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

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

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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. \(\frac{(180-x)}{2}\)

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

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

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

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

Originally posted by vksunder on 02 Mar 2009, 17:27.
Last edited by Bunuel on 17 Jul 2017, 06:44, edited 1 time in total.
Renamed the topic and edited the question.
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During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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New post 13 Jun 2010, 04:40
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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.
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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 26 Dec 2010, 02:41
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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.
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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 17 Aug 2010, 05:26
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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
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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 14 Jun 2010, 01:01
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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.

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.
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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 28 May 2014, 00:43
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Look at the diagram below:

Setting up the equation (We require to find value of s)

\(\frac{x}{40} + \frac{100-x}{60} = \frac{100}{s}\)

\(s = \frac{12,000}{x+200}\)

Answer = 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 Oct 2013, 16:23
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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)


Don't be scared of plugging in numbers. Sometimes it is just the most straightforward way to solve.

For this problem, assume Distance = D = 240
Then assume x = 50

Half of the distance = 120 @ 40mph = 3 hrs
The other half = 120 @ 60mph = 2 hrs

Then avg. speed = total distance / total time = 240 / 5 =48 (this is our target)

Answer choice (E) just replace x with 50.

So we get 240/5 = 48. Bingo.

This is our answer (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 28 May 2010, 06:48
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dimitri92 wrote:
I am really confused after looking at this question. It seems like both C and E, satisfy the conditions. What do you guys think. The OA is but I think C is good enough too

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
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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 26 Sep 2012, 13:25
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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.
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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 09 Dec 2014, 23:33
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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)


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.
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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 Mar 2009, 20:33
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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)


Alt method:

shortcut: avg speed = 1/(d1/v1 + d2/v2)

d1 = x*d/100; d2 = (1 - x/100)*d; v1 = 40, v2 = 60 --> 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 10 May 2016, 14:23
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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)



This problem can be solved by using a Rate-Time-Distance table. We are given that Francine traveled x percent of the distance at a rate of 40 mph.

Since we are working with percents, and 100% is the total distance percentage, we can say that (100 – x) percent = the percentage of the remaining distance. Thus we know that Francine traveled (100 – x) percent of the distance traveled, at a rate of 60 mph.

Since we are working with percents, we can choose a convenient number for the total distance driven; we'll use 100 miles.

Let’s fill in the table.

Image

Remember, time = distance/rate, so we use the entries from the chart to set up the times:

Time for x percent of the distance = x/40

Time for (100 – x) percent of the distance = (100 – x)/60

Finally, we must remember that average rate = total distance/total time. Our total distance is 100. The total time is the sum of the two expressions that we developed in the previous steps. Here is the initial setup:

100/[(x/40 + (100 – x)/60)]

Now work with the fractions in the denominator, getting a common denominator so that they can be added:

100/[(3x/120 + (200 – 2x)/120)]

100/[(200 + x)/120)]

This fraction division step requires that we invert and multiply:

100 x 120/(200 + x)

12,000/(200 + x)

Answer 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 24 Aug 2016, 11:08
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Mariwa wrote:
Why can't I apply the simple average method here?

Average Speed=40*X/100+60*(100-X)/100

That leads to answer C. It makes sense to me.

Please help!


The average speed is always the total distance, divided by the total time.

When you multiplied 40*x and 60*(100-x), that actually doesn't make sense, mathematically. You're multiplying a rate (40 miles) by a percent of distance (x% of the total distance). That doesn't give you anything meaningful. Always check before you do the math: what am I taking a percent of?

Here's the version of your formula that would work:

total distance = d
total time for first half of trip = 40/(x/100 * d)
total time for second half of trip = 60/((100-x)/100*d)

average speed = total distance/(total time for first half + total time for second half)

If you try writing that out, though, you'll see why picking numbers is actually the right choice here!
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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 Mar 2009, 20:17
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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)


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.
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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 23 Aug 2016, 13:39
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Why can't I apply the simple average method here?

Average Speed=40*X/100+60*(100-X)/100

That leads to answer C. It makes sense to me.

Please help!
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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 24 Aug 2016, 11:27
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Mariwa wrote:
ccooley wrote:
Mariwa wrote:
Why can't I apply the simple average method here?

Average Speed=40*X/100+60*(100-X)/100

That leads to answer C. It makes sense to me.

Please help!


The average speed is always the total distance, divided by the total time.

When you multiplied 40*x and 60*(100-x), that actually doesn't make sense, mathematically. You're multiplying a rate (40 miles) by a percent of distance (x% of the total distance). That doesn't give you anything meaningful. Always check before you do the math: what am I taking a percent of?

Here's the version of your formula that would work:

total distance = d
total time for first half of trip = 40/(x/100 * d)
total time for second half of trip = 60/((100-x)/100*d)

average speed = total distance/(total time for first half + total time for second half)

If you try writing that out, though, you'll see why picking numbers is actually the right choice here!


I got it thanks!

This is applicable to all problems concerning rates? No only the ones about speed right?


Any time a problem says 'average speed' or 'average rate', start by writing out 'total work' (or 'total distance') / 'total time' on your paper. Then, just fill in those two values as you work them out! That said, I can't remember ever seeing an official GMAT problem that asked for an average rate, rather than an average speed. But there's always a chance, and in that case, you'd do it exactly like this problem.
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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 17 Jul 2017, 05:07
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suntorytea wrote:
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.


How do you get 12000/(x + 2)?
You don't need to assume a value for d.

Average Speed = Total Distance / Total Time

Total Distance = d

Time 1 \(= \frac{D1}{S1} = \frac{(\frac{x}{100})*d}{40} = \frac{xd}{4000}\)

Time 2 \(= \frac{D2}{S2} = \frac{(1 - x/100)*d}{60} = \frac{(100 - x)d}{6000}\)

Average Speed \(= \frac{d}{\frac{xd}{4000} + \frac{(100-x)d}{6000}}\)

We take d common from the denominator and it cancels out with the d in the numerator

Average Speed \(= \frac{d}{d * (\frac{x}{4000} + \frac{(100-x)}{6000})}\)

Average Speed \(= \frac{1}{(\frac{x}{4000} + \frac{(100-x)}{6000})}\)

Average Speed \(= \frac{12000}{x + 200}\)
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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 09 Sep 2017, 18:56
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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. \(\frac{12,000}{(x+200)}\)


Answer: option E

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

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New post 28 Dec 2017, 06:24
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Bunuel, niks18, amanvermagmat

Quote:
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.


I hope you meant d as TOTAL DISTANCE

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


For above, I am getting:


\(t_2= \frac{distance_2}{speed_2}=\frac{d-xd}{60*100}\);

To be more precise, see VeritasPrepKarishma solution above:
x% of d = xd/100

then complimentary distance will be
(1-x)d/100 or (d-xd)/100

Can you advise what mistake I am making?
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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
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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:
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image2.PNG [ 25.84 KiB | Viewed 619 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] 10 Feb 2019, 23:52

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