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During a trip on an expressway, Don drove a total of x miles. His aver
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Updated on: 14 Oct 2019, 06:34
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During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5mile 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 xmile 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%
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Originally posted by ludwigvb on 27 Jan 2006, 17:35.
Last edited by Bunuel on 14 Oct 2019, 06:34, edited 2 times in total.
Renamed the topic, edited the question and added the OA.




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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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07 Mar 2013, 05:39
During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5mile 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 xmile 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.
Answer: E.
Hope it's clear.




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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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27 Jan 2006, 20:41
total dist = x miles
time take to clear 5 mile section at 30miles/hr = 5/30 = 1/6 hr
time taken to clear x5 miles section at 60 miles/hr = x5/60 hr
Total time = 1/6 + x5/60 = x+5/60 hr
Total time to clear x miles if he traveled at constant rate of 60 miles/hr = x/60 hr
Extra time = 5/60 = 1/12 hr
Percentage greater = (1/12)/(x/60) * 100% = 500/x %




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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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14 Aug 2006, 08:33
You were on the right track:
(x+5)/x = 1 + 5/x
So, increase as a ratio = 5/x
increase as percent = 500/x (%)
(E)



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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11 Dec 2010, 20:37
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% = (2015)/15 * 100% = 5/15 * 100% = 1/3 * 100% = 33 1/3 % Plugging x=15 into the choices: A) 8.5 %... nope B) 50%... nope C) x/12%.... 5/4 of 1%... nope D) 60/x%... 4%... nope. E) 500/x%... 500/15 = 100/3 = 33 1/3.. yay! Choose (E).



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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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 x5 miles in 60mph \(t2\)= \(5/30\) + \((x5)/60\) = \((x+5)/60\)
Percentage increase =\(100\)*\((t2  t1)/t1\) = \(100\) * \((x+5x)/x\)= \(500/x\) %



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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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
\(\frac{\frac{x}{60}\frac{x5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%\)
Ans: "E"



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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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%
500/x = 500/15 = 33.33%. Hence E



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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09 Mar 2013, 15:02
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 = \(\frac{(T1T2)}{T2}*100\) = 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



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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26 Jun 2013, 01:07
An alternative solution
Don’s time for 5 miles is \(\frac{5}{30}\)>> \(\frac{10}{60}\)
Don's time for X5 miles is \(\frac{X5}{60}\)
Total time is \(\frac{10}{60}\) + \(\frac{X5}{60}\) >>\(\frac{5+x}{60}\)
to find percentage change \(\frac{{5+xx}}{60}\) divided by \(\frac{X}{60}\) * 100
= \(\frac{500}{X}\) %
Answer E



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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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03 Jul 2013, 22:16
We can plug in any value for x because we can see that x is not tied to a specific value. Let us plug in a convenient value. 1. Assume x=10 miles. Total time taken when traveling at different speeds is 5/30 hr + x5 /60 hr. = 5/30 hr + 5/60 hr = 10 min + 5min = 15 min 2. Total time taken at 60 miles /hr = (5 + x5)/60 = 10/60 hr = 10 min 3. (1) greater than (2) by 50 %. 4. Substitute the assumed value of x in the choices . Only E gives 50 % B is not correct because if you change the value of x, the percentage increase will change.
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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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22 Jul 2015, 07:54
During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5mile 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 xmile 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%
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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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09 Feb 2017, 17:28
ludwigvb wrote: During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5mile 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 xmile 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% We are given that Don drives a total of x miles, his average speed on a 5mile section of the expressway was 30 mph, and his average speed for the remainder of the trip, or x  5, miles was 60 mph. Since time = distance/rate, the time for the first 5mile section is 5/30 = 1/6 of an hour and the time for the remainder of the trip is (x5)/60 hours. Thus, the total time is 1/6 + (x5)/60 = 10/60 + (x5)/60 = (x + 5)/60 hours. Had he traveled at a constant rate of 60 miles per hour for the entire trip, then his time would have been x/60 hours. We need to determine the percent by which (x + 5)/60 is greater than x/60. [(x + 5)/60  x/60]/(x/60) * 100% (5/60)/(x/60) * 100% 5/x * 100% 500/x% Answer: E
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