Class 11 Mathematics Sets MCQs Set 03

Question. If \( aN = \{ax : x \in N\} \) then \( 3N \cap 7N = \)
(a) \( 21N \)
(b) \( 10N \)
(c) \( 4N \)
(d) \( 5N \)
Answer: (a) \( 21N \)
Hint: \( 3N = \{3x : x \in N\}, 7N = \{7x : x \in N\} \implies 3N \cap 7N = 21N \).

Question. If \( A = \{1,2,3,4,5,6\}, B = \{1,2\} \), then \( \frac{A}{\left(\frac{A}{B}\right)} = \)
(a) \( A \)
(b) \( \phi \)
(c) \( A \cap B \)
(d) \( A \cup B \)
Answer: (c) \( A \cap B \)
Hint: \( \frac{A}{B} = A - B = \{3,4,5,6\} \). Then \( \frac{A}{\left(\frac{A}{B}\right)} = A - \{3,4,5,6\} = \{1,2\} = A \cap B \).

Question. If \( n(U) = 700, n(A) = 200, n(B) = 300, n(A \cap B) = 100 \), then \( n(A' \cap B') \) is equal to
(a) 400
(b) 240
(c) 300
(d) 500
Answer: (c) 300
Hint: \( n(A' \cap B') = n(A \cup B)' = n(U) - n(A \cup B) = 700 - (200 + 300 - 100) = 700 - 400 = 300 \).

Question. If \( n(U) = 48, n(A) = 28, n(B) = 33 \) and \( n(B-A) = 12 \), then \( n(A \cap B)^C \) is
(a) 27
(b) 28
(c) 29
(d) 30
Answer: (a) 27
Hint: \( n(A \cap B) = n(B) - n(B-A) = 33 - 12 = 21 \). \( n(A \cap B)^C = n(U) - n(A \cap B) = 48 - 21 = 27 \).

Question. If \( n(A \cap B^C) = 5, n(B \cap A^C) = 6, n(A \cap B) = 4 \) then the value of \( n(A \cup B) \) is
(a) 18
(b) 15
(c) 16
(d) 17
Answer: (b) 15
Hint: \( n(A) = n(A \cap B) + n(A \cap B^C) = 4 + 5 = 9 \). \( n(B) = n(A \cap B) + n(B \cap A^C) = 4 + 6 = 10 \). \( n(A \cup B) = n(A) + n(B) - n(A \cap B) = 9 + 10 - 4 = 15 \).

Question. Let \( n(A-B) = 25+X, n(B-A) = 2X \) and \( n(A \cap B) = 2X \). If \( n(A) = 2(n(B)) \) then 'X' is
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (b) 5
Hint: \( n(A) = n(A-B) + n(A \cap B) = 25 + X + 2X = 25 + 3X \). \( n(B) = n(B-A) + n(A \cap B) = 2X + 2X = 4X \). Given \( n(A) = 2n(B) \implies 25 + 3X = 2(4X) \implies 25 + 3X = 8X \implies 5X = 25 \implies X = 5 \).

Question. Of the members of three athletic teams in a school 21 are in the cricket team, 26 are in the hockey team and 29 are in the football team. Among them, 14 play hockey and cricket, 15 play hockey and football, and 12 play football and cricket. Eight play all the three games. The total number of members in the three athletic teams is
(a) 43
(b) 76
(c) 49
(d) 53
Answer: (a) 43
Hint: \( n(C \cup H \cup F) = n(C) + n(H) + n(F) - [n(C \cap H) + n(H \cap F) + n(F \cap C)] + n(C \cap H \cap F) \). \( = 21 + 26 + 29 - [14 + 15 + 12] + 8 = 76 - 41 + 8 = 43 \).

Question. If sets A and B have 3 and 6 elements each, then the minimum number of elements in \( A \cup B \) is
(a) 3
(b) 6
(c) 9
(d) 18
Answer: (b) 6
Hint: \( n(A \cup B) \ge \max\{n(A), n(B)\} = \max\{3, 6\} = 6 \).

Question. If \( n(U) = 60, n(A) = 21, n(B) = 43 \) then greatest value of \( n(A \cup B) \) and least value of \( n(A \cup B) \) are
(a) 60, 43
(b) 50, 36
(c) 70, 44
(d) 60, 38
Answer: (a) 60, 43
Hint: Greatest value of \( n(A \cup B) = n(U) = 60 \). Least value of \( n(A \cup B) = n(B) = 43 \).

Question. If \( A = \{x:x \text{ is a multiple of 4}\} \) and \( B = \{x:x \text{ is a multiple of 6}\} \) then \( A \cap B \) consists of all multiples of
(a) 16
(b) 12
(c) 8
(d) 4
Answer: (b) 12
Hint: \( A = \{4, 8, 12, 16, 20, 24, \dots\}, B = \{6, 12, 18, 24, \dots\} \implies A \cap B = \{12, 24, \dots\} \), which are multiples of 12.

Question. Two finite sets have \( m \) and \( n \) elements. The total number of subsets of the first set is 48 more than the total number of subsets of the second set. The values of \( m \) and \( n \) are
(a) 7,6
(b) 7,6
(c) 6, 4
(d) 7, 4
Answer: (c) 6, 4
Hint: \( 2^m - 2^n = 48 \implies m=6, n=4 \) as \( 64 - 16 = 48 \).

Question. If \( X = \{4^n - 3n - 1 : n \in N\} \) and \( Y = \{9(n-1) : n \in N\} \) then \( X \cup Y = \)
(a) X
(b) Y
(c) N
(d) \( \phi \)
Answer: (b) Y
Hint: \( 4^n - 3n - 1 \) is a multiple of 9. \( X \) contains elements which are multiples of 9. \( Y \) contains all multiples of 9. Thus \( X \subset Y \implies X \cup Y = Y \).

Question. If \( A = ]2, 4[ \) and \( B = [3, 5[ \) then \( A \cap B = \)
(a) [3, 4]
(b) [3, 4[
(c) ]3, 4[
(d) ]2, 5[
Answer: (b) [3, 4[
Hint: The common part between the interval \( (2, 4) \) and \( [3, 5) \) is \( [3, 4) \).

Question. Let A be the set of non-negative integers, I is the set of integers, B is the set of non-positive integers, E is the set of even integers and P is the set of prime numbers then
(a) I-A=B
(b) \( A \cap B = \phi \)
(c) \( E \cap P = \phi \)
(d) \( A \Delta B = I - \{0\} \)
Answer: (d) \( A \Delta B = I - \{0\} \)
Hint: \( A \Delta B = (A-B) \cup (B-A) = \{1,2,3,\dots\} \cup \{\dots, -3, -2, -1\} = I - \{0\} \).

Question. In a class of 100 students, 55 students have passed in Mathematics and 67 students have passed in physics. No of students who have passed in physics only is
(a) 22
(b) 33
(c) 10
(d) 45
Answer: (d) 45
Hint: \( n(M) = 55, n(P) = 67, n(M \cup P) = 100 \). \( n(M \cap P) = 55 + 67 - 100 = 22 \). \( n(P \text{ only}) = n(P) - n(M \cap P) = 67 - 22 = 45 \).

Question. 90 students take mathematics, 72 take science in a class of 120 students. If 10 take neither Mathematics nor science then number of students who take both the subjects is
(a) 52
(b) 110
(c) 162
(d) 100
Answer: (a) 52
Hint: \( n(M) = 90, n(S) = 72, n(M' \cap S') = 10 \). \( n(M \cup S) = 120 - 10 = 110 \). \( n(M \cap S) = 90 + 72 - 110 = 52 \).

Question. A set A has 3 elements and another set B has 6 elements. Then
(a) \( 3 \le n(A \cup B) \le 6 \)
(b) \( 3 \le n(A \cup B) \le 9 \)
(c) \( 6 \le n(A \cup B) \le 9 \)
(d) \( 0 \le n(A \cup B) \le 9 \)
Answer: (c) \( 6 \le n(A \cup B) \le 9 \)
Hint: \( \min\{n(A \cup B)\} = \max\{p, q\} \) and \( \max\{n(A \cup B)\} = p + q \).

Question. 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. then how many students were taking neither apple juice nor orange juice are
(a) 120
(b) 220
(c) 225
(d) 150
Answer: (c) 225
Hint: \( n(A' \cap B') = n(U) - n(A \cup B) = 400 - (100 + 150 - 75) = 400 - 175 = 225 \).

Question. Let A and B be two sets then \( (A \cup B)^C \cup (A^C \cap B) = \)
(a) \( A^C \)
(b) \( B^C \)
(c) \( \phi \)
(d) \( U \)
Answer: (a) \( A^C \)
Hint: \( (A \cup B)^C \cup (A^C \cap B) = (A^C \cap B^C) \cup (A^C \cap B) = A^C \cap (B^C \cup B) = A^C \cap U = A^C \)

Question. A set contains \( (2n + 1) \) elements. The number of subsets of this set containing more than n elements is equal to
(a) \( 2^{n-1} \)
(b) \( 2^n \)
(c) \( 2^{n+1} \)
(d) \( 2^{2n} \)
Answer: (d) \( 2^{2n} \)
Hint: Let the original set contains \( 2n+1 \) elements, then subsets of this set containing more than \( n \) elements means subsets containing \( (n+1) \) elements, \( (n+2) \) elements .... \( (2n+1) \) elements. Required number of subsets \( = {}^{2n+1}C_{n+1} + {}^{2n+1}C_{n+2} + \dots + {}^{2n+1}C_{2n+1} = \frac{1}{2} [ 2^{2n+1} ] = 2^{2n} \).

Question. From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the students has passed in at least one of the subject, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. Then the largest possible number that could have passed all three examinations is
(a) 16
(b) 14
(c) 18
(d) 15
Answer: (b) 14
Hint: \( n(M \cup P \cup C) = 50, n(M) = 37, n(P) = 24, n(C) = 43, n(M \cap P) \le 19, n(M \cap C) \le 29, n(P \cap C) \le 20 \).
\( n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(M \cap C) - n(P \cap C) + n(M \cap P \cap C) \)
\( 50 = 37 + 24 + 43 - [n(M \cap P) + n(M \cap C) + n(P \cap C)] + n(M \cap P \cap C) \)
\( \Rightarrow n(M \cap P \cap C) \le 19 + 29 + 20 - 54 = 14 \).
Therefore, the number of students is at most 14.

Question. Which is the simplified representation of \( (A^1 \cap B^1 \cap C) \cup (B \cap C) \cup (A \cap C) \) where A,B,C are subsets of set X
(a) A
(b) B
(c) C
(d) \( X \cap (A \cup B \cup C) \)
Answer: (c) C

Question. The set \( (A \cup B \cup C) \cap (A \cap B' \cap C')' \cap C' \) is equal to
(a) \( B \cap C' \)
(b) \( A \cap C \)
(c) \( B \cup C' \)
(d) \( A \cap C' \)
Answer: (a) \( B \cap C' \)
Hint: \( (A \cup B \cup C) \cap (A \cap B' \cap C')' \cap C' = (A \cup B \cup C) \cap (A' \cup B \cup C) \cap C' \)
\( = [(A \cap A') \cup (B \cup C)] \cap C' = (\phi \cup B \cup C) \cap C' = (B \cup C) \cap C' = (B \cap C') \cup (C \cap C') = B \cap C' \).

Question. If \( P = \{x \in R : f(x) = 0\} \) and \( Q = \{x \in R : g(x) = 0\} \) then \( P \cup Q \) is
(a) \( \{x \in R : f(x) + g(x) = 0\} \)
(b) \( \{x \in R : f(x) \cdot g(x) = 0\} \)
(c) \( \{x \in R : (f(x))^2 + (g(x))^2 = 0\} \)
(d) \( \{x \in R : x > 1\} \)
Answer: (b) \( \{x \in R : f(x) \cdot g(x) = 0\} \)
Hint: \( f(x) \cdot g(x) = 0 \Rightarrow \) either \( f(x) = 0 \) or \( g(x) = 0 \).

Question. Suppose \( A_1, A_2, ..., A_{30} \) are thirty sets each with five elements and \( B_1, B_2, ..., B_n \) are \( n \) sets each with three elements such that \( \bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S \). If each element of S belongs to exactly ten of the \( A_i \)'s and exactly 9 of the \( B_j \)'s, then the value of n is
(a) 15
(b) 135
(c) 45
(d) 90
Answer: (c) 45
Hint: \( S = \bigcup_{i=1}^{30} A_i \Rightarrow n(S) = \frac{1}{10}(5 \times 30) = 15 \).
Again, \( S = \bigcup_{j=1}^n B_j \Rightarrow n(S) = \frac{1}{9}(3 \times n) = \frac{n}{3} \).
Thus \( \frac{n}{3} = 15 \Rightarrow n = 45 \).

Question. If \( aN = \{ax / x \in N\} \) and \( bN \cap cN = dN \), where \( b, c \in N \) are relatively prime, then
(a) \( d = bc \)
(b) \( c = bd \)
(c) \( b = cd \)
(d) None of the options
Answer: (a) \( d = bc \)
Hint: \( bN = \{bx | x \in N\} \) = the set of positive integral multiples of \( b \) and \( cN = \{cx | x \in N\} \)= the set of positive integral multiples of \( c \).
\( \therefore bN \cap cN = \) the set of positive integral multiples of \( bc = bcN \) [\(\because b \) and \( c \) are relatively prime]. Hence, \( d = bc \).

Question. A survey show that in a city that 63% of the citizens like tea where as 76% like coffee. If x% like both tea and coffee, then
(a) \( x = 63 \)
(b) \( x = 39 \)
(c) \( 50 \le x \le 63 \)
(d) \( 39 \le x \le 63 \)
Answer: (d) \( 39 \le x \le 63 \)
Hint: Let the population of the city be 100. Let A denote tea and B denote coffee.
\( n(A) = 63, n(B) = 76 \).
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) and \( n(A \cup B) \le 100 \Rightarrow 63 + 76 - n(A \cap B) \le 100 \)
\( \Rightarrow 139 - n(A \cap B) \le 100 \Rightarrow 39 \le n(A \cap B) \rightarrow (1) \)
Also \( n(A \cap B) \le n(A) \) and \( n(A \cap B) \le n(B) \Rightarrow n(A \cap B) \le 63 \) and \( n(A \cap B) \le 76 \)
\( \Rightarrow n(A \cap B) \le 63 \rightarrow (2) \). From (1) and (2): \( 39 \le x \le 63 \).

Question. An investigator interviewed 100 students to determine their preferences for the three drinks: milk (M), coffee(C) and tea (T). He reported the following : 10 students had all the three drinks M,C,T; 20 had M and C only: 30 had C and T; 25 had M and T; 12 had M only; 5 had C only; 8 had T only. Then how many did not take any of the three drinks
(a) 20
(b) 3
(c) 36
(d) 42
Answer: (a) 20

Question. In a college of 300 students , every students reads 5 newspapers and every newspaper is read by 60 students. The number of newspapers is
(a) atleast 30
(b) atmost 20
(c) exactly 25
(d) atmost 10
Answer: (c) exactly 25
Hint: If \( n \) is the required number of newspapers then \( n \times 60 = 300 \times 5 \Rightarrow n = 25 \).

Question. In a class of 55 students the numbers of students studying different subjects are 23 in mathematics, 24 in physics, 19 in chemistry, 12 in mathematics and physics, 9 in mathematics and chemistry 7 in physics and chemistry and 4 in all the three subjects. the numbers of students who have taken exactly one subject is
(a) 6
(b) 13
(c) 16
(d) 22
Answer: (d) 22
Hint: \( n(M)=23, n(P)=24, n(C)=19, n(M \cap C)=9, n(P \cap C)=7, n(M \cap P \cap C)=4 \).
Number of students taking exactly one subject \( = n(M \cap P' \cap C') + n(P \cap M' \cap C') + n(C \cap M' \cap P') \).
\( n(M \cap P' \cap C') = n(M) - n(M \cap P) - n(M \cap C) + n(M \cap P \cap C) = 23 - 12 - 9 + 4 = 6 \).
\( n(P \cap M' \cap C') = n(P) - n(P \cap M) - n(P \cap C) + n(P \cap M \cap C) = 24 - 12 - 7 + 4 = 9 \).
\( n(C \cap M' \cap P') = n(C) - n(C \cap P) - n(C \cap M) + n(C \cap P \cap M) = 19 - 7 - 9 + 4 = 7 \).
Total exactly one \( = 6 + 9 + 7 = 22 \).

Question. Out of 800 boys in a school. 224 played cricket 240 played hockey and 336 played basketball. of the total, 64 played both basketball and hockey, 80 played cricket and basketball and 40 played cricket and hockey, 24 played all the three games. The numbers of boys who did not play any game is
(a) 128
(b) 216
(c) 240
(d) 160
Answer: (d) 160
Hint: \( n(C) = 224, n(H) = 240, n(B) = 336, n(H \cap B) = 64, n(B \cap C) = 80, n(H \cap C) = 40, n(C \cap H \cap B) = 24 \).
\( n(C^C \cap H^C \cap B^C) = n(U) - n(C \cup H \cup B) = 800 - [224 + 240 + 336 - 64 - 80 - 40 + 24] = 800 - 640 = 160 \).

Question. In a certain town 25% families own a phone and 15% own a car, 65% families own neither a phone nor a car. 2000 families own both a car and a phone. Consider the following statements in this regard.
i. 10% families own both a car and a phone
ii. 35% families own either a car or a phone
iii. 40,000 families live in the town.
Which of the above statements are correct?
(a) i and ii
(b) i and iii
(c) ii and iii
(d) i,ii and iii
Answer: (c) ii and iii
Hint: \( n(P) = 25\%, n(C) = 15\%, n(P^C \cap C^C) = 65\% \Rightarrow n(P \cup C)^C = 65\% \Rightarrow n(P \cup C) = 35\% \).
\( n(P \cup C) = n(P) + n(C) - n(P \cap C) \Rightarrow 35 = 25 + 15 - n(P \cap C) \Rightarrow n(P \cap C) = 5\% \).
Since \( 5\% \) of total \( = 2000 \Rightarrow \text{Total} = \frac{2000 \times 100}{5} = 40,000 \).
Statements (ii) and (iii) are correct.

Question. In a battle 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs the minimum value of x is
(a) 10
(b) 12
(c) 15
(d) 5
Answer: (a) 10
Hint: Minimum value of \( x = 100 - (30 + 20 + 25 + 15) = 100 - 90 = 10 \).

Assertion-Reason Type :

Note :
1) Statement-1 is true , Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
2) Statement-1 is true , Statement-2 is true, Statement-2 is not correct explanation for Statement-1.
3) Statement-1 is true , Statement-2 is false.
4) Statement-1 is false , Statement-2 is false.

Question. Assertion. If A={x:x is a prime number} and B={x: x \(\in\) N} then \( A \cap B \)= {x: x is a prime number}=A.
Reason. If \( A \subset B \) then \( A \cap B = A \).
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1

Question. Assertion. If A={2,4,7,10}, B={1,2,3,4} then A-B={1,3,7,10}.
Reason. A-B={x:x \( \notin \) A and x \( \in \) B}.
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4

Question. Assertion. A={x: 0<x<3, x \( \in \) R } and B = {x: 1 \( \le x \le \) 5 , x \( \in \) R } then A - B = { x : 0 <x< 1, x \( \in \) R }
Reason. \( A \Delta B = (A - B) \cup (B - A) \).
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2

Question. Assertion. If n(B)=3 then number of elements in power set of B= \( 2^3 = 8 \)
Reason. If n(A) = m then number of elements in power set of A = \( 2^m - 1 \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3

Question. Assertion. If X = {x : x is whole number }, Y = { x : x is natural number } then \( X \cup Y \) = { x : x is whole number } = X
Reason. If \( Y \subset X \) then \( Y \cup X = X \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1

Question. Assertion. If A and B are disjoint sets then A - B = A
Reason. If A and B are disjoint sets then \( A \cap B = \phi \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1

Question. Assertion. If n (A) = 5 then number of proper subsets of A= 31
Reason. If n (A) = m then number of proper subsets of B = \( 2^m - 1 \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1