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11. Higher Order Thinking Mark is training for a min triathlon. He rode his bike \( \frac{3}{4} \) mile, ran \( \frac{2}{4} \) mile, and swam \( \frac{1}{4} \) mile each day. How does the distance he biked in 3 days compare to the distance he swam in 3 days? In 5 days? In 6 days? Why?

Ask by Goodwin Nunez. in the United States
Mar 14,2025

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- **3 Days:** Biking = 3 × \( \frac{3}{4} = \frac{9}{4} \) miles; Swimming = 3 × \( \frac{1}{4} = \frac{3}{4} \) miles. Biking is 3 times swimming. - **5 Days:** Biking = 5 × \( \frac{3}{4} = \frac{15}{4} \) miles; Swimming = 5 × \( \frac{1}{4} = \frac{5}{4} \) miles. Biking is 3 times swimming. - **6 Days:** Biking = 6 × \( \frac{3}{4} = \frac{18}{4} = \frac{9}{2} \) miles; Swimming = 6 × \( \frac{1}{4} = \frac{6}{4} = \frac{3}{2} \) miles. Biking is 3 times swimming. **Why?** Each day, Mark bikes 3 times the distance he swims. This ratio remains consistent over any number of days because both biking and swimming distances are multiplied by the same number of days, maintaining the 3:1 ratio.

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In 3 days, Mark biked \( \frac{3}{4} \) mile per day, totaling \( \frac{3}{4} \times 3 = \frac{9}{4} \) miles biked. For swimming, he swam \( \frac{1}{4} \) mile per day, totaling \( \frac{1}{4} \times 3 = \frac{3}{4} \) miles. So after 3 days, Mark biked \( \frac{9}{4} \) miles, which is significantly more than the \( \frac{3}{4} \) miles he swam. If we consider 5 days, he biked \( \frac{3}{4} \times 5 = \frac{15}{4} \) miles, and swam \( \frac{1}{4} \times 5 = \frac{5}{4} \) miles. Again, the biking distance far exceeds the swimming distance. In 6 days, the pattern continues: biking \( \frac{3}{4} \times 6 = \frac{18}{4} \) miles and swimming \( \frac{1}{4} \times 6 = \frac{6}{4} \) miles. The biking distance always remains higher because he bikes more distance each day compared to the swimming. The consistent ratio of biking to swimming distances showcases how his triathlon training is structured to increase biking strength while maintaining a balanced overall fitness regimen!

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