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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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New post Updated on: 17 Jul 2017, 06:44
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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 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 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 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 13 Jun 2010, 23:38
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
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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 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 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 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 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 16 Jun 2014, 17:42
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)


Plug in is the best approach when you find variables in the choices.

1. The average speed is given by (d1+d2) / (d1/40 + d2/60). We see if we substitute 40 for d1 and 60 for d2 we get 50 as the average speed

2. Choice E gives the same value of average speed for the above assumed value of d1 i.e x
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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 31 Aug 2014, 06:38
Bunuel wrote:
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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

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

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

Answer: E.


How do i get (100-x)/150 ? I don't know how to solve the fraction.
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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 01 Sep 2014, 02:11
lou34 wrote:
Bunuel wrote:
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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

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

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

Answer: E.


How do i get (100-x)/150 ? I don't know how to solve the fraction.


\(\frac{(1-\frac{x}{100})*40}{60}=\frac{(\frac{100-x}{100})*40}{60}=\frac{100-x}{100*60}*40=\frac{100-x}{150}\)
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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 29 Nov 2015, 13:14
Given Info: Average Speed for some distance (x% of total distance) in a trip is given, and the average speed for the remaining distance is also given. We have to find average speed for entire trip

Interpreting the Problem: We are given average speed for some distances. Based on that we can calculate the total time taken for these distances, and then divide the total distance traveled by the total time calculated we will be able to calculate the average speed for the entire trip.

Solution:
Let the total distance be a

Distance travelled at a speed of 40 miles per hour = x%*a

Time taken to travel x% of total distance=\(\frac{xa}{100*40}\)

Time taken will be \(\frac{xa}{4000}\)

Distance traveled at a speed of 60 miles per hour = (100-x%)a.

Time taken to travel (100-x)% of total distance=\(\frac{(100-x)a}{100*60}\)

Time taken will be \(\frac{((100-x)a)}{6000}\)

Average speed= \(\frac{TotalDistanceTraveled}{TotalTimeTaken}\)

Total distance traveled =a

Total time taken = \(\frac{xa}{4000} + \frac{((100-x)a)}{6000}\)

Total time taken = \(\frac{xa+200a}{12000} = \frac{a(x+200)}{12000}\)

Average speed for entire trip = \(\frac{a}{(a(x+200)/12000)}\)

Average speed =\(\frac{12000}{(x+200)}\)

Hence, The answer is E

Key Takeaways:


Always remember that the average speed is total distance traveled/total time taken. It is never the average of the speeds for various distances.
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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
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 06 Jun 2016, 05:58
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)



Hello, I found assuming total distance to be 100 much easier.

D1= x and D2= 100-x

Average speed = Total distance /total time taken
= 100/(T1 + T2) where T1= D1/S1=x/40 and T2= D2/S2=(100-x)/60

or, Average Speed = 100/[(x/40)+(100-x)/60] or 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 23 Aug 2016, 13:39
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!
Re: During a trip, Francine traveled x percent of the total distance at an &nbs [#permalink] 23 Aug 2016, 13:39

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