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During a certain trip, Karl drove 3 hours at an average speed of 60

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During a certain trip, Karl drove 3 hours at an average speed of 60  [#permalink]

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New post 21 Mar 2019, 06:57
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During a certain trip, Karl drove 3 hours at an average speed of 60 m/h and drove the rest trip at an average speed of 45 m/h. If the average speed for the entire trip is 54 m/h, what is the total distance of the trip?

(A) 270

(B) 315

(C) 360

(D) 378

(E) 420

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Re: During a certain trip, Karl drove 3 hours at an average speed of 60  [#permalink]

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New post 21 Mar 2019, 08:21
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Average speed is a weighted average, weighted by the time spent at each speed. If Karl drove at 60 mph, then at 45 mph, and his overall average speed was 54 mph, he must have driven for more time at 60 mph, because 54 is closer to 60 than it is to 45. He spent 3 hours at 60mph, so he spent less than 3 hours at 45 mph, and his total distance was thus less than 3*60 + 3*45 = 315 miles. Only answer A makes any sense.

Of course one can calculate the answer exactly using algebra or weighted average methods (you'll find he spent 2 hours at 45 mph), but there's no reason to with these answer choices.
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Re: During a certain trip, Karl drove 3 hours at an average speed of 60  [#permalink]

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New post 24 Mar 2019, 18:27
rohan2345 wrote:
During a certain trip, Karl drove 3 hours at an average speed of 60 m/h and drove the rest trip at an average speed of 45 m/h. If the average speed for the entire trip is 54 m/h, what is the total distance of the trip?

(A) 270

(B) 315

(C) 360

(D) 378

(E) 420


Let t be the time (in hours) Karl drove at 45 mph. He drove 3 x 60 = 180 miles initially and then he drove 45t miles. The total time for the trip is (t + 3) hours. Thus, we can create the equation:

(180 + 45t)/(t + 3) = 54

180 + 45t = 54t + 162

18 = 9t

t = 2

Then, Karl drove 180 + 2*45 = 180 + 90 = 270 miles in total.

Answer: A
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Re: During a certain trip, Karl drove 3 hours at an average speed of 60   [#permalink] 24 Mar 2019, 18:27
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