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During a trip on an expressway, Don drove a total of x miles. His
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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 22 Jul 2015, 00:59, edited 1 time 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
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07 Mar 2013, 05:39
fozzzy wrote: 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 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
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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
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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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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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The general formula for % increase of A over B is:
100%*(AB)/B = 100%*(A/B1) = 100%(A/B)  100%



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Re: During a trip on an expressway, Don drove x miles.His average speed on
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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 x miles.His average speed on
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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 x miles.His average speed on
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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 x miles.His average speed on
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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
\(\frac{\frac{x}{60}\frac{x5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%\)
Ans: "E" The question is correct. It will be greater only as the time increases if the speed decreases. Note : Time to cross the 5 mile section increases if traveling at 30mph than at 60mph.
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Re: During a trip on an expressway, Don drove x miles.His average speed on
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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
\(\frac{\frac{x}{60}\frac{x5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%\)
Ans: "E" The question is correct. It will be greater only as the time increases if the speed decreases. Note : Time to cross the 5 mile section increases if traveling at 30mph than at 60mph.Agree!!! I was wrong in my interpretation. Thanks.
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Re: During a trip on an expressway, Don drove x miles.His average speed on
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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
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07 Mar 2013, 05:13
Whats the answer for this question and the best strategy for this question is to plug numbers or solve algebrically?
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Re: During a trip on an expressway, Don drove a total of x miles
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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
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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
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05 Aug 2013, 12:04
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?
The question is looking for how much longer it would take Don to complete the trip at slower speed + faster speed than just faster speed.
Formula for % increase of A over B:
100%*(AB)/B = 100%*(A/B1) = 100%(A/B)  100%
Percentage change in time = (t1t2)/t2 * 100%
I assume that we can set a number to x because we are looking for time and not distance.
x=10
30 miles/hour = .5 miles/minute t1=5/.5 t1=10
60 miles/hour = 1mile/minute t1=5/1 t1=5
So it took him 10 minutes to cover the first half and 5 minutes to cover the second half for a total of 15 minutes.
We are looking to compare how long it would take him to cover the distance of slow+fast vs. just fast. If he covered the entire 10 miles at 60 miles/hour he would have spent 10 minutes on the road.
Percentage change in time = (t1t2)/t2 * 100% (1510)/10 * 100% 5/10 * 100% .5*100% = 50%
He would have traveled 50% faster if the entire journey was completed at 60 miles/hour.
Now plug x into the answer choices to see which one matches.
500/x % 500/10 % 50%
ANSWER: E. 500/x%



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Re: During a trip on an expressway, Don drove a total of x miles
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07 Oct 2013, 09:14
You can always solve by pluggin in numbers. But a good student would know which number to pluggin I plugged in x =60 and it took me more than 2.5 min.
Bunuel's pick is the best number to pick from. I guess thats where the diff comes in Q 51 and the rest.



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During a trip on an expressway, Don drove a total of x miles
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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
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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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Re: During a trip on an expressway, Don drove a total of x miles
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03 Jul 2018, 05:51
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% hey niks18 here is my solution to the problem i got confused in the end, can you pls advise where did i go wrong thanks! Don`s distance X miles Ave speed on 5 miles section  30mph Ave speed for remainder of trip  60mph Time 1 = \(\frac{5}{30}\) Time 2 = \(\frac{x5}{60}\) Total Time = \(\frac{5}{30}\) + \(\frac{x5}{60}\) = \(\frac{5+x}{60}\) total speed = total distance / total time Total Average Speed = x / (50+x) /60 = \(\frac{x}{1} *\frac{60}{5+x}\) = \(\frac{60x}{5+x}\) Now question asks: 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? so total average speed / total average speed 60 for entire trip 60x/(5+x) / 60 =\(\frac{x}{5+x}\)
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