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 represents 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 represents 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 represents 2 shaded part 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 represents 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 represents 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 represents 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 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 × (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 fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (7/1) × (3/5)
= (7 × 3)/ (1 × 5)
= (21/5)
=4 1/2 (Mixed fraction)
(ii) 4 × (1/3)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4/1) × (1/3)
= (4 × 1)/ (1 × 3)
= (4/3)
= 1 1/3 (mixed fraction)
(iii) 2 × (6/7)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/1) × (6/7)
= (2 × 6)/ (1 × 7)
= (12/7) ( change it into mixed fraction) 1 5/7
(iv) 5 × (2/9)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/1) × (2/9)
= (5 × 2)/ (1 × 9)
= (10/9)
= 1 1/9
(v) (2/3) × 4
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/3) × (4/1)
= (2 × 4)/ (3 × 1)
= (8/3)
= 2 2/3
(vi) (5/2) × 6
Solution:-
By the rule Multiplication of fraction,
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 fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11/1) × (4/7)
= (11 × 4)/ (1 × 7)
= (44/7)
= 6 2/7 (mixed fraction)
viii) 20 × (4/5)
Solution:-
By the rule Multiplication of fraction,
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 fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (13/1) × (1/3)
= (13 × 1)/ (1 × 3)
= (13/3)
= 4 1/3
(x) 15 × (3/5)
Solution:-
By the rule Multiplication of fraction,
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:
(c) 7 × 2 ¼
Solution:-
First convert the given mixed fraction into improper fraction.
= 2 ¼ = 9/4
Now,
= 7 × (9/4)
= 63/4
= 15 ¾
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) × 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 – (2/5))
= (5-2)/5
= 3/5
∴ Total amount of water consumed by Pratap = 3/5 of 5 liters
= (3/5) × 5
= 3 liters
So, the total amount of water drank by Pratap is 3 liters