On a family vacation with his RV, Bill drove a total of x

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

On a family vacation with his RV, Bill drove a total of x

by AAPL » Mon Jul 22, 2019 5:53 am

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On a family vacation with his RV, Bill drove a total of \(x\) miles. He averaged 50 miles per hour for the entire trip except for a 10-mile section when he averaged only 20 miles per hour because of construction. His travel time for the \(x-\)mile trip was what percent greater than it would have been if he had been able to travel 50 miles per hour for the entire trip?

A. \(1.5\)%

B. \(15\)%

C. \(\frac{50}{x}\)%

D. \(\frac{300}{x}\)%

E. \(\frac{1500}{x}\)%

OA E

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Source: Beat The GMAT — Problem Solving | Mon Jul 22, 2019 5:53 am

swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

by swerve » Tue Jul 23, 2019 4:40 pm

Time would have taken if drove 50mph. \(\frac{x}{50}\)

Time taken when construction \(\frac{x-10}{50} + \frac{10}{20} \Rightarrow \frac{x+15}{50}\)

\(\frac{x+15}{50} - \frac{x}{50} = \frac{15}{50}\)

\(\frac{15}{50}\) divide by \(\frac{x}{50} \Rightarrow \frac{15}{x}\)

Multiplying by 100 to get percents \(\Rightarrow \frac{1500}{x}\) %. __E__

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue Jul 30, 2019 8:37 am

AAPL wrote:Veritas Prep

On a family vacation with his RV, Bill drove a total of \(x\) miles. He averaged 50 miles per hour for the entire trip except for a 10-mile section when he averaged only 20 miles per hour because of construction. His travel time for the \(x-\)mile trip was what percent greater than it would have been if he had been able to travel 50 miles per hour for the entire trip?

A. \(1.5\)%

B. \(15\)%

C. \(\frac{50}{x}\)%

D. \(\frac{300}{x}\)%

E. \(\frac{1500}{x}\)%

OA E

If Bill had traveled 50 mph for the entire trip, his time would have been x/50 hours. Since there was a 10-mile section where he had to travel 20 mph, his time for this section was 10/20 = 1/2 hour, and his time for the remaining (x - 10) miles was (x - 10)/50 = x/50 - 1/5 hours. Thus, his actual total time for the x-mile trip was 1/2 + x/50 - 1/5 = x/50 + 3/10 hours. We can determine the percent greater that this actual time is than what the time would have been if he had been able to travel 50 mph for the entire trip as: was:

[(x/50 + 3/10 - x/50)/(x/50)] * 100

[(3/10)/(x/50)] * 100

[150/(10x)] * 100

(15/x) *100

1500/x

Thus, the actual time was 1500/x percent greater than the time that he would have needed had he been able to travel 50 mph for the entire trip.

Answer: E

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