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During a trip on an expressway, Don drove a total of x miles. His aver
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16 Nov 2021, 08:21
16 Nov 2021, 08:21
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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%
A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%
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Joined: 10 Apr 2015
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Re: During a trip on an expressway, Don drove a total of x miles. His aver
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16 Nov 2021, 09:40
16 Nov 2021, 09:40
GeminiHeat wrote:
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%
A. 8.5%
B. 50%
C. x/12%
D. 60/x%
E. 500/x%
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
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During a trip on an expressway, Don drove a total of x miles. His aver
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16 Nov 2021, 11:14
16 Nov 2021, 11:14
With a question like this it will most likely be easier to think about it in terms of plugging numbers in; make the distance something simple, like x = 60 miles. I'll suggest making the time units minutes as that won't matter toward the final percentage difference in time.
Then if we travel at 60 miles/hr we go 1 min/mile, and at 30 miles/hr we go 2 min/mile.
Bearing that in mind, remember that Average Rate is Total Distance over Total Time; we'll need to calculate this for the altered value (where we change speeds), but it will be obvious for the non-altered value because if we go 60 miles that's simply 60 minutes. Therefore, t_1 = 60 minutes.
Take the average rate for the altered case: r_2 = 60 miles / t_2
We need to figure out what t_2 is: that will be 5 miles at 30 miles/hr, or 2 min/mi, so 10 minutes; add this to 55 miles at 1 min/mi or 55 minutes. That's 65 minutes total for t_2.
We're looking for the percentage change between t_2 and t_1. The simplest thing to do is simply t_2 / t_1 and figure out how much greater than 100% it is.
That is, (65-60)/60 * 100 = 5/60 * 100 for the percentage change. Now remember that we defined 60 miles as the x at the very beginning, so it's time to plug that x back in.
If we sub 60 for x, we get 500/x, which is Answer D.
Then if we travel at 60 miles/hr we go 1 min/mile, and at 30 miles/hr we go 2 min/mile.
Bearing that in mind, remember that Average Rate is Total Distance over Total Time; we'll need to calculate this for the altered value (where we change speeds), but it will be obvious for the non-altered value because if we go 60 miles that's simply 60 minutes. Therefore, t_1 = 60 minutes.
Take the average rate for the altered case: r_2 = 60 miles / t_2
We need to figure out what t_2 is: that will be 5 miles at 30 miles/hr, or 2 min/mi, so 10 minutes; add this to 55 miles at 1 min/mi or 55 minutes. That's 65 minutes total for t_2.
We're looking for the percentage change between t_2 and t_1. The simplest thing to do is simply t_2 / t_1 and figure out how much greater than 100% it is.
That is, (65-60)/60 * 100 = 5/60 * 100 for the percentage change. Now remember that we defined 60 miles as the x at the very beginning, so it's time to plug that x back in.
If we sub 60 for x, we get 500/x, which is Answer D.
