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During a trip on an expressway, Don drove a total of x miles

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During a trip on an expressway, Don drove a total of x miles. His [#permalink] New post 27 Jan 2006, 16:35
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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?

A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Jul 2015, 23:59, 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] New post 27 Jan 2006, 17:07
[5/30+((x-5)/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] New post 27 Jan 2006, 19:41
total dist = x miles

time take to clear 5 mile section at 30miles/hr = 5/30 = 1/6 hr

time taken to clear x-5 miles section at 60 miles/hr = x-5/60 hr

Total time = 1/6 + x-5/60 = 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] New post 27 Jan 2006, 22:57
Total distance X miles

Time taken to travel distance X at 60 mph X/60 …………..1

Time taken to travel distance X-5 at 30mph (X-5)/30 …………….2

Percentage of time taken with case2 compared to case 1.
( 1-2)/1. = 500%
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During a trip on an expressway, Don drove a total of x miles [#permalink] New post 14 Aug 2006, 07:29
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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, 04:20, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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 [#permalink] New post 14 Aug 2006, 07:33
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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 [#permalink] New post 14 Aug 2006, 08:12
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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 [#permalink] New post 14 Aug 2006, 09:11
The general formula for % increase of A over B is:

100%*(A-B)/B = 100%*(A/B-1) = 100%(A/B) - 100%
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Re: During a trip on an expressway, Don drove a total of x [#permalink] New post 07 Mar 2013, 04:13
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] New post 07 Mar 2013, 04:39
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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.

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] New post 09 Mar 2013, 14:02
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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
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Re: During a trip on an expressway, Don drove a total of x miles [#permalink] New post 26 Jun 2013, 00:07
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An alternative solution

Don’s time for 5 miles is \(\frac{5}{30}\)>> \(\frac{10}{60}\)

Don's time for X-5 miles is \(\frac{X-5}{60}\)

Total time is \(\frac{10}{60}\) + \(\frac{X-5}{60}\) >>\(\frac{5+x}{60}\)

to find percentage change \(\frac{{5+x-x}}{60}\) 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 [#permalink] New post 03 Jul 2013, 21:16
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 + x-5 /60 hr. = 5/30 hr + 5/60 hr = 10 min + 5min = 15 min
2. Total time taken at 60 miles /hr = (5 + x-5)/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 [#permalink] New post 05 Aug 2013, 11:04
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?

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%*(A-B)/B = 100%*(A/B-1) = 100%(A/B) - 100%

Percentage change in time = (t1-t2)/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 = (t1-t2)/t2 * 100%
(15-10)/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 [#permalink] New post 07 Oct 2013, 08:14
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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Re: During a trip on an expressway, Don drove a total of x miles [#permalink] New post 31 Oct 2014, 02:43
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Re: During a trip on an expressway, Don drove a total of x miles. His [#permalink] New post 21 Jul 2015, 10:13
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During a trip on an expressway, Don drove a total of x miles [#permalink] New post 22 Jul 2015, 06:54
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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%
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