This document contains solutions to multiple choice and multi-part questions about sets from a Maths Class 11 Chapter on sets. Some key details: - Questions cover topics like roster form, set-builder form, union, intersection, complement and power sets of sets. - Solutions provide step-by-step workings and explanations for finding subsets, set operations and properties of sets. - One question involves a Venn diagram to determine the number of individuals exposed to certain chemicals. - Another question uses set operations and properties to calculate the number of families buying different newspaper combinations in a town.

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Maths Class 11 Chapter 1 SETS Most Important Questions for Exams
SOLUTION OF ONE MARK QUESTIONS
Q1. Write the set A = {x : x is a letter of the word MATHEMATICS} in the roster form.
Sol 1. Distinct letters in the word ‘MATHEMATICS’ are :
M, A, T, H, E, I, C, S
⇒ In roster form, set A = {M, A, T, H, E, I, C, S}
Q2. Write the set {1, 4, 9, …, 100} in the set-builder form.
Sol 2. We observe that the elements of set A are the squares of all natural numbers less
than or equal to 10.
⇒ In set-builder form, set A = {x : x = n2
, where n ∈ N and n ≤ 10}
Q3. Find the value of X – Y if X = {a, b, c, d} and Y = {b, d, f, g}.
Sol 3. X = {a, b, c, d}, Y = {b, d, f, g}.
⇒ X – Y = {a, c}
Q4. If A = {-1, 0, 1}, then write the power set of A.
Imp Note : (i) The order in which the elements are written in a set makes no difference.
For example - A = {M, A, T, H, E, I, C, S} = {T, A, M, H, E, I, C, S}
(ii) The repetition of element has no effect.
For example - A = {M, A, T, H, E, M, A, T, I, C, S} = {M, A, T, H, E, I, C, S}

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Sol 4. Subsets of A are : 𝜙, {-1}, {0}, {1}, {-1, 0}, {0, 1}, {-1, 1}, {-1, 0, 1}
⇒ P(A) = { 𝝓, {-1}, {0}, {1}, {-1, 0}, {0, 1}, {-1, 1}, {-1, 0, 1}}
Q5. If A = {1, 3, 5, 7, 9}, how many elements has P(A)?
Sol 5. No. of elements in A, n(A) = 5
⇒ Total no. of subsets of set A = 2(No. of elements in A)
= 25
= 32
∴ Total no. of elements in P(A) = Total no. of subsets of set A = 32
SOLUTION OF TWO MARK QUESTIONS
Q6. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 6}, B = {1, 4, 5, 6, 7} and C = {3, 4, 8, 9}. Find
(A - B) ∪ C’.
Sol 6. U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {2, 3, 5, 6}, B = {1, 4, 5, 6, 7} and C = {3, 4, 8, 9}
⇒ A – B = {2, 3}
and C’ = {1, 2, 5, 6, 7}
⇒ (A - B) ∪ C’ = {1, 2, 3, 5, 6, 7}
Q7. Let A and B be two sets having 4 and 7 elements respectively. Then write the
(i) minimum number of elements that A ∪ B can have
(ii) maximum number of elements that A ∩ B can have
Sol 7. (i) Minimum number of elements that A ∪ B can have is 7 (greater no. b/w 4 and 7)
(ii) Maximum number of elements that A ∩ B can have is 4 (smaller no. b/w 4 and 7)

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Q8. Using properties of sets, prove that A ∪ (A ∩ B) = A.
Sol 8. To prove, A ∪ (A ∩ B) = A
L.H.S.
A ∪ (A ∩ B)
= (A ∪ A) ∩ (A ∪ B) [∵ ∪ is distributive over ∩]
= A ∩ (A ∪ B) [∵ A ∪ A = A]
⇒ A ∪ (A ∩ B) = A ∩ (A ∪ B) …(i)
Let a ∈ A,
⇒ a ∈ A ∪ B
⇒ a ∈ A ∩ (A ∪ B) [∵ a ∈ A and a ∈ A ∪ B]
⇒ a ∈ A ∪ (A ∩ B) [∵ from (i)]
∴ a ∈ A ⇒ a ∈ A ∪ (A ∩ B)
⇒ A ⊂ A ∪ (A ∩ B) …(ii)
Let b ∈ A ∪ (A ∩ B),
⇒ b ∈ A ∩ (A ∪ B) [∵ from (i)]
⇒ b ∈ A and b ∈ A ∪ B
⇒ b ∈ A and (b ∈ A or b ∈ B)
⇒ b ∈ A
∴ b ∈ A ∪ (A ∩ B) ⇒ b ∈ A
⇒ A ∪ (A ∩ B) ⊂ A …(iii)
From (ii) and (iii),
A ∪ (A ∩ B) = A
Hence proved.

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SOLUTION OF FOUR MARK QUESTIONS
Q9. From the adjoining venn diagram, write the value of the following :
(i) A’
(ii) B’
(iii) (A ∩ B)’
Sol 9. U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(i) A = {2, 4, 3, 0, 6}
⇒ A’ = {1, 5, 7, 8, 9}
(ii) B = {0, 3, 5, 6, 7}
⇒ B’ = {1, 2, 4, 8, 9}
(iii) A ∩ B = {3, 0, 6}
(A ∩ B)’ = {1, 2, 4, 5, 7, 8, 9}
Q10. If U = {x : x ∈ N and x ≤ 10}, A = {x : x is prime and x ≤ 10} and B = {x : x is a factor of 24}
then verify the following result
(i) A – B = A ∩ B’
(ii) (A ∪ B)’ = A’ ∩ B’
(iii) (A ∩ B)’ = A’ ∪ B’
(iv) A ∪ (B - A) = (A ∪ B)
Sol 10. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
A = {2, 3, 5, 7}
A B
1 8
2 3
4
5
6 7
9
0
U

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and B = {1, 2, 3}
(i) B’ = {4, 5, 6, 7, 8, 9, 10}
A – B = {5, 7} …(i)
A ∩ B’ = {5, 7} …(ii)
From (i) & (ii),
A – B = A ∩ B’
Hence verified.
(ii) A’ = {1, 4, 6, 8, 9, 10}
A ∪ B = {1, 2, 3, 5, 7} …(iii)
(A ∪ B)’ = {4, 6, 8, 9, 10} …(iv)
A’ ∩ B’ = {4, 6, 8, 9, 10} …(v)
From (iv) & (v),
(A ∪ B)’ = A’ ∩ B’
Hence verified.
(iii) A ∩ B = {2, 3}
(A ∩ B)’ = {1, 4, 5, 6, 7, 8, 9, 10} …(vi)
A’ ∪ B’ = {1, 4, 5, 6, 7, 8, 9, 10} …(vii)
From (vi) & (vii),
(A ∩ B)’ = A’ ∪ B’
Hence verified.
(iv) B - A = {1}
A ∪ (B - A) = {1, 2, 3, 5, 7} …(viii)
From (iii) & (viii),
A ∪ (B - A) = (A ∪ B)
Hence verified.

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Q11. On the real axis, if A = [0, 3) and B = [2, 6], then find the following
(i) A’
(ii) A ∪ B
(iii) A ∩ B
(iv) A - B
Sol 11. A = [0, 3)
B = [2, 6]
(i) A’ = (-∞, 0) ∪ [3, ∞)
(ii) A ∪ B = [0, 6]
(iii) A ∩ B = [2, 3)
(iv) A – B = [0, 2)
Q12. 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.
Sol 12. Let U = Set of all surveyed students,
A = Set of students drinking Apple Juice
and O = Set of students drinking Orange Juice
then, n(U) = 400, n(A) = 100, n(O) = 150 and n(A ∩ O) = 75
To find : n(A’ ∩ O’)
+∞−∞ 0 3
+∞−∞ 0 2 6
+∞−∞ 0 3
+∞−∞ 0 6
+∞−∞ 0 2 3
+∞−∞ 0 2

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now, n(A ∪ O) = n(A) + n(O) – n(A ∩ O)
= 100 + 150 – 75
= 250 – 75 = 175
⇒ n(A ∪ O) = 175
⇒ n(A’ ∪ O’) = n(U) - n(A ∪ O)
= 400 – 175 = 225
SOLUTION OF SIX MARK QUESTIONS
Q13. There are 200 individuals with a skin disorder, 120 has 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
(iii) Chemical C1 or Chemical C2
Sol 13. Let U = Set of individuals suffering from the skin disorder,
A = Set of individuals exposed to the chemical C1
and B = Set of individuals exposed to the chemical C2
then, n(U) = 200, n(A) = 120, n(B) = 50 and n(A ∩ B) = 30
To find : (i) n(A − B)
(ii) n(B − A)
(iii) n(A ∪ B)
(i) From the Venn diagram, we have
A = (A – B) ∪ (A ∩ B)
A BU
A - B B - A
(A ∩ B)

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⇒ n (A) = n(A – B) + n(A ∩ B) [Since A – B and A ∩ B are disjoint sets]
⇒ n (A – B) = n (A) – n (A ∩ B)
= 120 – 30 = 90
Hence, the number of individuals exposed to chemical C1 but not to chemical C2 is 90.
(ii) From the Venn diagram, we have
B = (B – A) ∪ (A ∩ B)
⇒ n (B) = n (B – A) + n (A ∩ B) [Since B – A and A ∩ B are disjoint sets]
⇒ n(B – A) = n (B) – n (A ∩ B)
= 50 – 30 = 20
Hence, the number of individuals exposed to chemical C2 but not to chemical C1 is 20.
(iii) n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
= 120 + 50 – 30
= 170 – 30 = 140
The number of individuals exposed either to chemical C1 or to chemical C2 is 140.
Q14. A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these
medals went to a total of 58 men and only three men got medals in all the three sports. How
many received medals in exactly two of the three sports?
Sol 14. Let F = Set of men who received medals in football,
B = Set of men who received medals in basketball
and C = Set of men who received medals in cricket
then, n(F) = 38, n(B) = 15, n(C) = 20 n (F ∪ B ∪ C) = 58 and n(F ∩ B ∩ C) = 3
⇒ n(F ∪ B ∪ C) = n(F) + n(B) + n(C) - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) + n(F ∩ B ∩ C)
Putting values,
58 = 38 + 15 + 20 - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) + 3

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F B
U
C
a
b c
d
⇒ 58 = 76 - n(F ∩ B) - n(B ∩ C) - n(F ∩ C)
⇒ n(F ∩ B) + n(B ∩ C) + n(F ∩ C) = 76 – 58 = 18
⇒ n(F ∩ B) + n(B ∩ C) + n(F ∩ C) = 18 …(i)
Consider the Venn diagram,
Here, a = n(F ∩ B) = number of men who got medals in football and basketball only,
b = n(F ∩ C) = number of men who got medals in football and cricket only,
c = n(B ∩ C) = number of men who got medals in basket ball and cricket only
and d = n(F ∩ B ∩ C) = number of men who got medal in all the three
Putting values in (i), we get
a + d + b + d + c + d = 18
⇒ a + b + c + 3d = 18
⇒ a + b + c + 3(3) = 18 [∵ n(F ∩ B ∩ C) = d = 3]
⇒ a + b + c + 9 = 18
⇒ a + b + c = 18 – 9
⇒ a + b + c = 9
∴ The number of people who got medals in exactly two of the three sports is 9.
Q15. In a town of 10,000 families, it was found that 40% families buy newspaper A, 20%
families buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3%
buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, find the no. of
families which buy

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A
U
Bd
f
g
e
C
a b
c
(i) A only
(ii) B only
(iii) exactly two newspapers
(iv) A and C but not B
(v) exactly one newspaper
(vi) atleast one of A, B and C
(vii) none of A, B and C.
Sol 14. Let U = Set of families in a town,
A = Set of families buying newspaper A,
B = Set of families buying newspaper B
and C = Set of families buying newspaper C
then, n(U) = 10,000
n(A) = 40% of 10,000 =
40
100
× 10,000 = 4,000
n(B) = 20% of 10,000 =
20
100
× 10,000 = 2,000
n(C) = 10% of 10,000 =
10
100
× 10,000 = 1,000
n(A ∩ B) = 5% of 10,000 =
5
100
× 10,000 = 500
n(B ∩ C) = 3% of 10,000 =
3
100
× 10,000 = 300
n(A ∩ C) = 4% of 10,000 =
4
100
× 10,000 = 400
n(A ∩ B ∩ C) = 2% of 10,000 =
2
100
× 10,000 = 200

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Consider the Venn diagram,
Here, a = no. of families buying newspaper A only,
b = no. of families buying newspaper B only,
c = no. of families buying newspaper C only,
d = no. of families buying newspaper A and B but not C,
e = no. of families buying newspaper B and C but not A,
f = no. of families buying newspaper A and C but not B
and g = no. of families buying all the three newspapers
⇒ g = n(A ∩ B ∩ C) = 200
now, n(A ∩ B) = d + g
⇒ 500 = d + 200 [Putting values of n(A ∩ B) and g]
⇒ d = 500 – 200 = 300
n(B ∩ C) = e + g
⇒ 300 = e + 200 [Putting values of n(B ∩ C) and g]
⇒ e = 300 – 200 = 100
n(A ∩ C) = f + g
⇒ 400 = f + 200 [Putting values of n(A ∩ C) and g]
⇒ f = 400 – 200 = 200
n(A) = a + d + f + g
⇒ 4,000 = a + 300 + 200 + 200 [Putting values of n(A), d, f and g]
⇒ a = 4,000 - 300 - 200 - 200
⇒ a = 4,000 – 700 = 3,300
n(B) = b + d + e + g
⇒ 2,000 = b + 300 + 100 + 200 [Putting values of n(B), d, e and g]
⇒ b = 2,000 - 300 - 100 - 200

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⇒ b = 2,000 – 600 = 1,400
n(C) = c + e + f + g
⇒ 1,000 = c + 100 + 200 + 200 [Putting values of n(C), e, f and g]
⇒ c = 1,000 - 100 - 200 - 200
⇒ c = 1,000 – 500 = 500
(i) A only = a = 3,300
(ii) B only = a = 1,400
(iii) Exactly two newspapers = d + e + f = 300 + 100 + 200 = 600
(iv) A and C but not B = f = 200
(v) Exactly one newspaper = a + b + c = 3,300 + 1,400 + 500 = 5,200
(vi) Atleast one of A, B and C, n(A ∪ B) = a + b + c + d + e + f + g
= 3,300 + 1,400 + 500 + 300 + 100 + 200 + 200
= 6,000
(vii) None of A, B and C = n(U) – n(A ∪ B) = 10,000 – 6,000 = 4,000