<Question>
The value of $y$ varies inversely as $\sqrt x$ and when $x=24$, $y=15$. What is $x$ when $y=3$?
</Question>

<think>
<Plan_1>
Okay, let's see. The problem says that y varies inversely as the square root of x. Hmm, inverse variation, right? So I remember that inverse variation means that as one quantity increases, the other decreases proportionally. The general formula for inverse variation is y = k/x, where k is the constant of variation. But in this case, it's not just x; it's the square root of x. So I think the formula here should be y = k / sqrt(x). Let me confirm that.
Yes, if y varies inversely as sqrt(x), then y is equal to some constant divided by sqrt(x).

Alright, so the problem gives me that when x is 24, y is 15. I need to find x when y is 3. So first, maybe I should find the constant k using the first set of values, and then use that k to find the new x when y is 3.

Let me write down the equation with the first pair of values.

<prompt_1.1> Use the given values of x = 24 and y = 15 to find the constant of variation k. </prompt_1.1>
-----

Wait, let me think again. The problem says y varies inversely as sqrt(x). So the relationship is y = k / sqrt(x). So when y is 3, we can plug into the same equation. But we need to remember that k is the same constant, right? Because the variation is with respect to the same x and y.

<prompt_1.2> Verify the solution using the inverse variation relationship and confirm the result. </prompt_1.2>
-----

Let me just check if the answer makes sense intuitively. Since y is inversely proportional to sqrt(x), if y decreases from 15 to 3, that's a decrease by a factor of 15/3 = 5. So sqrt(x) should increase by the same factor.
Therefore, sqrt(x) becomes 5 times larger. So sqrt(x) = 5*sqrt(24). Then x would be 25*24, which is 600. Yeah, that matches. So that's another way to see it.

Alternatively, if I think about the product y*sqrt(x). Since y varies inversely as sqrt(x), the product y*sqrt(x) should be constant. So initially, y=15 and x=24.
So 15*sqrt(24) = k. Then when y=3, sqrt(x) should be k/3 = (15*sqrt(24))/3 = 5*sqrt(24). Therefore, sqrt(x) = 5*sqrt(24), so x = 25*24 = 600. Same result. So that's consistent.

Another way to check is to compute sqrt(24).

<prompt_1.3> Perform additional verification by calculating the square roots and recalculating y. </prompt_1.3>

Alternatively, maybe someone might make a mistake by thinking it's direct variation, but the problem clearly states inverse variation. So if someone did y = k*sqrt(x), then they would get a different answer, but the problem says y varies inversely, so that's not the case here.

Alternatively, maybe miscalculating the square roots or arithmetic. Let me check sqrt(24) again.
24 is 4*6, so sqrt(4*6) = 2*sqrt(6). Then 5*2*sqrt(6) is 10*sqrt(6). Squared is 100*6=600.

I think that's solid. So the answer is 600.
</Plan_1>

<Execution_1>
<execution_1.1>
So 15 = k / sqrt(24). To find k, I can rearrange this equation: k = 15 * sqrt(24). Let me compute that. But maybe I can leave it in exact form instead of using a decimal approximation to keep it precise.

Wait, sqrt(24) can be simplified. sqrt(24) is sqrt(4*6) which is 2*sqrt(6). So k = 15 * 2 * sqrt(6) = 30*sqrt(6). Okay, so k is 30*sqrt(6).
</execution_1.1>

<execution_1.2>
So let's do that step by step.

Given x = 24, y = 15. So plugging into the equation:

15 = k / sqrt(24)

So to find k, we can rearrange this:

k = 15 * sqrt(24)

Let me compute sqrt(24). Hmm, sqrt(24) is equal to sqrt(4*6) which is 2*sqrt(6). So sqrt(24) = 2*sqrt(6). Therefore, k = 15 * 2 * sqrt(6) = 30*sqrt(6). So k is 30 times the square root of 6.
</execution_1.2>

<execution_1.3>
which is approximately 4.899. Then 15*4.899 is approximately 73.485. When y is 3, sqrt(x) would be 73.485 / 3 ≈ 24.495, and x would be (24.495)^2 ≈ 600.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, when y is 3, we can use the same k to find the new x. So plugging into the equation:

<prompt_2.1> Use the constant of variation k to find the new x when y = 3. </prompt_2.1>
-----

Based on execution_1.2:
Now, we need to find x when y = 3. So plug into the equation again:

<prompt_2.2> Use the equation y = k / sqrt(x) to find x when y = 3. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
3 = k / sqrt(x). But we know k is 30*sqrt(6), so substituting that in:

3 = (30*sqrt(6)) / sqrt(x)

Hmm, now I need to solve for x. Let me rearrange this equation. Multiply both sides by sqrt(x):

3*sqrt(x) = 30*sqrt(6)

Then divide both sides by 3:

sqrt(x) = (30*sqrt(6)) / 3

Simplify 30 divided by 3, which is 10:

sqrt(x) = 10*sqrt(6)

Now, to solve for x, square both sides:

x = (10*sqrt(6))^2

Calculating that, 10 squared is 100, and (sqrt(6))^2 is 6. So:

x = 100 * 6 = 600

So x should be 600 when y is 3.
</execution_2.1>

<execution_2.2>
3 = k / sqrt(x)

We know k is 30*sqrt(6), so:

3 = (30*sqrt(6)) / sqrt(x)

Let me solve for sqrt(x). Multiply both sides by sqrt(x):

3*sqrt(x) = 30*sqrt(6)

Then divide both sides by 3:

sqrt(x) = (30*sqrt(6)) / 3

Simplify 30/3 to 10:

sqrt(x) = 10*sqrt(6)

Now, to find x, square both sides:

x = (10*sqrt(6))^2

Calculating that:

10 squared is 100, and (sqrt(6))^2 is 6. So 100*6 = 600. Therefore, x = 600.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, let me check if that makes sense. Let's verify the steps again to make sure I didn't make a mistake.

<prompt_3.1> Recheck the calculations to verify the solution. </prompt_3.1>
-----

Based on execution_2.1, execution_2.2:
But let me just make sure that when we squared 10*sqrt(6), we did that correctly.

<prompt_3.2> Verify the squaring of 10*sqrt(6) to ensure the calculation is correct. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
First, inverse variation with sqrt(x) means y = k / sqrt(x). Correct. Then when x=24, y=15. So plugging in, 15 = k / sqrt(24). Then k = 15*sqrt(24). Which is 15*2*sqrt(6) = 30*sqrt(6). That seems right.

Then, when y=3, 3 = 30*sqrt(6) / sqrt(x). So solving for sqrt(x): sqrt(x) = 30*sqrt(6)/3 = 10*sqrt(6). Then x is (10*sqrt(6))^2. Which is 100*6=600.
</execution_3.1>

<execution_3.2>
(10*sqrt(6))^2 = 10^2 * (sqrt(6))^2 = 100 * 6 = 600.
</execution_3.2>
</Execution_3>

<Final_answer>
Therefore, when $y=3$, $x$ is $\boxed{600}$.
</Final_answer>
</think>