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During a trip, Francine traveled x percent of the total

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Senior Manager
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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
bingo
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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
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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Re: During a trip, Francine traveled x percent of the total   [#permalink] 16 Jun 2014, 16:42
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