Nice question!!

Given: The ratio of \(y\) to \(c\) is equal to the ratio of \(x\) to \(b\).

We can rewrite this as: \(\frac{y}{c}=\frac{x}{b}\)

Cross multiply to get: \(yb = xc\)

Let's solve this equation for \(b\) by dividing both sides by \(y\) to get: \(b = \frac{xc}{y}\)

Now that we've solved the equation for \(b\), we can rewrite Quantity B to get:
Quantity A: \(3c\)
Quantity B: \(2(\frac{xc}{y})\)

Simplify:
Quantity A: \(3c\)
Quantity B: \(\frac{2xc}{y}\)

Since we are told that \(c\) is POSITIVE, we can safely divide both quantities by \(c\) to get:
Quantity A: \(3\)
Quantity B: \(\frac{2x}{y}\)

Let's also divide both quantities by \(2\) to get:
Quantity A: \(\frac{3}{2}\)

Quantity B: \(\frac{x}{y}\)

Finally, let's apply the given fact that \(0 < y < 4\) and \(x > 4\).
We can MINIMIZE the value of Quantity B by minimizing the numerator (x) and maximizing by the denominator (y)
Since x and y are integers, the minimum value of x is 5, and the maximum value of y is 3
So the MINIMIZE value of Quantity B is 5/3, which is greater than Quantity A (3/2)

If the MINIMUM value of Quantity B is greater than Quantity A, we can conclude that Quantity B will always be greater than Quantity A

Answer: B

Cheers,
Brent

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