During a trip on an expressway, Don drove a total of...

[swerve]

During a trip on an expressway, Dron 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 x-mile trip was what percent greater than it would have been if he has 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%

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

[regor60]

During a trip on an expressway, Dron 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 x-mile trip was what percent greater than it would have been if he has 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%

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

Use Distance = Rate x Time. Call the total distance X per the problem.

Call the total time for the slower overall trip T. Call the time for the trip at 60 mph T1.

Transcribing the slower trip: X [distance] = 5 [30 mph distance] + 60 mph*(T-1/6) [time spent at 60mph is total time minus time spent at 30mph]

So X = 5+ 60T-10 = 60T-5

Now rearrange to find T since the question wants to know something about the times: T = (X+5)/60

Transcribe the faster trip: X [same distance] = 60 mph*T1 > Solve for T1 = X/60

The difference between these times is T1-T = (X+5)/60 - X/60 = 5/60

The question asks for the percentage of the faster time this difference is. First find the decimal equivalent:

(5/60)/T1 = (5/60)/(X/60) = (5/60)*(60/X) = 5/X

5/X is the decimal answer. To put in terms of percent, multiply by 100 = [spoiler]500/X, E[/spoiler]

[GMATWisdom]

During a trip on an expressway, Dron 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 x-mile trip was what percent greater than it would have been if he has 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%

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

In the first case he has traveled first 5 miles at a speed of 30mph and in the second case
he travells same 5 miles at the speed of 60 miles per hour
therefore the the difference in travel time would be 5/30 - 5/60 = 5/60 = 1/12 hours.
travel time for distance X miles at the rate of 60 miles per hour = X/ 60 hours
Hence percentage increase in time = (1/12)*100/(X/60)= 500/X
Hence option E

[Scott@TargetTestPrep]

During a trip on an expressway, Dron 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 x-mile trip was what percent greater than it would have been if he has 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%

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

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

[Brent@GMATPrepNow]

During a trip on an expressway, Dron 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 x-mile trip was what percent greater than it would have been if he has 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%

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

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