# NCERT Solutions for Class 7 Maths Exercise 2.2 Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Maths Exercise 2.2 Chapter 2 Fractions and Decimals is available in downloadable PDF format. Multiplication of fractions and multiplication of a fraction by a whole number are the two topics covered in this exercise of NCERT Solutions for Class 7 Maths Chapter 2. From this exercise, students can learn how to multiply the fractions by practising more number of problems provided in NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals. Our expert tutors prepare these solutions with detailed explanations.

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 2 Fractions and Decimals â€“ Exercise 2.2

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### Access answers to NCERT Solutions for Class 7 Maths Chapter 2 â€“ Fractions and Decimals Exercise 2.2

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

(i) 2 Ã— (1/5) (ii) 2 Ã— Â½ (iii) 3 Ã— (2/3) (iv) 3 Ã— Â¼

Solution:-

(i) 2 Ã— (1/5) represents the addition of 2 figures, each represents 1 shaded part out of the given 5 equal parts.

âˆ´ 2 Ã— (1/5) is represented by fig (d).

(ii) 2 Ã— Â½ represents the addition of 2 figures, each represents 1 shaded part out of the given 2 equal parts.

âˆ´ 2 Ã— Â½ is represented by fig (b).

(iii) 3 Ã— (2/3) represents the addition of 3 figures, each represents 2 shaded part out of the given 3 equal parts.

âˆ´ 3 Ã— (2/3) is represented by fig (a).

(iii) 3 Ã— Â¼ represents the addition of 3 figures, each represents 1 shaded part out of the given 4 equal parts.

âˆ´ 3 Ã— Â¼ is represented by fig (c).

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

(i) 3 Ã— (1/5) = (3/5) (ii) 2 Ã— (1/3) = (2/3) (iii) 3 Ã— (3/4) = 2 Â¼

Solution:-

(i) 3 Ã— (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.

âˆ´ 3 Ã— (1/5) = (3/5) is represented by fig (c).

(ii) 2 Ã— (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.

âˆ´ 2 Ã— (1/3) = (2/3) is represented by fig (a).

(iii) 3 Ã— (3/4) represents the addition of 3 figures, each represents 3 shaded part out of the given 4 equal parts and 2 Â¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.

âˆ´ 3 Ã— (3/4) = 2 Â¼ is represented by fig (b).

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

(i) 7 Ã— (3/5)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (7/1) Ã— (3/5)

= (7 Ã— 3)/ (1 Ã— 5)

= (21/5)

=

(ii) 4 Ã— (1/3)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4/1) Ã— (1/3)

= (4 Ã— 1)/ (1 Ã— 3)

= (4/3)

=

(iii) 2 Ã— (6/7)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/1) Ã— (6/7)

= (2 Ã— 6)/ (1 Ã— 7)

= (12/7)

=

(iv) 5 Ã— (2/9)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/1) Ã— (2/9)

= (5 Ã— 2)/ (1 Ã— 9)

= (10/9)

=

(v) (2/3) Ã— 4

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/3) Ã— (4/1)

= (2 Ã— 4)/ (3 Ã— 1)

= (8/3)

=

(vi) (5/2) Ã— 6

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/2) Ã— (6/1)

= (5 Ã— 6)/ (2 Ã— 1)

= (30/2)

= 15

(vii) 11 Ã— (4/7)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (11/1) Ã— (4/7)

= (11 Ã— 4)/ (1 Ã— 7)

= (44/7)

=

(viii) 20 Ã— (4/5)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (20/1) Ã— (4/5)

= (20 Ã— 4)/ (1 Ã— 5)

= (80/5)

= 16

(ix) 13 Ã— (1/3)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13/1) Ã— (1/3)

= (13 Ã— 1)/ (1 Ã— 3)

= (13/3)

=

(x) 15 Ã— (3/5)

Solution:-

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (15/1) Ã— (3/5)

= (15 Ã— 3)/ (1 Ã— 5)

= (45/5)

= 9

(i) Â½ of the circles in box (a) (b) 2/3 of the triangles in box (b)

(iii) 3/5 of the squares in the box (c)

Solution:-

(i) From the question,

We may observe that there are 12 circles in the given box. So, we have to shade Â½ of the circles in the box.

âˆ´ 12 Ã— Â½ = 12/2

= 6

So we have to shade any 6 circles in the box.

(ii) From the question,

We may observe that there are 9 triangles in the given box. So, we have to shade 2/3 of the triangles in the box.

âˆ´ 9 Ã— (2/3) = 18/3

= 6

So we have to shade any 6 triangles in the box.

(iii) From the question,

We may observe that there are 15 squares in the given box. So, we have to shade 3/5 of the squares in the box.

âˆ´ 15 Ã— (3/5) = 45/5

= 9

So we have to shade any 9 squares in the box.

5. Find:

(a) Â½ of (i) 24 (ii) 46

Solution:-

(i) 24

We have,

= Â½ Ã— 24

= 24/2

= 12

(ii) 46

We have,

= Â½ Ã— 46

= 46/2

= 23

(b) 2/3 of (i) 18 (ii) 27

Solution:-

(i) 18

We have,

= 2/3 Ã— 18

= 2 Ã— 6

= 12

(ii) 27

We have,

= 2/3 Ã— 27

= 2 Ã— 9

= 18

(c) Â¾ of (i) 16 (ii) 36

Solution:-

(i) 16

We have,

= Â¾ Ã— 16

= 3 Ã— 4

= 12

(ii) 36

We have

= Â¾ Ã— 36

= 3 Ã— 9

= 27

(d) 4/5 of (i) 20 (ii) 35

Solution:-

(i) 20

We have,

= 4/5 Ã— 20

= 4 Ã— 4

= 16

(ii) 35

We have,

= 4/5 Ã— 35

= 4 Ã— 7

= 28

6. Multiply and express as a mixed fraction:

(a) 3 Ã—

Solution:-

First convert the given mixed fraction into improper fraction.

== 26/5

Now,

= 3 Ã— (26/5)

= 78/5

=

(b) 5 Ã— 6 Â¾

Solution:-

First convert the given mixed fraction into improper fraction.

= 6 Â¾ = 27/4

Now,

= 5 Ã— (27/4)

= 135/4

= 33 Â¾

(c) 7 Ã— 2 Â¼

Solution:-

First convert the given mixed fraction into improper fraction.

= 2 Â¼ = 9/4

Now,

= 7 Ã— (9/4)

= 63/4

= 15 Â¾

(d) 4 Ã—

Solution:-

First convert the given mixed fraction into improper fraction.

== 19/3

Now,

= 4 Ã— (19/3)

= 76/3

=

(e) 3 Â¼ Ã— 6

Solution:-

First convert the given mixed fraction into improper fraction.

= 3 Â¼ = 13/4

Now,

= (13/4) Ã— 6

= (13/2) Ã— 3

= 39/2

= 19 Â½

(f) Ã— 8

Solution:-

First convert the given mixed fraction into improper fraction.

== 17/5

Now,

= (17/5) Ã— 8

= 136/5

=

7. Find:

(a) Â½ of (i) 2 Â¾ (ii)

Solution:-

(i) 2 Â¾

First convert the given mixed fraction into improper fraction.

= 2 Â¾ = 11/4

Now,

= Â½ Ã— 11/4

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= Â½ Ã— (11/4)

= (1 Ã— 11)/ (2 Ã— 4)

= (11/8)

=

(ii)

First convert the given mixed fraction into improper fraction.

== 38/9

Now,

= Â½ Ã— (38/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= Â½ Ã— (38/9)

= (1 Ã— 38)/ (2 Ã— 9)

= (38/18)

= 19/9

=

(b) 5/8 of (i) (ii)

Solution:-

(i)

First convert the given mixed fraction into improper fraction.

== 23/6

Now,

= (5/8) Ã— (23/6)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) Ã— (23/6)

= (5 Ã— 23)/ (8 Ã— 6)

= (115/48)

=

(ii)

First convert the given mixed fraction into improper fraction.

== 29/3

Now,

= (5/8) Ã— (29/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) Ã— (29/3)

= (5 Ã— 29)/ (8 Ã— 3)

= (145/24)

=

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed 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 = 2/5 of 5 liters

= (2/5) Ã— 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 â€“ water consumed by Vidya)

= (1 â€“ (2/5))

= (5-2)/5

= 3/5

âˆ´ Total amount of water consumed by Pratap = 3/5 of 5 liters

= (3/5) Ã— 5

= 3 liters

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

### Access Other Exercises of NCERT Solutions for Class 7 Maths Chapter 2 â€“ Fractions and Decimals

Exercise 2.1 Solutions

Exercise 2.3 Solutions

Exercise 2.4 Solutions

Exercise 2.5 Solutions

Exercise 2.6 Solutions

Exercise 2.7 Solutions