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During a trip, Francine traveled x percent of the total dist

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During a trip, Francine traveled x percent of the total dist [#permalink] New post 26 Aug 2011, 12:28
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During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)

[Reveal] Spoiler: Solution
First, we should assume x = 50, both distances are the same. To find the average speed over the same distance, the equation is: 2*s1*s2/(s1+s2). In this case, that's 2*40*60/100, which is 48.

So, plug 50 back into the choices for x, and look for 48... E works.


OPEN DISCUSSION OF THIS QUESTION IS HERE: during-a-trip-francine-traveled-x-percent-of-the-total-94933.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Jun 2014, 10:09, edited 1 time in total.
Renamed the topic and edited the question.
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Re: During a trip, Francine traveled x percent of the total dist [#permalink] New post 26 Aug 2011, 16:09
distance time

xd/100 40 mph

(100-x)(d/100) 60 mph


Average speed = total distance/ total time

= d/ \((xd/4000)+(((100-x)d)/6000)\)

= 1000 /\(((x/4) + ((100-x)/6))\)

= 12000/(x+200)

Answer is E.
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Re: During a trip, Francine traveled x percent of the total dist [#permalink] New post 26 Aug 2011, 23:39
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Total time taken = \(\frac{xd}{100*40} + \frac{(100-x)d}{100*60}\)

\(= \frac{d(200+x)}{12000}\)

Avg speed = \(\frac{d*12000}{d(200+x)}\)

\(= \frac{12000}{x+200}\)
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Re: During a trip, Francine traveled x percent of the total dist [#permalink] New post 27 Aug 2011, 05:28
Say the total distance is 100 miles and x is 60

60 miles w/ 40m/h => 60/40 = 3/2 hours
40 miles w/ 60m/h => 40/60 = 2/3 hours

In total = 3/2 + 2/3 = 13/6 hours
Total distance = 100

Avg. speed: 100/(13/6) = 10 * (6/13) = 600/13

Check the answers by plugging in x = 60 and see that for E it is 12000/260, cancel with 2 and 10 and get 600/13 => E it is
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Re: During a trip, Francine traveled x percent of the total dist [#permalink] New post 27 Aug 2011, 11:01
Should be E.

Let distance be 1.

Then avg speed will be 1/ (x/40+1-x/60) = 1200/x+200
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Re: During a trip, Francine traveled x percent of the total dist [#permalink] New post 30 Jun 2014, 09:51
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Re: During a trip, Francine traveled x percent of the total dist [#permalink] New post 30 Jun 2014, 10:11
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During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?

A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

Algebraic approach:

\(Average \ speed=\frac{distance}{total \ time}\), let's assume \(distance=40\) (distance \(d\) will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.

Francine traveled \(x\) percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}\);

Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}\);

\(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\).

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: during-a-trip-francine-traveled-x-percent-of-the-total-94933.html
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Re: During a trip, Francine traveled x percent of the total dist   [#permalink] 30 Jun 2014, 10:11
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