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Airplane A flew against a headwind a distance of 900 miles

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Airplane A flew against a headwind a distance of 900 miles [#permalink] New post 03 May 2012, 16:06
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Question Stats:

75% (03:49) correct 25% (01:25) wrong based on 12 sessions
Airplane A flew against a headwind a distance of 900 miles at an average speed of (s - 50) miles per hour. Airplane B fl‡ew 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

I love this question. I solved in this way:
(s - 50) * (x + \frac{3}{2}) = 900
(s + 50)* (x) = 900
Being x the number of hours.

We solve both equations and we get:
2x^2 + 3x - 27 = 0
(2x + 9)(x - 3) = 0
ANSWER: x = 3

This solution took time. Is there a faster way to solve it?

Source: http://www.gmathacks.com
[Reveal] Spoiler: OA

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Re: Airplane A ‡flew against a headwind a distance of 900 miles [#permalink] New post 03 May 2012, 18:59
I did almost the same. it took me 4 minutes. really not acceptable for GMAT... :(
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Re: Airplane A ‡flew against a headwind a distance of 900 miles [#permalink] New post 03 May 2012, 19:23
akrish1982 wrote:
I did almost the same. it took me 4 minutes. really not acceptable for GMAT... :(


The OE says that a faster way is picking numbers, using the numbers of the choices. But that strategy is time consuming also. Maybe this is not a GMAT like question (because of this time issue). What do you think Bunuel?
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Re: Airplane A flew against a headwind a distance of 900 miles [#permalink] New post 03 May 2012, 21:05
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metallicafan wrote:
Airplane A flew against a headwind a distance of 900 miles at an average speed of (s - 50) miles per hour. Airplane B fl‡ew 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

I love this question. I solved in this way:
(s - 50) * (x + \frac{3}{2}) = 900
(s + 50)* (x) = 900
Being x the number of hours.

We solve both equations and we get:
2x^2 + 3x - 27 = 0
(2x + 9)(x - 3) = 0
ANSWER: x = 3

This solution took time. Is there a faster way to solve it?

Source: http://www.gmathacks.com


Time taken by A \frac{900}{s-50};
Time taken by B \frac{900}{s+50};

We are told that the time for A was 3/2 hours longer: \frac{900}{s-50}-\frac{900}{s+50}=\frac{3}{2} --> \frac{900(s+50-s+50)}{s^2-2,500}=\frac{3}{2} --> \frac{300*100}{s^2-2,500}=\frac{1}{2} --> s^2-2,500=600*100 --> s^2=62,500 --> s=250.

Time taken by B \frac{900}{250+50}=3.

Answer: D.
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Re: Airplane A flew against a headwind a distance of 900 miles [#permalink] New post 04 May 2012, 20:33
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metallicafan wrote:
Airplane A flew against a headwind a distance of 900 miles at an average speed of (s - 50) miles per hour. Airplane B fl‡ew 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

I love this question. I solved in this way:
(s - 50) * (x + \frac{3}{2}) = 900
(s + 50)* (x) = 900
Being x the number of hours.

We solve both equations and we get:
2x^2 + 3x - 27 = 0
(2x + 9)(x - 3) = 0
ANSWER: x = 3

This solution took time. Is there a faster way to solve it?

Source: http://www.gmathacks.com


You can club number picking with algebra to get a faster solution.
As Bunuel did above:

\frac{900}{s-50}-\frac{900}{s+50}=\frac{3}{2}

Multiply both sides by 2 to get:
\frac{1800}{s-50}-\frac{1800}{s+50}=3

Now try values for s which will give you integers with a difference of 3.

First guess: s = 150
You get 18 - 9. You need to go higher to get smaller values so that difference between them is lesser i.e. 3

Second guess: s = 250
You get 9 - 6
So s must be 250.

Time taken = 900/(250 + 50) = 3 hrs
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Re: Airplane A flew against a headwind a distance of 900 miles   [#permalink] 04 May 2012, 20:33
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