Question : A varies directly as the positive square root of B, and inversely as the cube of C. If A = 15, when B = 27 and C = 2, then find B when A = 9 and C = 2.
Option 1: $\frac{281}{42}$
Option 2: $\frac{243}{25}$
Option 3: $\frac{264}{37}$
Option 4: $\frac{275}{51}$
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Correct Answer: $\frac{243}{25}$
Solution :
Let $A$ = $\frac{k\sqrt{B}}{C^3}$
Substituting $A$ = 15, $B$ = 27, and $C$ = 2, we get:
15 = $\frac{k\sqrt{27}}{2^3}$
⇒ $k=\frac{120}{3\sqrt{3}}$
⇒ $k=\frac{40}{\sqrt{3}}$
Now, $B = (\frac{AC^3}{k})^2$
Substituting $A$ = 9 and $C$ = 2, we get:
⇒ $B = (\frac{9\times 2^3}{\frac{40}{\sqrt{3}}})^2$
⇒ $B = \frac{243}{25}$
Hence, the correct answer is $\frac{243}{25}$.
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