Question

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

Answer 1

anybody know??

Answer 2

1:00 PM?
8:00PM?
9:00PM?
16:00PM

Answer 3

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.

Answer 4

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.