NCERT Solutions for Class 7 Maths Exercise 2.2 Chapter 2 Fractions and Decimals are 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.
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals – Exercise 2.2
Access Answers to NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals Exercise 2.2
1. Which of the drawings (a) to (d) show:
(i) 2 × (1/5) (ii) 2 × ½ (iii) 3 × (2/3) (iv) 3 × ¼
Solution:-
(i) 2 × (1/5) represents the addition of 2 figures, each representing 1 shaded part out of the given 5 equal parts.
∴ 2 × (1/5) is represented by fig (d).
(ii) 2 × ½ represents the addition of 2 figures, each representing 1 shaded part out of the given 2 equal parts.
∴ 2 × ½ is represented by fig (b).
(iii) 3 × (2/3) represents the addition of 3 figures, each representing 2 shaded parts out of the given 3 equal parts.
∴ 3 × (2/3) is represented by fig (a).
(iii) 3 × ¼ represents the addition of 3 figures, each representing 1 shaded part out of the given 4 equal parts.
∴ 3 × ¼ is represented by fig (c).
2. Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 × (1/5) = (3/5) (ii) 2 × (1/3) = (2/3) (iii) 3 × (3/4) = 2 ¼
Solution:-
(i) 3 × (1/5) represents the addition of 3 figures, each representing 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.
∴ 3 × (1/5) = (3/5) is represented by fig (c).
(ii) 2 × (1/3) represents the addition of 2 figures, each representing 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.
∴ 2 × (1/3) = (2/3) is represented by fig (a).
(iii) 3 × (3/4) represents the addition of 3 figures, each representing 3 shaded parts 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 × (3/4) = 2 ¼ is represented by fig (b).
3. Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 7 × (3/5)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (7/1) × (3/5)
= (7 × 3)/ (1 × 5)
= (21/5)
=
(ii) 4 × (1/3)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4/1) × (1/3)
= (4 × 1)/ (1 × 3)
= (4/3)
=
(iii) 2 × (6/7)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/1) × (6/7)
= (2 × 6)/ (1 × 7)
= (12/7)
=
(iv) 5 × (2/9)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/1) × (2/9)
= (5 × 2)/ (1 × 9)
= (10/9)
=
(v) (2/3) × 4
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/3) × (4/1)
= (2 × 4)/ (3 × 1)
= (8/3)
=
(vi) (5/2) × 6
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/2) × (6/1)
= (5 × 6)/ (2 × 1)
= (30/2)
= 15
(vii) 11 × (4/7)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11/1) × (4/7)
= (11 × 4)/ (1 × 7)
= (44/7)
=
(viii) 20 × (4/5)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (20/1) × (4/5)
= (20 × 4)/ (1 × 5)
= (80/5)
= 16
(ix) 13 × (1/3)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (13/1) × (1/3)
= (13 × 1)/ (1 × 3)
= (13/3)
=
(x) 15 × (3/5)
Solution:-
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (15/1) × (3/5)
= (15 × 3)/ (1 × 5)
= (45/5)
= 9
4. Shade:
(i) ½ 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 ½ of the circles in the box.
∴ 12 × ½ = 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.
∴ 9 × (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.
∴ 15 × (3/5) = 45/5
= 9
So we have to shade any 9 squares in the box.
5. Find:
(a) ½ of (i) 24 (ii) 46
Solution:-
(i) 24
We have,
= ½ × 24
= 24/2
= 12
(ii) 46
We have,
= ½ × 46
= 46/2
= 23
(b) 2/3 of (i) 18 (ii) 27
Solution:-
(i) 18
We have,
= 2/3 × 18
= 2 × 6
= 12
(ii) 27
We have,
= 2/3 × 27
= 2 × 9
= 18
(c) ¾ of (i) 16 (ii) 36
Solution:-
(i) 16
We have,
= ¾ × 16
= 3 × 4
= 12
(ii) 36
We have
= ¾ × 36
= 3 × 9
= 27
(d) 4/5 of (i) 20 (ii) 35
Solution:-
(i) 20
We have,
= 4/5 × 20
= 4 × 4
= 16
(ii) 35
We have,
= 4/5 × 35
= 4 × 7
= 28
6. Multiply and express as a mixed fraction:
(a) 3 × 
Solution:-
First, convert the given mixed fraction into an improper fraction.
=
= 26/5
Now,
= 3 × (26/5)
= 78/5
=
(b) 5 × 6 ¾
Solution:-
First, convert the given mixed fraction into an improper fraction.
= 6 ¾ = 27/4
Now,
= 5 × (27/4)
= 135/4
= 33 ¾
(c) 7 × 2 ¼
Solution:-
First, convert the given mixed fraction into an improper fraction.
= 2 ¼ = 9/4
Now,
= 7 × (9/4)
= 63/4
= 15 ¾
(d) 4 × 
Solution:-
First, convert the given mixed fraction into an improper fraction.
=
= 19/3
Now,
= 4 × (19/3)
= 76/3
=
(e) 3 ¼ × 6
Solution:-
First, convert the given mixed fraction into an improper fraction.
= 3 ¼ = 13/4
Now,
= (13/4) × 6
= (13/2) × 3
= 39/2
= 19 ½
(f) 
× 8
Solution:-
First, convert the given mixed fraction into an improper fraction.
=
= 17/5
Now,
= (17/5) × 8
= 136/5
=
7. Find:
(a) ½ of (i) 2 ¾ (ii) 
Solution:-
(i) 2 ¾
First, convert the given mixed fraction into an improper fraction.
= 2 ¾ = 11/4
Now,
= ½ × 11/4
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (11/4)
= (1 × 11)/ (2 × 4)
= (11/8)
=
(ii)
First, convert the given mixed fraction into an improper fraction.
=
= 38/9
Now,
= ½ × (38/9)
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ½ × (38/9)
= (1 × 38)/ (2 × 9)
= (38/18)
= 19/9
=
(b) 5/8 of (i) 
(ii) 
Solution:-
(i)
First, convert the given mixed fraction into an improper fraction.
=
= 23/6
Now,
= (5/8) × (23/6)
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (23/6)
= (5 × 23)/ (8 × 6)
= (115/48)
=
(ii)
First, convert the given mixed fraction into an improper fraction.
=
= 29/3
Now,
= (5/8) × (29/3)
By the rule Multiplication of fractions,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/8) × (29/3)
= (5 × 29)/ (8 × 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 litres
Amount of water consumed by Vidya = 2/5 of 5 litres
= (2/5) × 5
= 2 litres
So, the total amount of water drank by Vidya is 2 litres
(ii) From the question, it is given that,
Amount of water in the water bottle = 5 litres
Then,
Amount of water consumed by Pratap = (1 – water consumed by Vidya)
= (1 – (2/5))
= (5-2)/5
= 3/5
∴ The total amount of water consumed by Pratap = 3/5 of 5 litres
= (3/5) × 5
= 3 litres
So, the total amount of water drank by Pratap is 3 litres
Access Other Exercises of NCERT Solutions for Class 7 Maths Chapter 2 – 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
Also, explore –
NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7
NCERT Solutions
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