It is currently 25 Jul 2017, 09:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 10 Mar 2008
Posts: 360
During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

02 Mar 2009, 17:27
10
KUDOS
88
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

71% (03:32) correct 29% (02:47) wrong based on 2289 sessions

### HideShow timer Statistics

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.
Manager
Joined: 01 Mar 2009
Posts: 50
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

02 Mar 2009, 20:17
1
This post was
BOOKMARKED
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,
Manager
Joined: 26 Dec 2008
Posts: 58
Schools: Booth (Admit R1), Sloan (Ding R1), Tuck (R1)
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

02 Mar 2009, 20:33
2
KUDOS
4
This post was
BOOKMARKED
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
Senior Manager
Joined: 10 Mar 2008
Posts: 360
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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?
Manager
Joined: 01 Mar 2009
Posts: 50
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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.
Senior Manager
Joined: 10 Mar 2008
Posts: 360
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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?
Manager
Joined: 01 Mar 2009
Posts: 50
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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)
VP
Joined: 05 Mar 2008
Posts: 1469
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

28 May 2010, 06:48
6
KUDOS
1
This post was
BOOKMARKED
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
Manager
Joined: 05 Mar 2010
Posts: 217
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

13 Jun 2010, 01:26
Is there any clear way of getting correct answer without plugging in different numbers ????????????????
_________________

Success is my Destiny

Math Expert
Joined: 02 Sep 2009
Posts: 40376
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

13 Jun 2010, 04:40
18
KUDOS
Expert's post
25
This post was
BOOKMARKED
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: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}$$;

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

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

_________________
Manager
Joined: 05 Mar 2010
Posts: 217
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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
_________________

Success is my Destiny

Math Expert
Joined: 02 Sep 2009
Posts: 40376
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

14 Jun 2010, 01:01
9
KUDOS
Expert's post
5
This post was
BOOKMARKED
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.
_________________
Manager
Joined: 05 Mar 2010
Posts: 217
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

14 Jun 2010, 22:09
Thanks .......its clear now
_________________

Success is my Destiny

Intern
Affiliations: NYSSA
Joined: 07 Jun 2010
Posts: 34
Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate - Thomson Reuters
WE 2: Analyst - TIAA CREF
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

16 Jun 2010, 10:52
1
KUDOS
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: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}$$;

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

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

Hope it's clear.

Def. the best method. I tried plugging numbers and got in trouble doing that.
Manager
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 68
WE 1: 6 years - Consulting
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

17 Aug 2010, 05:26
8
KUDOS
9
This post was
BOOKMARKED

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
_________________

Consider giving Kudos if my post helps in some way

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 326
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

26 Dec 2010, 02:41
15
KUDOS
13
This post was
BOOKMARKED
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.

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Director
Joined: 22 Mar 2011
Posts: 611
WE: Science (Education)
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

26 Sep 2012, 13:25
5
KUDOS
3
This post was
BOOKMARKED
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.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Senior Manager
Joined: 23 Oct 2010
Posts: 383
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

03 Oct 2013, 16:23
2
KUDOS
1
This post was
BOOKMARKED
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.

Gimme Kudos! I need those GMAT Club tests!!
Senior Manager
Joined: 13 May 2013
Posts: 468
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

### Show Tags

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)
Re: During a trip, Francine traveled x percent of the total distance at an   [#permalink] 07 Nov 2013, 10:37

Go to page    1   2   3    Next  [ 51 posts ]

Similar topics Replies Last post
Similar
Topics:
6 Frederique traveled x/8 of the total distance of a trip 4 27 Feb 2017, 03:58
8 During a trip, Yogi Bear traveled 62.5% of the total distance at an 10 07 Feb 2016, 10:18
Tim traveled 40 percent of the distance of his trip alone, continued a 8 09 Jan 2015, 06:59
22 On his trip to Alaska, Joe traveled y percent of the total 11 01 Oct 2016, 00:23
10 During a trip, Francine traveled x percent of the total dist 6 30 Jun 2014, 11:11
Display posts from previous: Sort by

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

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.