CBSE Class 11 Maths Revision Notes Chapter 1

Q.1 Draw appropriate Venn diagram for each of the following:

a) A’ ∩ B’.
b) A’.

Ans

a) A’ ∩ B’ = (A ∪ B)’

b) A’

Q.2 In a survey of 200 students of a school it was found that 120 study mathematics, 90 study physics and 70 study chemistry, 40 study mathematics and physics, 30 study physics and chemistry, 50 study chemistry and mathematics and 20 study none of these subjects. Find the number of students who study all three subjects.

Ans

Q.3 If S and T are two sets such that S has 21 elements. T has 32 elements and S ∩ T has 11 elements. How many elements does S ∪ T has?

Ans

Q.4 Fill in the blanks to make the following statements true.

1) A ∪ A’ = ____
2) Φ ∪ A’ = _____
3) R – Q = _____, where R is the set of real numbers and Q is the set of rational numbers.
4) A ∩ A’ = ______
5) U ∩ A = _____
6) A ∩ A = _____

Ans

Q.5 Taking U = {x : x is natural number} write the complements of the following sets:

1. A = {x : x is an even natural number}
2. B = {x : 2x + 5 = 9}
3. C = {x : x ≥ 7}
4. D = {x : x is a prime number}

Ans

1. A’ = {x : x is an odd natural number}
2. B’ = {x : x ∈ N, x ≠ 2}
3. C’ = {x : x ∈ N and x < 7}
4. D’ = {x : x is a composite number}

Q.6 Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Show that (A ∪ B)’ = A’ ∩ B’.

Ans

Q.7 If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13} and C = {11, 13, 15}, find

(i) (A ∩ B) ∩ (B ∩ C)
(ii) A ∩ (B ∪ C)
(iii) Are (i) and (ii) equal?

Ans

Q.8 State True or False justify your answers

(i) {2,3,4,5} and{3,6} are disjoint sets
(ii) {a,e,i,o,u} and {x : x is a vowel} are disjoint sets
(iii) {2,6,10} and {3,7,11} are disjoint sets

Ans

Q.9 In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Find the number of persons who read neither.

Ans

Q.10 If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, find (A ∪ B)^c.

Ans

Given A = {1, 2, 3, 4}, B = {2, 4, 6, 8}
A ∪ B = {1, 2, 3, 4, 6, 8}(A ∪ B)^c = {5, 7, 9}

Q.11 State whether A = B or not.

(a) If A = {2, 4, 6, 8, 10} and B = {x : x is positive even integer and x ≤ 10}.
(b) A = {7, 3} and B = {x : x is a solution of x² + 5x + 6 = 0}.

Ans

Q.12 Consider the sets Φ, A = {1, 3, 9} and B = {1, 5, 9, 11}. Fill in with Φ, ⊂ or ⊄.

1) Φ _____ B
2) A _____ B

Ans

1) Φ ⊂ B as Φ is a subset of every set.
2) A ⊄ B as 3 ∈ A but 3 ∉ B.

Q.13 a) Write the set {1, 4, 9, …,100} in set builder form.
b) Write the Set E = {x: x is a letter of the word ‘MATHEMATICS’}

Ans

a) A = {x: x = y², y is a natural number less than or equal to 10}
OR
A = {x: x = y², y ∈ N and 1 ≤ y ≤ 10}

b) {M,A,T,H,E,I,C,S}

Q.14 Which type of set is the set of odd natural numbers divisible by 2?

Ans

Null set, as there is no odd natural number which is divisible by 2.

Q.15 If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find

a) A ∪ B
b) A ∩ B

Ans

Q.16 If A = {1, 2, 3}, B = {3, 4, 5}, find A – B and B – A. Is A – B = B – A?

Ans

Set of elements of A that are not in B: A – B = {1, 2}
Set of elements of B that are not in A: B – A = {4, 5}
Thus , A – B ≠ B – A

Q.17 Write the set {3, 6, 9, 12, …} in set builder form.

Ans

A = {3x : x ∈ N}

Q.18 Define disjoint sets.

Ans

Two sets A and B are said to be disjoint if A ∩ B = Φ.
E.g. {a, b, c, d} and {e, f, g, h} are disjoint sets.

Q.19 If A = {x : x is a prime number ∀ x ∈ N}, then find A^c.

Ans

A^c = {x : x is not a prime number ∀ x ∈ N}

Q.20 If A = {1, 2, 3}, then find P(A).

Ans

P(A) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}}

Q.21 If A = {a, b, c}, B = {b, e, f}, find A – B.

AnsA – B = {a, c}

Q.22 Define the difference of sets.

Ans

The difference of sets A and B, denoted by A-B, is the set of all those elements of A which do not belongs to B, i.e.,
A – B = {x : x ∈ A and x ∉ B}

Q.23 If A = {x : x ∈ N and x < 100} and B = {y : y ∈ N and y is a multiple of 2 less than 100}, find A ∪ B.

Ans

A ∪ B = A = {x : x ∈ N and x < 100}

Q.24 Write the set A = {x : x ∈ N and x² < 25} in roster form.

Ans

{1, 2, 3, 4}

Q.25 Find the pair of equal sets from the following sets:
A = {0}, B = {x : x > 10 and x < 4}, C = {x : x – 3 = 0} and D = {x : x is a positive root of equation x² – 2x – 3}

Ans

Here A = {0}, B = Φ, C = {3}, D = {3}
Thus C = D

Q.26 Define finite and infinite sets.

Ans

Finite Set: A set which is empty or contains finite number of elements is known as a finite set.
Infinite Set: A set that contains infinite number of elements is known as an infinite set.

Q.27 Write the following intervals in set builder form:

(i) (–7, 0)
(ii) [–2, 3)

Ans

(i) (–7, 0) = {x : x ∈ R and –7 < x < 0}.
(ii) [-2, 3) = {x : x ∈ R and –2 ≤ x < 3}.

Q.28 Define symmetric difference of two sets.

Ans

Let A and B be two sets, then symmetric difference denoted as A Δ B, is defined as follows:
A Δ B = (A – B) ∪ (B – A) = {x : x ∉ A ∩ B}
The Venn diagram is shown below:

Q.29 If A,B and C are any three sets, then prove that A ∩ (BΔC) = (A ∩ B) Δ (A ∩ C).

Ans

Here,
A ∩ (BΔC) = A ∩ [(B – C) ∪ (C – B)]
= [A ∩ (B – C)] ∪ [A ∩ (C – B)]
= [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)]
= (A ∩ B) Δ (A ∩ C)

Q.30 Fill in the blanks to make each of the following a true statement:

(i) A ∪ A’ = _____.
(ii) A ∩ A’ = _____.

Ans

(i) A ∪ A’ = U (Universal set)
(ii) A ∩ A’ = Φ (No element is common in both the sets)

Q.31

$\text{If}\mathrm{A}\cap \mathrm{B}=\mathrm{\varphi },\text{then find\hspace{0.17em}}\mathrm{n}(\mathrm{A}\cup \mathrm{B})\text{.}$

Ans

We know, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
If A ∩ B = Φ, then n(A ∩ B) = 0.
This implies that,
n(A ∪ B) = n(A) + n(B) – 0
or, n(A ∪ B) = n(A) + n(B)

Q.32 In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics?

Ans

Let M denote the set of mathematics teacher and P denote the set of physics teacher.
Given that, n(M ∪ P) = 20
n(M ∩ P) = 4
n(M) = 12
We wish to determine n(P)
We have,
n(M ∪ P) = n(M) + n(P) – n(M ∩ P)
we obtain
20 = 12 + n (P) – 4
⇒ 8 + n(P) = 20
Thus, n(P) = 12
Hence, 12 teachers teach physics.

Q.33 In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play at least one of the two games. How many students like to play both cricket and football?

Ans

$$\begin{gathered} \text{Let}\mathrm{X}\text{be the set of students who like to play cricket} \\ \text{and}\mathrm{Y}\text{be the set of students who like to play football.} \\ \text{Then}\mathrm{X}\cup \mathrm{Y}\text{is the set of students who like to play} \\ \text{at least one game.} \\ \text{and}\mathrm{X}\cap \mathrm{Y}\text{is the set of students who like to play both games.} \\ \text{Given,}\mathrm{n}\left(\mathrm{X}\right)=24, \\ \mathrm{n}\left(\mathrm{Y}\right)=16, \\ \mathrm{n}(\mathrm{X}\cup \mathrm{Y})=35, \\ \mathrm{n}(\mathrm{X}\cap \mathrm{Y})=? \\ \text{Using the formula,} \\ \mathrm{n}(\mathrm{X}\cup \mathrm{Y})=\mathrm{n}\left(\mathrm{X}\right)+\mathrm{n}\left(\mathrm{Y}\right)–\mathrm{n}(\mathrm{X}\cap \mathrm{Y}) \\ \mathrm{we}\mathrm{get} \\ 35=24+16–\mathrm{n}(\mathrm{X}\cap \mathrm{Y}) \\ \Rightarrow \mathrm{n}(\mathrm{X}\cap \mathrm{Y})=40–35 \\ \Rightarrow \mathrm{n}(\mathrm{X}\cap \mathrm{Y})=5 \\ \text{Thus,}5\text{students like to play both games.} \end{gathered}$$

Q.34 In a group of 70 Students, 37 like coffee, 52 like tea and each student likes at least one of the two drinks. How many students like both coffee and tea?

Ans

$$\begin{gathered} \text{Let}\mathrm{X}\text{be the set of students who like to take coffee} \\ \text{and}\mathrm{Y}\text{be the set of students who like to take tea.} \\ \text{Then XUY is the set of students who like at least} \\ \text{one of the two drinks and} \\ \mathrm{X}\cap \mathrm{Y}\text{is the set of students who like both coffee and tea.} \\ \text{Given,}\mathrm{n}\left(\mathrm{X}\right)=37 \\ \mathrm{n}\left(\mathrm{Y}\right)=52 \\ \mathrm{n}(\mathrm{X}\cup \mathrm{Y})=70,\mathrm{n}(\mathrm{X}\cap \mathrm{Y})=? \\ \text{Using the formula,} \\ \mathrm{n}(\mathrm{X}\cup \mathrm{Y})=\mathrm{n}\left(\mathrm{X}\right)+\mathrm{n}\left(\mathrm{Y}\right)–\mathrm{n}(\mathrm{X}\cap \mathrm{Y}) \\ \mathrm{we}\mathrm{get}, \\ 70=37+52–\mathrm{n}(\mathrm{X}\cap \mathrm{Y}) \\ \Rightarrow \mathrm{n}(\mathrm{X}\cap \mathrm{Y})=89–70 \\ \Rightarrow \mathrm{n}(\mathrm{X}\cap \mathrm{Y})=19 \\ \text{Thus,}19\text{students like both coffee and tea.} \end{gathered}$$

Q.35 Mr. X has five friends. He wishes to arrange a dinner party on some occasion. In how many ways he can invite his friends on dinner party by inviting at least one of them.

Ans

Let a, b, c, d, e be the friends of Mr. X
A = {a,b,c,d,e}
The number of all possible subsets of set A = 2⁵ = 32
[The no. of subset of a set having n elements is 2ⁿ]
This includes a null set that indicates not inviting any friend.
The number of ways in which he can invite his friends = 2⁵ = 32
Hence, the number of ways in which he can invite at least one of his friends = 2⁵ – 1
= 32 – 1
= 31

Q.36 In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics only?

Ans

Let M denote the set of mathematics teacher and P denote the set of physics teacher.
Given that,

$$\begin{gathered} \mathrm{n}(\mathrm{M}\mathrm{U}\mathrm{P})=20 \\ \mathrm{n}(\mathrm{M}\cap \mathrm{P})=4,\mathrm{n}\left(\mathrm{M}\right)=12 \\ \text{We wish to determine}\mathrm{n}\left(\mathrm{P}\right) \\ \text{We have,} \\ \mathrm{n}(\mathrm{M}\cup \mathrm{P})=\mathrm{n}\left(\mathrm{M}\right)+\mathrm{n}\left(\mathrm{P}\right)–\mathrm{n}(\mathrm{M}\cap \mathrm{P}) \end{gathered}$$

We obtain
20 = 12 + n (P) – 4
⇒ 8 + n(P) = 20
Thus, n(P) = 12
⇒ The number of teachers teach physics = 12
Since the number of teachers teach both physics and mathematics = 4
∴ Number of teachers teach physics only = 12 – 4
= 8

Q.37 In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.

Ans

Let U denote the set of surveyed students and A denote the set of students taking apple juice and B denote the set of students taking orange juice.
Then n(U) = 400
n(A) = 100
n(B) = 150 and

$$\begin{gathered} \mathrm{n}(\mathrm{A}\cap \mathrm{B})=75 \\ \text{Using the formula,} \\ \mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)–\mathrm{n}(\mathrm{A}\cap \mathrm{B}) \\ \text{we get} \\ \Rightarrow \mathrm{n}(\mathrm{A}\cup \mathrm{B})=100+150–75 \\ \Rightarrow \mathrm{n}(\mathrm{A}\cup \mathrm{B})=175 \\ \text{Also,}\mathrm{n}(\mathrm{A}\cup \mathrm{B}{)}^{‘}=\mathrm{n}(\mathrm{U})–\mathrm{n}(\mathrm{A}\cup \mathrm{B}) \\ \Rightarrow \mathrm{n}(\mathrm{A}\cup \mathrm{B}{)}^{\mathrm{r}}=400–175 \\ \Rightarrow \mathrm{n}(\mathrm{A}\cup \mathrm{B}{)}^{‘}=225 \\ \text{Hence,}225\text{students were taking neither apple juice nor orange juice.} \end{gathered}$$

Q.38 A market research group conducted a survey of 2000 consumers and reported that 1350 consumers like beverage A and 750 consumers like beverage B, what is the least number that must have liked both beverages?

Ans

$$\begin{gathered} \text{Let}\mathrm{A}\text{be the set of consumers who like beverage}\mathrm{A} \\ \text{and}\mathrm{B}\text{be the set of consumers who like beverage}\mathrm{B}\text{.} \\ \text{Then, AUB is the set of consumers who like least one of the two drink and} \\ \mathrm{A}\cap \mathrm{B}\text{is the set of consumers who like both beverages.} \\ \mathrm{Given}, \\ \mathrm{n}\left(\mathrm{A}\right)=1350 \\ \mathrm{n}\left(\mathrm{B}\right)=750 \\ \text{For}\mathrm{n}(\mathrm{A}\cap \mathrm{B})\text{to be least,}\mathrm{n}(\mathrm{A}\cup \mathrm{B})\text{is maximum.} \\ \text{So, maximum values of}\mathrm{n}(\mathrm{A}\cup \mathrm{B})\text{is}2000 \\ \mathrm{n}(\mathrm{A}\cup \mathrm{B})\ge \mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)–\mathrm{n}(\mathrm{A}\cap \mathrm{B}) \\ 2000\ge 1350+750–\mathrm{n}(\mathrm{A}\cap \mathrm{B}) \\ \Rightarrow 2000\ge 2100–\mathrm{n}(\mathrm{A}\cap \mathrm{B}) \\ \Rightarrow \mathrm{n}(\mathrm{A}\cap \mathrm{B})\ge 2100–2000 \\ \Rightarrow \mathrm{n}(\mathrm{A}\cap \mathrm{B})\ge 100 \\ \text{The least number of consumers who like both beverages is}100\text{.} \end{gathered}$$

Q.39 There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C₁, 50 to chemical C₂ and 30 to both the chemicals C₁ and C₂. Find the number of individuals exposed to

(i) Chemical C₁ but not chemical C₂
(ii) Chemical C₂ but not chemical C₁

Ans

Let U denote the universal set consisting of individuals suffering from the skin disorder.
A denote the set of individuals exposed to chemical C_1.
B denote the set of individuals exposed to chemical C₂.
Here, n(U) = 200,
n(A) = 120, n(B) = 50 and

$$\begin{gathered} \mathrm{n}(\mathrm{A}\cap \mathrm{B})=30 \\ \text{(i) we have,} \\ \mathrm{n}(\mathrm{A}–\mathrm{B})=\mathrm{n}\left(\mathrm{A}\right)–\mathrm{n}(\mathrm{A}\cap \mathrm{B}) \\ \Rightarrow \mathrm{n}(\mathrm{A}–\mathrm{B})=120–30 \\ \Rightarrow \mathrm{n}(\mathrm{A}–\mathrm{B})=90 \\ \text{Hence, the number of individuals exposed to chemical} \\ {\mathrm{C}}_1\text{but not to chemical}{\mathrm{C}}_2\text{is}90\text{.} \\ \text{(ii) we have,} \\ \mathrm{n}(\mathrm{B}–\mathrm{A})=\mathrm{n}\left(\mathrm{B}\right)–\mathrm{n}(\mathrm{A}\cap \mathrm{B}) \\ \Rightarrow \mathrm{n}(\mathrm{B}–\mathrm{A})=50–30 \\ \Rightarrow \mathrm{n}(\mathrm{B}–\mathrm{A})=20 \\ \text{Hence, the number of individuals exposed to chemical} \\ {\mathrm{C}}_2\text{but not to chemical}{\mathrm{C}}_1\text{is}20. \end{gathered}$$

Q.40 In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find the number of people who read at least one of the newspapers.

Ans

$$\begin{gathered} \text{Let}\mathrm{H}\text{denote the people who read newspaper}\mathrm{H}\text{.} \\ \text{I denote the people who read newspaper I.} \\ \text{T denote the people who read newspaperT.} \\ \mathrm{n}\left(\mathrm{H}\right)=25 \\ \mathrm{n}\left(\mathrm{I}\right)=26 \\ \mathrm{n}\left(\mathrm{T}\right)=26 \\ \mathrm{n}(\mathrm{H}\cap \mathrm{I})=9 \\ \mathrm{n}(\mathrm{H}\cap \mathrm{T})=11 \\ \mathrm{n}(\mathrm{T}\cap \mathrm{I})=8 \\ \mathrm{n}(\mathrm{H}\cap \mathrm{I}\cap \mathrm{T})=3 \\ \text{Since}\mathrm{n}(\mathrm{H}\cup \mathrm{T}\cup \mathrm{I})\text{denote the number of people who read at least one of} \\ \text{the newspapers, we have} \\ \mathrm{n}(\mathrm{H}\cup \mathrm{T}\cup \mathrm{I})=\mathrm{n}\left(\mathrm{H}\right)+\mathrm{n}\left(\mathrm{T}\right)+\mathrm{n}\left(\mathrm{I}\right)–\mathrm{n}(\mathrm{H}\cap \mathrm{I})–\mathrm{n}(\mathrm{H}\cap \mathrm{T})–\mathrm{n}(\mathrm{T}\cap \mathrm{I})+\mathrm{n}(\mathrm{H}\cap \mathrm{T}\cup \mathrm{I}) \\ =25+26+26–9–11–8+3 \\ =77–28+3 \\ =80–28 \\ =52 \\ \text{Hence, the number of people who read at least one of the newspapers is}52\text{.} \end{gathered}$$