The question states that a function of two variables varies directly with $x$ and inversely with $y$. The general form for such a function is:

$f(x, y) = k \frac{x}{y}$

where $k$ is a constant to be determined from the given conditions.

Step 1: Use the given information to find $k$

From the problem:

• When $x = 0$ and $y = 1$, $f = 15$,
• When $x = 1$ and $y = 15$, $f = 2$.

Using the first condition: $f(0, 1) = 15 \implies 15 = k \frac{0}{1} = 0$ This shows that there's no meaningful information from the first condition, likely because any term multiplied by zero cancels out.

Now using the second condition: $f(1, 15) = 2 \implies 2 = k \frac{1}{15}$ Solving for $k$: $k = 2 \times 15 = 30$

Step 2: Determine the function when $x = 5$ and $y = 3$

Now that we have $k = 30$, we can use the function to find $f$ when $x = 5$ and $y = 3$: $f(5, 3) = 30 \frac{5}{3} = 30 \times \frac{5}{3} = 50$

So, the value of the function when $x = 5$ and $y = 3$ is $50$. However, none of the options provided (A = 10, B = 2, C = 5, D = 15) match this result, suggesting there might be an issue with the options. Would you like to clarify or explore this further?

Here are some relative questions you might consider:

1. Could there be a misprint in the provided options?
2. Can we check for other variations of the function form?
3. How would the result change if the function had additional constraints?
4. What if the values for $x$ or $y$ were different?
5. Is there an alternative interpretation of the inverse relationship?

Tip: When solving direct and inverse variation problems, always substitute known values carefully to avoid confusion from undefined conditions (like dividing by zero).