Solving the Problem

Method 1 (choosing numbers)

I argue that choosing numbers is the easier approach here. The given information tells us that $S = \frac{k}{t}$. Let's say that $k=12$, $T=3$, and $S=4$. If we plug those values into the given equation, everything checks out:

$$4 = \frac{12}{3}$$

It then says that $S$ increases by $50 \%$, so the new $S$ value is $1.5 \times 4 = 6$. We know $k$ can't change because it's constant, so the question becomes what happens to $T$? Let's go back to our equation:

$$6 = \frac{12}{T}$$

$$T = 2$$

So $T$ starts out as $3$ and then goes to $2$. Using the percent decrease formula, we get the following:

$$\frac{3-2}{3} \times 100 = 33 \frac{1}{3} \%$$

Thus, the correct answer is B.

Method 2 (algebra)

The given information tells us that $k$ is constant, so to make our lives easier, we should not think of the equation as $S = \frac{k}{T}$, but rather as $k = ST_{1}$. We labeled the $T$ with the $1$ to denote it as the initial $T$ value.

The given information then tells us that $S$ increases by $50$ percent, which can be represented by $1.5S$, causing $T_{1}$ to change as a result. We'll call that new value $T_{2}$. The new equation is thus $k = 1.5ST_{2}$. If we set the two equations equal to each other, we get the following:

$$ST_{1} = 1.5ST_{2} $$

The $S$ cancels on both sides to give us...

$$T_{1} = 1.5T_{2}$$

If we divide both sides by $T_{2}$, we get...

$$\frac{T_{1}}{T_{2}} = 1.5 = \frac{3}{2}$$

So $T_{1}$ is $3$ and $T_{2}$ is $2$. Using the percent decrease formula, we get the following:

$$\frac{3-2}{3} \times 100 = 33 \frac{1}{3} \%$$

Thus, the correct answer is B.