Let \(S=\{x:x \text{ is a positive multiple of }3 \text{ less than }100\}\), \[ P=\{x:x \text{ is a prime less than }20\} \] Then, \(n(S)+n(P)\) is (a) 34 (b) 31 (c) 33 (d) 30 Solution Positive multiples of \(3\) less than \(100\): \[ 3,6,9,\ldots,99 \] Number of elements in \(S\): \[ \frac{99}{3}=33 \] Therefore, \[ […]

If A and B are two sets, then \(A \cap (A \cup B)\) equals (a) \(A\) (b) \(B\) (c) \(\phi\) (d) \(A\cap B\) Solution Using absorption law, \[ A\cap(A\cup B)=A \] Answer \[ \boxed{A} \] Correct option: (a) Next Question / Full Exercise

The set \((A \cup B \cup C) \cap (A \cap B′ \cap C′)′ \cup C′\) is equal to (a) \(B \cap C’\) (b) \(A \cap C\) (c) \(B \cup C’\) (d) \(A \cap C’\) Solution \[ (A\cup B\cup C)\cap(A\cap B’\cap C’)’\cup C’ \] Using De Morgan’s law, \[ (A\cap B’\cap C’)’=A’\cup B\cup C \] Therefore,

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 value of \(m\) and \(n\) respectively are: (a) \(7,6\) (b) \(5,1\) (c) \(6,3\) (d) \(8,7\) Solution Number of subsets of a set having \(m\)

Each set \(X_r\) contains 5 elements and each set \(Y_r\) contains 2 elements and \[ \bigcup_{r=1}^{20}X_r=S=\bigcup_{r=1}^{n}Y_r \] If each element of \(S\) belongs to exactly 10 of the \(X_r\)’s and to exactly 4 of the \(Y_r\)’s, then \(n\) is (a) 10 (b) 20 (c) 100 (d) 50 Solution Total element occurrences in all \(X_r\)’s: \[

If sets A and B are defined as \[ A=\{(x,y):y=\frac1x,\ 0\ne x\in R\} \] \[ B=\{(x,y):y=-x,\ x\in R\} \] then (a) \(A\cap B=A\) (b) \(A\cap B=B\) (c) \(A\cap B=\phi\) (d) \(A\cup B=A\) Solution For intersection, \[ \frac1x=-x \] \[ 1=-x^2 \] \[ x^2=-1 \] There is no real value of \(x\) satisfying this equation. Therefore,

A survey shows that 63% of the people watch a News channel whereas 76% watch another channel. If \(x\%\) of the people watch both channel, then (a) \(x=35\) (b) \(x=63\) (c) \(39\le x\le63\) (d) \(x=39\) Solution Let \[ n(A)=63\%,\qquad n(B)=76\% \] Using, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] \[ 100\ge63+76-x \] \[ 100\ge139-x \] \[ x\ge39

If \[ X=\{8^n-7n-1:n\in N\} \] and \[ Y=\{49n-49:n\in N\} \] Then, (a) \(X\subset Y\) (b) \(Y\subset X\) (c) \(X=Y\) (d) \(X\cap Y=\phi\) Solution Consider \[ 8^n=(1+7)^n \] Using binomial expansion, \[ 8^n=1+7n+\text{terms containing }7^2 \] Therefore, \[ 8^n-7n-1 \] is divisible by \[ 49 \] Hence every element of \(X\) is a multiple of \(49\).

Let \(F_1\) be the set of all parallelograms, \(F_2\) the set of all rectangles, \(F_3\) the set of all rhombuses, \(F_4\) the set of all squares and \(F_5\) the set of all trapeziums in a plane. Then \(F_1\) may be equal to (a) \(F_2\cap F_3\) (b) \(F_3\cap F_4\) (c) \(F_2\cup F_3\) (d) \(F_2\cup F_3\cup F_4\cup

The set \((A \cup B’) \cup (B \cap C)\) is equal to (a) \(A’ \cup B \cup C\) (b) \(A’ \cup B\) (c) \(A’ \cup C’\) (d) \(A’ \cap B\) Solution \[ (A\cup B’)\cup(B\cap C) \] \[ =A\cup[B’\cup(B\cap C)] \] Using distributive law, \[ =B’\cup(B\cap C) \] \[ =(B’\cup B)\cap(B’\cup C) \] \[ =U\cap(B’\cup C)