Airplane A flew against a headwind a distance of 900 miles

Question

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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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: https://www.gmathacks.com

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Official Answer

D

Bunuel · Accepted Answer

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.

akrish1982 · Answer

I did almost the same. it took me 4 minutes. really not acceptable for GMAT...

metallicafan · Answer

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?

KarishmaB · Answer

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

jlgdr · Answer

I backsolved this way

900/s-50 - 900/s+50 = 3/2

900 (1-s/50 - 1/s+50) = 3/2

1-s/50 +1/s+50 = 1/600

Now, let's try with D

900 / s+50 = 3

So s+50 is 300

So s = 250

1/200 - 1/300 = 1/600

Yes it works, actually one can just do it really quick with smaller numbers
Remember that 1/2 - 1/3 = 1/6 this should already be familiar

So D is the correct answer

Hope it helps
Cheers
J

EMPOWERgmatRichC · Answer

Hi All,

This question has a couple of great "pattern-matching" shortcuts built into it (and the answers). While some Algebra is required to solve it, how YOU choose to do the Algebra will dictate how long this question takes to solve.

We're told a number of Facts about two airplanes
1) They both travel 900 miles
2) One travels (S - 50) miles/hour and the other travels (S + 50) miles/hour
3) The slower plane takes an extra 1.5 hours to travel that distance.

We're asked to figure out the TIME it took the FASTER plane to travel that distance.

Here are the "shortcuts":
1) The answers are "nice" numbers, in half-hour increments, which means we'll likely be dealing with nice "ROUND" numbers for the speeds and "easy" times of the two planes. 
2) We can rewrite the two speeds of the planes as X and (X+100); this is considerably easier than using (S - 50) and (S + 50). Since the question is NOT asking about the speeds, we won't have to "undo" any of this later.

Here are the two equations that we can set up from this info:

Slow plane:
900 miles = (X miles/hour)(T + 1.5 hours)

Fast plane:
900 miles = (X + 100 miles/hour)(T hours)

Remember how the answer choices implied that we would be dealing with ROUND numbers for the times and speeds? Now consider the speed of the FASTER plane (X + 100)....What would be a nice ROUND number for X, that when added to 100, divides EVENLY into 900?.....

How about X = 200....? What happens to the "math" when X = 200.....?

900 = (300)(T)
3 hours = T

Will X = 200 and T = 3 "fit" into the equation for the slower plane? Let's check:

900 = (200)(3 + 1.5)
900 = (200)(4.5)

This MATCHES PERFECTLY!!!

Thus, the TIME of the FASTER plane MUST be 3 hours.

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

TarPhi · Answer

We solve both equations and we get:
2x2+3x−27=02x2+3x−27=0
(2x+9)(x−3)=0(2x+9)(x−3)=0
ANSWER: x=3x=3

How did you get this??

I solved it by Bunuel's method and got the answer!

finisher009 · Answer

Step-by-Step Solution using Backsolving:
1. Analyze the relationship: Airplane A (headwind) is slower than Airplane B (tailwind). The time difference is 1.5 hours. The equations are:
 • Time_A = 900 / (s - 50)
 • Time_B = 900 / (s + 50)
 • Time_A = Time_B + 1.5

2. Pick an answer choice to test. Let’s start with a “nice” number like D) 3 hours.
 • Assume Airplane B’s time (Time_B) is 3 hours.

3. Calculate based on this assumption:
 • If Time_B is 3 hours, then Speed_B = Distance / Time = 900 / 3 = 300 mph.
 • Since Speed_B = s + 50, the plane’s speed in still air (s) is 300 - 50 = 250 mph.
 • Now, find Speed_A, which is s - 50. So, Speed_A = 250 - 50 = 200 mph.
 • Calculate Time_A = Distance / Speed_A = 900 / 200 = 4.5 hours.

4. Check if it matches the problem: The problem states that Airplane A’s trip took 1.5 hours longer than Airplane B’s.
 • Our calculated Time_A (4.5 hours) is exactly 1.5 hours more than our assumed Time_B (3 hours).
 • This matches perfectly. We’ve found the correct answer.

Final Answer: The correct answer is D) 3.

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

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