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During a trip, Francine traveled x percent of the total [#permalink]
02 Mar 2009, 16:27
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00:00
A
B
C
D
E
Difficulty:
55% (hard)
Question Stats:
70% (03:40) correct
30% (02:39) wrong based on 1096 sessions
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)
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)
Need to figure out (average speed=total distance/total time)
xy+(100y-xy) = total distance (xy/40)+((100y-xy)/60) = total time
take total distance divided by total time you get average speed Answer: E
If wrong, let me know what assumptions were wrong. Thanks,
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)
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) _________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
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)
Let's say the total mileage = 100 and x = 40% If x = 40 then it takes 1 hour at 40MPH Therefore, at 60MPH it takes another hour to go the rest of the distance
Add the numbers: 2r = 100 r = 50MPH
c. 300-40/5 = 52 e. 12000/240 = 50 of course B could be the answer as well
So if we were to reverse the numbers and set x = 60 then the answer is E. Sometimes you have to try different sets of numbers because initially two answers can be correct.
3/2r + 2/3r = 100 13/6 r = 100 r = 600/13 only E would win
Re: Aren't E and C both answers ? [#permalink]
13 Jun 2010, 03:40
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Expert's post
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hardnstrong wrote:
Is there any clear way of getting correct answer without plugging in different numbers ????????????????
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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);
Re: Aren't E and C both answers ? [#permalink]
13 Jun 2010, 22:38
Where does it say that total distance is 40 Distance is given as x % of total distance we only have speed, which is 40m/ph and 60m/ph _________________
Re: Aren't E and C both answers ? [#permalink]
14 Jun 2010, 00:01
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Expert's post
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hardnstrong wrote:
Where does it say that total distance is 40 Distance is given as x % of total distance we only have speed, which is 40m/ph and 60m/ph
Nowhere it's said that the total distance is 40 miles. I should have written this more clearly: distance \(d\) will cancel out from the equation (edited the earlier post to clear this). So we can assume distance to be some number. I chose 40 as it's easy for calculation.
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: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*d}{40}=\frac{dx}{40*100}\);
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);
Re: Aren't E and C both answers ? [#permalink]
16 Jun 2010, 09:52
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Bunuel wrote:
hardnstrong wrote:
Where does it say that total distance is 40 Distance is given as x % of total distance we only have speed, which is 40m/ph and 60m/ph
Nowhere it's said that the total distance is 40 miles. I should have written this more clearly: distance \(d\) will cancel out from the equation (edited the earlier post to clear this). So we can assume distance to be some number. I chose 40 as it's easy for calculation.
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: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*d}{40}=\frac{dx}{40*100}\);
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);
Re: Aren't E and C both answers ? [#permalink]
17 Aug 2010, 04:26
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Easier approach:
Formula
:When an body covers m part of journey at speed p and next n part of the journey at speed q then the Average speed of the total journey is: (m+n)*pq / (np+mq).
Using above formula: initial part of journey =x remaining part 100-x (since x is in percent) m+n=100
so we have => 100*40*60/x*60+(100-x)40 -> solves to 12000/x+200 -E is the Answer _________________
Consider giving Kudos if my post helps in some way
Re: Aren't E and C both answers ? [#permalink]
26 Dec 2010, 01:41
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a nice quick way of solving this question in under a min.
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 = 48.
So, plug 50 back into the choices for x, and look for 48... E works. _________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
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?
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?
Re: During a trip, Francine traveled x percent of the total [#permalink]
26 Sep 2012, 12:25
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vksunder wrote:
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)
If x = 0, the answer should give 60, as this would mean that Francine traveled the whole distance with the average speed of 60. We should choose between C and E. Similarly, for x = 100, the answer should give 40. We are still left with choices C and E.
For x = 50, let's say Francine traveled 2 * 120 miles, 120 with 40 mph and the other 120 miles with 60 mph. The average speed would be 240/(120/40 + 120/60) = 240/(3+2) = 48. Only E gives the correct answer (12,000/250 = 48, while (300 - 50)/5 = 50).
Answer E.
After I posted my reply, I saw that dimitri92 used a similar approach. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
gmatclubot
Re: During a trip, Francine traveled x percent of the total
[#permalink]
26 Sep 2012, 12:25
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