\displaystyle \textbf{Question 1: } \text{For any set }A,\ (A')'\text{ is equal to}
\displaystyle \text{(a) }A'\qquad \text{(b) }A\qquad \text{(c) }\phi\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the complement of the complement of a set is the set itself.}
\displaystyle \therefore (A')'=A
\displaystyle \text{Hence, the correct option is (b) }A.
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\displaystyle \textbf{Question 2: } \text{Let }A\text{ and }B\text{ be two sets in the same universal set. Then, } \\ A-B=
\displaystyle \text{(a) }A\cap B\qquad \text{(b) }A'\cap B\qquad \text{(c) }A\cap B'\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{We know that }A-B=A\cap B'.
\displaystyle \text{Hence, the correct option is (c) }A\cap B'.
\\

\displaystyle \textbf{Question 3: } \text{The number of subsets of a set containing }n\text{ elements is}
\displaystyle \text{(a) }n\qquad \text{(b) }2^n-1\qquad \text{(c) }n^2\qquad \text{(d) }2^n
\displaystyle \text{Answer:}
\displaystyle \text{A set containing }n\text{ elements has }2^n\text{ subsets.}
\displaystyle \text{Hence, the correct option is (d) }2^n.
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\displaystyle \textbf{Question 4: } \text{For any two sets }A\text{ and }B,\ A\cap(A\cup B)=
\displaystyle \text{(a) }A\qquad \text{(b) }B\qquad \text{(c) }\phi\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Using the absorption law, }A\cap(A\cup B)=A.
\displaystyle \text{Hence, the correct option is (a) }A.
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\displaystyle \textbf{Question 5: } \text{If }A=\{1,3,5,B\}\text{ and }B=\{2,4\},\text{ then}
\displaystyle \text{(a) }4\in A\qquad \text{(b) }\{4\}\subset A\qquad \text{(c) }B\subset A\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 4\notin A,\ \{4\}\not\subset A\text{ and }B=\{2,4\}\not\subset A.
\displaystyle \text{Hence, the correct option is (d) none of these.}
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\displaystyle \textbf{Question 6: } \text{The symmetric difference of }A\text{ and }B\text{ is not equal to}
\displaystyle \text{(a) }(A-B)\cap(B-A)\qquad \text{(b) }(A-B)\cup(B-A)
\displaystyle \text{(c) }(A\cup B)-(A\cap B)\qquad \text{(d) }[(A\cup B)-A]\cup[(A\cup B)-B]
\displaystyle \text{Answer:}
\displaystyle \text{The symmetric difference of }A\text{ and }B\text{ is }(A-B)\cup(B-A).
\displaystyle \text{But }(A-B)\cap(B-A)=\phi.
\displaystyle \text{Hence, the correct option is (a).}
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\displaystyle \textbf{Question 7: } \text{The symmetric difference of }A=\{1,\ 2,\ 3\}\text{ and }B=\{3,\ 4,\ 5\}\text{ is}
\displaystyle \text{(a) }\{1,\ 2\}\qquad \text{(b) }\{1,\ 2,\ 4,\ 5\}\qquad \text{(c) }\{4,\ 3\}\qquad \text{(d) }\{2,\ 5,\ 1,\ 4,\ 3\}
\displaystyle \text{Answer:}
\displaystyle A-B=\{1,\ 2\}\text{ and }B-A=\{4,\ 5\}
\displaystyle \therefore (A-B)\cup(B-A)=\{1,\ 2,\ 4,\ 5\}
\displaystyle \text{Hence, the correct option is (b).}
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\displaystyle \textbf{Question 8: } \text{For any two sets }A\text{ and }B,\ (A-B)\cup(B-A)=
\displaystyle \text{(a) }(A-B)\cup A\qquad \text{(b) }(B-A)\cup B
\displaystyle \text{(c) }(A\cup B)-(A\cap B)\qquad \text{(d) }(A\cup B)\cap(A\cap B)
\displaystyle \text{Answer:}
\displaystyle \text{We know that }(A-B)\cup(B-A)=(A\cup B)-(A\cap B).
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 9: } \text{Which of the following statement is false?}
\displaystyle \text{(a) }A-B=A\cap B'\qquad \text{(b) }A-B=A-(A\cap B)
\displaystyle \text{(c) }A-B=A-B'\qquad \text{(d) }A-B=(A\cup B)-B
\displaystyle \text{Answer:}
\displaystyle A-B=A\cap B'\text{ but }A-B'=A\cap B.
\displaystyle \text{Hence, }A-B=A-B'\text{ is false.}
\displaystyle \text{Therefore, the correct option is (c).}
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\displaystyle \textbf{Question 10: } \text{For any three sets }A,\ B\text{ and }C
\displaystyle \text{(a) }A\cap(B-C)=(A\cap B)-(A\cap C)
\displaystyle \text{(b) }A\cap(B-C)=(A\cap B)-C
\displaystyle \text{(c) }A\cup(B-C)=(A\cup B)\cap(A\cup C')
\displaystyle \text{(d) }A\cup(B-C)=(A\cup B)-(A\cup C)
\displaystyle \text{Answer:}
\displaystyle A\cup(B-C)=A\cup(B\cap C')=(A\cup B)\cap(A\cup C').
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 11: } \text{Let }A=\{x:x\in R,\ x>4\}\text{ and } \\ B=\{x\in R:x<5\}.\text{ Then, }A\cap B=
\displaystyle \text{(a) }(4,\ 5]\qquad \text{(b) }(4,\ 5)\qquad \text{(c) }[4,\ 5)\qquad \text{(d) }[4,\ 5]
\displaystyle \text{Answer:}
\displaystyle A\cap B=\{x:x\in R,\ 4<x<5\}=(4,\ 5).
\displaystyle \text{Hence, the correct option is (b).}
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\displaystyle \textbf{Question 12: } \text{Let }U\text{ be the universal set containing }700\text{ elements. If }A,\ B\text{ are subsets of }U
\displaystyle \text{such that }n(A)=200,\ n(B)=300\text{ and }n(A\cap B)=100.\text{ Then, }n(A'\cap B')=
\displaystyle \text{(a) }400\qquad \text{(b) }600\qquad \text{(c) }300\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle n(A'\cap B')=n((A\cup B)')=n(U)-n(A\cup B)
\displaystyle =700-[200+300-100]=300
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 13: } \text{Let }A\text{ and }B\text{ be two sets such that }n(A)=16,\ n(B)=14,
\displaystyle n(A\cup B)=25.\text{ Then, }n(A\cap B)\text{ is equal to}
\displaystyle \text{(a) }30\qquad \text{(b) }50\qquad \text{(c) }5\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle n(A\cap B)=n(A)+n(B)-n(A\cup B)
\displaystyle =16+14-25=5
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 14: } \text{If }A=\{1,\ 2,\ 3,\ 4,\ 5\},\text{ then the number of proper subsets of }A\text{ is}
\displaystyle \text{(a) }120\qquad \text{(b) }30\qquad \text{(c) }31\qquad \text{(d) }32
\displaystyle \text{Answer:}
\displaystyle \text{Number of subsets of }A=2^5=32.
\displaystyle \text{Number of proper subsets }=32-1=31.
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 15: } \text{In set-builder method the null set is represented by}
\displaystyle \text{(a) }\{\}\qquad \text{(b) }\phi\qquad \text{(c) }\{x:x\neq x\}\qquad \text{(d) }\{x:x=x\}
\displaystyle \text{Answer:}
\displaystyle \text{No element satisfies }x\neq x.
\displaystyle \text{Therefore, }\{x:x\neq x\}\text{ represents the null set.}
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 16: } \text{If }A\text{ and }B\text{ are two disjoint sets, then }n(A\cup B)\text{ is equal to}
\displaystyle \text{(a) }n(A)+n(B)\qquad \text{(b) }n(A)+n(B)-n(A\cap B)
\displaystyle \text{(c) }n(A)+n(B)+n(A\cap B)\qquad \text{(d) }n(A)n(B)
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ and }B\text{ are disjoint, }n(A\cap B)=0.
\displaystyle \therefore n(A\cup B)=n(A)+n(B).
\displaystyle \text{Hence, the correct option is (a).}
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\displaystyle \textbf{Question 17: } \text{For two sets }A\cup B=A\text{ iff}
\displaystyle \text{(a) }B\subseteq A\qquad \text{(b) }A\subseteq B\qquad \text{(c) }A\neq B\qquad \text{(d) }A=B
\displaystyle \text{Answer:}
\displaystyle A\cup B=A\text{ iff every element of }B\text{ is already in }A.
\displaystyle \therefore B\subseteq A
\displaystyle \text{Hence, the correct option is (a).}
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\displaystyle \textbf{Question 18: } \text{If }A\text{ and }B\text{ are two sets such that }n(A)=70,\ n(B)=60,
\displaystyle n(A\cup B)=110,\text{ then }n(A\cap B)\text{ is equal to}
\displaystyle \text{(a) }240\qquad \text{(b) }50\qquad \text{(c) }40\qquad \text{(d) }20
\displaystyle \text{Answer:}
\displaystyle n(A\cap B)=n(A)+n(B)-n(A\cup B)
\displaystyle =70+60-110=20
\displaystyle \text{Hence, the correct option is (d).}
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\displaystyle \textbf{Question 19: } \text{If }A\text{ and }B\text{ are two given sets, then }A\cap(A\cap B)'\text{ is equal to}
\displaystyle \text{(a) }A\qquad \text{(b) }B\qquad \text{(c) }\phi\qquad \text{(d) }A\cap B'
\displaystyle \text{Answer:}
\displaystyle A\cap(A\cap B)'=A\cap(A'\cup B')
\displaystyle =(A\cap A')\cup(A\cap B')=\phi\cup(A\cap B')=A\cap B'
\displaystyle \text{Hence, the correct option is (d).}
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\displaystyle \textbf{Question 20: } \text{If }A=\{x:x\text{ is a multiple of }3\}\text{ and }B=\{x:x\text{ is a multiple of }5\},
\displaystyle \text{then }A-B\text{ is}
\displaystyle \text{(a) }A\cap B\qquad \text{(b) }A\cap \overline{B}\qquad \text{(c) }\overline{A}\cap B\qquad \text{(d) }\overline{A\cap B}
\displaystyle \text{Answer:}
\displaystyle A-B\text{ means elements of }A\text{ which are not in }B.
\displaystyle \therefore A-B=A\cap \overline{B}
\displaystyle \text{Hence, the correct option is (b).}
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\displaystyle \textbf{Question 21: } \text{In a city }20\%\text{ of the population travels by car, }50\%\text{ travels by bus}
\displaystyle \text{and }10\%\text{ travels by both car and bus. Then, persons travelling by car or bus is}
\displaystyle \text{(a) }80\%\qquad \text{(b) }40\%\qquad \text{(c) }60\%\qquad \text{(d) }70\%
\displaystyle \text{Answer:}
\displaystyle \text{Required percentage}=20\%+50\%-10\%=60\%
\displaystyle \text{Hence, the correct option is (c).}
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\displaystyle \textbf{Question 22: } \text{If }A\cap B=B,\text{ then}
\displaystyle \text{(a) }A\subseteq B\qquad \text{(b) }B\subseteq A\qquad \text{(c) }A=\phi\qquad \text{(d) }B=\phi
\displaystyle \text{Answer:}
\displaystyle A\cap B=B\text{ means every element of }B\text{ is also an element of }A.
\displaystyle \therefore B\subseteq A
\displaystyle \text{Hence, the correct option is (b).}
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\displaystyle \textbf{Question 23: } \text{An investigator interviewed }100\text{ students to determine the preference of three drinks:}
\displaystyle \text{milk, coffee and tea. The investigator reported that }10\text{ students take all three drinks;}
\displaystyle 20\text{ students take milk and coffee; }25\text{ students take milk and tea;}
\displaystyle 20\text{ students take coffee and tea; }12\text{ students take milk only;}
\displaystyle 5\text{ students take coffee only and }8\text{ students take tea only. Then the number of students}
\displaystyle \text{who did not take any of three drinks is}
\displaystyle \text{(a) }10\qquad \text{(b) }20\qquad \text{(c) }25\qquad \text{(d) }30
\displaystyle \text{Answer:}
\displaystyle \text{Students taking exactly milk and coffee}=20-10=10
\displaystyle \text{Students taking exactly milk and tea}=25-10=15
\displaystyle \text{Students taking exactly coffee and tea}=20-10=10
\displaystyle \text{Students taking at least one drink}=12+5+8+10+15+10+10=70
\displaystyle \text{Students taking none}=100-70=30
\displaystyle \text{Hence, the correct option is (d).}
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\displaystyle \textbf{Question 24: } \text{Two finite sets have }m\text{ and }n\text{ elements. Then number of elements in}
\displaystyle \text{the  power set of first set is }48\text{ more than the total number of elements in power set}
\displaystyle \text{of the second set. Then, the values of }m\text{ and }n\text{ are:}
\displaystyle \text{(a) }7,\ 6\qquad \text{(b) }6,\ 3\qquad \text{(c) }6,\ 4\qquad \text{(d) }7,\ 4
\displaystyle \text{Answer:}
\displaystyle \text{Number of elements in the power sets are }2^m\text{ and }2^n.
\displaystyle \therefore 2^m=2^n+48
\displaystyle \text{Checking the options, for }m=6,\ n=4,\text{ we get }2^6=64\text{ and }2^4+48=16+48=64.
\displaystyle \text{Hence, the correct option is (c) }6,\ 4.
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\displaystyle \textbf{Question 25: } \text{In a class of }175\text{ students the following data shows the number of student}
\displaystyle \text{opting students opting one or more subjects. Mathematics }100;\text{ Physics }70;\text{ Chemistry }40;
\displaystyle \text{Mathematics and Physics }30;\text{ Mathematics and Chemistry }28;
\displaystyle \text{Physics and Chemistry }23;\text{ Mathematics, Physics and Chemistry }18.
\displaystyle \text{How many students have offered Mathematics alone?}
\displaystyle \text{(a) }35\qquad \text{(b) }48\qquad \text{(c) }60\qquad \text{(d) }22
\displaystyle \text{Answer:}
\displaystyle \text{Students offering Mathematics alone}
\displaystyle =n(M)-n(M\cap P)-n(M\cap C)+n(M\cap P\cap C)
\displaystyle =100-30-28+18=60
\displaystyle \text{Hence, the correct option is (c) }60.
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\displaystyle \textbf{Question 26: } \text{Suppose }A_1,\ A_2,\ldots,A_{30}\text{ are thirty sets each having }5\text{ elements}
\displaystyle \text{and }B_1,\ B_2,\ldots,B_n\text{ are }n\text{ sets each with }3\text{ elements. Let}
\displaystyle \bigcup_{i=1}^{30}A_i=\bigcup_{j=1}^{n}B_j=S\text{ and each element of }S\text{ belongs to exactly }10
\displaystyle \text{of the }A_i\text{'s and exactly }9\text{ of the }B_j\text{'s, then }n\text{ is equal to}
\displaystyle \text{(a) }15\qquad \text{(b) }3\qquad \text{(c) }45\qquad \text{(d) }35
\displaystyle \text{Answer:}
\displaystyle n(S)=\frac{1}{10}\sum_{i=1}^{30}n(A_i)=\frac{1}{10}\times30\times5=15
\displaystyle \text{Also, }n(S)=\frac{1}{9}\sum_{j=1}^{n}n(B_j)=\frac{1}{9}\times3n=\frac{n}{3}
\displaystyle \therefore \frac{n}{3}=15
\displaystyle \Rightarrow n=45
\displaystyle \text{Hence, the correct option is (c) }45.
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\displaystyle \textbf{Question 27: } \text{Two finite sets have }m\text{ and }n\text{ elements. The number of subsets of the}
\displaystyle \text{first set is }112\text{ more than that of the second. The values of }m\text{ and }n\text{ are respectively}
\displaystyle \text{(a) }4,\ 7\qquad \text{(b) }7,\ 4\qquad \text{(c) }4,\ 4\qquad \text{(d) }7,\ 7
\displaystyle \text{Answer:}
\displaystyle 2^m=2^n+112
\displaystyle \text{Checking the options, for }m=7,\ n=4,\text{ we get }2^7=128\text{ and }2^4+112=16+112=128.
\displaystyle \text{Hence, the correct option is (b) }7,\ 4.
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\displaystyle \textbf{Question 28: } \text{For any two sets }A\text{ and }B,\ A\cap(A\cup B)'\text{ is equal to}
\displaystyle \text{(a) }A\qquad \text{(b) }B\qquad \text{(c) }\phi\qquad \text{(d) }A\cap B
\displaystyle \text{Answer:}
\displaystyle A\cap(A\cup B)'=A\cap(A'\cap B')
\displaystyle =(A\cap A')\cap B'=\phi\cap B'=\phi
\displaystyle \text{Hence, the correct option is (c) }\phi.
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\displaystyle \textbf{Corrected Question 29: } \text{The set }(A\cup B')'\cup B\text{ is equal to}
\displaystyle \text{(a) }A'\cup B\cup C\qquad \text{(b) }A'\cup B
\displaystyle \text{(c) }A'\cup C'\qquad \text{(d) }A'\cap B
\displaystyle \text{Answer:}
\displaystyle (A\cup B')'=A'\cap B
\displaystyle \therefore (A'\cap B)\cup B=B
\displaystyle \text{Since }B\subseteq A'\cup B,\text{ the matching option is (b).}
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\displaystyle \textbf{Question 30: } \text{Let }F_1\text{ be the set of all parallelograms, }F_2\text{ the set of all rectangles,}
\displaystyle F_3\text{ the set of all rhombuses, }F_4\text{ the set of all squares and }F_5\text{ the set of trapeziums}
\displaystyle \text{in a plane. Then }F_1\text{ may be equal to}
\displaystyle \text{(a) }F_2\cap F_3\qquad \text{(b) }F_3\cap F_4\qquad \text{(c) }F_2\cup F_3\qquad \text{(d) }F_2\cup F_3\cup F_4\cup F_1
\displaystyle \text{Answer:}
\displaystyle F_2,\ F_3\text{ and }F_4\text{ are all subsets of }F_1.
\displaystyle \therefore F_2\cup F_3\cup F_4\cup F_1=F_1
\displaystyle \text{Hence, the correct option is (d).}
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