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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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27 Feb 2017, 04:04
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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02 May 2017, 00:23
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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29 May 2017, 01:26
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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29 Jun 2017, 09:38
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 )

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

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15 Jul 2017, 12:19
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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16 Jul 2017, 16:37
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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17 Jul 2017, 05:07
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Expert's post
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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17 Jul 2017, 05:19
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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17 Jul 2017, 06:27
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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24 Aug 2017, 12:17
Expert's post
Top Contributor
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)

[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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09 Sep 2017, 18:56
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. $$\frac{12,000}{(x+200)}$$

Check solution attached
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Re: During a trip, Francine traveled x percent of the total distance at an   [#permalink] 09 Sep 2017, 18:56

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