Airplane A flew against a headwind a distance of 900 miles

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?

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

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

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:

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

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!

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