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During a trip, Francine traveled x percent of the total distance at an

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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 27 Feb 2017, 03:04
Shortcut: x =0 answer should be 60 and x =100 answer should be 40. After checking the option we can see that Option C and E could be the answer.
Now avg speed = 1/(d1/v1 + d2/v2). Since distance part is coming in denominator, x will be in the denominator. Hence Option 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 01 May 2017, 23:23
Consider x% as 50% means 1/2 of total distance covered with an average speed of 40 mph and another half at 60.
Average speed will be 2 *40*60 /(40 + 60) =48

putting x=50 back in option E gives 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 29 May 2017, 00:26
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)


1. Since the distance traveled is given in terms of percent, assume the total distance traveled as 100 units.
2. A distance of x at 40 miles per hour and a distance of 100-x at 60 miles per hour
3. Average speed = 100 / total time taken
4. Total time taken = x/40 + (100-x)/60
5. Substituting (4) in (3) we get, Average speed= 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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New post 29 Jun 2017, 08:38
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)


Though I tried to solve it by formula.. But really if we try to apply our mind, we can easily solve the problem without solving using complex x and y ...

So the solution goes like this..
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

x% with 40 miles/hr
(100-x) with 60 miles/hr

So, if we assume x as 0% i.e. whole distance is traveled with speed of 60 miles/hr.
Now start checking options .
A. 90
B. 15
C. 60 (May be correct).. Always check all options when we are solving through options and assuming the values.
D. 600/115
E. 60

So we can eliminate A,B,D.

So C or E is correct.

Lets assume x as 100% i.e. whole distance is traveled with 40 miles/hr.
Now start checking option C and E.
C. 200/5 = 40
E. 12000/300 = 40

So this was wrong assumption..

Lets assume x= 50% i.e. the average speed = 2x40x60/(40+60) = 48 miles/hr

Lets check the option C & E again
C. 250/5 = 50 miles /hr (It is wrong )
E. 12000/250 = 48 miles/hr (correct option )


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 15 Jul 2017, 11:19
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)



t1 = x/40
t2 = 100-x/60

t1 + t2 = 200 + x/120

average spped: 100 * 120/200+x
=12000/200+x

hope this helps ..
cheers ....
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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 Jul 2017, 15:37
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.
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Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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suntorytea wrote:
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.


How do you get 12000/(x + 2)?
You don't need to assume a value for d.

Average Speed = Total Distance / Total Time

Total Distance = d

Time 1 \(= \frac{D1}{S1} = \frac{(\frac{x}{100})*d}{40} = \frac{xd}{4000}\)

Time 2 \(= \frac{D2}{S2} = \frac{(1 - x/100)*d}{60} = \frac{(100 - x)d}{6000}\)

Average Speed \(= \frac{d}{\frac{xd}{4000} + \frac{(100-x)d}{6000}}\)

We take d common from the denominator and it cancels out with the d in the numerator

Average Speed \(= \frac{d}{d * (\frac{x}{4000} + \frac{(100-x)}{6000})}\)

Average Speed \(= \frac{1}{(\frac{x}{4000} + \frac{(100-x)}{6000})}\)

Average Speed \(= \frac{12000}{x + 200}\)
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Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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New post 17 Jul 2017, 04:19
Time taken = Distance Travelled/Speed = x/40 + 100-x/60 = 60x + 4000 - 40x/2400 = 20x+4000/2400 = x+200/120.

Average speed = Total distance/Total time taken = 120*100/x+200 = 12000/x+200. Ans - 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 17 Jul 2017, 05:27
suntorytea wrote:
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.



hi

whenever you find any PS or DS pertaining to percents, you can safely assume 100 as total. i can guarantee it will save you time and will prevent you from messy calculations ...
i have found this rule in many books ..

thanks `
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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 24 Aug 2017, 11:17
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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. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)


I like to begin with a word equation:
Average speed = (total distance)/(total time)
For this question, let's let the total distance = D

Next, observe that: total time = (time spent driving 40 mph) + (time spent driving 60 mph)

time spent driving 40 mph = distance/speed
Aside: distance driven = (x/100)(D)
So, time spent driving 40 mph = (x/100)(D)/40


time spent driving 60 mph = distance/speed
Aside: if x% of the distance was driven at 40 mph, then the distance driven at 60 mph = [(100-x)/100](D)
So, time spent driving 60 mph = [(100-x)/100](D)/60


Here comes the awful algebra ...

Total time = (x/100)(D)/40 + [(100-x)/100](D)/60

Simplify ...
Total time = xD/4000 + [100D-xD]/6000
Total time = 3xD/12000 + [200D-2xD]/12000
Total time = (xD+200D)/12000

And finally,
Average speed = (total distance)/(total time)
= D/[(xD+200D)/12000]
= (12000D)/(xD+200D)
= (12000)/(x+200)

Answer:
[Reveal] Spoiler:
E


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

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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. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)


Answer: option 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 11 Dec 2017, 10:55
This is an extremely easy problem and you can solve it within one minute if you apply correct strategy.

First thing to know is that on average speeds problems weight is time. It means on which leg you spend more time traveling your average speed will be closer to the speed at which you traveled that leg.

If you spend the same amount of time on each leg your average speed will be average of your actual speeds, regardless of how much distance you traveled at each leg.

Let’s say first you traveled for 3 hours at 40 miles per hour. And then you traveled for 3 hours at 60 m/h. Because you spend equal amount of time on each leg your average speed will be 50 miles per hour. Which is (60 + 40)/2 = 50 miles per hour. You need to internalize this concept in order to do this kind of problems fast.

To come back to our problem. Lets make x equal to 50 percent. Because distance of each leg is equal, time spent on the first leg will be greater than the time spent on the second leg. Mean of 40 and 60 is 50. But because Francie spent more time at 40 mph, her average speed for entire trip will be closer to 40 mph than to 60 mph. In other words her average speed will be between 40 and 50 mph.

Now plug the 50 into the answer choices

A) (180 – 50)/2 = 65 incorrect, right answer must be less than 50
B) (50 + 60)/4 = 25 impossible, average speed will always be between 40 and 60. Avergae speed will never be less than 40 or greater than 60
C) (300-50)/5 = 50 incorrect, this could only be true if times spent on each leg were equal.
D) 600/(115-50) = 600/65 which is very close to 10, an impossible answer
E) 1200/50+200 = 48 this is the only possible answer
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During a trip, Francine traveled x percent of the total distance at an [#permalink]

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Bunuel, niks18, amanvermagmat

Quote:
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.


I hope you meant d as TOTAL DISTANCE

Quote:
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);


For above, I am getting:


\(t_2= \frac{distance_2}{speed_2}=\frac{d-xd}{60*100}\);

To be more precise, see VeritasPrepKarishma solution above:
x% of d = xd/100

then complimentary distance will be
(1-x)d/100 or (d-xd)/100

Can you advise what mistake I am making?
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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 Dec 2017, 10:17
adkikani wrote:
Bunuel, niks18, amanvermagmat

Quote:
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.


I hope you meant d as TOTAL DISTANCE

Quote:
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);


For above, I am getting:


\(t_2= \frac{distance_2}{speed_2}=\frac{d-xd}{60*100}\);

To be more precise, see VeritasPrepKarishma solution above:
x% of d = xd/100

then complimentary distance will be
(1-x)d/100 or (d-xd)/100
Can you advise what mistake I am making?


Hi adkikani

i think you are making simple calculation errors. I would suggest write down your approach and try to get the answer. For the moment forget about other's method and focus on you approach and calculation steps.

d-dx/100 = (100d-dx)/100 =>(100-x)d/100. I am not sure how you got the highlighted portion
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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 Dec 2017, 18:35
1st part: t1 = (x/100)*d/40
2nd part: t2 = ((100-x)/100)*d/60

For average speed: Total distance / total time taken
= d/(t1+t2)
On substituting the values of t1 & t2 you get 120*100/d*(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 29 Dec 2017, 08:13
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.


Hello Bunuel,

here is my solution:

let the first part of distance be X1
Speed = 40 mph
Time = t1
the first part distance done --> x1= 40*t1
----> T1 = x1/40

the rest distance let be x2
Spped = 60
Time = t2
the rest part of distance done -->x2= 60*t2
T2 = x2/60

now total distance / total time = average speed

40t+60t / x/40+x/60 = 12000t/5 <---- this is the answer i got

Can you please explain what is wrong in my solution ? Isn`t my logic correct? :) I always have hard time solving such kind of questons with many variables.

Thank you!
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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 Dec 2017, 09:57
dave13 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.


Hello Bunuel,

here is my solution:

let the first part of distance be X1
Speed = 40 mph
Time = t1
the first part distance done --> x1= 40*t1
----> T1 = x1/40

the rest distance let be x2
Spped = 60
Time = t2
the rest part of distance done -->x2= 60*t2
T2 = x2/60

now total distance / total time = average speed

40t+60t / x/40+x/60 = 12000t/5 <---- this is the answer i got

Can you please explain what is wrong in my solution ? Isn`t my logic correct? :) I always have hard time solving such kind of questons with many variables.

Thank you!


hi dave13

you have used 4 different variable in you solution t1, t2, x1 & x2. Finally while calculating you are using only 2 variable t & x ( see the highlighted part) which is wrong.
you need to have another equation connecting all your variables.
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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 Dec 2017, 10:25
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2 :?
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During a trip, Francine traveled x percent of the total distance at an [#permalink]

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New post 29 Dec 2017, 10:33
dave13 wrote:
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2 :?


Hi dave13

what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?
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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 30 Dec 2017, 06:40
niks18 wrote:
dave13 wrote:
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2 :?


Hi dave13

what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?


Hi niks18, and Bunuel

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2) a minor correcton

I solved denominator seperatly but the point is that i didnt indicate x as x1 and x2 and t as t1 and t2 - was it a must ?

so i did this 3x+2x/120 = 5x/120 is total total time

than numerotor calculated 40t+60t = 100t is a total distance

average speed = total distance / total time

100t / (5x/120) = 100t/1 *120/5x = 12000t/5x this is how i solved it

you are saying i need additional equation for linking variables, but i seperatly expressed time and distance for one part and seperatly expressed time and distance for second part

so i i linked all these variables through formula of finding average speed = total distance divided by total time.

if so which additional equation am i missing and why ? :?

many thanks! :)
Re: During a trip, Francine traveled x percent of the total distance at an   [#permalink] 30 Dec 2017, 06:40

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