Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Jun 2016, 15:22

### 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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 23 Oct 2010
Posts: 386
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 21

Kudos [?]: 269 [0], given: 73

Re: During a trip, Francine traveled x percent of the total [#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: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 46

Kudos [?]: 473 [1] , given: 355

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

### Show Tags

03 Oct 2013, 16:23
1
KUDOS
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!!
Intern
Joined: 15 Mar 2013
Posts: 23
WE: Information Technology (Consulting)
Followers: 0

Kudos [?]: 18 [0], given: 6

### Show Tags

30 Oct 2013, 09:03
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
Senior Manager
Joined: 13 May 2013
Posts: 472
Followers: 2

Kudos [?]: 131 [0], given: 134

Re: During a trip, Francine traveled x percent of the total [#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)
Senior Manager
Joined: 29 Oct 2013
Posts: 289
Concentration: Finance
Schools: Cornell AMBA'17
GMAT 1: 750 Q V46
GPA: 3.7
WE: Corporate Finance (Retail Banking)
Followers: 13

Kudos [?]: 312 [0], given: 189

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

### Show Tags

27 May 2014, 13:14
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).
_________________

Please consider giving 'kudos' if you like my post and want to thank

My journey V46 and 750 -> my-journey-to-46-on-verbal-750overall-171722.html#p1367876

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1858
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 36

Kudos [?]: 1562 [1] , given: 193

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

### Show Tags

28 May 2014, 00:43
1
KUDOS
Look at the diagram below:

Setting up the equation (We require to find value of s)

$$\frac{x}{40} + \frac{100-x}{60} = \frac{100}{s}$$

$$s = \frac{12,000}{x+200}$$

Attachments

spe.jpg [ 23.02 KiB | Viewed 1718 times ]

_________________

Kindly press "+1 Kudos" to appreciate

Manager
Joined: 27 Apr 2014
Posts: 53
GMAT 1: 710 Q47 V40
Followers: 0

Kudos [?]: 2 [0], given: 20

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

### Show Tags

11 Jun 2014, 06:41
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

Hard problem - Would take 3 mins to solve
_________________

Kudos my back and I Kudos your back

Senior Manager
Joined: 17 Dec 2012
Posts: 433
Location: India
Followers: 22

Kudos [?]: 352 [0], given: 14

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

### Show Tags

16 Jun 2014, 17:42
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
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com

Classroom Courses in Chennai

Intern
Joined: 04 Jun 2014
Posts: 49
Followers: 0

Kudos [?]: 2 [0], given: 5

During a trip, Francine traveled x percent of the total [#permalink]

### Show Tags

31 Aug 2014, 06:38
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: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}$$;

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

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

How do i get (100-x)/150 ? I don't know how to solve the fraction.
Math Expert
Joined: 02 Sep 2009
Posts: 33499
Followers: 5929

Kudos [?]: 73439 [0], given: 9902

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

### Show Tags

01 Sep 2014, 02:11
Expert's post
lou34 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: $$t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}$$;

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

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

How do i get (100-x)/150 ? I don't know how to solve the fraction.

$$\frac{(1-\frac{x}{100})*40}{60}=\frac{(\frac{100-x}{100})*40}{60}=\frac{100-x}{100*60}*40=\frac{100-x}{150}$$
_________________
Manager
Joined: 07 Apr 2014
Posts: 147
Followers: 1

Kudos [?]: 19 [0], given: 81

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

### Show Tags

12 Sep 2014, 22:36
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)

I have used plug in method..
let the total distance be 100 & X=10

time for travelling x = 1/4 ; time for travelling remaining = 90/60 = 3/2

t= t1+t2

1/4 +3/2 = 7/4

D=ST ==> 100 = (7/4)s

s= 400/7

apply x= 10 in options .. close call between C & E but E Wins...
Intern
Joined: 17 May 2012
Posts: 49
Followers: 0

Kudos [?]: 8 [0], given: 126

During a trip, Francine traveled x percent of the total [#permalink]

### Show Tags

07 Oct 2014, 04:32
The approach I followed is to assume total distance as 100 miles.
Then x% of total distance is x miles, at the speed of 40 miles per hour and the remaining distance is 100-x.

Formula is Speed = Total Distance/Total Time

T1 = X/40
T2 = 100-X/50
T1 + T2 = X+200/120

_________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
Manager
Joined: 30 Mar 2013
Posts: 137
Followers: 0

Kudos [?]: 38 [0], given: 196

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

### Show Tags

08 Oct 2014, 04:00
I assumed x=20.
so remaining distance =80.
first part= s=d/t. put in speed as 40 (given in the question), distance= 20. find out time. It comes out to be 1/2.

Second part distance= 80. speed =60. Time will now be t=4/3

average speed for the whole journey is total distance / total time.

That's 100/(11/6)

simplifies to 600/11, which is equal to option E. Start with option C first and see if your target answer (in this case, 600/11) is less than or greater than C. Then decide whether you need to go higher or lower.

Kudos if I helped. thanks!
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6666
Location: Pune, India
Followers: 1828

Kudos [?]: 11099 [3] , given: 218

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

### Show Tags

09 Dec 2014, 23:33
3
KUDOS
Expert's post
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)

Responding to a pm:
Quote:

is this a good formula for different distances? so if i learn just these two (first one given in my pm above), then I can pretty much solve anything...is that right? how will this formula change for three different average speeds?

Again, I do not encourage the use of formulas. You will need to learn many formulas to cover various different scenarios and even then you can not cover all.

Say, overall distance is 100. So, distance covered at speed 40 is x. So distance covered at speed 60 will be 100-x

Avg Speed = Total Distance/Total Time $$= \frac{100}{\frac{x}{40} + \frac{100-x}{60}}= \frac{100*40*60}{60x + 40(100-x)}$$ (same as given formula)

You might have to take one step extra here but it makes much more sense than learning up every formula you come across and then getting confused whether the formula will work in a particular situation or not.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 04 Oct 2013
Posts: 162
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Followers: 3

Kudos [?]: 93 [0], given: 54

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

### Show Tags

10 Dec 2014, 23:57
Let Francine travel 40 miles/hr for 1 hour and the rest of the distance at 60 miles/hr for another 1 hour.

40-------------50--------------60

Using time of travel as weight (1:1), the weighted average is calculated as 50 miles/hr.

x = 40% as Francine covers 40 miles of 100 miles

Plugging x as 40/100, option E yields 50 miles/hr.

Manager
Joined: 26 Dec 2011
Posts: 122
Schools: HBS '18, IIMA
Followers: 0

Kudos [?]: 55 [0], given: 39

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

### Show Tags

17 May 2015, 00:00
xyz21 wrote:
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

CORRECTION :

avg speed = d/(d1/v1 + d2/v2)
_________________

Thanks,

Intern
Joined: 18 Aug 2012
Posts: 10
GMAT 1: 730 Q50 V39
Followers: 0

Kudos [?]: 4 [0], given: 1

During a trip, Francine traveled x percent of the total [#permalink]

### Show Tags

29 Nov 2015, 13:14
Given Info: Average Speed for some distance (x% of total distance) in a trip is given, and the average speed for the remaining distance is also given. We have to find average speed for entire trip

Interpreting the Problem: We are given average speed for some distances. Based on that we can calculate the total time taken for these distances, and then divide the total distance traveled by the total time calculated we will be able to calculate the average speed for the entire trip.

Solution:
Let the total distance be a

Distance travelled at a speed of 40 miles per hour = x%*a

Time taken to travel x% of total distance=$$\frac{xa}{100*40}$$

Time taken will be $$\frac{xa}{4000}$$

Distance traveled at a speed of 60 miles per hour = (100-x%)a.

Time taken to travel (100-x)% of total distance=$$\frac{(100-x)a}{100*60}$$

Time taken will be $$\frac{((100-x)a)}{6000}$$

Average speed= $$\frac{TotalDistanceTraveled}{TotalTimeTaken}$$

Total distance traveled =a

Total time taken = $$\frac{xa}{4000} + \frac{((100-x)a)}{6000}$$

Total time taken = $$\frac{xa+200a}{12000} = \frac{a(x+200)}{12000}$$

Average speed for entire trip = $$\frac{a}{(a(x+200)/12000)}$$

Average speed =$$\frac{12000}{(x+200)}$$

Key Takeaways:

Always remember that the average speed is total distance traveled/total time taken. It is never the average of the speeds for various distances.
Senior Manager
Joined: 07 Dec 2014
Posts: 276
Followers: 1

Kudos [?]: 30 [0], given: 0

During a trip, Francine traveled x percent of the total [#permalink]

### Show Tags

29 Nov 2015, 14:45
1/[x/40+(1-x)/60]=120/(x+2)=12000/(x+200)

Last edited by gracie on 10 May 2016, 14:12, edited 2 times in total.
VP
Joined: 09 Jun 2010
Posts: 1288
Followers: 3

Kudos [?]: 85 [0], given: 697

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

### Show Tags

10 May 2016, 08:44
percentage problem , pick the specific number of 100
suppose the road is 100 meter
Manager
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 124
Followers: 7

Kudos [?]: 24 [0], given: 2

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

### Show Tags

10 May 2016, 14:23
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)

This problem can be solved by using a Rate-Time-Distance table. We are given that Francine traveled x percent of the distance at a rate of 40 mph.

Since we are working with percents, and 100% is the total distance percentage, we can say that (100 – x) percent = the percentage of the remaining distance. Thus we know that Francine traveled (100 – x) percent of the distance traveled, at a rate of 60 mph.

Since we are working with percents, we can choose a convenient number for the total distance driven; we'll use 100 miles.

Let’s fill in the table.

Remember, time = distance/rate, so we use the entries from the chart to set up the times:

Time for x percent of the distance = x/40

Time for (100 – x) percent of the distance = (100 – x)/60

Finally, we must remember that average rate = total distance/total time. Our total distance is 100. The total time is the sum of the two expressions that we developed in the previous steps. Here is the initial setup:

100/[(x/40 + (100 – x)/60)]

Now work with the fractions in the denominator, getting a common denominator so that they can be added:

100/[(3x/120 + (200 – 2x)/120)]

100/[(200 + x)/120)]

This fraction division step requires that we invert and multiply:

100 x 120/(200 + x)

12,000/(200 + x)

_________________

Jeffrey Miller
Jeffrey Miller

Re: During a trip, Francine traveled x percent of the total   [#permalink] 10 May 2016, 14:23

Go to page   Previous    1   2   3    Next  [ 41 posts ]

Similar topics Replies Last post
Similar
Topics:
7 During a trip, Yogi Bear traveled 62.5% of the total distance at an 10 09 Aug 2015, 15:41
18 On his trip to Alaska, Joe traveled y percent of the total 10 31 Oct 2012, 11:46
9 During a trip, Francine traveled x percent of the total dist 6 26 Aug 2011, 13:28
8 During a trip on an expressway, Don drove a total of x miles. His aver 4 26 Sep 2006, 09:31
During a trip on an expressway, Don drove a total of x miles. His 0 21 Jul 2015, 11:13
Display posts from previous: Sort by