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Intern  Joined: 20 Oct 2005
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During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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56 00:00

Difficulty:   85% (hard)

Question Stats: 62% (03:04) correct 38% (03:13) wrong based on 826 sessions

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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%

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59561
Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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17
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.

Hope it's clear.
CEO  Joined: 07 Jul 2004
Posts: 3422
Location: Singapore
Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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6
5
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 %
##### General Discussion
Manager  Joined: 26 Jun 2006
Posts: 124
Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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1
1
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)
Kaplan GMAT Instructor Joined: 21 Jun 2010
Posts: 142
Location: Toronto
Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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1
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%

= (20-15)/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  [#permalink]

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2
Time when droven x miles in 60 mph $$t1$$= $$x/60$$
time when droven 5 miles in 30mph and x-5 miles in 60mph $$t2$$= $$5/30$$ + $$(x-5)/60$$ = $$(x+5)/60$$

Percentage increase =$$100$$*$$(t2 - t1)/t1$$ = $$100$$ * $$(x+5-x)/x$$= $$500/x$$ %
Retired Moderator Joined: 20 Dec 2010
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Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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1
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

$$\frac{\frac{x}{60}-\frac{x-5}{60}}{\frac{x}{60}}*100 = \frac{500}{x}%$$

Ans: "E"
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Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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1
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
Manager  Joined: 08 Dec 2012
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Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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1
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
Director  Joined: 29 Nov 2012
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Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#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 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

= $$\frac{500}{X}$$ %

Director  S
Joined: 17 Dec 2012
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Location: India
Re: During a trip on an expressway, Don drove a total of x miles. His aver  [#permalink]

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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. His aver  [#permalink]

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

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ludwigvb wrote:
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%

We are given that Don drives a total of x miles, his average speed on a 5-mile 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 5-mile section is 5/30 = 1/6 of an hour and the time for the remainder of the trip is (x-5)/60 hours.

Thus, the total time is 1/6 + (x-5)/60 = 10/60 + (x-5)/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%

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