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During a trip on an expressway, Don drove a total of x miles. His [#permalink]

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27 Jan 2006, 17:35

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During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

During a trip on an expressway, Don drove a total of x miles [#permalink]

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14 Aug 2006, 08:29

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During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A. 8.5% B. 50% C. x/12% D. 60/x% E. 500/x%

Last edited by Bunuel on 07 Mar 2013, 05:20, edited 1 time in total.

Renamed the topic, edited the question and added the OA.

During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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11 Dec 2010, 18:11

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During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile seciton was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been if he had travelled at a constant rate of 60 miles per hour for the entire trip

Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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11 Dec 2010, 20:37

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Yellow22 wrote:

During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile seciton was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been if he had travelled at a constant rate of 60 miles per hour for the entire trip

A) 8.5 % B) 50% C) x/12% D) 60/x% E) 500/x%

Long word problem with a variable in the choices - perfect situation for picking numbers.

Let's let x=15 to keep things simple.

So, he drove 5 miles at 30mph and 10 miles at 60mph.

time = dist/rate, so he spent 5/30 + 10/60 = 1/6 + 1/6 = 1/3 of an hour = 20 minutes total.

If he had travelled at 60mph for the entire 15 miles, his time would have been 15/60 = 1/4 hour = 15 minutes.

The question is "His travel time for the x mile trip was what percent greater", so we use the percent change formula:

% change = (amount of change / original amount) * 100%

Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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22 Feb 2011, 15:28

Time when droven x miles in 60 mph \(t1\)= \(x/60\) time when droven 5 miles in 30mph and x-5 miles in 60mph \(t2\)= \(5/30\) + \((x-5)/60\) = \((x+5)/60\)

Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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22 Feb 2011, 15:49

I think the question should be:

During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile section was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent smaller than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip

Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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22 Feb 2011, 16:24

fluke wrote:

I think the question should be:

During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile section was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip

Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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22 Feb 2011, 16:31

maddy2u wrote:

fluke wrote:

I think the question should be:

During a trip on an expressway, Don drove x miles.His average speed on a certain 5 mile section was 30 miles per hour and his average speed of the remainder ofthe trip was 60 miles per hour. His travel time for the x mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip

Re: During a trip on an expressway, Don drove x miles.His average speed on [#permalink]

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08 Mar 2011, 19:39

Let x = 15 5 miles @ 30 mph time = 1/6 10 miles @ 60 mph time = 1/6 Total time = 1/3 15 miles @ 60 mph, time = 1/4 % change (1/3 - 1/4) /1/4 = 1/3 or 100/3% = 33.33%

Whats the answer for this question and the best strategy for this question is to plug numbers or solve algebrically?

During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

A. 8.5% B. 50% C. x/12% D. 60/x% E. 500/x%

For such questions I personally prefer plug in method. For this particular problem this method should give you the answer in less than a minute.

Say x = 5 miles (so no remainder of the trip).

Time to cover x = 5 miles at 30 miles per hour = (time) = (distance)/(rate) = 5/30 = 1/6 hours = 10 minutes. Time to cover x = 5 miles at 60 miles per hour = (time) = (distance)/(rate) = 5/60 = 1/12 hours = 5 minutes.

(Or simply, half rate will result in doubling the time.)

So, we can see that the time to cover x = 5 miles at 30 miles per hour (10 minutes) is 100% greater than the time to cover x = 5 miles at 60 miles per hour (5 minutes).

Now, plug x = 5 miles into the answer choices to see which one yields 100%. Only answer E works.

Re: During a trip on an expressway, Don drove a total of x miles [#permalink]

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09 Mar 2013, 15:02

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I assumed x=10

time taken for 5 miles at 30mph = 10min time taken for the remaining 5 miles at 60mph = 5min total time taken to travel 10miles = 15min = T1

If travelled at 60mph constantly for the entire 10 miles then total time taken = 10min = T2

Percentage change in time = \(\frac{(T1-T2)}{T2}*100\) = 50%

Now this is just one of the answers, it is supposed to change if the total distance changes, as 5 miles becomes different fraction of the total distance (so answer must contain the total distance x, thereby eliminating choice A & B).

By plugging in with answer choices to see which results in 50% will get the answer i.e E