Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 26 Jul 2010
Posts: 24

A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
27 Jan 2011, 16:35
Question Stats:
64% (02:21) correct 36% (02:02) wrong based on 393 sessions
HideShow timer Statistics
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C? A. 13 B. 13.5 C. 14 D. 14.5 E. 15
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 56277

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
27 Jan 2011, 17:28
pgmat wrote: A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
13 13.5 14 14.5 15
Can some one help me solve this problem using weighted averages? Why cannot we use gallons as weights? Thanks.
Source: GMAT club problem Average miles per gallon equals to total miles covered/total gallons used (miles/gallons), so if the distance from A to B is \(2x\) miles and from B to C is \(x\) miles then we'll have: \(average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5\). Answer: B. Now, \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) and as we are asked to determine average miles per gallon then exactly miles per gallon should be the weights and gallons used for 1st and 2nd parts of the trip should be the values, so in this case the weighted average formula will give the same expression as the one above. Hope it's clear. P.S.
_________________




Intern
Joined: 26 Jul 2010
Posts: 24

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
27 Jan 2011, 18:07
excellent explanation. Thank you.



Senior Manager
Joined: 21 Dec 2010
Posts: 405

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
28 Jan 2011, 00:02
yes thats fairly simple. the formula for avg. rate of the two rates r1 and r2 is avg= (2r1*r2)/(r1+r2). apply this to anything, speed , work done, mileage , provided the work done using the two rates is the same.



Senior Manager
Joined: 08 Nov 2010
Posts: 313
WE 1: Business Development

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
29 Jan 2011, 10:10
good job. thanks for that. i like these kind of questions.
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9442
Location: Pune, India

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
29 Jan 2011, 20:58
pgmat wrote: A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
13 13.5 14 14.5 15
Can some one help me solve this problem using weighted averages? Why cannot we use gallons as weights? Thanks.
Source: GMAT club problem Note: Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled should they be gallons of fuel used...) look at the units. Average required is \(\frac{miles}{gallon}\). So you are trying to find the weighted average of two quantities whose units must be \(\frac{miles}{gallon}\). \(C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}\) \(C_{avg}, C_1, C_2  \frac{miles}{gallon}\) So \(W_1\) and \(W_2\) should be in gallon to get: \(\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)\) Food for thought: If I sold 10 apples at a profit of 10% and 15 oranges at a profit of 20%, what was my overall profit%?
_________________
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 >



Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 82
Concentration: Strategy, General Management
GPA: 3.6
WE: Consulting (Computer Software)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
15 Jan 2012, 07:29
Karishma,
Why is this not working ?
12*(2/3) + 18 *(1/3) /1
14??



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9442
Location: Pune, India

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
16 Jan 2012, 00:59
shankar245 wrote: Karishma,
Why is this not working ?
12*(2/3) + 18 *(1/3) /1
14?? We need to find the average miles per gallon. Average miles/gallon = Total miles/Total gallons What is the distance travelled (i.e. miles)? "traveled twice as many miles from Town A to Town B as it did from Town B to Town C" We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d. What is the total gallons used? Fuel used to go from A to B = 2d/12 Fuel used to go from B to C = d/18 So Average miles/gallon = (2d + d)/(2d/12 + d/18) Go back to my post above. It explains that you have to be mindful of how to use weighted averages. You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon. If we change the question and make it: A certain car used twice as many gallons from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C? Now you can use 12*(2/3) + 18 *(1/3)/1 I hope it makes sense now.
_________________
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 >



Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 82
Concentration: Strategy, General Management
GPA: 3.6
WE: Consulting (Computer Software)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
16 Jan 2012, 10:41
Thanks karishma.. Guess to excel in gmat you need pay attention to every single word in the question before jumpin on answer it



Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 133
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
17 Jan 2012, 05:24
Ans is B Given d_ab = 2*d_bc let d_ab = d and d_bc = x so d=2x
for average miles per gallon = (d+x)/((d/12)+(x/18)) = 13.5 (formula avg speed = total distance/ total time)



Intern
Joined: 06 Apr 2012
Posts: 1
GMAT Date: 09012012

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
26 Dec 2012, 09:46
How about using a plugin techenique? since we have the car gas milages (miles/gallon) are 12mpg and 18 mpg LCM of 12 and 18=36 Let's say distance between B and C is 36 miles Distance between A and B =72 So for traveling from A to B , the required gallons of gas is =72 miles/(12 miles per gallon)=6 gallons ( note : cancell the miles) For travelling from B to C, the required gallons of gas is = 36 miles/ ( 18 miles per gallon) = 2 gallons therefore avg miles per gallon for the total trip is = (total distance)/(total gallons) = (36+72) miles/(8) gallons = 108/8=13.5 miles per gallon.



Manager
Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 108
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
21 Nov 2014, 07:17
pgmat wrote: A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13 B. 13.5 C. 14 D. 14.5 E. 15 Let, Distance between Town A to Town B = D1 Distance between Town B to Town C = D2
Avg miles/gallon from Town A to Town B = p = 12 Avg miles/gallon from Town B to Town C = q = 18
Given : D1 = 2 * D2
Therefore Required Average can be found out by :
\(Req. Avg = \frac{(D1+D2)pq}{[D1*q+D2*p]}\)
\(Req. Avg = \frac{3*D2*12 * 18}{[2*D2*18 + D2*12]}\)
\(Req. Avg = \frac{12*18*3*D2}{48*D2}\)
Req Avg = 13.5 miles/gallon
Answer is B



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
14 Jan 2015, 01:41
Let distance A to B = 2, then distance B to C = 1 Overall Average \(= \frac{Total distance}{Total consumption} = \frac{2+1}{\frac{2}{12} + \frac{1}{18}} = \frac{3}{4} * 18 = 13.5\) Answer = B
_________________
Kindly press "+1 Kudos" to appreciate



VP
Joined: 07 Dec 2014
Posts: 1206

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
24 Dec 2015, 10:01
average miles per gallon=total miles/total gallons let total miles=3m total gallons=2m/9 3m/(2m/9)=13.5 miles per gallon



Manager
Joined: 13 Dec 2013
Posts: 150
Location: United States (NY)
Concentration: General Management, International Business
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
25 Apr 2017, 10:17
VeritasPrepKarishma wrote: shankar245 wrote: Karishma,
Why is this not working ?
12*(2/3) + 18 *(1/3) /1
14?? We need to find the average miles per gallon. Average miles/gallon = Total miles/Total gallons What is the distance travelled (i.e. miles)? "traveled twice as many miles from Town A to Town B as it did from Town B to Town C" We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d. What is the total gallons used? Fuel used to go from A to B = 2d/12 Fuel used to go from B to C = d/18 So Average miles/gallon = (2d + d)/(2d/12 + d/18) Go back to my post above. It explains that you have to be mindful of how to use weighted averages. You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon. If we change the question and make it: A certain car used twice as many gallons from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C? Now you can use 12*(2/3) + 18 *(1/3)/1 I hope it makes sense now. Shouldn't the weights be the distances???



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9442
Location: Pune, India

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
25 Apr 2017, 22:42
Cez005 wrote: VeritasPrepKarishma wrote: shankar245 wrote: Karishma,
Why is this not working ?
12*(2/3) + 18 *(1/3) /1
14?? We need to find the average miles per gallon. Average miles/gallon = Total miles/Total gallons What is the distance travelled (i.e. miles)? "traveled twice as many miles from Town A to Town B as it did from Town B to Town C" We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d. What is the total gallons used? Fuel used to go from A to B = 2d/12 Fuel used to go from B to C = d/18 So Average miles/gallon = (2d + d)/(2d/12 + d/18) Go back to my post above. It explains that you have to be mindful of how to use weighted averages. You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon. If we change the question and make it: A certain car used twice as many gallons from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C? Now you can use 12*(2/3) + 18 *(1/3)/1 I hope it makes sense now. Shouldn't the weights be the distances??? No. Since you need to average miles/gallon, the weights will be in gallons (the denominator). It is obvious from the formula of weighted average: \(Cavg = \frac{(C1*w1 + C2*w2)}{(w1 + w2)}\) Also check this post: https://www.veritasprep.com/blog/2014/1 ... averages/
_________________
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 >



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6923
Location: United States (CA)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
29 Apr 2017, 09:03
pgmat wrote: A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13 B. 13.5 C. 14 D. 14.5 E. 15 We are given that a certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. Thus, if the car traveled d miles from Town B to Town C, it traveled 2d miles from Town A to Town B. We are also given that from Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. Thus, the number of gallons used from Town A to Town B is 2d/12 = d/6 and the number of gallons used from Town B to Town C is d/18. Let’s now determine the overall average miles per gallon, using the following formula: average = (total distance)/(total gallons) average = (d + 2d)/(d/6 + d/18) average = 3d/(4d/18) = (3d x 18)/4d = 54/4 = 13.5 mpg Answer: B
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Intern
Joined: 23 Aug 2016
Posts: 48

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
02 May 2017, 14:06
Choosing smart numbers: A to B = 36 miles B to C = 18 miles From A to B at 12mpg = 36 miles/12mpg = 3 gallons used From B to C at 18mpg = 18 miles/18mpg = 1 gallon use Total Gallons = 4 Average mpg = Total Miles/Total Gallons = (36+18)/4 = 54/4 = 13.5, Answer B.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14563
Location: United States (CA)

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
09 Oct 2018, 23:11
Hi All, We're told that a certain car traveled TWICE as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. We're asked for the average MILES PER GALLON that the car achieved on its trip from Town A through Town B to Town C. This question can be solved in a couple of different ways, including by TESTing VALUES. To start, we should look to pick distances that work well with the given miles/gallon data that we have (re: 12 miles/gallon and 18 miles/gallon). Choosing 36 miles for the first part of the trip and 18 miles for the second part of the trip will make the math fairly easy. Town A to Town B: 36 miles traveled at 12 miles/gallon > 36/12 = 3 gallons used to travel 36 miles Town B to Town C: 18 miles traveled at 18 miles/gallon > 18/18 = 1 gallon used to travel 18 miles Total miles traveled = 36 + 18 = 54 miles Total gallons used = 3 + 1 = 4 gallons Average miles/gallon = 54/4 = 13.5 Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/



VP
Joined: 09 Mar 2018
Posts: 1002
Location: India

Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
Show Tags
08 Feb 2019, 10:38
pgmat wrote: A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13 B. 13.5 C. 14 D. 14.5 E. 15 So i plugged in here A>B twice than B>C 105 average miles per gallon that the car achieved on its trip from Town A through Town B to Town C = Total Distance/Gallons consumed per distance = 15 / [10/12 + 5/18] = 15 *12*18/{240} = 13.5 B
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.




Re: A certain car traveled twice as many miles from Town A to Town B as it
[#permalink]
08 Feb 2019, 10:38






