A man drives 150 km to the seashore in 3 hours 20 min. He returns from the shore to the starting point in 4 hours 10 min. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$, in kilometres per hour, by

### Concept & Average Rate Average rate (speed) is always calculated as the Total Distance divided by the Total Time. Do not average the two speeds. $$Average\ Speed = \frac{Total\ Distance}{Total\ Time}$$ ### Step-by-Step Solution * **Step 1: Calculate the average rate for the trip going.** * Distance = $150$ km. * Time going = $3$ hours $20$ min = $3 + \frac{20}{60} = 3 \frac{1}{3} = \frac{10}{3}$ hours. * Rate going = $\frac{150}{10/3} = \frac{150 \times 3}{10} = 45$ km/hr. * **Step 2: Calculate the average rate for the entire trip ($r$).** * Total Distance = $150$ km (going) + $150$ km (returning) = $300$ km. * Total Time = $3$ hr $20$ min + $4$ hr $10$ min = $7$ hr $30$ min = $7.5$ hours = $\frac{15}{2}$ hours. * Average rate $r$ = $\frac{300}{15/2} = \frac{300 \times 2}{15} = 40$ km/hr. * **Step 3: Find the difference.** * The question asks how much the rate going ($45$ km/hr) exceeds the average rate $r$ ($40$ km/hr). * Difference = $45 - 40 = 5$ km/hr. ### Exam Strategy & Shortcut Convert minutes to fractions of hours immediately to keep calculations clean. Work out each distinct piece of the puzzle (rate going, then total rate) sequentially rather than trying to build one giant formula. ### Common Pitfall A common mistake is finding the return rate (which is $36$ km/hr) and averaging the two rates to find $r$ ($\frac{45 + 36}{2} = 40.5$ km/hr), which would lead to an incorrect difference of $4.5$ km/hr. ### Final Answer Therefore, the correct answer is **5**.