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During a trip, Francine traveled x percent of the total distance at an

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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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New post 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  [#permalink]

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New post 08 Jan 2018, 18:54
rate * time = distance

40_____ x/40_______________ = x

60_____ (100-x)/60_________ = 100-x

(100)/(x/40+[100-x]/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  [#permalink]

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New post 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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);

\(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}\);

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

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

\(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}\);

thank you :-)

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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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New post 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  [#permalink]

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New post 22 Jul 2018, 01:30
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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. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)


Assume total distance = 100 miles
so, x percent = 100*(x/100) = x, the rest is 100-x
Time 1 = x/40 [Time * speed = Total distance, i.e, TS = D, T = D/S]
Time 2 = (100-x)/60

Now, average speed = Total Distance/ Total time

100/[(x/40)+(100-x/60)]
= 12000/(x+200)
Ans. E
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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New post 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 : 100-x since (100-x)% of 100 will be 100-x
Speed = 60mph
Time taken = (100-x)/60

Avg speed= total distance/ total time
So Total distance = 100-x+x =100
Avg speed = 100/[(x+40) +(100-x/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  [#permalink]

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New post 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  [#permalink]

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New post 10 Feb 2019, 23:52
1
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
image2.PNG [ 25.84 KiB | Viewed 537 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  [#permalink]

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New post 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  [#permalink]

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New post 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:

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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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New post 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. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
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 100-x

Avg Speed = Total Distance/Total Time \(= \frac{100}{\frac{x}{40} + \frac{100-x}{60}}= \frac{100*40*60}{60x + 40(100-x)}\) (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  [#permalink]

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New post 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  [#permalink]

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New post 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.

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Re: During a trip, Francine traveled x percent of the total distance at an   [#permalink] 27 Oct 2019, 19:39

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