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During a trip, Francine traveled x percent of the total

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During a trip, Francine traveled x percent of the total [#permalink] New post 02 Mar 2009, 16:27
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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. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)
[Reveal] Spoiler: OA
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Re: Average speed [#permalink] New post 02 Mar 2009, 19:17
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.
Thanks,
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Re: Average speed [#permalink] New post 02 Mar 2009, 19:33
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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: Average speed [#permalink] New post 03 Mar 2009, 13:42
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: Average speed [#permalink] New post 03 Mar 2009, 14:19
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: Average speed [#permalink] New post 03 Mar 2009, 14:56
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: Average speed [#permalink] New post 03 Mar 2009, 15:15
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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Aren't E and C both answers ? [#permalink] New post 28 May 2010, 02:07
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)
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Re: Aren't E and C both answers ? [#permalink] New post 28 May 2010, 05:48
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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

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: Aren't E and C both answers ? [#permalink] New post 13 Jun 2010, 00:26
Is there any clear way of getting correct answer without plugging in different numbers ????????????????
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Re: Aren't E and C both answers ? [#permalink] New post 13 Jun 2010, 03:40
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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}.

Answer: E.
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Re: Aren't E and C both answers ? [#permalink] New post 13 Jun 2010, 22:38
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: Aren't E and C both answers ? [#permalink] New post 14 Jun 2010, 00:01
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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: Aren't E and C both answers ? [#permalink] New post 14 Jun 2010, 21:09
Thanks .......its clear now
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Re: Aren't E and C both answers ? [#permalink] New post 16 Jun 2010, 09:52
Bunuel wrote:
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.



Def. the best method. I tried plugging numbers and got in trouble doing that.
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Re: Aren't E and C both answers ? [#permalink] New post 17 Aug 2010, 04:26
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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: Aren't E and C both answers ? [#permalink] New post 26 Dec 2010, 01:41
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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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RATE DISTANCE [#permalink] New post 26 Sep 2012, 10:46
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)
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Re: RATE DISTANCE [#permalink] New post 26 Sep 2012, 11:38
Expert's post
ishanand 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)


Merging similar topics. Please refer to the solutions above.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: During a trip, Francine traveled x percent of the total [#permalink] New post 26 Sep 2012, 12:25
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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. (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   [#permalink] 26 Sep 2012, 12:25
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