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)

For any two sets A and B, \(A \cap (A \cup B)’\) is equal to (a) \(A\) (b) \(B\) (c) \(\phi\) (d) \(A\cap B\) Solution \[ A\cap(A\cup B)’ \] Using De Morgan’s law, \[ =A\cap(A’\cap B’) \] \[ =(A\cap A’)\cap B’ \] \[ =\phi\cap B’ \] \[ =\phi \] Answer \[ \boxed{\phi} \] Correct option:

Two finite sets have \(m\) and \(n\) elements. The number of subsets of the first set is 112 more than that of the second. The values of \(m\) and \(n\) are respectively (a) \(4,7\) (b) \(7,4\) (c) \(4,4\) (d) \(7,7\) Solution Number of subsets of a set having \(m\) elements: \[ 2^m \] Number of

Suppose \(A_1, A_2,\ldots,A_{30}\) are thirty sets each having 5 elements and \(B_1, B_2,\ldots,B_n\) 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 (a) 15 (b) 3 (c) 45

In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics \(=100\) Physics \(=70\) Chemistry \(=40\) Mathematics and Physics \(=30\) Mathematics and Chemistry \(=28\) Physics and Chemistry \(=23\) Mathematics, Physics and Chemistry \(=18\) How many students have offered Mathematics alone? (a) 35 (b) 48 (c) 60