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

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Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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03 Jul 2018, 07:18
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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!

Dons 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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Joined: 09 Mar 2016
Posts: 1234
Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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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!

Dons 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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Joined: 20 Feb 2015
Posts: 795
Concentration: Strategy, General Management
Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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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

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)%
Intern
Joined: 17 Jun 2018
Posts: 7
Re: During a trip on an expressway, Don drove a total of x miles  [#permalink]

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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%
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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