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Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.8 | Set 1
  • Last Updated : 13 Jan, 2021
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Question 1. If A and B are two sets such that n (A ∪ B) = 50, n (A) = 28 and n (B) = 32, find n (A ∩ B).

Solution:

n(A∪B) = 50
n(A) = 28
n(B) = 32
We know the formula, n(A∪B) = n(A) + n(B) – n(A∩B)
Putting the values we get
50 = 28 + 32 – n(A∩B)
50 = 60 – n(A∩B)
–10 = – n(A∩B)
n(A∩B) = 10

Question 2. If P and Q are two sets such that P has 40 elements, P ∪ Q has 60 elements and P ∩ Q has 10 elements, how many elements does Q have?

Solution:

n(P) = 40
n(P ∪ Q)= 60
n(P ∩ Q) =10
We know, n(P∪Q) = n(P) + n(Q) – n(P∩Q)
Putting the values we get
60 = 40 + n(Q)–10
60 = 30 + n(Q)
N(Q) = 30
Q has 30 elements.

Question 3. In a school, there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics, and 4 teach physics and mathematics. How many teach physics?

Solution:



Teachers teaching physics or math = 20
Teachers teaching physics and math = 4
Teachers teaching maths = 12
Let teachers who teach physics be ‘n(P)’ and for Maths be ‘n(M)’
20 teachers who teach physics or math = n (P ∪ M) = 20
4 teachers who teach physics and math = n (P ∩ M) = 4
12 teachers who teach maths = n (M) = 12
We know the formula
n (P ∪ M) = n (M) + n (P) – n (P ∩ M)
Putting the values we get,
20 = 12 + n (P) – 4
20 = 8 + n (P)
n (P) =12
There are 12 physics teachers.

Question 4. In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many like both coffee and tea?

Solution:

Total number of people = 70
Number of people who like Coffee = n(C) = 37
Number of people who like Tea = n(T) = 52
Total number = n(C∪T) = 70
Person who likes both would be n(C∩T)
We know as formula
n(C∪T) = n(C) + n(T) – n(C ∩ T)
Putting the values we get
70 = 37 + 52 – n (C ∩ T)
70 = 89 – n (C ∩ T)
n(C∩T)=19

Question 5. Let A and B be two sets such that: n (A) = 20, n (A ∪ B) = 42 and n (A ∩ B) = 4. Find

(i) n (B)

Solution:

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
(As we know the formula)
Putting the values we get
42 = 20 + n (B) – 4
42 = 16 + n (B)
n(B) = 26
n(B) = 26

(ii) n (A – B)

Solution:

We know, the formula
n(A – B) = n (A ∪ B) – n (B)
Putting the values we get
n(A – B) = 42 – 26
= 16

(iii) n (B – A)

n(B – A) = n(B) – n(A ∩ B)
Putting the values we get
n(B – A) = 26 – 4 = 22
n(B–A) = 22



Question 6. A survey shows that 76% of the Indians like oranges, whereas 62% like bananas. What percentage of the Indians like both oranges and bananas?

Solution:

According to question
People who like oranges = 76%
People who like bananas = 62%
Let assume that people who like oranges be n(O)
Let assume that people who like bananas be n(B)
Total number of people who like oranges or bananas = n (O ∪ B) = 100
People who like both oranges and bananas = n (O ∩ B)
We know the formula
n(O∪B) = n(O) + n(B) – n(O ∩ B)
Substituting the values we get
100 = 76 + 62 – n (O ∩ B)
100 = 138 – n (O ∩ B)
n(O∩B) = 38

Question 7. In a group of 950 persons, 750 can speak Hindi, and 460 can speak English. Find:

(i) How many can speak both Hindi and English.

(ii) How many can speak Hindi only.

(iii) how many can speak English only.

Solution:

Let, total number of people be n (P) = 950
People who can speak English n (E) = 460
People who can speak Hindi n (H) = 750
Given in the question:
(i) People who can speak both Hindi and English = n (H ∩ E)
We know the formula
n(P) = n(E) + n(H) – n(H ∩ E)
Putting the values
950 = 460 + 750 – n (H ∩ E)
950 = 1210 – n(H∩E)
n(H ∩ E) = 260
Number of people who can speak both English and Hindi are 260.

(ii) H is disjoint union of n(H–E) and n(H ∩ E).
(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))
H = n (H–E) ∪ n (H ∩ E)
n (H) = n (H–E) + n (H ∩ E)
750 = n(H – E) + 260
n(H–E) = 490
490 people can speak only Hindi.

(iii) How many can speak English only. (According to Question)
E is disjoint union of n (E–H) and n (H ∩ E)
(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))
E = n(E–H) ∪ n(H ∩ E).
n(E) = n(E–H) + n(H ∩ E).
460 = n(H – E) + 260
n H–E) = 460 – 260 = 200
200 people can speak only English.

Question 8. A survey shows that 76% of the Indians like oranges, whereas 62% bananas. What percentage of Indians like both oranges and bananas?

Solution:

Let assume that n(p) denote total percentage of Indian. 
n(o) denotes the percentage of who like oranges.(let)
n(B) denotes the percentage of who like banana. (let)
n(p) = 100 n(O)=76 n(B)=62
We need to find n(O∩B)
n(p) = n(O)+n(B) – n(O∩B)
100 = 76 + 62 – n(O∩B)
n(O∩B) = 138 – 100
= 38

Question 9. In a group of 950 people, 750 can speak Hindi, and 460  can speak English. Find

(i) How many can speak both Hindi and English?

(ii) How many can speak Hindi only?

(iii) How many can speak English only?

Solution:

(i) Let assume that n(P) denote that number of person.
n(H) assume that number of people who speak Hindi.
n(E) assume that number of people who speak English.
n(p) = 950 , n(H) = 750 and n(E) = 460
we need to find n(H∩E)
According to formula n(P) = n(H) + n(E) – n(H∩E)
950 = 750 + 460 – n(H∩E)
n(H∩E) = 260
(ii) Here we can say that H is disjoint set union of H-E & H∩E 
that implies that H = (H – E) ∪ (H ∩ E)
n(H) = n(H – E) + n(H∩E)
750 = n(H – E) – 260
n(H – E) = 490
(iii) E = (E – H) ∪ ( H ∩ E)
n(E) = n(E-H) + n(H ∩ E)
n(E-H) = 460 – 260
= 200

Question 10. In a group of  50 people, 14 drink tea but not coffee, and 30 drink tea. Find:

(i) How many drink tea and coffee both?

(ii) How many drink coffee but not tea?

(i) Let assume that, n(p) denote the total number of persons
n(T) denote the number of person who drink tea
n(C) denote the number of person who drink coffee
We need to find n(T∩C)
T is clear;y disjoint union of T-C and T∩C
T = (T – C) ∪ (T ∩ C)
30 = 14 + n(T ∩ C)
n(T ∩ C) = 16
(ii) Here we need to find C-T
According to the formula: n(P) = n(C) + n(T) – n(T ∩ C)
50 = n(C) + 14
n(C) = 36
Now , C is the disjoint union of C-T and (T ∩ C)
C = (C-T) ∪ (T ∩ C)
n(C) = n(C-T) + n(C ∩ T)
n(C- T) = 36 – 20
= 20

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