Suppose \(A_1, A_2, \dots, A_{30}\) are thirty sets each having 5 elements and \(B_1, B_2, \dots, B_n\) are \(n\) sets each with 3 elements. Let \(\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S\) and each element of \(S\) belongs to exactly 10 of the \(A_i\)'s and exactly 9 of the \(B_j\)'s. Then \(n\) is equal to
15
3
45
35
Two finite sets have \(m\) and \(n\) elements. The number of subsets of the first set is 112 more than that of the second set. The values of \(m\) and \(n\) are, respectively,
4, 7
7, 4
4, 4
7, 7
The set \((A \cap B') \cup (B \cap C)\) is equal to
\(A' \cup B \cup C\)
\(A' \cup B\)
\(A' \cup C'\)
\(A' \cap B\)
Let \(F_1\) be the set of parallelograms, \(F_2\) the set of rectangles, \(F_3\) the set of rhombuses, \(F_4\) the set of squares and \(F_5\) the set of trapeziums in a plane. Then \(F_1\) may be equal to
\(F_2 \cap F_3\)
\(F_3 \cap F_4\)
\(F_2 \cup F_5\)
\(F_2 \cup F_3 \cup F_4 \cup F_1\)
Let \(S\) = set of points inside the square, \(T\) = set of points inside the triangle and \(C\) = set of points inside the circle. If the triangle and circle intersect each other and are contained in a square, then
\(S \cap T \cap C = \varnothing\)
\(S \cup T \cup C = C\)
\(S \cup T \cup C = S\)
\(S \cup T = S \cap C\)
Let \(R\) be set of points inside a rectangle of sides \(a\) and \(b\) (\(a,b>1\)) with two sides along the positive direction of \(x\)-axis and \(y\)-axis. Then
\(R=\{(x,y):0\le x\le a,\;0\le y\le b\}\)
\(R=\{(x,y):0\le x
\(R=\{(x,y):0\le x\le a,\;0< y< b\}\)
\(R=\{(x,y):0< x< a,\;0< y< b\}\)
In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is
0
25
35
45
In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
210
290
180
260
If \(X=\{8^n-7n-1\mid n\in \mathbb{N}\}\) and \(Y=\{49n-49\mid n\in \mathbb{N}\}\). Then
\(X \subset Y\)
\(Y \subset X\)
\(X = Y\)
\(X \cap Y = \varnothing\)
A survey shows that 63% of the people watch a News Channel whereas 76% watch another channel. If \(x\%\) of the people watch both channels, then
\(x = 35\)
\(x = 63\)
\(39 \le x \le 63\)
\(x = 39\)
If sets \(A\) and \(B\) are defined as \(A=\{(x,y)\mid y=\tfrac{1}{x},\;0\ne x\in\mathbb{R}\}\) and \(B=\{(x,y)\mid y=-x,\;x\in\mathbb{R}\}\), then
\(A \cap B = A\)
\(A \cap B = B\)
\(A \cap B = \varnothing\)
\(A \cup B = A\)
If \(A\) and \(B\) are two sets, then \(A \cap (A \cup B)\) equals
\(A\)
\(B\)
\(\varnothing\)
\(A \cap B\)
If \(A=\{1,3,5,7,9,11,13,15,17\},\; B=\{2,4,\dots,18\}\) and \(\mathbb{N}\) the set of natural numbers is the universal set, then \(A' \cup (A \cup B) \cap B'\) is
\(\varnothing\)
\(\mathbb{N}\)
\(A\)
\(B\)
Let \(S=\{x\mid x\text{ is a positive multiple of }3\text{ less than }100\}\). \(P=\{x\mid x\text{ is a prime number less than }20\}\). Then \(n(S)+n(P)\) is
34
31
33
30
If \(X\) and \(Y\) are two sets and \(X'\) denotes the complement of \(X\), then \(X \cap (X \cup Y)'\) is equal to
\(X\)
\(Y\)
\(\varnothing\)
\(X \cap Y\)