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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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02 Mar 2009, 17:27
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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. $$\frac{12,000}{(x+200)}$$
[Reveal] Spoiler: OA

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

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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

If wrong, let me know what assumptions were wrong.
Thanks,

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

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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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03 Mar 2009, 14:42
Thanks for the explanation guys.

I have one question for you: Can we pick a variable for total distance and solve the problem? What prompted you to pick 100 for total distance?

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

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03 Mar 2009, 15:19
vksunder wrote:
Thanks for the explanation guys.

I have one question for you: Can we pick a variable for total distance and solve the problem? What prompted you to pick 100 for total distance?

I arrived at 100 as in 100%. Question provides x percent at 40mph. Therefore 100-x is for 60mph. Picking a variable is also possible.

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

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03 Mar 2009, 15:56
I picked a variable: Y for total distance and my answer turns out to be - 120y/(2y+x)

Rate Time Distance
40 -- 40/x -- x
60 -- 60/(y-x) -- (y - x)

Total Dist/Total time = x+(y-x) / 40/x + 60/(y-x) = 120y/(2y+x)

Do you know where I'm messing up?

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

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03 Mar 2009, 16:15
vksunder wrote:
I picked a variable: Y for total distance and my answer turns out to be - 120y/(2y+x)

Rate Time Distance
40 -- 40/x -- x
60 -- 60/(y-x) -- (y - x)

Total Dist/Total time = x+(y-x) / 40/x + 60/(y-x) = 120y/(2y+x)

Do you know where I'm messing up?

Time should be:
x/40 and (y-x)/60

and replace y with 100 which helps you convert two unknown variables (x and y) into only one (x)

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

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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
[Reveal] Spoiler:
E
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

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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13 Jun 2010, 01:26
Is there any clear way of getting correct answer without plugging in different numbers ????????????????
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Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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13 Jun 2010, 04:40
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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: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}$$;

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

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

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

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

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

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

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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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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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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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).

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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02 Mar 2013, 10:50
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
bingo
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Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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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.

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

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07 Nov 2013, 10:37
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 distance at an [#permalink]

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27 May 2014, 13:14
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).
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Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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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)

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

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

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Re: During a trip, Francine traveled x percent of the total distance at an   [#permalink] 28 May 2014, 00:43

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