Question 5·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

Count the original total revenue as $1$, or $100\%$. Let $s$ be the original share from standard tickets. The theater sells only two kinds of tickets, so the student-ticket share is $1-s$. Both shares are parts of the original total.

The number sold in each category stays the same, so each category’s revenue changes by the same percent as its ticket price. A $15\%$ increase gives standard tickets the multiplier $1.15$, while a $10\%$ decrease leaves student tickets with the multiplier $0.90$. Their adjusted revenues, measured against the original total, are $1.15s$ and $0.90(1-s)$.

A $4\%$ increase makes the new total $1.04$ times the original total of $1$. The two adjusted categories must add to that total, so $1.15s+0.90(1-s)=1.04$. Adding the two price-change percents instead would ignore how much revenue each category started with.

Type $1.15x_{1}+0.90(1-x_{1})\sim1.04$ in Desmos, using $x_{1}$ for $s$ and $\sim$ to tell Desmos to solve rather than graph. Under PARAMETERS, it shows $x_{1}=0.56$. So standard tickets brought in $56\%$ of the original revenue, not necessarily the new revenue.

Add $1.15x_{1}$ in Desmos. It shows $0.644$, the standard-ticket revenue after its price increase, measured against the original total of $1$. That makes $64.4\%$ a tempting stopping point, but it still uses the original total as its whole.

The question asks for a share after the adjustment, so divide the adjusted standard revenue by the adjusted total, $1.04$. Type $100\frac{1.15x_{1}}{1.04}$; Desmos shows $61.923076\ldots$. Rounded to the nearest $0.01\%$, standard tickets provided $61.92\%$ of the theater’s adjusted revenue. Choice A.