If A = {(a, b) : 4a = 5b; a, b in {1, 2, 3, d | vedclass

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(B) Given the equation $4a = 5b$,we can write $a = \frac{5}{4}b$. Since $a$ must be an integer,$b$ must be a multiple of $4$. Given $b \in \{1, 2, 3,

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$6$

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(B) Given the equation $4a = 5b$,we can write $a = \frac{5}{4}b$. Since $a$ must be an integer,$b$ must be a multiple of $4$. Given $b \in \{1, 2, 3, \dots, 30\}$,the possible values for $b$ are $4, 8, 12, 16, 20, 24, 28$. For each $b$,we calculate $a = \frac{5}{4}b$: If $b = 4, a = 5$. If $b = 8, a = 10$. If $b = 12, a = 15$. If $b = 16, a = 20$. If $b = 20, a = 25$. If $b = 24, a = 30$. If $b = 28, a = 35$ (which is not in the set $\{1, 2, \dots, 30\}$). Thus,the valid ordered pairs $(a, b)$ are $(5, 4), (10, 8), (15, 12), (20, 16), (25, 20), (30, 24)$. The total number of such pairs is $6$.

If $A = \{(a, b) : 4a = 5b; a, b \in \{1, 2, 3, \dots, 30\}\}$,then the number of such ordered pairs $(a, b)$ is

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