If (a/3 + 1, b − 2/3) = (5/3, 1/3), Find the Values of a and b If \( \left(\frac{a}{3}+1,\ b-\frac{2}{3}\right)=\left(\frac{5}{3},\ \frac{1}{3}\right) \), Find the Values of \(a\) and \(b\) Question If \[ \left(\frac{a}{3}+1,\ b-\frac{2}{3}\right)=\left(\frac{5}{3},\ \frac{1}{3}\right) \] find the values of \(a\) and \(b\). Solution Since two ordered pairs are equal, their corresponding components are […]

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 (a) 0 (b) 25 (c) 35 (d) 45 Solution Let \[ n(C)=25 \] \[ n(T)=20 \] \[ n(C\cap T)=10 \] Using the formula, \[

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 both. Then the number of persons who read neither is (a) 210 (b) 290 (c) 180 (d) 260 Solution Let \[ n(H)=450 \] \[ n(E)=300 \] \[ n(H\cap E)=200 \] Using the formula, \[ n(H\cup E)=n(H)+n(E)-n(H\cap E) \] \[ =450+300-200

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