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by moneyman » Sat May 10, 2008 6:18 am
Should I pick numbers for this question ? What is the best strategy ?

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 an 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 travelled at a constant rate of 60 miles per hour for the entire trip ?

8.5%
50%
x/12 %
60/x %
500/x %

Ans E

Diatance = Time*Rate

For the 5 mile section Time=1/6 or 10 mins

For the remainder Distance= (x-5) Rate=60 So Time = (x-5)/60 or (x-5)mins

Total time = (x-5)mins + 10 mins

Option 2

Distance = x Rate=60 so Time = x/60 or x mins

This comes down to how much more is x+5 > x ?

I know I am missing something here. PLease help!! Exam is on Monday
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by VP_Jim » Sat May 10, 2008 5:48 pm
Hi moneyman,

An easier way to look at this problem is to plug in numbers, so how about using 20 for X? With X = 20, Don took 10 minutes if he traveled 5 miles of the total distance at 30 miles/hour, plus 15 minutes for the rest of the trip at 60 miles/hour, for a total of 25 minutes.

However, if he traveled the entire distance at 60 miles/hour, it would only have taken him 20 minutes.

Thus, 25 minutes - 20 minutes = a difference of 5 minutes, and 5/20 = 25% (the difference between the two options). Now, find the answer that gives you this percentage, by plugging in the original number you chose for X (which is 20) and you'll see that the correct answer is choice E.

One last word of advice, no matter how good you are at algebra, you're better at arithmetic; so plug in whenever possible.

Good luck on your exam!
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by moneyman » Sat May 10, 2008 9:16 pm
Thanks a lot Jim..I am very confortable with using numbers but I thought I could save some time by using this approach..but guess it dint work quite well..
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by CITI29 » Sun May 11, 2008 9:24 pm
Time taken with two seperate speeds:
5/30 +x-5/60 =x+5/60

Time if the speed 60miles/hr was constant:
x/60

Difference in two times : x+5/60 -x/60 =5/60

%age (5/60) / x/60 *100
=500/x

[E]
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by gkumar » Wed Oct 21, 2009 6:01 am
I'd just assume x = 5.

Then for 30 miles per hour on a 5 mile road, he averaged 30 miles per hour. The remaining road is 0 miles so the total average speed is 30 miles per hour.

If he traveled 60 miles per hour on that 5 mile road, then his average speed would be 60 miles per hour. This is double the 30 miles per hour rate.

So I plug in 5 in answer choices to see if I get 100% anywhere. Voila, E is answer.
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by Scott@TargetTestPrep » Tue Dec 12, 2017 7:13 am
moneyman wrote:Should I pick numbers for this question ? What is the best strategy ?

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 an 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 travelled at a constant rate of 60 miles per hour for the entire trip ?

8.5%
50%
x/12 %
60/x %
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%

Answer:E

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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