Solutions for Class 7 Mathematics – Chapter 2 Fractions and Decimals Exercise 2.2

NCERT Solutions for Class 7 Mathematics – Chapter 2: Fractions and Decimals (Exercise 2.2)

Welcome to Digital Pipal Academy! In this post, we present detailed solutions for Exercise 2.2 from Class 7 General Mathematics – Fractions and Decimals. These solutions will help students understand the concepts of reciprocals, division of fractions, and real-life applications of fractions and decimals.

SCERT Solution for Class 7 Maths Chapter 2 Fractions and Decimals

Chapter Name	Solution Link
Exercise 2.1	Click Here
Exercise 2.2	Click Here
Exercise 2.3	Click Here
Exercise 2.4	Click Here
Exercise 2.5	Click Here

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Exercise 2.2 – Solutions

Question 1. Find the reciprocal -

Solution:

$6$ → Reciprocal = $\frac{1}{6}$

$\frac{1}{2}$ → Reciprocal = $2$

$\frac{3}{5}$ → Reciprocal = $\frac{5}{3}$

$1$ → Reciprocal = $1$

$2$ → Reciprocal = $\frac{1}{2}$

Question 2. Evaluate:

A.
(i) $6 \div \frac{3}{8}$
(ii) $31 \div \frac{2}{3}$
(iii) $51 \div \frac{17}{3}$
(iv) $4 \div \left( \frac{3}{4} \right)$
(v) $3 \div \left( 2 \frac{1}{4} \right)$

Solution:

(i) $6 \div \frac{3}{8}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$$6 \div \frac{3}{8} = 6 \times \frac{8}{3}$$ $$= \frac{6 \times 8}{3} = \frac{48}{3} = 16$$

(ii) $31 \div \frac{2}{3}$

$$31 \div \frac{2}{3} = 31 \times \frac{3}{2}$$ $$= \frac{31 \times 3}{2} = \frac{93}{2} = 46.5$$

(iii) $51 \div \frac{17}{3}$

$$51 \div \frac{17}{3} = 51 \times \frac{3}{17}$$ $$= \frac{51 \times 3}{17} = \frac{153}{17} = 9$$

(iv) $4 \div \frac{3}{4}$

$$4 \div \frac{3}{4} = 4 \times \frac{4}{3}$$ $$= \frac{4 \times 4}{3} = \frac{16}{3} = 5\frac{1}{3}$$

(v) $3 \div \left( 2\frac{1}{4} \right)$

Convert the mixed fraction $2\frac{1}{4}$ into an improper fraction:

$$2\frac{1}{4} = \frac{9}{4}$$

Now, divide:

$$3 \div \frac{9}{4} = 3 \times \frac{4}{9}$$ $$= \frac{3 \times 4}{9} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3}$$

 

B. 

(i) $2\frac{1}{4} \div 3$
(ii) $\frac{60}{7} \div 15$
(iii) $5\frac{1}{3} \div 4$
(iv) $4\frac{1}{3} \div 3$
(v) $4\frac{3}{7} \div 7$

Solution:

(i) $2\frac{1}{4} \div 3$

Convert the mixed fraction to an improper fraction:

$$2\frac{1}{4} = \frac{9}{4}$$

Now, divide:

$$\frac{9}{4} \div 3 = \frac{9}{4} \times \frac{1}{3}$$ $$= \frac{9 \times 1}{4 \times 3} = \frac{9}{12} = \frac{3}{4}$$

(ii) $\frac{60}{7} \div 15$

$$\frac{60}{7} \div 15 = \frac{60}{7} \times \frac{1}{15}$$ $$= \frac{60 \times 1}{7 \times 15} = \frac{60}{105}$$

Simplify:

$$= \frac{4}{7}$$

(iii) $5\frac{1}{3} \div 4$

Convert the mixed fraction to an improper fraction:

$$5\frac{1}{3} = \frac{16}{3}$$

Now, divide:

$$\frac{16}{3} \div 4 = \frac{16}{3} \times \frac{1}{4}$$ $$= \frac{16 \times 1}{3 \times 4} = \frac{16}{12} = \frac{4}{3} = 1\frac{1}{3}$$

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

Convert the mixed fraction to an improper fraction:

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

Now, divide:

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

(v) $4\frac{3}{7} \div 7$

Convert the mixed fraction to an improper fraction:

$$4\frac{3}{7} = \frac{31}{7}$$

Now, divide:

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

C. 

(i) $3\frac{1}{6} \div 2\frac{1}{3}$
(ii) $5\frac{2}{3} \div 4\frac{1}{4}$
(iii) $11\frac{7}{13} \div 4\frac{2}{13}$
(iv) $3\frac{5}{6} \div 2\frac{4}{5}$

Solution:

(i) $3\frac{1}{6} \div 2\frac{1}{3}$

Convert mixed fractions to improper fractions:

$$3\frac{1}{6} = \frac{19}{6}, \quad 2\frac{1}{3} = \frac{7}{3}$$

Now, divide:

$$\frac{19}{6} \div \frac{7}{3} = \frac{19}{6} \times \frac{3}{7}$$ $$= \frac{19 \times 3}{6 \times 7} = \frac{57}{42}$$

Simplify:

$$= \frac{19}{14} = 1\frac{5}{14}$$

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(ii) $5\frac{2}{3} \div 4\frac{1}{4}$

Convert mixed fractions to improper fractions:

$$5\frac{2}{3} = \frac{17}{3}, \quad 4\frac{1}{4} = \frac{17}{4}$$

Now, divide:

$$\frac{17}{3} \div \frac{17}{4} = \frac{17}{3} \times \frac{4}{17}$$ $$= \frac{17 \times 4}{3 \times 17} = \frac{68}{51}$$

Simplify:

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

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(iii) $11\frac{7}{13} \div 4\frac{2}{13}$

Convert mixed fractions to improper fractions:

$$11\frac{7}{13} = \frac{150}{13}, \quad 4\frac{2}{13} = \frac{54}{13}$$

Now, divide:

$$\frac{150}{13} \div \frac{54}{13} = \frac{150}{13} \times \frac{13}{54}$$ $$= \frac{150 \times 13}{13 \times 54} = \frac{150}{54}$$

Simplify:

$$= \frac{25}{9} = 2\frac{7}{9}$$

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(iv) $3\frac{5}{6} \div 2\frac{4}{5}$

Convert mixed fractions to improper fractions:

$$3\frac{5}{6} = \frac{23}{6}, \quad 2\frac{4}{5} = \frac{14}{5}$$

Now, divide:

$$\frac{23}{6} \div \frac{14}{5} = \frac{23}{6} \times \frac{5}{14}$$ $$= \frac{23 \times 5}{6 \times 14} = \frac{115}{84}$$

Simplify:

$$= 1\frac{31}{84}$$

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Question 3. (i) Divide $\left( \frac{8}{15} \text{ of } \frac{3}{4} \right)$ by $2\frac{3}{2}$.

Solution:

$\left( \frac{8}{15} \times \frac{3}{4} \right) \div 2\frac{3}{2}$
$= \frac{24}{60} \div \frac{7}{2}$
$= \frac{2}{5} \times \frac{2}{7}$
$= \frac{4}{35}$

(ii) Divide $1 \frac{13}{22}$ by the reciprocal of $\frac{6}{11}$.

Solution:

$1 \frac{13}{22} \div \frac{11}{6}$
$= \frac{35}{22} \times \frac{6}{11}$
$= \frac{210}{242}$
$= \frac{105}{121}$

(iii) Divide the product of $\frac{1}{3}$ and $\frac{2}{5}$ by the product of $\frac{3}{7}$ and $\frac{2}{5}$.

Solution:

$\left( \frac{1}{3} \times \frac{2}{5} \right) \div \left( \frac{3}{7} \times \frac{2}{5} \right)$
$= \frac{2}{15} \div \frac{6}{35}$
$= \frac{2}{15} \times \frac{35}{6}$
$= \frac{70}{90}$
$= \frac{7}{9}$

(iv) Product of two numbers is $1 \frac{1}{2}$. If one of them is $\frac{9}{14}$, then find the other number.

Solution:

Let the other number be $x$.

Given:

$$x \times \frac{9}{14} = 1 \frac{1}{2}$$

Convert to improper fraction:

$$1 \frac{1}{2} = \frac{3}{2}$$

Solving for $x$:

$$x = \frac{3}{2} \div \frac{9}{14}$$

Multiply by the reciprocal:

$$x = \frac{3}{2} \times \frac{14}{9}$$ $$x = \frac{3 \times 14}{2 \times 9} = \frac{42}{18}$$

Simplify:

$$x = \frac{7}{3}$$

Thus, the other number is $\frac{7}{3}$ or $2 \frac{1}{3}$.

(v) A car covers 240 km in $3 \frac{1}{3}$ hours. How much distance will the car cover in 1 hour?

Solution:

To find the distance covered in 1 hour, we divide the total distance by the total time.

Given:
Total distance = 240 km
Total time = $3 \frac{1}{3}$ hours
Convert mixed fraction to improper fraction:
$3 \frac{1}{3} = \frac{10}{3}$ hours

Now,
Distance covered in 1 hour =

$$\frac{240}{\frac{10}{3}}$$ $$= 240 \times \frac{3}{10}$$ $$= \frac{240 \times 3}{10} = \frac{720}{10} = 72 \text{ km}$$

The car will cover 72 km in 1 hour.

Question 4: Area of a rectangle is 24 cm². If the length of the rectangle is $6 \frac{2}{3}$ cm, then what is its breadth?

Solution:

Breadth of the rectangle = $\frac{\text{Area}}{\text{Length}}$

= $\frac{24}{6 \frac{2}{3}}$

= $\frac{24}{\frac{20}{3}}$

= $24 \times \frac{3}{20}$

= $\frac{72}{20}$

= 3.6 cm

Question 5. A ribbon of $12 \frac{1}{2}$ m long is divided into 10 equal parts. What is the length of its parts?

Solution:

Length of each part = $\frac{\text{Total length}}{\text{Number of parts}}$

= $\frac{12 \frac{1}{2}}{10}$

= $\frac{25}{2} \div 10$

= $\frac{25}{2} \times \frac{1}{10}$

= $\frac{25}{20}$

= 1.25 m

Question 6: $\frac{3}{4}$ parts of a container is filled with water. If this volume of water is divided equally into $\frac{1}{8}$ parts, how many containers will be required to keep the whole volume of it?

Solution:

Number of containers required =
$\frac{\text{Total volume}}{\text{Each part}}$

= $\frac{3}{4} \div \frac{1}{8}$
= $\frac{3}{4} \times \frac{8}{1}$
= $\frac{3 \times 8}{4 \times 1}$
= $\frac{24}{4}$
= 6 containers

Question 7. How many $\frac{1}{2}$ are there in $\frac{6}{4}$? Illustrate with the help of a diagram.

Solution:

To find how many $\frac{1}{2}$ are there in $\frac{6}{4}$, we divide $\frac{6}{4}$ by $\frac{1}{2}$.

$$\text{Required number} = \frac{6}{4} \div \frac{1}{2}$$

Using the division rule for fractions:

$$\frac{6}{4} \times \frac{2}{1} = \frac{6 \times 2}{4 \times 1} = \frac{12}{4} = 3$$

Thus, there are 3 halves in $\frac{6}{4}$.

Question 8. An aeroplane covers 200 km in $\frac{1}{5}$ hours. How far will it fly in 5 hours?

Solution:

Distance covered in $\frac{1}{5}$ hours = 200 km

Speed = $200 \times 5 = 1000$ km/hour

Distance covered in 5 hours = $1000 \times 5 = 5000$ km

Question 9. A boy covers $5 \frac{1}{8}$ km in $1 \frac{1}{4}$ hours on a bicycle. If the boy maintains the same speed, how far will he cover in 1 hour?

Solution:

Given,
Total distance = $5 \frac{1}{8}$ km
Total time = $1 \frac{1}{4}$ hours

Now,
The distance covered in 1 hour

$$= \left( \frac{41}{8} \div \frac{5}{4} \right) \text{ km}$$ $$= \left( \frac{41}{8} \times \frac{4}{5} \right) \text{ km}$$ $$= \frac{164}{40} \text{ km}$$ $$= \frac{41}{10} \text{ km}$$ $$= 4.1 \text{ km}$$

The boy will cover 4.1 km in 1 hour.

Question 10. Four options are given to the question below. Choose the correct answer—If $A + B = 1$ and $A - B = \frac{2}{3}$, then the fractions A and B are—

(i) A = $\frac{5}{6}$, B = $-\frac{3}{6}$
(ii) A = $\frac{2}{3}$, B = $\frac{1}{3}$
(iii) A = $\frac{5}{6}$, B = $-\frac{1}{6}$
(iv) A = $\frac{3}{5}$, B = $\frac{1}{5}$

Solution:

We are given the equations:

$$A + B = 1$$ $$A - B = \frac{2}{3}$$

Step 1: Add Both Equations

$$(A + B) + (A - B) = 1 + \frac{2}{3}$$ $$2A = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$$

Step 2: Solve for A

$$A = \frac{5}{6}$$

Step 3: Substitute A into the First Equation

$$\frac{5}{6} + B = 1$$

Step 4: Solve for B

$$B = 1 - \frac{5}{6} = \frac{6}{6} - \frac{5}{6} = \frac{1}{6}$$

Final Answer:

The values of A and B are $A = \frac{5}{6}$ and $B = \frac{1}{6}$.

Correct Option:

From the given options, the correct answer is (iii) A = $\frac{5}{6}$, B = $\frac{1}{6}$.

Understanding fractions and decimals is crucial for problem-solving in mathematics. Practice these solutions to strengthen your concepts and improve accuracy.

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