anonymous
  • anonymous
Mark went rock climbing on Saturday on Mt. Rockytop. He started at 10:00 a.m. and climbed at the rate of 4 miles per hour. His friend Paul began climbing at noon, climbing at a rate of 5 miles per hour. At what time would Paul catch up to Mark
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anybody know??
anonymous
  • anonymous
1:00 PM? 8:00PM? 9:00PM? 16:00PM
anonymous
  • anonymous
Let t=Mark's time spent going up the mountain. Then Mark's distance up the mountain is 4t. (rate * time=distance). Since Paul's started two hours later, t-2 is the amount of time that Paul spends going up the mountain. (He is 2 hours behind Mark) So Paul's distance up the mountain is 5(t-2). Set the two distances equal: 4t=5t-10. So t=10. Since t is the amount of time it took Mark to go up the mountain, we can say that t=0 is at 10:00AM, when he started. So t=10 is 8:00PM. And to check, after 10 hours, Mark is 4*10, or 40 miles up the mountain. And after 8 hours, Paul is 5*8=40 miles up the mountain.

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anonymous
  • anonymous
10 hours after Mark started is 8:00PM, and 8 hours after Paul started is also 8:00PM. Therefore, at 8:00PM, they are at the same height.

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