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During a trip, Francine traveled x percent of the total

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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 02 Mar 2013, 09:50
let the whole distance be 100, and x=40

then we got that the 1st distance took 1 h ( distance =40%*100=40 . time = distance/speed =40/40=1)
the 2nd distance also took 1 h (distance =100-40=60 ; time = 60/60=1)

so, average speed = total distance/total time =100/2 =50


lets plug in answer choice E
12000/ (40+200)=50
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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 03 Oct 2013, 15:23
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)


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: Aren't E and C both answers ? [#permalink] New post 30 Oct 2013, 08:03
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.



Quick and easy! Thank you
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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 07 Nov 2013, 09:37
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?

Average speed = total distance/total time.

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

d1 = (percentage traveled on first leg of journey)*d/100(where 100 represents total distance)
d2 = (100%-percentage traveled on first leg of journey)/100(where 100 represents total distance)*d

v1=average speed for first portion of journey
v2=average speed for second portion of journey

In this problem, what is d?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)
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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 27 May 2014, 12:14
I think the easiest way to do this problem is to assume distance is 100.

So X= distance traveled at 40mph and (100-x)=distance traveled at 60mph

Let A = average speed

Time required to travel total distance = time required to travel x + time required to travel (100-x)
100/A = x/40 + (100-x)/60 ===> A = 12,000 / ( x + 200).
Answer E
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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 27 May 2014, 23:43
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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 [#permalink] New post 11 Jun 2014, 05:41
Pick x = 0
Avg speed we need it then 60

Pick x= 100
Avg speed we need is 40

We are now down to C & E

Pick x= 50
Avg speed we need is 48

E is the right answer

Hard problem - Would take 3 mins to solve
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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 16 Jun 2014, 16:42
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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During a trip, Francine traveled x percent of the total [#permalink] New post 31 Aug 2014, 05:38
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 [#permalink] New post 01 Sep 2014, 01:11
Expert's post
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 [#permalink] New post 12 Sep 2014, 21:36
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)



I have used plug in method..
let the total distance be 100 & X=10

time for travelling x = 1/4 ; time for travelling remaining = 90/60 = 3/2

t= t1+t2

1/4 +3/2 = 7/4

D=ST ==> 100 = (7/4)s



s= 400/7

apply x= 10 in options .. close call between C & E but E Wins...
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During a trip, Francine traveled x percent of the total [#permalink] New post 07 Oct 2014, 03:32
The approach I followed is to assume total distance as 100 miles.
Then x% of total distance is x miles, at the speed of 40 miles per hour and the remaining distance is 100-x.

Formula is Speed = Total Distance/Total Time

T1 = X/40
T2 = 100-X/50
T1 + T2 = X+200/120

Thus, Speed = 100/(X+200/120) and that leads to the answer 12,000/(x+200)


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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 08 Oct 2014, 03:00
I assumed x=20.
so remaining distance =80.
first part= s=d/t. put in speed as 40 (given in the question), distance= 20. find out time. It comes out to be 1/2.

Second part distance= 80. speed =60. Time will now be t=4/3

average speed for the whole journey is total distance / total time.

That's 100/(11/6)

simplifies to 600/11, which is equal to option E. Start with option C first and see if your target answer (in this case, 600/11) is less than or greater than C. Then decide whether you need to go higher or lower.

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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 09 Dec 2014, 22:33
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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 [#permalink] New post 10 Dec 2014, 22:57
Let Francine travel 40 miles/hr for 1 hour and the rest of the distance at 60 miles/hr for another 1 hour.

40-------------50--------------60

Using time of travel as weight (1:1), the weighted average is calculated as 50 miles/hr.

x = 40% as Francine covers 40 miles of 100 miles

Plugging x as 40/100, option E yields 50 miles/hr.

Answer: E
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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 16 May 2015, 23:00
xyz21 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)


Alt method:

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

d1 = x*d/100; d2 = (1 - x/100)*d; v1 = 40, v2 = 60 --> E



CORRECTION :

avg speed = d/(d1/v1 + d2/v2)
Re: During a trip, Francine traveled x percent of the total   [#permalink] 16 May 2015, 23:00

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