SOLUTION: At 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces. His horse travels at 10 miles per hour. A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose

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Question 62466: At 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces. His horse travels at 10 miles per hour. A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose of giving an additional package for the first Pony Express rider to deliver. The second rider’s horse travels at 18 miles per hour. At what time will the second Pony Express rider catch up to the first Pony Express rider. (Assume that the horses maintain the same rate of speed and that no hindrances affect either horse or rider).
Found 2 solutions by ankor@dixie-net.com, checkley71:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
At 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces. His horse travels at 10 miles per hour. A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose of giving an additional package for the first Pony Express rider to deliver. The second rider’s horse travels at 18 miles per hour. At what time will the second Pony Express rider catch up to the first Pony Express rider. (Assume that the horses maintain the same rate of speed and that no hindrances affect either horse or rider).
:
We know that when the fast horse catches up with the slow horse the will have
traveled the same distance.
Let t = time required for the fast horse to catch up
Then (t+3) = time the 1st horse has traveled when he is caught
:
Dist = speed * time
:
2nd horse dist = 1st horse dist
18t = 10(t+3)
18t = 10t + 30
18t - 10t = 30
8t = 30
t = 30/8
t = 3.75 hrs or 3 hrs 45 min
:
Add this to 9 AM: 12:45 PM when they are together
:
Check using distance:
18(3.75) = 10(6.75)
67.5 = 67.5 miles

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
DISTANCE=RATE*TIME SO
D=10T FOR THE FIRST RIDER
FOR THE SECOND RIDER WE HAVE
D=18(T-3) SEEING AS THE DISTANCES ARE EQUAL WE HAVE
10T=18(T-3)
10T=18T-54
10T-18T=-54
-8T=-54
T=-54/-8
T=6.75 HOURS AFTER 6:00 AM WHICH MAKES IT 6+6.75 OR 12:45 P.M.