Pricing

PP1 (Shorter) Quant Section 1 (Medium) Q12

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.