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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 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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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 miles. His [#permalink]
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27 Jan 2006, 18:07
[5/30+((x5)/60)x/60]/(x/60)
My answer is 500/x%



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Re: During a trip on an expressway, Don drove a total of x miles. His [#permalink]
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27 Jan 2006, 20:41
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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 [#permalink]
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27 Jan 2006, 23:57
Total distance X miles
Time taken to travel distance X at 60 mph X/60 â€¦â€¦â€¦â€¦..1
Time taken to travel distance X5 at 30mph (X5)/30 â€¦â€¦â€¦â€¦â€¦.2
Percentage of time taken with case2 compared to case 1.
( 12)/1. = 500%
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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 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%
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.



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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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v1rok wrote: 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)
Thank you but I still can not catch it.
(x+5)/x = x/x +5/x = 1 + 5/x, but then were why we remove 1?



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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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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
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 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% = (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 [#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 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 [#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 \(\frac{\frac{x}{60}\frac{x5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%\) Ans: "E"
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Re: During a trip on an expressway, Don drove x miles.His average speed on [#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
\(\frac{\frac{x}{60}\frac{x5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%\)
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 [#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
\(\frac{\frac{x}{60}\frac{x5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%\)
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 [#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%
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 [#permalink]
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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 [#permalink]
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07 Mar 2013, 05:39
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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 [#permalink]
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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 = \(\frac{(T1T2)}{T2}*100\) = 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 [#permalink]
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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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