 Try These solving word problems that involve fractions and decimals. Try These problems help you apply your knowledge to real-world scenarios.

NCERT Solutions App for Class 7

There are plenty of online resources, interactive games, and practice exercises that can make learning fractions and decimals more engaging. Create flashcards with fraction and decimal representations on one side and the corresponding numerical values on the other. This can help reinforce your memory. Teaching someone else what you’ve learned can solidify your own understanding.

Collaborate with classmates to discuss concepts, solve problems together, and clarify doubts. As you feel more confident, attempt practice test papers or sample question papers. This simulates exam conditions and helps you manage time.

7 Mathematics Chapter 2 Fractions and Decimals all exercises solutions are given below in Hindi and English Medium. Fractions and decimals might take time to master. Stay patient and maintain a positive attitude towards learning. You can download it without any login or password.

Important Questions on Class 7 Maths Chapter 2

Lipika reads a book for 1 and 3/4 hours everyday. she reads the entire book in 6 days. how many hours in all were required by her to read the book.

Time taken by Lipika to read a book = 1 and 3/4 hours. She reads entire book in 6 days. Now, hours taken by her to read the entire book = (1 and 3/4) x 6 = 7/4 x 6 = 21/2 Thus, 10 and 1/2 hours were required by her to read the book.

A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 and 3/4 litres of petrol?

In 1 litre of pertrol, car covers the distance = 16 km In litres of petrol, car covers the distance = 2 and 3/4 of 16 km = 11/4 x 16 = 44 km Thus, the car will cover 44 km distance.

Which is greater: 0.5 or 0.05

Shyam bought 5 kg 300 g apples and 3 kg 250 g mangoes. sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. who bought more fruits.

Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g On comparing the quantity of fruits, 8 kg 550 g < 8 kg 950 g Therefore, Sarala bought more fruits.

How much less is 28 km than 42.6 km?

We have to find the difference of 42.6 km and 28 km. Difference = 42.6 – 28.0 = 14.6 km Therefore 14.6 km less is 28 km than 42.6 km.

In Class 7 Mathematics Chapter 2 Fractions and Decimals, we will learn how to convert a number into decimal and decimal into fraction. Normally we represent fraction in the following ways: 1. Proper Fraction: A proper fraction is a fraction that represents a part of a whole. In which the numerator is always smaller than the denominator. 2. Improper Fraction: An improper fraction is a combination of whole and a proper fraction. Note: When two proper fractions are multiplied, the product is less than each of the fractions. Or, we say the value of the product of two proper fractions is smaller than each of the two fractions. Download Class 7 Maths App in English or Class 7 Maths App in Hindi for offline use.

We also go through the multiplication of a number with fractions. To multiply a whole number with a proper or an improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator same. But to multiply a mixed fraction to a whole number, first convert the mixed fraction to an improper fraction and then multiply.

Now the NCERT solutions for 2024-25 and answers of class 7 Maths are available in both the mode Online as well as Downloadable form for the convenience of students. Remember, practice and consistent effort are key to mastering fractions and decimals. Don’t rush through the material; focus on understanding the concepts thoroughly. The new updates is uploaded frequently based on latest NCERT Books.

How can students easily prepare chapter 2 of class 7th Maths?

No, students can’t skip chapter 2 (Fractions and Decimals) of class 7th Maths. Chapter 2 of class 7th Maths is a very useful chapter because this chapter is helpful in higher maths classes also. This chapter is a very interesting chapter.

How many exercises are there in chapter 2 of class 7th Maths?

As per new syllabus and latest textbook issued for session 2024-25, chapter 2 of class 7th Maths has five exercises. The fifth exercise (exercise 2.5) has the highest number of questions and the third exercise (exercise 2.3) has the least number of questions (4 questions).

What are the sub-topics under the topic Fractions and Decimals (chapter 2 of grade 7th Maths)?

The sub-topics under the topic Fractions and Decimals (chapter 2 of grade 7th Maths) are: 1. Fractions. 2. Greater than, less than in fractions 3. Addition and Subtraction of fractions. 4. Addition and Subtraction of fractions-statement questions 5. Multiplication of fractions. 6. Multiplication of fractions-statement questions. 7. Division of fractions. 8. Reciprocal of fractions. 9. Decimal numbers. 10. Greater than, less than in decimals. 11. Addition of decimals. 12. Multiplication of decimals. 13. Multiplication of decimals-statement questions. 14. Division of decimals. 15. Division of decimals-statement questions.

Is there any need to practice extra questions of chapter 2 of class 7th Maths?

No, there is no need to practice extra questions of chapter 2 of class 7th Maths. Questions of NCERT are enough for practice. If students want to practice more questions of chapter 2 of class 7th Maths, they can refer to R.S. Aggarwal.

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NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals

NCERT solutions for class 7 maths chapter 2 fractions and decimals comprise exercises based on proper, improper, and mixed fractions as well as their addition and subtraction. Additionally, comparison of fractions, equivalent fractions, representation of fractions on the number line, and ordering of fractions are some of the key topics covered in this chapter.

Fractions represent a part of a whole and it is expressed in the form of a numerator and a denominator. Decimals are the numerical representation of fractions where the denominator is a power of ten like 10, 100, 1000, etc. The fractions and decimals are significantly important in performing arithmetic operations and hence, should be studied nicely. NCERT solutions class 7 maths chapter 2 fractions and decimals cover all the important concepts based on the properties of these numbers in detail with examples.

The first few exercises in these class 7 maths chapter 2 fractions and decimals explain the addition and subtraction of fractions and decimals along with the revision of fractions and decimals concepts studied in the previous classes with suitable examples. Sample problems covered in the class 7 maths NCERT solutions chapter 2 fractions and decimals are sufficient for students to gain an in-depth understanding of applying arithmetic operations on fractions and decimals. You can find some of these in the exercises given below.

• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.1
• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.2
• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.3
• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.4
• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.5
• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.6
• NCERT Solutions Class 7 Maths Chapter 2 Ex 2.7

NCERT Solutions for Class 7 Maths Chapter 2 PDF

NCERT solutions for class 7 maths are well-curated to provide the optimal coverage of the complete math syllabus. To obtain the best results in exams students can easily plan their preparation and revision with these competent resources. These solutions are also available for free pdf download as given below.

☛ Download Class 7 Maths NCERT Solutions Chapter 2

NCERT Class 7 Maths Chapter 2   Download PDF

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

With regular practice of the above exercises, students will gain the right approach required for solving arithmetic operations on fractions and decimals. The practice of these exercises will provide an insight into the type of questions and their step-wise solutions. The chapter-wise detailed analysis of NCERT Solutions Class 7 Maths Chapter 2 fractions and decimals are given below.

• Class 7 Maths Chapter 2 Ex 2.1 - 8 Questions
• Class 7 Maths Chapter 2 Ex 2.2 - 8 Questions
• Class 7 Maths Chapter 2 Ex 2.3 - 8 Questions
• Class 7 Maths Chapter 2 Ex 2.4 - 4 Questions
• Class 7 Maths Chapter 2 Ex 2.5 - 9 Questions
• Class 7 Maths Chapter 2 Ex 2.6 - 5 Questions
• Class 7 Maths Chapter 2 Ex 2.7 - 6 Questions

☛ Download Class 7 Maths Chapter 2 NCERT Book

Topics Covered: Addition and subtraction of fractions and decimals, multiplication of fractions and decimals, multiplication of fractions with whole numbers, multiplication of fractions by fractions, division of fraction and decimals, division of a fraction by another fraction, multiplication of decimal numbers by 10, 100 and 1000 and division of decimals by 10, 100 and 1000. Other topics included in these class 7 maths NCERT solutions chapter 2 are the comparison of fractions, multiplication of fractions with whole numbers, and division of decimals with whole numbers.

Total Questions: Class 7 maths chapter 2 fractions and decimals consists of a total of 48 questions of which 20 are easy, 5 are moderate and 23 are long answer type questions.

List of Formulas in NCERT Solutions Class 7 Maths Chapter 2

NCERT solutions class 7 maths chapter 2 covers lots of important concepts based on fractions and decimals that are crucial for strengthening the math foundation of students. Some important concepts in NCERT solutions for class 7 maths chapter 2 are about:

• Comparison of Fractional Values
• Multiplication of Fractions with Whole Numbers
• Division of Whole Numbers with Fractions

Important Questions for Class 7 Maths NCERT Solutions Chapter 2

NCERT Solutions for Class 7 Maths Video Chapter 2

FAQs on NCERT Solutions Class 7 Maths Chapter 2

What is the importance of ncert solutions class 7 maths chapter 2 fractions and decimals.

NCERT Solutions Class 7 Maths Chapter 2 deals with fractions and decimals and their concepts that help to enhance the student’s fundamental knowledge. This enables them to score well in their examinations as NCERT solutions are highly competent to deliver the in-depth conceptual knowledge for a strong mathematical foundation.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 7 Maths Fractions and Decimals?

Most questions in the NCERT Solutions Class 7 Maths Chapter 2 fractions and decimals are important. Questions based on the comparison of fractions and decimals, their addition and subtraction, along with multiplication and division are required to be practiced regularly while studying. Proper and Improper fractions are one topic that needs regular practice.

What are the Important Topics Covered in NCERT Solutions Class 7 Maths Chapter 2?

The important sub-topics covered in NCERT Solutions Class 7 Maths Chapter 2 are addition and subtraction of fractions and decimals, multiplication of fractions and decimals, multiplication of fractions with whole numbers, multiplication of fractions by fractions, division of fraction and decimals, division of whole numbers by fractions, reciprocal of fraction, division of a fraction by a whole number, division of a fraction by another fraction, multiplication of decimal numbers by 10, 100 and 1000 and division of decimals by 10, 100 and 1000. Fractions and decimals chapter 2 is very well structured to include all the important sub-topics for students to learn better for final exams.

How Many Questions are there in NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals?

There are a total of 48 questions in NCERT class 7 maths chapter 2 fractions and decimals. Out of these 48 questions, 30 are well-structured to give an in-detailed understanding of the concepts and context of the theory. These 30 questions can be further categorized into long answers, moderate level, and easy ones. Students are required to plan their practice time well for each subcategory as they need time and practice.

What are the Important Formulas in NCERT Solutions Class 7 Maths Chapter 2?

The important concepts covered in the NCERT Solutions class 7 maths chapter 2 are based on fractions and decimals comparison, multiplication, and division like multiplication of fractions with whole numbers, multiplication of fractions by fractions, division of fraction and decimals, division of whole numbers by fractions, reciprocal of fraction, division of a fraction by a whole number, division of a fraction by another fraction, etc. These concepts are also important for students to acquire a clear understanding of applying arithmetic operators on fractions and decimals.

Why Should I Practice Class 7 Maths NCERT Solutions fractions and decimals chapter 2?

Practicing NCERT Solutions Class 7 Maths fractions and decimals chapter 2 can help students to improve their mathematical proficiency and skills. These solutions are well designed after thorough research to help students get the best learning outcome. While practicing the fractions and decimals operations, ensure to use all the tips mentioned in the NCERT solutions.

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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

September 9, 2019 by phani

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

• Class 7 Maths Fractions and Decimals Exercise 2.1 
• Class 7 Maths Fractions and Decimals Exercise 2.2
• Class 7 Maths Fractions and Decimals Exercise 2.3
• Class 7 Maths Fractions and Decimals Exercise 2.4
• Class 7 Maths Fractions and Decimals Exercise 2.5
• Class 7 Maths Fractions and Decimals Exercise 2.6
• Class 7 Maths Fractions and Decimals Exercise 2.7

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