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During a trip on an expressway, Don drove a total of x miles

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Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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New post 03 Jul 2018, 07:18
1
dave13 wrote:
ludwigvb 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%



hey niks18

here is my solution to the problem :-) i got confused in the end, can you pls advise where did i go wrong :-) thanks! :-)

Don`s distance X miles
Ave speed on 5 miles section - 30mph
Ave speed for remainder of trip - 60mph


Time 1 = \(\frac{5}{30}\)

Time 2 = \(\frac{x-5}{60}\)

Total Time = \(\frac{5}{30}\) + \(\frac{x-5}{60}\) = \(\frac{5+x}{60}\)

total speed = total distance / total time

Total Average Speed = x / (50+x) /60 = \(\frac{x}{1} *\frac{60}{5+x}\) = \(\frac{60x}{5+x}\)

Now question asks: 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?

so total average speed / total average speed 60 for entire trip

60x/(5+x) / 60 =\(\frac{x}{5+x}\) :?


Hi dave13

You are asked % of "Time" but you are trying to find out average speeds. I suggest you read the question few times more and try to break it. Find out the time take to travel the entire x miles had the speed been constant at 60mph.

If you are still stuck let me know.
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Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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New post 03 Jul 2018, 08:12
niks18 wrote:
dave13 wrote:
ludwigvb 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%



hey niks18

here is my solution to the problem :-) i got confused in the end, can you pls advise where did i go wrong :-) thanks! :-)

Don`s distance X miles
Ave speed on 5 miles section - 30mph
Ave speed for remainder of trip - 60mph


Time 1 = \(\frac{5}{30}\)

Time 2 = \(\frac{x-5}{60}\)

Total Time = \(\frac{5}{30}\) + \(\frac{x-5}{60}\) = \(\frac{5+x}{60}\)

total speed = total distance / total time

Total Average Speed = x / (50+x) /60 = \(\frac{x}{1} *\frac{60}{5+x}\) = \(\frac{60x}{5+x}\)

Now question asks: 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?

so total average speed / total average speed 60 for entire trip

60x/(5+x) / 60 =\(\frac{x}{5+x}\) :?


Hi dave13

You are asked % of "Time" but you are trying to find out average speeds. I suggest you read the question few times more and try to break it. Find out the time take to travel the entire x miles had the speed been constant at 60mph.

If you are still stuck let me know.



niks18 thank you :)

i got it. just one thing i dont get

if this is total time in case car travelled 5 miles section at 30 mph. \(\frac{5+x}{60}\) that means that it took more time to do the whole distance compared the case if average speed during whole trip would be 60 mph.


so I divide x/60 by \(\frac{5+x}{60}\) ...but i see in other solutions vice versa ...why ? :?


when we need to calculate percetage we chhose smaller value in numerator and bigger value in denominator ... for example what percetage is 4 of 6 hence i divide 4 by 6 and result multiply by 100% :? :)
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Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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New post 03 Jul 2018, 10:20
ludwigvb 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%



average speed on the 5 mile section =30 miles per hour =5/t1
t1=5/30 =1/6 hour

where t1 is the time taken by Don to cover the 5 mile section

let t2 be the time taken by don to cover the remaining
then ,
t2=x-5/60
total time taken t=t1+t2=1/6+(x-5)/60=(x+5)/60
additional time taken =5/60=1/12

If Don had driven at 60 miles per hour t3=x/60

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?

ratio = (1/12)*100/(x/60)=(500/x)%
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Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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New post 04 Jul 2018, 20:46
At the beginning, I spent a lot of time on trying to figure out the time it takes for the part before the 5 miles section, because I think his journey contains 3 part--the first section before 5 miles section, 5 miles section and the remaining section. The question is not given any information regarding to the time, distance or speed for the first section. But later on, I realize that we don't even need the any information about the first section. lol

Original time it takes to complete the trip: X/60 hour
t1: time needs to complete 5 miles section with speed 30 miles / hour, which is 5/30 hour
t2: time needs to complete 5 miles section with speed 60 miles / hour, which is 5/60 hour
The actual time it takes to complete the trip = original time - t2 + t3 = x/60 - 5/60 + 5/30 = (X+5)/60
Percentage Change:
[(x+5)/60-x/60]/(x/60) = 500/x%
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Re: During a trip on an expressway, Don drove a total of x miles &nbs [#permalink] 04 Jul 2018, 20:46

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