Question 315·Hard·Math

Let $x$ be 2023 revenue, in millions of dollars. A profit margin is profit as a percent of revenue, so the $22\%$ margin makes 2023 profit $0.22x$ million.

Revenue increased by $18\%$, so 2024 revenue is $118\%$, or $1.18$ times, the 2023 revenue: $\text{2024 revenue}=1.18x$. Using $0.18x$ would count only the increase, not the whole new revenue.

The $20\%$ margin is taken of 2024 revenue, not of 2023 profit. Each year’s profit margin applies to that year’s revenue. So $\text{2024 profit}=0.20(1.18x)$.

The 2024 profit was $1.8 million greater, so subtract the 2023 profit from the 2024 profit: $0.20(1.18x)-0.22x=1.8$. Reversing the subtraction would describe a loss.

Type $0.20(1.18x_1)-0.22x_1\sim1.8$ in Desmos. Use $x_1$ for the unknown and $\sim$ to ask Desmos to find its value; plain $x$ is treated as a graph variable. Under PARAMETERS, Desmos shows $x_1=112.5$. That’s already to the nearest tenth, so the company’s 2023 revenue was $112.5 million. Choice C.

For each $1 million of 2023 revenue, the $18\%$ increase gives $1.18 million of 2024 revenue. Type $0.20\cdot1.18$; Desmos returns $0.236$. That means $0.236 million of 2024 profit per original revenue million.

The 2023 margin gave $0.22 million of profit per original revenue million. Type $0.236-0.22$; Desmos returns $0.016$. That’s the extra profit per original revenue million, not a change in profit margin.

Divide the $1.8 million extra profit by $0.016 million of extra profit per original revenue million: $x=\frac{1.8}{0.016}=112.5$. Type $1.8/0.016$; Desmos returns $112.5$. So 2023 revenue was $112.5 million. Choice C.