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NCERT Solutions for Class 7 Maths Chapter 2 Fractions And Decimals Ex 2.3

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NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.3 - FREE PDF Download

NCERT Solutions for Class 7 Fractions and Decimals Exercise 2.3  is a critical part of the curriculum that introduces the essential concepts of addition and subtraction of decimals. This exercise focuses on developing accuracy in aligning decimal points to ensure correct calculations.  Class 7 maths chapter 2 exercise 2.3 solutions focuses on the addition and subtraction of decimals, a crucial skill for solving real-life problems involving money, measurement, and data. Ex 2.3 Class 7 emphasizes the importance of understanding the fundamental principles of fraction multiplication to handle more complex mathematical problems in future studies. Mastery of these topics is essential for scoring well in exams and developing a strong foundation in fractions and decimals. Access the NCERT Solutions for Class 7 Maths here.

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Table of Content
1. NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.3 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 2 Exercise 2.3 Class 7 | Vedantu
3. Access NCERT Solutions for Maths Class 7 Chapter 2 - Fractions and Decimals
    3.1Exercise - 2.3
4. Conclusion
5. NCERT Solution Class 7 Maths Chapter 2 - Fractions and Decimals Other Exercises
6. CBSE Class 7 Maths Chapter 2 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 7 Maths
8. Important Related Links for NCERT Class 7 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 2 Exercise 2.3 Class 7 | Vedantu

  • Class 7 Maths Exercise 2.1 Solutions for Chapter 2 Fractions and Decimals likely cover concepts related to multiplying two or more fractions by multiplying the numerators and the denominators.

  • Understand how to simplify fractions before or after multiplication to make calculations easier and results simpler.

  • Gain proficiency in converting mixed numbers to improper fractions for multiplication and then converting the result back to a mixed number.

  • Apply multiplication of fractions to solve practical problems, enhancing comprehension of real-world applications and improving problem-solving skills.

  • There are 4 fully solved questions in class 7th maths chapter 2 Fractions and Decimal.

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

Exercise - 2.3

1. Find:

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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


2. Find the reciprocal of each of the following fractions. Classify the reciprocals as a proper fraction, improper fractions and whole numbers.

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

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

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

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

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

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

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

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

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

(iv) $\dfrac{\text{6}}{\text{5}}$ 

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

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

(v) $\dfrac{\text{12}}{\text{7}}$ 

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

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

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

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

Reciprocal of $\dfrac{\text{1}}{\text{8}}\text{=8}\to \text{Whole number}$

(vii) $\dfrac{\text{1}}{\text{11}}$ 

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

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


3. Find:

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans:  Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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


4. Find:

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

Ans:  Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans:  Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans: Calculating the value,

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

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

Ans:  Calculating the value,

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

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

Ans:  Calculating the value,

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

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

Ans: Calculating the value,

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


Conclusion

Class 7 Maths Chapter 2.3, "Fractions and Decimals," emphasizes the importance of careful alignment of decimal points to ensure accurate calculations. By practising these skills, students will be better equipped to tackle more complex mathematical problems in higher classes. Ex 2.3 Class 7 focuses on the concept of Multiplication of Fractions, which is essential for understanding how to deal with parts of a whole in various mathematical and real-life contexts. The comprehensive approach ensures that students gain confidence and proficiency in handling fractional operations, preparing them for future academic success.

NCERT Solution Class 7 Maths Chapter 2 - Fractions and Decimals Other Exercises

Exercises

Number of Questions

Exercise 2.1

8 Questions & Solutions

Exercise 2.2

8 Questions & Solutions

Exercise 2.4

5 Questions & Solutions

Exercise 2.5

6 Questions & Solutions



CBSE Class 7 Maths Chapter 2 Other Study Materials



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.




Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.


FAQs on NCERT Solutions for Class 7 Maths Chapter 2 Fractions And Decimals Ex 2.3

1. What topics are covered in Class 7 Maths Chapter 2 Exercise 2.3 solutions?

The exercise covers the addition and subtraction of decimals, focusing on aligning decimal points accurately to ensure correct calculations. It includes practice questions and word problems to apply these concepts in real-life scenarios.

2. How do you add decimals in exercise 2.3 class 7?

To add decimals, align the decimal points of the numbers, then add them as you would whole numbers. Ensure the decimal point in the result is placed directly below the decimal points of the numbers being added.

3. How is subtracting decimals explained in class 7 maths 2.3?

According to class 7 maths 2.3, similar to addition, align the decimal points of the numbers. Subtract as you would whole numbers, borrowing if necessary. Place the decimal point in the result directly below the decimal points of the numbers being subtracted.

4. What are some examples of word problems in class 7 maths exercise 2.3?

Word problems in this exercise typically involve real-life scenarios such as money transactions, measurements, and data handling, requiring the addition or subtraction of decimals to find the solution.

5. Why is aligning decimal points important in exercise 2.3 class 7 maths?

Aligning decimal points ensures that each digit is in the correct place value, which is crucial for accurate addition and subtraction of decimals. Misalignment can lead to incorrect results.

6. Are there any tips for solving class 7 maths ex 2.3?

Yes, always double-check the alignment of decimal points before performing the operation. Simplify the decimals if possible, and practice regularly to gain confidence and accuracy in decimal calculations.

7. How does class 7th maths exercise 2.3 solutions help in real-life applications?

Understanding the addition and subtraction of decimals is essential for various real-life applications, such as managing finances, measuring quantities accurately, and interpreting data. This exercise provides a strong foundation for these practical skills.