1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

RS Aggarwal - Mathematics

1. Sets

Exercise 1A

1 A 1 B 1 C 1 D 1 E 1 F 1 G 1 H 1 I 1 J 1 K 1 L 1 M 2 3 A 3 B 3 C 3 D 3 E 3 F 3 G 3 H 3 I 3 J 4 A 4 B 4 C 4 D 4 E 4 F 4 G 4 H 5 6

Exercise 1B

1 A 1 B 1 C 1 D 1 E 1 F 1 G 1 H 1 I 2 3 A 3 B 3 C 3 D 3 E 4 5 A 5 B 5 C 5 D 5 E 5 F 5 G 5 H 5 I 5 J 5 K 6

Exercise 1C

1 A 1 B 1 C 1 D 1 E 1 F 1 G 1 H 1 I 1 J 1 K 2 3 4 5 6 7 8 9 10 11 12 13 14

Exercise 1D

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Exercise 1E

1 2 3 4 5 6 7 8

Exercise 1F

Exercise 1G

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Exercise 1H

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Q. 6

If A and B are tw

Class 11th RS Aggarwal - Mathematics 1. Sets

Answer :

Given: n(A) = 54, n(B) = 39, n(B – A) = 13

Using the hint

n(B) = n(B – A) + n(A ∩ B)

⇒ 39 = 13 + n(A ∩ B)

⇒ n(A ∩ B) = 39 – 13

⇒ n(A ∩ B) = 26

Visualizing the hint given,

We know that n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ n(A ∪ B) = 54 + 39 – 26

⇒ n(A ∪ B) = 67

Hence n(A ∪ B) = 67

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