document.write( "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). \n" ); document.write( "
Algebra.Com's Answer #43222 by ankor@dixie-net.com(22740)\"\" \"About 
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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).
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\n" ); document.write( "We know that when the fast horse catches up with the slow horse the will have
\n" ); document.write( "traveled the same distance.
\n" ); document.write( "Let t = time required for the fast horse to catch up
\n" ); document.write( "Then (t+3) = time the 1st horse has traveled when he is caught
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\n" ); document.write( "Dist = speed * time
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\n" ); document.write( "2nd horse dist = 1st horse dist
\n" ); document.write( "18t = 10(t+3)
\n" ); document.write( "18t = 10t + 30
\n" ); document.write( "18t - 10t = 30
\n" ); document.write( "8t = 30
\n" ); document.write( "t = 30/8
\n" ); document.write( "t = 3.75 hrs or 3 hrs 45 min
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\n" ); document.write( "Add this to 9 AM: 12:45 PM when they are together
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\n" ); document.write( "Check using distance:
\n" ); document.write( "18(3.75) = 10(6.75)
\n" ); document.write( "67.5 = 67.5 miles\r
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