1. **Stating the problem:** A car travels from point A to point B at an unknown speed $v$ km/h (unloaded), and returns from B to A at 60 km/h (loaded). The average speed for the entire round trip is 70 km/h. We need to find the speed $v$ when the car was unloaded. 2. **Formula for average speed:** The average speed for a round trip when distances are equal but speeds differ is given by the harmonic mean: $$\text{Average speed} = \frac{2 \times v_1 \times v_2}{v_1 + v_2}$$ where $v_1$ and $v_2$ are the speeds for each leg. 3. **Applying the formula:** Let $v_1 = v$ (unloaded speed) and $v_2 = 60$ km/h (loaded speed). The average speed is 70 km/h, so: $$70 = \frac{2 \times v \times 60}{v + 60}$$ 4. **Solving for $v$:** Multiply both sides by $(v + 60)$: $$70(v + 60) = 120v$$ Distribute 70: $$70v + 4200 = 120v$$ Subtract $70v$ from both sides: $$4200 = 120v - 70v$$ $$4200 = 50v$$ Divide both sides by 50: $$v = \frac{4200}{50}$$ Simplify: $$v = 84$$ 5. **Answer:** The speed of the car when unloaded was **84 km/h**.