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A train traveled 500 miles at x mph and arrived one hour early. The tr
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13 Feb 2022, 04:00
13 Feb 2022, 04:00
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A train traveled 500 miles at x mph and arrived one hour early. The train would have arrived exactly on time if it had traveled at what speed in miles per hour?
(A) \(x + 1 \)
(B) \(\frac{500x}{500 + x}\)
(C) \(\frac{500}{x + 1}\)
(D) \(\frac{500}{500 + x}\)
(E) \(\frac{x}{x + 500}\)
(A) \(x + 1 \)
(B) \(\frac{500x}{500 + x}\)
(C) \(\frac{500}{x + 1}\)
(D) \(\frac{500}{500 + x}\)
(E) \(\frac{x}{x + 500}\)
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Re: A train traveled 500 miles at x mph and arrived one hour early. The tr
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13 Feb 2022, 12:17
13 Feb 2022, 12:17
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Carcass wrote:
A train traveled 500 miles at x mph and arrived one hour early. The train would have arrived exactly on time if it had traveled at what speed in miles per hour?
(A) \(x + 1 \)
(B) \(\frac{500x}{500 + x}\)
(C) \(\frac{500}{x + 1}\)
(D) \(\frac{500}{500 + x}\)
(E) \(\frac{x}{x + 500}\)
(A) \(x + 1 \)
(B) \(\frac{500x}{500 + x}\)
(C) \(\frac{500}{x + 1}\)
(D) \(\frac{500}{500 + x}\)
(E) \(\frac{x}{x + 500}\)
There are two parts to this scenario: the ACTUAL trip and the HYPOTHETICAL trip.
We're told that the HYPOTHETICAL trip would take 1 hour longer than the ACTUAL trip.
So we can start with the following word equation: (time of ACTUAL trip) + 1 = time of HYPOTHETICAL trip
Let k = the speed traveled during the hypothetical trip
time = distance/speed, which means we can plug our values into the word equation to get: (500/x) + 1 = 500/k
Multiply both sides of the equation by kx to get: 500k + kx = 500x
Factor the left side: k(500 + x) = 500x
Divide both sides of the equation by (500 + x) to get: \(k = \frac{500x}{500+x}\)
Answer: B
gmatclubot
Re: A train traveled 500 miles at x mph and arrived one hour early. The tr [#permalink]
13 Feb 2022, 12:17
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