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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...

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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...

by AAPL » Tue Oct 13, 2020 4:05 am

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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%

OA E
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Re: During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the..

by Scott@TargetTestPrep » Sat Oct 17, 2020 7:19 am
AAPL wrote: ↑
Tue Oct 13, 2020 4:05 am
 GMAT Prep

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%

OA E
Solution:

We are given that Don drove 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 was 5/30 = 1/6 of an hour, and the time for the remainder of the trip was (x-5)/60 hours.

Thus, the total time was 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. We use the percent change formula: (New - Old)/Old * 100%.

[(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. His average speed on a certain 5-mile section of the..

by deloitte247 » Fri Oct 23, 2020 10:38 pm
Total distance covered = x miles
Average speed on a 5-mile section = 30 mph
Average speed on remaining (x-5) miles = 60 mph
$$Time\ taken\ =\frac{dis\tan ce\ covered}{Average\ speed}$$
$$Time\ taken\ for\ 5-mile\ stretch=\frac{5}{30}=\frac{1}{6}hour$$
For the remaining portion of the distance
$$Time=\frac{x-5}{60}$$
$$Time\ taken=\frac{1}{6}+\frac{x-5}{60}=\frac{10+x-5}{60}=\frac{x-5}{60}$$
Had Don traveled at an average speed of 60mph for the entire trip, the time taken = x/60 hours
$$\%\ by\ which\ \frac{\left(x+5\right)}{60}is\ greater\ than\ \frac{x}{60}=>$$ $$\frac{\left(\frac{\left(x+5\right)}{60}-\frac{x}{60}\right)}{\frac{x}{60}}\cdot100=\frac{500}{x}\%\ \ \ \left(Answer\ =\ option\ E\right)$$
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