Question

Let $A=\left\{2,3,4,5,..,30\right\}$ and $\text{'}\cong \text{'}$ be an equivalence relation $A\times A$, defined by $\left(a,b\right)\cong \left(c,d\right)$, if and only if $ad=bc$. Then the number of ordered pair $\left(4,3\right)$ is equal to:

• A. $7$
• B. $5$
• C. $6$
• D. $8$

Answer 1

The correct option is A

$7$

Explanation for the correct option:

Determine the number of ordered pairs:

Given $A=\left\{2,3,4,5,..,30\right\}$where $A\times A$is defined by $\left(a,b\right)\cong \left(c,d\right)$

Hence, $\left(a,b\right)\cong \left(c,d\right)$ implies that it satisfies the reflexive, symmetric, and transitive conditions.

Given $ad=bc$ and the ordered pair is $\left(4,3\right)$

Considering the given $\left(a,b\right)\cong \left(c,d\right)$

Hence, $\left(4,3\right)=\left(c,d\right)$

$\begin{array}{rcl}4d& =& 3c\\ \frac{4}{3}& =& \frac{c}{d}\\ & & \text{or}\\ \frac{4}{3}& =& \frac{a}{b}\end{array}$

$b$ must be multiple of $3$, since the ordered pair is $\left(4,3\right)$

$\therefore b=\left\{3,6,9,12,15,18,21,24,27,30\right\}$

Thus if $a=\frac{4}{3}b$, then $a=\left\{4,8,12,16,20,24,28,32,36,40\right\}$

On applying condition $2\le A\le 30$, we can say that $a=\left\{4,8,12,16,20,24,28\right\}$ is possible set value.

Therefore, the ordered pair of $\left(a,b\right)$ is given as: $\left(a,b\right)=\left(4,3\right),\left(8,6\right),\left(12,9\right),\left(16,12\right),\left(20,15\right),\left(24,18\right),\left(28,21\right)$

That is $7$ ordered pairs

Hence, option (A) is the correct answer.