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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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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]
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 [#permalink]
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