NCERT Solutions for exercise 2.2 class 7 maths

 

NCERT Solutions for Class 7 Maths Exercise 2.2

NCERT Solutions for exercise 2.2 class 7 maths Chapter 2 Fractions and Decimals in simple solutions are given here. exercise 2.2 class 7 maths of NCERT Solutions for Maths Class 7 Chapter 2 contains all the topics containing the basic information of Fractions. exercise 2.2 class 7 maths provide students the study of fractions including proper, improper and mixed fractions as well as their addition and subtraction. NCERT Solutions for exercise 2.2 class 7 maths provides necessary material for solving different range of questions that test the students capability of understanding the concepts. It also helps the students of CBSE Class 7 Maths students

Question 1. 

Which of the drawings (a) to (d) show:

(i) 2 × \((\dfrac{1}{5})\) (ii) 2 × \((\dfrac{1}{2})\) (iii) 3 × \((\dfrac{2}{3}\) (iv) 3 × \((\dfrac{1}{4})\)

Solution:-

(i) 2 × \((\dfrac{1}{5})\) represents the addition of 2 figures, each represents 1 shaded part out of the given 5 equal parts.

∴ 2 × \((\dfrac{1}{5})\) is represented by fig (d).

(ii) 2 × \((\dfrac{1}{2})\) represents the addition of 2 figures, each represents 1 shaded part out of the given 2 equal parts.

∴ 2 × \((\dfrac{1}{2})\) is represented by fig (b).

(iii) 3 × \((\dfrac{2}{3}\) represents the addition of 3 figures, each represents 2 shaded part out of the given 3 equal parts.

∴ 3 × \((\dfrac{2}{3}\) is represented by fig (a).

(iii) 3 × \((\dfrac{1}{4})\) represents the addition of 3 figures, each represents 1 shaded part out of the given 4 equal parts.

∴ 3 × \((\dfrac{1}{4})\) is represented by fig (c).

Question 2. 

Some pictures (a) to (c) are given below. Tell which of them show:

(i) 3 × \((\dfrac{1}{5})\)= \((\dfrac{3}{5})\) (ii) 2 × \((\dfrac{1}{3})\) = \((\dfrac{2}{3})\) (iii) 3 × \((\dfrac{3}{4})\) = 2 \((\dfrac{1}{4})\)

Solution:-

(i) 3 × \((\dfrac{1}{5})\) represents the addition of 3 figures, each represents 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.

∴ 3 × \((\dfrac{1}{5})\) = \((\dfrac{3}{5})\) is represented by fig (c).

(ii) 2 × \((\dfrac{1}{3})\) represents the addition of 2 figures, each represents 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.

∴ 2 × \((\dfrac{1}{3})\) = \((\dfrac{2}{3})\) is represented by fig (a).

(iii) 3 × \((\dfrac{3}{4})\)  represents the addition of 3 figures, each represents 3 shaded part out of the given 4 equal parts and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.

∴ 3 × \((\dfrac{3}{4})\) = 2 \((\dfrac{1}{4})\) is represented by fig (b).

Question 3. 

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

(i) 7 × \((\dfrac{3}{5})\)

Solution:-

= \((\dfrac{7 \times 3}{1 \times 5 })\) 

= \((\dfrac{21}{5 })\) 

(ii) 4 × \((\dfrac{1}{3 })\) 

Solution:-

4 × \((\dfrac{1}{3 })\)

= \((\dfrac{4}{1 }) \times (\dfrac{1}{3 })\) 

= \(\dfrac{(4 × 1)}{1 × 3 }\)

= \((\dfrac{4}{3 })\)

(iii) 2 × \((\dfrac{6}{7 })\) 

Solution:-

= \((\dfrac{2}{1 }) \times (\dfrac{6}{7 })\) 

= \(\dfrac{(2 × 6)}{1 × 7}\)

= \((\dfrac{12}{7 })\)

(iv) 5 × \((\dfrac{2}{9 })\)

Solution:-

= \((\dfrac{5}{1 }) \times (\dfrac{2}{9 })\) 

= \(\dfrac{(5 × 2)}{1 × 9}\)

= \((\dfrac{10}{9 })\)

(v) (2/3) × 4

Solution:-

= \((\dfrac{2}{3 }) \times (\dfrac{4}{1 })\) 

= \(\dfrac{(2 × 4)}{3 × 1}\)

= \((\dfrac{8}{3 })\)

(vi) \((\dfrac{5}{2})\) × 6

Solution:-

= \((\dfrac{5}{2}) \times (\dfrac{6}{1 })\) 

= \(\dfrac{(5 × 6)}{2 × 1}\)

= \((\dfrac{30}{2})\)

=15

(vii) 11 × \((\dfrac{4}{7})\) 

Solution:-

= \((\dfrac{11}{1}) \times (\dfrac{4}{7})\) 

= \(\dfrac{(11 × 4)}{1 × 7}\)

= \((\dfrac{44}{7})\)

(viii) 20 × \((\dfrac{4}{5})\)

Solution:-

= \((\dfrac{20}{1}) \times (\dfrac{4}{5})\) 

= \(\dfrac{(20 × 4)}{1 × 5}\)

= \((\dfrac{80}{5})\)

= 16

(ix) 13 × \((\dfrac{1}{3})\)

Solution:-

= \((\dfrac{13}{1}) \times (\dfrac{1}{3})\) 

= \(\dfrac{(13 × 1)}{1 × 3}\)

= \((\dfrac{13}{3})\)

(x) 15 × \((\dfrac{3}{5})\)

Solution:-

= \((\dfrac{15}{1}) \times (\dfrac{3}{5})\) 

= \(\dfrac{(15 × 3)}{1 × 5}\)

= \((\dfrac{45}{5})\)

= 9

Question 4. 

Shade:

(i) \(\dfrac{1}{2}\) of the circles in box (a)  (ii) \(\dfrac{2}{3}\) of the triangles in box (b) (iii) \(\dfrac{3}{5}\) of the squares in the box (c)

Solution:-

(i) We may observe that there are 12 circles in the given box. So, we have to shade \(\dfrac{1}{2}\) of the circles in the box. ∴ 12 × \(\dfrac{1}{2}\) = \(\dfrac{12}{2}\)=6 So we have to shade any 6 circles in the box. (ii) We may observe that there are 9 triangles in the given box. So, we have to shade \(\dfrac{2}{3}\) of the triangles in the box. ∴ 9 × \(\dfrac{2}{3}\) = \(\dfrac{18}{3}\)= 6 So we have to shade any 6 triangles in the box. (iii) We may observe that there are 15 squares in the given box. So, we have to shade \(\dfrac{3}{5}\) of the squares in the box. ∴ 15 × \(\dfrac{3}{5}\) = \(\dfrac{45}{5}\) = 9 So we have to shade any 9 squares in the box.

Question 5. 

Find:

(a) \(\dfrac{1}{2}\) of (i) 24 (ii) 46

Solution:-

(i) 24

= \(\dfrac{1}{2}\) × 24

= \(\dfrac{24}{2}\)

= 12

(ii) 46

= \(\dfrac{1}{2}\) × 46

= \(\dfrac{46}{2}\)

= 23

(b) \(\dfrac{2}{3}\) of (i) 18 (ii) 27

Solution:-

(i) 18

= \(\dfrac{2}{3}\) × 18

= 2 × 6

= 12

(ii) 27

= \(\dfrac{2}{3}\) × 27

= 2 × 9

= 18

(c) \(\dfrac{3}{4}\) of (i) 16 (ii) 36

Solution:-

(i) 16

= \(\dfrac{3}{4}\) × 16

= 3 × 4

= 12

(ii) 36

= \(\dfrac{3}{4}\) × 36

= 3 × 9

= 27

(d) \(\dfrac{4}{5}\) of (i) 20 (ii) 35

Solution:-

(i) 20

= \(\dfrac{4}{5}\) × 20

= 4 × 4

= 16

(ii) 35

= \(\dfrac{4}{5}\) × 35

= 4 × 7

= 28

Question 6. 

Multiply and express as a mixed fraction:

(a) 3 × 5\(\dfrac{1}{5}\)

Solution:-

= 3 × \(\dfrac{26}{5}\)

= \(\dfrac{78}{5}\)

\(15\dfrac{3}{5}\)

(b) 5 × 6 \(\dfrac{3}{4}\)

Solution:-

= 6 \(\dfrac{3}{4}\) = \(\dfrac{27}{4}\)

= 5 × \(\dfrac{27}{4}\)

= \(\dfrac{135}{4}\)

= 33 \(\dfrac{3}{4}\)

(c) 7 × 2 \(\dfrac{1}{4}\)

Solution:-

= 2 \(\dfrac{1}{4}\) = \(\dfrac{9}{4}\)

= 7 × \(\dfrac{9}{4}\)

= \(\dfrac{63}{4}\)

= 15 \(\dfrac{3}{4}\)

(d) 4 × \(6\dfrac{1}{3}\)

Solution:-

= 4 × \(\dfrac{19}{3}\)

= \(\dfrac{76}{3}\)

=\(25\dfrac{1}{3}\)

(e) 3 \(\dfrac{1}{4}\) × 6

Solution:-

= 3 \(\dfrac{1}{4}\) = \(\dfrac{13}{4}\)

= (\(\dfrac{13}{4}\)) × 6

= (\(\dfrac{13}{2}\)) × 3

= \(\dfrac{39}{2}\)

= \(19\dfrac{1}{2}\)

(f) \(3\dfrac{2}{5}\) × 8

Solution:-

=\(3\dfrac{2}{5}\) × 8

= (\(\dfrac{17}{5}\)) × 8

= \(\dfrac{136}{5}\)

\(27\dfrac{1}{5}\)

Question 7. 

Find:

(a) \(\dfrac{1}{2}\) of (i) 2 \(\dfrac{3}{4}\) (ii) 4\(\dfrac{2}{9}\)

Solution:-

(i) 2 \(\dfrac{3}{4}\)

First convert the given mixed fraction into improper fraction.

= 2 \(\dfrac{3}{4}\) = \(\dfrac{11}{4}\)

Now,

=\(\dfrac{1}{2}\) × \(\dfrac{11}{4}\)

=\(\dfrac{1}{2} \times \dfrac{1}{2}\)

=\(\dfrac{1 \times 11}{2 \times 4}\)

= \(\dfrac{11}{8}\)

=\(1\dfrac{3}{8}\)

(ii)4\(\dfrac{2}{9}\)

First convert the given mixed fraction into improper fraction.

=4\(\dfrac{2}{9}\)= \(\dfrac{38}{9}\)

Now,

= \(\dfrac{1}{2} \times \dfrac{38}{9}\)

= \(\dfrac{1 × 38}{2 × 9}\)

=  \(\dfrac{ 38}{18}\)

=  \(\dfrac{ 19}{9}\)

= \(2\dfrac{ 1}{9}\)

(b)   \(\dfrac{ 5}{8}\) of (i) \(3\dfrac{ 5}{6}\)  (ii)   \(9\dfrac{ 2}{3}\)

Solution:-

(i)

=\(\dfrac{ 5}{8}\) × \(3\dfrac{ 5}{6}\) 

=\(\dfrac{ 5}{8}\) × \(\dfrac{ 23}{6}\) 

= \(\dfrac{ 5 \times 23}{8 \times 6}\)

= \(\dfrac{ 115}{8 \times 48}\)

=\(2\dfrac{ 19}{48}\)

(ii)\(9\dfrac{ 2}{3}\)

=\(\dfrac{5}{8} \times \) \(\dfrac{ 29}{3}\)

=\(\dfrac{5 \times28}{8\times 3} \)

=\(\dfrac{145 }{ 24} \)

=\(6\dfrac{1}{ 24} \)

Question 8. 

Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed \((\dfrac{2}{5})\) of the water. Pratap consumed the remaining water.

(i) How much water did Vidya drink?

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

Solution:-

(i) From the question, it is given that,

Amount of water in the water bottle = 5 liters

Amount of water consumed by Vidya = \((\dfrac{2}{5})\) of 5 liters

= \((\dfrac{2}{5})\) × 5

= 2 liters

So, the total amount of water drank by Vidya is 2 liters

(ii) From the question, it is given that,

Amount of water in the water bottle = 5 liters

Then,

Amount of water consumed by Pratap = (1 – water consumed by Vidya)

= (1 – \((\dfrac{2}{5})\))

= \((\dfrac{5-2}{5})\)

= \((\dfrac{3}{5})\)

∴ Total amount of water consumed by Pratap = \((\dfrac{3}{5})\) of 5 liters

= \((\dfrac{3}{5})\) × 5

= 3 liters

So, the total amount of water drank by Pratap is 3 liters.