The problem states that $a$ varies directly as the square of $b$ and inversely as $c$. We are given that $a = 2$ when $b = 4$ and $c = 24$. We need to find the value of the constant of variation $k$ and then find the value of $a$ when $b = 9$ and $c = 27$.

1. Problem Description

The problem states that $a$ varies directly as the square of $b$ and inversely as $c$. We are given that $a = 2$ when $b = 4$ and $c = 24$. We need to find the value of the constant of variation $k$ and then find the value of $a$ when $b = 9$ and $c = 27$.

2. Solution Steps

(a) To find the value of $k$, we use the given relationship:

$a = \frac{kb^2}{c}$.

Substitute the given values $a=2$, $b=4$, and $c=24$ into the equation:

$2 = \frac{k(4^2)}{24}$.

$2 = \frac{16k}{24}$.

$2 = \frac{2k}{3}$.

Multiply both sides by 3:

$6 = 2k$.

Divide both sides by 2:

$k = 3$.

(b) Now, we need to find the value of $a$ when $b = 9$ and $c = 27$. We use the formula $a = \frac{kb^2}{c}$ with $k=3$, $b=9$, and $c=27$.

$a = \frac{3(9^2)}{27}$.

$a = \frac{3(81)}{27}$.

$a = \frac{243}{27}$.

$a = 9$.

3. Final Answer

(a) $k = 3$

(b) $a = 9$