Codebreaker2k wrote:Airplane A flew against a headwind a distance of 900 miles at
an average speed of (s - 50) miles per hour. Airplane B flew
the same route in the opposite direction with a tailwind and
traveled the same distance at an average speed of (s + 50)
miles per hour. If Airplane A's trip took 1.5 hours longer than
Airplane B's trip, how many hours did Airplane B's trip take?
(A) 1.5
(B) 2
(C) 2.5
(D) 3
(E) 3.5
B's rate - A's rate = (s+50) - (s-50) = 100.
Thus, the difference between B's rate and A's rate is 100 miles per hour.
We can plug in the answers, which represent B's time.
Since A's time is 1.5 hours longer, the answer choices imply the following times:
B = 1.5 hours, A = 3 hours.
B = 2 hours, B = 3.5 hours.
B = 2.5 hours, A = 4 hours.
B = 3 hours, A = 4.5 hours.
B = 3.5 hours, A = 5 hours.
Look for combinations that divide EASILY into 900.
The times in red -- 1.5, 3, and 4.5 -- seem the most viable, implying the following rates:
900/1.5 = 600 miles per hour.
900/3 = 300 miles per hour.
900/4.5 = 200 miles per hour.
The required rate difference (100 miles per hour) is yielded by the times in blue (3 and 4.5).
The correct answer is
D.
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