Sets and Relations: MCQ test, Set 6

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

If $A = \{t, y\}$ , then $A \times A =$
A $\{(t, y), (y, t)\}$ B $\{(t, t), (t, y), (y, t), (y, y)\}$ C $\{(t, t), (y, y)\}$ D $\{(t, y)\}$

Answer and solution

Answer: (B) $\{(t, t), (t, y), (y, t), (y, y)\}$

Pair each element of $A$ with every element of $A$ (itself included).

$A \times A = \{(t, t), (t, y), (y, t), (y, y)\}$
If $A = \{1, 11, 12\}$ and $B = \{c, p\}$ , then $A \times B =$
A $\{(1, c), (11, p)\}$ B $\{(c, 1), (c, 11), (c, 12), (p, 1), (p, 11), (p, 12)\}$ C $\{(1, c), (1, p), (11, c), (11, p)\}$ D $\{(1, c), (1, p), (11, c), (11, p), (12, c), (12, p)\}$

Answer and solution

Answer: (D) $\{(1, c), (1, p), (11, c), (11, p), (12, c), (12, p)\}$

$A \times B = \{(a, b) : a \in A, b \in B\}$

$= \{(1, c), (1, p), (11, c), (11, p), (12, c), (12, p)\}$
If $A = \{7, 9\}$ and $B = \{p, q\}$ , then $B \times A =$
A $\{(7, p), (7, q), (9, p), (9, q)\}$ B $\{(p, 7), (q, 9)\}$ C $\{(p, 7), (p, 9), (q, 7), (q, 9)\}$ D $\{(p, 7), (p, 9)\}$

Answer and solution

Answer: (C) $\{(p, 7), (p, 9), (q, 7), (q, 9)\}$

$B \times A = \{(b, a) : b \in B, a \in A\}$

$= \{(p, 7), (p, 9), (q, 7), (q, 9)\}$
If $A = \{3, 5, 9\}$ and $B = \{t, x\}$ , which of the following belongs to $A \times B$ ?
A $(t, x)$ B $(t, 5)$ C $(5, t)$ D $(5, 9)$

Answer and solution

Answer: (C) $(5, t)$

$(a, b) \in A \times B$ only when $a \in A$ and $b \in B$ .

$5 \in A$ and $t \in B$ , so $(5, t) \in A \times B$ .
If $n(A) = 9$ and $n(B) = 4$ , then $n(A \times B) =$
A $13$ B $45$ C $72$ D $36$

Answer and solution

Answer: (D) $36$

$n(A \times B) = n(A) \times n(B) = mn$

$= 9 \times 4 = 36$
If $n(A \times B) = 20$ and $n(B) = 2$ , then $n(A) =$
A $10$ B $22$ C $40$ D $18$

Answer and solution

Answer: (A) $10$

$n(A \times B) = n(A) \times n(B)$

$20 = n(A) \times 2$

$n(A) = 20 \div 2 = 10$
If $A = \{4, 5\}$ and $B = \{6, 7, 9, 11\}$ , then $n(A \times B) =$
A $16$ B $6$ C $8$ D $256$

Answer and solution

Answer: (C) $8$

$n(A) = 2, n(B) = 4$

$n(A \times B) = 2 \times 4 = 8$
If $A = \{3, 10, 11\}$ and $B = \{q, y\}$ , which of the following is a relation from $A$ to $B$ ?
A $\{(y, 10), (11, y)\}$ B $\{(10, y), (11, y)\}$ C $\{(q, 11), (y, 10)\}$ D $\{(10, p), (10, y)\}$

Answer and solution

Answer: (B) $\{(10, y), (11, y)\}$

A relation from $A$ to $B$ is a subset of $A \times B$ .

Only $\{(10, y), (11, y)\}$ has every pair in $A \times B$ .
If $A = \{4, 10\}$ and $B = \{p, q, t\}$ , which of the following is NOT a relation from $A$ to $B$ ?
A $\{(4, q)\}$ B $\{(4, z), (4, q)\}$ C $\{(4, p), (4, q), (4, t)\}$ D $\{(10, p), (10, t)\}$

Answer and solution

Answer: (B) $\{(4, z), (4, q)\}$

A relation from $A$ to $B$ is a subset of $A \times B$ .

$z \notin B$ , so $(4, z) \notin A \times B$ .
If $A = \{b, t, x\}$ and $B = \{4, 8\}$ , which of the following is a relation from $B$ to $A$ ?
A $\{(4, 8), (8, 4)\}$ B $\{(t, x), (x, b)\}$ C $\{(b, 4), (t, 8)\}$ D $\{(4, t), (8, x)\}$

Answer and solution

Answer: (D) $\{(4, t), (8, x)\}$

A relation from $B$ to $A$ is a subset of $B \times A$ .

In $\{(4, t), (8, x)\}$ every first element is in $B$ and every second element is in $A$ .
If $A = \{2\}$ and $B = \{1, 2, 6, 12\}$ , the number of relations from $A$ to $B$ is
A $8$ B $16$ C $4$ D $32$

Answer and solution

Answer: (B) $16$

$n(A \times B) = 1 \times 4 = 4$

$\text{Number of relations} = 2^{4} = 16$
If $A = \{1, 7, 11\}$ , the number of relations from $A$ to $A$ is
A $64$ B $9$ C $512$ D $8$

Answer and solution

Answer: (C) $512$

$n(A \times A) = 3 \times 3 = 9$

$\text{Number of relations} = 2^{9} = 512$
If $n(A) = 4$ and there are 65536 relations from $A$ to $B$ , then $n(B) =$
A $32768$ B $16$ C $16384$ D $4$

Answer and solution

Answer: (D) $4$

$2^{4 \times n(B)} = 65536 = 2^{16}$

$4 \times n(B) = 16$

$n(B) = 4$
If $R = \{(b, 8), (b, 9), (z, 2)\}$ , which of the following is true?
A $bR9$ B $9Rb$ C $zR8$ D $bR2$

Answer and solution

Answer: (A) $bR9$

$aRb$ means $(a, b) \in R$ .

$(b, 9) \in R$ , so $bR9$ .
If $R = \{(y, 5), (y, 8), (y, 10)\}$ , which of the following is false?
A $pR5$ B $yR10$ C $yR5$ D $yR8$

Answer and solution

Answer: (A) $pR5$

$aRb$ means $(a, b) \in R$ .

$(p, 5) \notin R$ , so $pR5$ is false.
If $R = \{(a, 1), (a, 12), (z, 1), (z, 4)\}$ , the set of all $b$ with $zRb$ is
A $\{1, 4, 12\}$ B $\{1, 12\}$ C $\{z\}$ D $\{1, 4\}$

Answer and solution

Answer: (D) $\{1, 4\}$

$zRb$ means $(z, b) \in R$ .

Such pairs: $(z, 1)$ , $(z, 4)$ .
If $A = \{a, c\}$ and $B = \{a, p\}$ , the universal relation from $A$ to $B$ is
A $\{(a, a), (c, p)\}$ B $\{(a, a), (a, p), (c, a), (c, p)\}$ C $\{(a, a), (a, c), (p, a), (p, c)\}$ D $\phi$

Answer and solution

Answer: (B) $\{(a, a), (a, p), (c, a), (c, p)\}$

$A \times B \subseteq A \times B$ , so $A \times B$ is a relation from $A$ to $B$ : the universal relation.

$A \times B = \{(a, a), (a, p), (c, a), (c, p)\}$
If $A = \{7, 10\}$ and $B = \{c, p, z\}$ , the null relation from $A$ to $B$ is
A $\phi$ B $\{(7, c), (7, p), (7, z), (10, c), (10, p), (10, z)\}$ C $\{0\}$ D $\{(7, c)\}$

Answer and solution

Answer: (A) $\phi$

$\phi \subseteq A \times B$ , so $\phi$ is a relation from $A$ to $B$ : the null (void) relation.
If $n(A) = 12$ and $n(B) = 9$ , the number of ordered pairs in the universal relation from $A$ to $B$ is
A $216$ B $0$ C $108$ D $21$

Answer and solution

Answer: (C) $108$

The universal relation is $A \times B$ .

$n(A \times B) = 12 \times 9 = 108$
If $A = \{4, 10, 11\}$ and $B = \{a, q\}$ , the domain of the null relation from $A$ to $B$ is
A $\phi$ B $\{4, 10, 11\}$ C $\{a, q\}$ D $\{0\}$

Answer and solution

Answer: (A) $\phi$

The null relation $\phi$ has no ordered pairs.

So its domain is $\phi$ .
Which of the following relations has domain $\{10, 12\}$ and range $\{p, y\}$ ?
A $\{(p, 10), (y, 10), (y, 12)\}$ B $\{(7, p), (12, p)\}$ C $\{(10, p), (10, y), (12, y)\}$ D $\{(7, p), (10, y), (12, y)\}$

Answer and solution

Answer: (C) $\{(10, p), (10, y), (12, y)\}$

Domain $=$ first elements, Range $=$ second elements.

For $\{(10, p), (10, y), (12, y)\}$ : Domain $= \{10, 12\}$ , Range $= \{p, y\}$ .
$R = \{(2, c), (2, x), (8, c), (11, c)\}$ is a relation from $A = \{2, 8, 10, 11\}$ to $B = \{c, x\}$ . Which element of $A$ is not in the domain of $R$ ?
A $8$ B $10$ C $11$ D $2$

Answer and solution

Answer: (B) $10$

Domain of $R$ $=$ set of first elements of the pairs in $R$ .

$\text{Domain} = \{2, 8, 11\}$

$10 \in A$ but it is not a first element of any pair.
If $R = \{(2, b), (2, t), (5, b), (8, b)\}$ , the range of $R^{-1}$ is
A $\{b, t\}$ B $\{2, 5\}$ C $\{2, 5, 8, b, t\}$ D $\{2, 5, 8\}$

Answer and solution

Answer: (D) $\{2, 5, 8\}$

$R^{-1} = \{(b, 2), (t, 2), (b, 5), (b, 8)\}$

$\text{Range of } R^{-1} = \{2, 5, 8\}$

It is the domain of $R$ .
If $R = \{(c, 3), (q, 10), (z, 3)\}$ , which ordered pair belongs to $R^{-1}$ ?
A $(10, q)$ B $(q, 10)$ C $(c, 3)$ D $(7, c)$

Answer and solution

Answer: (A) $(10, q)$

$R^{-1}$ : interchange the two elements of every pair of $R$ .

$(q, 10) \in R$ , so $(10, q) \in R^{-1}$ .
$R = \{(2, a), (8, a), (8, b)\}$ is a relation from $A = \{2, 8, 11\}$ to $B = \{a, b, x\}$ . Then $R^{-1}$ is a relation from
A $A$ to $B$ B $B$ to $B$ C $B$ to $A$ D $A$ to $A$

Answer and solution

Answer: (C) $B$ to $A$

$R^{-1} = \{(a, 2), (a, 8), (b, 8)\}$

First elements are in $B$ , second elements in $A$ .

$\therefore R^{-1} \subseteq B \times A$

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MCQ test, Set 6