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

BTGmoderatorDC wrote: ↑

Fri Sep 23, 2022 4:50 pm

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

Source: GMAT Prep

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.
Total time = (time spent driving 30 mph) + (time spent driving 60 mph)
time = distance/speed
So, total driving time = 5/30 + (x - 5)/60
= 10/60 + (x - 5)/60
= (10 + x - 5)/60
= (x + 5)/60

Hypothetically speaking, Don could have driven the entire x miles at a speed of 60 mph
time = distance/speed
Total driving time = x/60

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?
In other word: (x + 5)/60 is what percent greater than x/60?

Percentage = 100[(x + 5)/60 - x/60]/(x/60)
= 100[5/60]/(x/60)
= 100(5/60)(60/x)
= 500/x

Answer: E

BTGmoderatorDC wrote: ↑

Fri Sep 23, 2022 4:50 pm

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

Source: GMAT Prep

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