January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 10 Mar 2008
Posts: 305

During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
Updated on: 17 Jul 2017, 05:44
Question Stats:
74% (02:28) correct 26% (02:43) wrong based on 3379 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. \(\frac{(180x)}{2}\) B. \(\frac{(x+60)}{4}\) C. \(\frac{(300x)}{5}\) D. \(\frac{600}{(115x)}\) E. \(\frac{12,000}{(x+200)}\)
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by vksunder on 02 Mar 2009, 16:27.
Last edited by Bunuel on 17 Jul 2017, 05:44, edited 1 time in total.
Renamed the topic and edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 52294

During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
13 Jun 2010, 03:40
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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1\frac{x}{100})*40}{60}=\frac{100x}{150}\); \(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100x}{150}=\frac{x+200}{300}\); \(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\). Answer: E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: 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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 301

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
26 Dec 2010, 01:41
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.




Intern
Joined: 01 Mar 2009
Posts: 31

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) E. 12,000/(x+200) Need to figure out (average speed=total distance/total time) xy+(100yxy) = total distance (xy/40)+((100yxy)/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,



Manager
Joined: 26 Dec 2008
Posts: 56
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, 19:33
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) 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



VP
Joined: 05 Mar 2008
Posts: 1399

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
28 May 2010, 05:48
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 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. 30040/5 = 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: 173

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
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
_________________
Success is my Destiny



Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
14 Jun 2010, 00:01
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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1\frac{x}{100})*d}{60}=\frac{(100x)d}{60*100}\); \(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100x)d}{60*100}=\frac{d(x+200)}{12000}\); \(Average \ speed=\frac{distance}{total \ time}=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{x+200}\). Answer: E. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: 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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 57
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, 04:26
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 100x (since x is in percent) m+n=100 so we have => 100*40*60/x*60+(100x)40 > solves to 12000/x+200 E is the Answer
_________________
Consider giving Kudos if my post helps in some way



Director
Joined: 22 Mar 2011
Posts: 600
WE: Science (Education)

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
26 Sep 2012, 12:25
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) 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.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



SVP
Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
03 Oct 2013, 15:23
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) 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. This is our answer (E) Gimme Kudos! I need those GMAT Club tests!!



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1823
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
27 May 2014, 23:43
Look at the diagram below: Setting up the equation (We require to find value of s) \(\frac{x}{40} + \frac{100x}{60} = \frac{100}{s}\) \(s = \frac{12,000}{x+200}\) Answer = E
Attachments
spe.jpg [ 23.02 KiB  Viewed 22284 times ]
_________________
Kindly press "+1 Kudos" to appreciate



Director
Joined: 17 Dec 2012
Posts: 625
Location: India

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
16 Jun 2014, 16:42
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) 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 Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Intern
Joined: 04 Jun 2014
Posts: 47

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
31 Aug 2014, 05:38
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: \(t_2= \frac{distance_2}{speed_2}= \frac{(1\frac{x}{100})*40}{60}=\frac{100x}{150}\);\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100x}{150}=\frac{x+200}{300}\); \(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\). Answer: E. How do i get (100x)/150 ? I don't know how to solve the fraction.



Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
01 Sep 2014, 01:11
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: \(t_2= \frac{distance_2}{speed_2}= \frac{(1\frac{x}{100})*40}{60}=\frac{100x}{150}\);\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100x}{150}=\frac{x+200}{300}\); \(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\). Answer: E. How do i get (100x)/150 ? I don't know how to solve the fraction. \(\frac{(1\frac{x}{100})*40}{60}=\frac{(\frac{100x}{100})*40}{60}=\frac{100x}{100*60}*40=\frac{100x}{150}\)
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: 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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8795
Location: Pune, India

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
09 Dec 2014, 22:33
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) 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 100x Avg Speed = Total Distance/Total Time \(= \frac{100}{\frac{x}{40} + \frac{100x}{60}}= \frac{100*40*60}{60x + 40(100x)}\) (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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 18 Aug 2012
Posts: 10

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
29 Nov 2015, 12:14
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 = (100x%)a. Time taken to travel (100x)% of total distance=\(\frac{(100x)a}{100*60}\) Time taken will be \(\frac{((100x)a)}{6000}\) Average speed= \(\frac{TotalDistanceTraveled}{TotalTimeTaken}\) Total distance traveled =a Total time taken = \(\frac{xa}{4000} + \frac{((100x)a)}{6000}\) 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 =\(\frac{12000}{(x+200)}\) Hence, The answer is E 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.



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
10 May 2016, 13:23
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) E. 12,000/(x+200) This problem can be solved by using a RateTimeDistance 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) Answer is E.
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 14 Feb 2016
Posts: 61

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
06 Jun 2016, 04:58
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. (180x)/2 B. (x+60)/4 C. (300x)/5 D. 600/(115x) E. 12,000/(x+200) Hello, I found assuming total distance to be 100 much easier. D1= x and D2= 100x Average speed = Total distance /total time taken = 100/(T1 + T2) where T1= D1/S1=x/40 and T2= D2/S2=(100x)/60 or, Average Speed = 100/[(x/40)+(100x)/60] or 12000/(x+200)
_________________
It is not who I am underneath but what I do that defines me.



Intern
Joined: 05 Apr 2016
Posts: 30

Re: During a trip, Francine traveled x percent of the total distance at an
[#permalink]
Show Tags
23 Aug 2016, 12:39
Why can't I apply the simple average method here?
Average Speed=40*X/100+60*(100X)/100
That leads to answer C. It makes sense to me.
Please help!




Re: During a trip, Francine traveled x percent of the total distance at an &nbs
[#permalink]
23 Aug 2016, 12:39



Go to page
1 2 3
Next
[ 47 posts ]



