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by thephoenix » Sat May 29, 2010 10:51 am
Selma and Miriam are racing their sailboats from Miami,Florida to Georgetown, Grand Cayman. The loser must buy the
first round of drinks at their favorite beach bar. At 1 P.M., Selma passes a cruise ship that is anchored 160 miles
away from Georgetown. Selma is traveling at 20 miles per hour. At 2 P.M., Miriam passes the anchored cruise ship.
Miriam is traveling at 24 miles per hour. If they continue traveling at the same rates, when will Miriam overtake
Selma?

A. 5.45 P.M.
B. 6 P.M.
C. 6:30 P.M.
D. 7 P.M.
E. 8:30 P.M.
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by mbaonmind » Sat May 29, 2010 11:38 am
thephoenix wrote:Selma and Miriam are racing their sailboats from Miami,Florida to Georgetown, Grand Cayman. The loser must buy the
first round of drinks at their favorite beach bar. At 1 P.M., Selma passes a cruise ship that is anchored 160 miles
away from Georgetown. Selma is traveling at 20 miles per hour. At 2 P.M., Miriam passes the anchored cruise ship.
Miriam is traveling at 24 miles per hour. If they continue traveling at the same rates, when will Miriam overtake
Selma?

A. 5.45 P.M.
B. 6 P.M.
C. 6:30 P.M.
D. 7 P.M.
E. 8:30 P.M.
1:00 PM -> Selma crosses the ship
2:00 PM -> Miriam crosses the ship.

Miriam will cross Selma after T hours, such that
Miriam Distance traveled in T hours = Miriam Distance traveled in T+1 hours
24T = 20(T+1)
T = 5.

Thus Answer is C -> 7:00 PM
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by indiantiger » Sat May 29, 2010 12:47 pm
Lets call the cruise ships locations as X.

First Selma passes the cruise ship at 1pm

Selma's speed (s) = 20 m/h

Second Miriam passes the point X at 2pm,
Miriam's speed (m) = 24m/h

At 2 pm the distance covered by selma will be = 20m/h*1hour = 20 miles.

So the distance gap between Selma and Miriam is 20 miles (lets call it as g).

Total relative speed = (m - s) = 24-20 = 4m/h

Total time taken for Miriam to catch up => g/(m-s) = 20/4 = 5 Hours (7pm)
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by thephoenix » Sat May 29, 2010 6:54 pm
oa [spoiler]7:00 pm[/spoiler]
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