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by Codebreaker2k » Thu Apr 18, 2013 1:16 pm
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

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by BenMiller » Thu Apr 18, 2013 1:33 pm
Ok... relax!

This problem is designed to stress you out by diving into what would appear to be a lot of complex algebra. Instead, reposition the perspective on the important information and don't let the GMAT dictate HOW you solve this problem.

Important details:
*100 mph differential between A and B's speeds
*Since all the answers are wholes or halves, the speed will divide fairly evenly into 900.
*Since all answers are low numbers, the speeds are relatively high.

If we took, say, 100 and 200 as the speeds of the planes, we could easily test these:
A - 900/100 = 9hrs
B - 900/200 = 4.5hrs
Difference is too large, so each must have been going faster.

Here, we're NOT concerned with solving the problem but with understanding *about* where the answer lies.

Trying the next easiest pair of numbers to divide, 200 and 300, we get:
a - 900/200 = 4.5hrs
B - 900/300 = 3hrs
Difference is 1.5, so the answer is D - 3hrs!

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by [email protected] » Thu Apr 18, 2013 6:02 pm
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?
Total time taken by A = (distance covered)/(average speed) = 900/(s - 50) hours
Total time taken by B = (distance covered)/(average speed) = 900/(s + 50) hours

So, 900/(s - 50) - 900/(s + 50) = 1.5
Now, plugging number for s such that satisfy the above equation is a great idea.
As a similar solution has been posted by Ben, I'm posting the algebraic solution which is not that much complicated for this particular problem.

So, 900/(s - 50) - 900/(s + 50) = 1.5
--> [900(s + 50) - 900(s - 50)]/[(s - 50)(s + 50)] = 1.5
--> [900*50 + 900*50]/(sÂ² - 50Â²) = 1.5
--> (sÂ² - 50Â²) = 2*900*50/(1.5) = 2*900*50/(3/2) = 1200*50
--> sÂ² = 1200*50 + 50*50 = 50(1200 + 50) = 50*1250 = 50*50*25 = (5*50)Â²
--> s = 5*50 - 250

So, time taken by B = 900/(250 + 50) = 900/300 = 3

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by GMATGuruNY » Thu Apr 18, 2013 7:31 pm
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).