1. **Problem:** Henri purchased a stereo for 400 and sold it for 450. Find the markup as a percentage of the purchase price. 2. **Formula:** Markup percentage = \( \frac{\text{Selling Price} - \text{Purchase Price}}{\text{Purchase Price}} \times 100 \% \) 3. **Calculation:** \[ \text{Markup} = 450 - 400 = 50 \] \[ \text{Markup Percentage} = \frac{50}{400} \times 100 = 12.5\% \] 4. **Answer:** Henri's markup is 12.5% of the purchase price. --- 1. **Problem:** The speedboat travels twice as fast as the yacht. The speedboat covers 240 miles in one hour less than the yacht takes to cover 150 miles. Find the speeds and times for both. 2. **Let:** - Yacht speed = \( s \) mph - Speedboat speed = \( 2s \) mph - Yacht time = \( t \) hours - Speedboat time = \( t - 1 \) hours 3. **Using distance = speed \( \times \) time:** For yacht: \( 150 = s \times t \Rightarrow t = \frac{150}{s} \) For speedboat: \( 240 = 2s \times (t - 1) \) 4. **Substitute \( t \) from yacht into speedboat equation:** \[ 240 = 2s \left( \frac{150}{s} - 1 \right) = 2s \left( \frac{150 - s}{s} \right) = 2(150 - s) \] 5. **Simplify:** \[ 240 = 300 - 2s \] \[ 2s = 300 - 240 = 60 \] \[ s = 30 \text{ mph (yacht speed)} \] 6. **Find times:** \[ t = \frac{150}{30} = 5 \text{ hours (yacht time)} \] \[ t - 1 = 5 - 1 = 4 \text{ hours (speedboat time)} \] 7. **Speedboat speed:** \[ 2s = 2 \times 30 = 60 \text{ mph} \] **Final answers:** - Yacht speed = 30 mph, time = 5 hours - Speedboat speed = 60 mph, time = 4 hours