Important Questions For Class 11 Maths Chapter 1 Sets are given here to help the students with their exam preparation for the academic year of 2020-2021. Students can refer to these important questions of Sets to learn different methods of solving the problems easily.
This chapter has simple questions related to the basic operations involving sets. You can refer to the stepwise solutions of all the important questions of this chapter here along with the practice questions at the bottom.
Important Questions & Answers For Class 11 Maths Chapter 1 Sets
Q. 1: Write the following sets in the roster form.
(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}
(ii) C = {x : x2 + 7x – 8 = 0, x ∈ R}
Solution:
(i) 2x – 1 is always an odd number for all positive integral values of x since 2x is an even number.
In particular, 2x – 1 is an odd number for x = 1, 2, … , 9.
Therefore, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(ii) x2 + 7x – 8 = 0
(x + 8) (x – 1) = 0
x = – 8 or x = 1
Therefore, C = {– 8, 1}
Q. 2: Write the following sets in roster form:
(i) A = {x : x is an integer and –3 ≤ x < 7}
(ii) B = {x : x is a natural number less than 6}
Solution:
(i) A = {x : x is an integer and –3 ≤ x < 7}
Integers are …-5, -4, -3, -2, -2, 0, 1, 2, 3, 4, 5, 6, 7, 8,…..
A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {x : x is a natural number less than 6}
Natural numbers are 1, 2, 3, 4, 5, 6, 7, ……
B = {1, 2, 3, 4, 5}
Q. 3: Given that N = {1, 2, 3, …, 100}, then
(i) Write the subset A of N, whose elements are odd numbers.
(ii) Write the subset B of N, whose elements are represented by x + 2, where x ∈ N.
Solution:
(i) A = {x | x ∈ N and x is odd}
A = {1, 3, 5, 7, …, 99}
(ii) B = {y | y = x + 2, x ∈ N}
1 ∈ N, y = 1 + 2 = 3
2 ∈ N, y = 2 + 2 = 4, and so on.
Therefore, B = {3, 4, 5, 6, … , 100}
Q. 4: Let X = {1, 2, 3, 4, 5, 6}. If n represent any member of X, express the following as sets:
(i) n ∈ X but 2n ∉ X
(ii) n + 5 = 8
(iii) n is greater than 4
Solution:
(i) For X = {1, 2, 3, 4, 5, 6}, it is given that n ∈ X, but 2n ∉ X.
Let, A = {x | x ∈ X and 2x ∉ X}
Now, 1 ∉ A as 2.1 = 2 ∈ X
2 ∉ A as 2.2 = 4 ∈ X
3 ∉ A as 2.3 = 6 ∈ X
But 4 ∈ A as 2.4 = 8 ∉ X
5 ∈ A as 2.5 = 10 ∉ X
6 ∈ A as 2.6 = 12 ∉ X
Therefore, A = {4, 5, 6}
(ii) Let B = {x | x ∈ X and x + 5 = 8}
Here, B = {3} as x = 3 ∈ X and 3 + 5 = 8 and there is no other element belonging to X such that x + 5 = 8.
(iii) Let C = {x | x ∈ X, x > 4}
Therefore, C = {5, 6}
Q. 5: Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.
Find A′, B′, A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′∩ B′.
Solution:
Given,
U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}
A′ = {1, 4, 5, 6}
B′ = { 1, 2, 6 }.
Hence, A′ ∩ B′ = { 1, 6 }
Also, A ∪ B = { 2, 3, 4, 5 }
(A ∪ B)′ = { 1, 6 }
Therefore, ( A ∪ B )′ = { 1, 6 } = A′ ∩ B′
Q. 6: Use the properties of sets to prove that for all the sets A and B, A – (A ∩ B) = A – B
Solution:
A – (A ∩ B) = A ∩ (A ∩ B)′ (since A – B = A ∩ B′)
= A ∩ (A′ ∪ B′) [by De Morgan’s law)
= (A∩A′) ∪ (A∩ B′) [by distributive law]
= φ ∪ (A ∩ B′)
= A ∩ B′ = A – B
Hence, proved that A – (A ∩ B) = A – B.
Q. 7: Let U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}, find
(i) A′ ∪ (B ∩ C′)
(ii) (B – A) ∪ (A – C)
Solution:
Given,
U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}
(i) A′ = {1, 3, 5, 7}
C′ = {3, 5, 6}
B ∩ C′ = {3, 5}
A′ ∪ (B ∩ C′) = {1, 3, 5, 7}
(ii) B – A = {3, 5}
A – C = {6}
(B – A) ∪ (A – C) = {3, 5, 6}
Q. 8: In a class of 60 students, 23 play hockey, 15 play basketball,20 play cricket and 7 play hockey and basketball, 5 play cricket and basketball, 4 play hockey and cricket, 15 do not play any of the three games. Find
(i) How many play hockey, basketball and cricket
(ii) How many play hockey but not cricket
(iii) How many play hockey and cricket but not basketball
Solution:
Venn diagram of the given data is:
15 students do not play any of three games.
n(H ∪ B ∪ C) = 60 – 15 = 45
n(H ∪ B ∪ C) = n(H) + n(B) + n(C) – n(H ∩ B) – n(B ∩ C) – n(C ∩ H) + n(H ∩ B ∩ C)
45 = 23 + 15 + 20 – 7 – 5 – 4 + d
45 = 42 + d
d = 45- 42 = 3
Number of students who play all the three games = 3
Therefore, the number of students who play hockey, basketball and cricket = 3
a + d = 7
a = 7 – 3 = 4
b + d = 4
b = 4 – 3 = 1
a + b + d + e = 23
4 + 1 + 3 + e = 23
e = 15
Similarly, c = 2, g =14, f = 6
Number of students who play hockey but not cricket = a + e
= 4 + 15
= 19
Number of students who play hockey and cricket but not basketball = b = 1
Q. 9: Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.
Solution:
Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}
U = { 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8}
A U B = {2, 3, 4, 5, 6, 7, 8}
(A U B)’ = {1, 9}
Q. 10: In a survey of 600 students in a school, 150 students were found to be drinking Tea and 225 drinking Coffee, 100 were drinking both Tea and Coffee. Find how many students were drinking neither Tea nor Coffee.
Solution:
Given,
Total number of students = 600
Number of students who were drinking Tea = n(T) = 150
Number of students who were drinking Coffee = n(C) = 225
Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100
n(T U C) = n(T) + n(C) – n(T ∩ C)
= 150 + 225 -100
= 375 – 100
= 275
Hence, the number of students who are drinking neither Tea nor Coffee = 600 – 275 = 325
Practice Questions For Class 11 Maths Chapter 1 Sets
- Let A, B and C be sets, then show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
- Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed:
- in English and Mathematics but not in Science
- in Mathematics and Science but not in English
- in Mathematics only
- in more than one subject only
- Two finite sets have m and n elements, respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The values of m and n respectively are: (A) 7, 6 (B) 5, 1 (C) 6, 3 (D) 8, 7
- Let A and B be two sets, if A ∩ X = B ∩ X = φ and A U X = B U X for some set X, prove that A =B.
- Let P be the set of prime numbers and let S = {t | 2t – 1 is a prime}. Prove that S ⊂ P.
- If A and B are subsets of the universal set U, then show that:
- (i) A ⊂ A ∪ B
- (ii) A ⊂ B ⇔ A ∪ B = B
- (iii) (A ∩ B) ⊂ A
- A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20} B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.
- In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 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:
- (a) The number of families which buy newspaper A only.
- (b) The number of families which buy none of A, B and C
- If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
- If X= { a, b, c, d } and Y = { f, b, d, g}, find: (i) X – Y (ii) Y – X (iii) X ∩ Y
- Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science,6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed
(i) in English and Mathematics but not in Science
(ii) in Mathematics and Science but not in English
(iii) in Mathematics only
(iv) in more than one subject only - Let F1 be the set of parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane. Then F1may be equal to
(a) F2 ∩F3
(b) F3 ∩F4
(c) F2 u Fs
(d) F2 ∪ F3 ∪ F4 ∪ F1 - If X= {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by
(i) 4n
(ii) n + 6
(iii) n/2
(iv) n-1