In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus is (a) 80% (b) 40% (c) 60% (d) 70% Solution Let \[ n(C)=20\% \] \[ n(B)=50\% \] \[ n(C\cap B)=10\% \] Using the formula, \[ n(C\cup […]

If \(A=\{x:x \text{ is a multiple of }3\}\) and \(B=\{x:x \text{ is a multiple of }5\}\), then \(A-B\) is (a) \(A\cap B\) (b) \(A\cap \overline{B}\) (c) \(\overline{A}\cap \overline{B}\) (d) \(\overline{A}\cap B\) Solution By definition, \[ A-B \] means the elements which belong to \(A\) but do not belong to \(B\). Therefore, \[ A-B=A\cap \overline{B} \]

If A and B are two given sets, then \(A \cap (A \cap B)^c\) is equal to (a) \(A\) (b) \(B\) (c) \(\Phi\) (d) \(A \cap B^c\) Solution \[ A\cap(A\cap B)^c \] Using De Morgan’s law, \[ =A\cap(A^c\cup B^c) \] \[ =(A\cap A^c)\cup(A\cap B^c) \] \[ =\Phi\cup(A\cap B^c) \] \[ =A\cap B^c \] Answer \[

If A and B are two sets such that \[ n(A)=70,\quad n(B)=60,\quad n(A\cup B)=110 \] Then, \(n(A\cap B)\) is equal to (a) 240 (b) 50 (c) 40 (d) 20 Solution Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] Substituting the values, \[ 110=70+60-n(A\cap B) \] \[ 110=130-n(A\cap B) \] \[ n(A\cap B)=130-110 \] \[ =20

For two sets \(A \cup B = A\) iff (a) \(B\subseteq A\) (b) \(A\subseteq B\) (c) \(A\ne B\) (d) \(A=B\) Solution If \[ A\cup B=A \] then every element of \(B\) is already present in \(A\). Therefore, \[ B\subseteq A \] Answer \[ \boxed{B\subseteq A} \] Correct option: (a) Next Question / Full Exercise

If A and B are two disjoint sets, then \(n(A \cup B)\) is equal to (a) \(n(A)+n(B)\) (b) \(n(A)+n(B)-n(A\cap B)\) (c) \(n(A)+n(B)+n(A\cap B)\) (d) \(n(A)n(B)\) Solution For disjoint sets, \[ A\cap B=\Phi \] Therefore, \[ n(A\cap B)=0 \] Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] \[ =n(A)+n(B)-0 \] \[ =n(A)+n(B) \] Answer \[ \boxed{n(A)+n(B)}

In set-builder method the null set is represented by (a) \(\{\}\) (b) \(\Phi\) (c) \(\{x:x\ne x\}\) (d) \(\{x:x=x\}\) Solution A null set contains no element. In set-builder form, \[ \{x:x\ne x\} \] means the set of all \(x\) such that \(x\) is not equal to itself. This is impossible for any element. Therefore, the set

Let A and B be two sets such that \[ n(A)=16,\quad n(B)=14,\quad n(A\cup B)=25 \] Then, \(n(A\cap B)\) is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] Substituting the values, \[ 25=16+14-n(A\cap B) \] \[ 25=30-n(A\cap B) \] \[ n(A\cap B)=30-25 \]

Let A and B be two sets such that \[ n(A)=16,\quad n(B)=14,\quad n(A\cup B)=25 \] Then, \(n(A\cap B)\) is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution Using the formula, \[ n(A\cup B)=n(A)+n(B)-n(A\cap B) \] Substituting the values, \[ 25=16+14-n(A\cap B) \] \[ 25=30-n(A\cap B) \] \[ n(A\cap B)=30-25 \]

Let U be the universal set containing 700 elements. If A, B are sub-sets of U such that \[ n(A)=200,\quad n(B)=300,\quad n(A\cap B)=100 \] Then, \[ n(A’ \cap B’)= \] (a) 400 (b) 600 (c) 300 (d) none of these Solution Using De Morgan’s law, \[ A’\cap B’=(A\cup B)’ \] Therefore, \[ n(A’\cap B’)=n(U)-n(A\cup B)