Question 170·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

The fraction describes distance, not time. Let $d$ be the first stretch's distance in kilometers. Type $d=42\left(\frac{2}{3}\right)$ in Desmos; it shows $d=28$. So the cyclist rides $28$ kilometers at the first speed.

Subtract the first stretch from the full route: $42-28=14$. So the cyclist rides $14$ kilometers at the second speed.

Divide the first stretch's distance by its speed: $\frac{28}{28}=1\text{ hour}$. Kilometers divided by kilometers per hour leaves hours.

Divide the remaining distance by the second speed: $\frac{14}{16}=\frac78\text{ hour}$. Don't round this to a full hour; that would change the trip's total time.

Resting still uses time, even though it adds no distance. An hour has $60$ minutes, so convert the rest: $\frac{15}{60}=\frac14\text{ hour}$.

For the entire trip, total elapsed time includes both riding stretches and the rest. Type $t=\frac{d}{28}+\frac{42-d}{16}+\frac{15}{60}$ in Desmos, where $t$ is the total time in hours. It shows $t=2.125$, or $\frac{17}{8}$ with the fraction button.

Average speed is total distance divided by total time. Multiply by $\frac{1000\text{ meters}}{1\text{ kilometer}}$ and $\frac{1\text{ hour}}{3600\text{ seconds}}$ so kilometers and hours cancel. Type $\frac{42}{t}\cdot\frac{1000}{3600}$; Desmos shows $5.490196\ldots$, or $\frac{280}{51}$ with the fraction button. The cyclist's average speed for the entire trip is $\frac{280}{51}$ meters per second. Choice D.