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Re: During a trip, Francine traveled x percent of the total distance at an
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30 Dec 2017, 13:23
Thanks SO much niks18 for a fantastic explanation! Everything is clear! Highly appreciated !



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Re: During a trip, Francine traveled x percent of the total distance at an
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08 Jan 2018, 18:54
rate * time = distance
40_____ x/40_______________ = x
60_____ (100x)/60_________ = 100x
(100)/(x/40+[100x]/60)
after simplifying we get choice E.
A good idea is to assume the distance is 100.



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Re: During a trip, Francine traveled x percent of the total distance at an
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18 Apr 2018, 14:17
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. The solution with \(d\): \(Average \ speed=\frac{distance}{total \ time}\). 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: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*d}{40}=\frac{dx}{40*100}\); Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1\frac{x}{100})*d}{60}=\frac{(100x)d}{60*100}\); \(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100x)d}{60*100}=\frac{d(x+200)}{12000}\); \(Average \ speed=\frac{distance}{total \ time}=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{x+200}\). Answer: E. Hope it's clear. Hi Bunuel can you expain step by step(with comments) how from adding up two TIMES you get this \(\frac{d(x+200)}{12000}\) \(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100x)d}{60*100}=\frac{d(x+200)}{12000}\); thank you D.



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Re: During a trip, Francine traveled x percent of the total distance at an
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20 Jun 2018, 21:57
dimitri92 wrote: 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. Hello  anyone can help me explain the rationale of the formula in bold ?



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Re: During a trip, Francine traveled x percent of the total distance at an
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22 Jul 2018, 01:30
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. \(\frac{(180x)}{2}\)
B. \(\frac{(x+60)}{4}\)
C. \(\frac{(300x)}{5}\)
D. \(\frac{600}{(115x)}\)
E. \(\frac{12,000}{(x+200)}\) Assume total distance = 100 miles so, x percent = 100*(x/100) = x, the rest is 100x Time 1 = x/40 [Time * speed = Total distance, i.e, TS = D, T = D/S] Time 2 = (100x)/60 Now, average speed = Total Distance/ Total time 100/[(x/40)+(100x/60)] = 12000/(x+200) Ans. E
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Re: During a trip, Francine traveled x percent of the total distance at an
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02 Sep 2018, 00:56
i just assumed total distance to be 100. We can assume distance here because we are only bothered about x% and remaining distance. The total distance can be assumed here in this case as distance will cancel out So x% of 100 at 40 mph . Distance covered will be x Time taken will be x/40 Remaining distance : 100x since (100x)% of 100 will be 100x Speed = 60mph Time taken = (100x)/60 Avg speed= total distance/ total time So Total distance = 100x+x =100 Avg speed = 100/[(x+40) +(100x/60)] Simplify and you will get the answer
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Re: During a trip, Francine traveled x percent of the total distance at an
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16 Nov 2018, 01:02
dimitri92 wrote: 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. its nowhere mentioned the distances are same?



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Re: During a trip, Francine traveled x percent of the total distance at an
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10 Feb 2019, 23:52
Solution:Here, we can solve the question in two steps: Step 1: Find the total time taken (T). Step 2: Find the average speed of the entire trip. (Avg Speed) Finding total time taken (T). Let’s take the total distance as “D”. Attachment:
image2.PNG [ 25.84 KiB  Viewed 620 times ]
So, the correct answer option is “E”.
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Re: During a trip, Francine traveled x percent of the total distance at an
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03 Sep 2019, 08:47
This is a poorly worded gmat question. It says "x percent" if we put in the formula say 6% of the distance then it nonsensically becomes wrong.



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Re: During a trip, Francine traveled x percent of the total distance at an
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13 Sep 2019, 18:20
Hi All, We're told that 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. We're asked, in terms of X, what was Francine's AVERAGE SPEED was for the entire trip. This question can be approached in an number of different ways  and can be solved rather easily by TESTing VALUES. IF.... Total Distance = 100 miles and X = 40.... 40% of 100 = 40 miles, so Francine traveled 40 miles at 40 miles/hour > 1 hour traveled The rest = 60 miles, so Francine traveled 60 miles at 60 miles/hour > 1 hour traveled Total Distance traveled = 100 miles Total Time traveled = 2 hours Average Speed = 100/2 = 50 miles/hour So, we're looking for an answer that equals 50 when X = 40. There's only one answer that matches... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: During a trip, Francine traveled x percent of the total distance at an
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20 Oct 2019, 14:19
VeritasKarishma wrote: 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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) E. 12,000/(x+200) Responding to a pm: Quote: is this a good formula for different distances? so if i learn just these two (first one given in my pm above), then I can pretty much solve anything...is that right? how will this formula change for three different average speeds?
Again, I do not encourage the use of formulas. You will need to learn many formulas to cover various different scenarios and even then you can not cover all. Say, overall distance is 100. So, distance covered at speed 40 is x. So distance covered at speed 60 will be 100x Avg Speed = Total Distance/Total Time \(= \frac{100}{\frac{x}{40} + \frac{100x}{60}}= \frac{100*40*60}{60x + 40(100x)}\) (same as given formula) You might have to take one step extra here but it makes much more sense than learning up every formula you come across and then getting confused whether the formula will work in a particular situation or not. can i apply the alligation formula in this question



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Re: During a trip, Francine traveled x percent of the total distance at an
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27 Oct 2019, 15:04
Can someone help with one aspect:
In other posts, people have been saying that the average speed is 48 m.p.h by showing this math:
(2)(40)(60)/(40+60)
How is 2 X 40 X 60 the total distance? Where do the individual numbers come from?
Thanks



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Re: During a trip, Francine traveled x percent of the total distance at an
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27 Oct 2019, 19:39
Hi mborski, In this question, the Average Speed will depend on the total distance that the car travels at each of the two speeds (40 miles/hour and 60 miles/hour). The ONLY time that the Average Speed will be exactly 48 miles/hour is when the car travels the SAME distance at each of those two speeds. Under all other situations, the Average Speed will be something OTHER than 48 miles/hour. GMAT assassins aren't born, they're made, Rich
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Re: During a trip, Francine traveled x percent of the total distance at an
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