NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.2: FREE PDF Download
Class 7 Maths Chapter 2, "Fractions and Decimals," aims to strengthen students' understanding and use of fractions and decimals in different scenarios. Exercise 2.2 provides a variety of problems designed to improve computational skills and deepen conceptual knowledge. This section is essential as it consolidates learning through carefully crafted questions, helping students to master fractions and decimals. By working through these exercises, students enhance their problem-solving abilities, preparing them for more advanced mathematical concepts in the future.
Glance on NCERT Solutions Maths Chapter 2 Ex 2.2 Class 7 | Vedantu
Chapter 2, Exercise 2.2 of your Class 7 Maths textbook likely deals with multiplying fractions. This is a foundational concept that builds on your understanding of fractions from previous chapters.
In multiplying fractions, you don't simply multiply the numerators and the denominators separately. Instead, you multiply the numerator of the first fraction with the numerator of the second fraction, and the denominator of the first fraction with the denominator of the second.
Exercise 2.2 would likely contain practice problems related to multiplying different fractions. These problems might involve whole numbers multiplied by fractions as well.
Exercise 2.2 Class 7 maths NCERT Solutions has overall 8 Questions with 2 fill-in-the-blanks.
Topics Covered in Class 7th Maths Chapter 2 Exercise 2.2
Multiplication of a fraction by a fraction
Value of Products
Access PDF for Maths NCERT Chapter 2 Fractions and Decimals Exercise 2.2 Class 7
Exercise 2.2
1.Find:
(i) $\dfrac{\text{1}}{\text{4}}\,\text{of}$
$\dfrac{\text{1}}{\text{4}}$
$\dfrac{\text{3}}{\text{5}}$
$\dfrac{\text{4}}{\text{3}}$
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.
(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,
$\dfrac{\text{1}}{\text{7}}\,\text{of}\,\dfrac{\text{2}}{\text{9}}\text{=}\dfrac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{10}}\text{=}\dfrac{\text{1 }\!\!\times\!\!\text{ 3}}{\text{7 }\!\!\times\!\!\text{ 10}}\text{=}\dfrac{3}{70}$
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.
Given: Saili plants \[4\] saplings in a row where the distance between two adjacent saplings $=\dfrac{3}{4}$m
The number of gaps in saplings \[=\text{ }3\]
Hence,
The distance between the first and the last saplings$\text{=3 }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{4}}\text{=}\dfrac{\text{9}}{\text{4}}\text{m=2}\dfrac{\text{1}}{\text{4}}\text{m}$
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 $\dfrac{\text{8}}{\text{15}}\,\text{is}\,\dfrac{\text{8}}{\text{15}}$
Conclusion
The "NCERT Solutions for Class 7 Exercise 2.2 Fractions and Decimals" by Vedantu covers essential concepts of adding, subtracting, multiplying, and dividing fractions. It's important to understand how to convert mixed fractions to improper fractions and find the least common multiple (LCM) for addition and subtraction of unlike fractions. Focus on mastering these operations as they form the foundation for more advanced math topics.
NCERT Solution Class 7 Maths Chapter 2 - Fractions and Decimals Other Exercises
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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.
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FAQs on NCERT for Class 7 Maths Chapter 2 - Fractions and Decimals Exercise 2.2
1. What are like fractions and unlike fractions?
Like Fractions: Fractions that have the same denominator. For example, 2/5 and 3/5 are like fractions because they both have the denominator 5.
Unlike Fractions: Fractions that have different denominators. For example, 2/3 and 4/7 are unlike fractions because their denominators are different.
2. How do you add or subtract like fractions?
To add or subtract like fractions, simply add or subtract the numerators and keep the same denominator. For example:
3/7+2/7=3+2 /7=5/7
6/9−4/9=6−4 /9=2/9