The key is that what's known from the given info is that the RELATIVE change in number of cheetahs was greater than the RELATIVE change in number of pandas, but what the Quantities are comparing are the ABSOLUTE increases.

Say that the number of cheetahs is C and number of pandas is P, 5 years ago.
\(\frac{C }{ P} = \frac{1}{3}\)
\(3C = P\)

Say that the increase in the number of cheetahs = X and increase in the number of pandas = Y
\(\frac{C + X}{ P + Y }= \frac{1}{2}\)
\(2C+2X = 3C+Y\)
\(2X = C+Y\)

Since the info doesn't specify what C or P's actual values are, we're left with an equation with multiple values of X and Y that could be solutions.

To demonstrate:
Say C = 2. Since P = 3C, P = 6.
\(2X = C+Y = 2+Y\) <-- subbing in C = 2

If X = Y, then \(2X = 2+X\). So, X and Y = 2.
Number of cheetahs now = 2+2 = 4, number of pandas now = 6+2 = 8. Ratio now is 4/8 = 1/2, which fits the given info.

OR If X = 1, then \(2 = 2+Y\). So, Y = 0 < X.
Number of cheetahs now = 2+1 = 3, number of pandas now = 6. Ratio now is 3/6 = 1/2, which also fits the given info.

There are relationships of Quantity A and B that could fit the given info. Hence, the answer is D.