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A trip of 900 miles would have taken 2 hour less if the average speed

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A trip of 900 miles would have taken 2 hour less if the average speed  [#permalink]

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New post 26 Sep 2018, 05:02
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A trip of 900 miles would have taken 2 hour less if the average speed for the trip had been greater by 10 miles per hour. What was the average speed for the trip?

(A) 40 miles per hour
(B) 45 miles per hour
(C) 60 miles per hour
(D) 75 miles per hour
(E) 90 miles per hour

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A trip of 900 miles would have taken 2 hour less if the average speed  [#permalink]

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New post 27 Sep 2018, 15:57
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Bunuel wrote:
A trip of 900 miles would have taken 2 hour less if the average speed for the trip had been greater by 10 miles per hour. What was the average speed for the trip?

(A) 40 miles per hour
(B) 45 miles per hour
(C) 60 miles per hour
(D) 75 miles per hour
(E) 90 miles per hour


Dear GMATPrepNow

Is the OA correct?

Let v=90...........t=10 hrs...........Total distance =900

If v= 100 (increased by 10)..........t=8 hrs (less by 2 hrs).............Total distance =800 (the total distance must match in both cases which does not happen under the conditions given by the question.

Thanks Brent
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A trip of 900 miles would have taken 2 hour less if the average speed  [#permalink]

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New post 28 Sep 2018, 08:01
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Bunuel wrote:
A trip of 900 miles would have taken 2 hour less if the average speed for the trip had been greater by 10 miles per hour. What was the average speed for the trip?

(A) 40 miles per hour
(B) 45 miles per hour
(C) 60 miles per hour
(D) 75 miles per hour
(E) 90 miles per hour


I agree with Mo2men!

Theere is no correct Answer provided!

In my calculation the time is: 14.454 and the speed: 62.266.... (Only possible with the help of a calculator)
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Re: A trip of 900 miles would have taken 2 hour less if the average speed  [#permalink]

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New post 28 Sep 2018, 09:48
Top Contributor
Bunuel wrote:
A trip of 900 miles would have taken 2 hour less if the average speed for the trip had been greater by 10 miles per hour. What was the average speed for the trip?

(A) 40 miles per hour
(B) 45 miles per hour
(C) 60 miles per hour
(D) 75 miles per hour
(E) 90 miles per hour


Let's start with a "word equation"

Travel time at regular speed = Travel time at FASTER speed + 2 hours
Let x = REGULAR speed
So, x + 10 = FASTER speed

Travel time = distance/speed

We get: 900/x = 900/(x + 10)+ 2
Multiply both sides of the equation by x to get: 900 = 900x/(x + 10) + 2x
Multiply both sides of the equation by (x + 10) to get: 900(x + 10) = 900x + 2x(x + 10)
Expand: 900x + 9000 = 900x + 2x² + 20x
Subtract 900x from both sides: 9000 = 2x² + 20x
Subtract 9000 from both sides: 0 = 2x² + 20x - 9000
Divide both sides by 2 to get: 0 = x² + 10x - 4500
Factor to get....hmm, this quadratic equation does not factor nicely (with nice integer solutions, as the 5 answer choices suggest)

Cheers,
Brent
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Re: A trip of 900 miles would have taken 2 hour less if the average speed  [#permalink]

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New post 28 Sep 2018, 10:02
Bunuel Is there a typo somewhere, especially in the answer choices? All attempts to solve this question made me blank out. Thanks.

Bunuel wrote:
A trip of 900 miles would have taken 2 hour less if the average speed for the trip had been greater by 10 miles per hour. What was the average speed for the trip?

(A) 40 miles per hour
(B) 45 miles per hour
(C) 60 miles per hour
(D) 75 miles per hour
(E) 90 miles per hour
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Re: A trip of 900 miles would have taken 2 hour less if the average speed &nbs [#permalink] 28 Sep 2018, 10:02
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A trip of 900 miles would have taken 2 hour less if the average speed

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