text: A certain car traveled twice as many miles from Town A : GMAT Problem Solving (PS)
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  # A certain car traveled twice as many miles from Town A
  
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  A certain car traveled twice as many miles from Town A [#permalink]
  
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  21 Feb 2012, 13:58
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  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
  
  MGMAT 1 Q. 11
  The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  21 Feb 2012, 14:15
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  docabuzar 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
  
  MGMAT 1 Q. 11
  The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
  
  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:
  $$average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5$$.
  
  P.S. As you can see weighted average concept is exactly what we used here.
  
  Hope it's clear.
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  21 Feb 2012, 14:40
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  Lets pick numbers:
  
  Distance A to B: 36 miles (2x) --> 3 gallons
  Distance B to C: 18 miles (x) --> 1 gallon
  
  Average: $$\frac{(36+18)}{4}$$ = 13.5
  
  Hence, B
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  21 Feb 2012, 20:58
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  docabuzar 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
  
  MGMAT 1 Q. 11
  The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
  
  If the concept you do not understand is why (12*2 + 18*1)/3 doesn't work, here you go:
  
  'Distances traveled' (i.e. ratio of 2:1) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.
  
  If we change the question and make it:
  A certain car used [highlight]twice as many gallons[/highlight] 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 + 18 *1)/3
  
  Why?
  
  Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or 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:
  
  $$\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)$$
  
  Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.
  
  We know that Average miles/gallon = Total miles/Total gallons
  
  [highlight]Takeaway: The weights have to be the denominator units of the average.[/highlight]
  
  So what do we do in this question?
  
  What is the distance travelled (i.e.total 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. Total distance must be (2d + 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)
  
  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.
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  Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 17 Jan 2012 Posts: 41 GMAT 1: 610 Q43 V31 Followers: 1 Kudos [?]: 146 [0], given: 16 Re: A certain car traveled twice as many miles from Town A [#permalink] ### Show Tags 22 Feb 2012, 03:10 Many Thanks. Just to make myself clearer, we can also say that: Case 1. Car moves A to B @ 12mph and covers 2 miles B to C @ 18 mph and covers 1 mile We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing). Case 2. Car moves A to B @ 12mph for 2 hrs B to C @ 18 mph for 1 hr We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here! Math Expert Joined: 02 Sep 2009 Posts: 36554 Followers: 7078 Kudos [?]: 93169 [0], given: 10553 Re: A certain car traveled twice as many miles from Town A [#permalink] ### Show Tags 22 Feb 2012, 03:50 docabuzar wrote: Many Thanks. Just to make myself clearer, we can also say that: Case 1. Car moves A to B @ 12mph and covers 2 miles B to C @ 18 mph and covers 1 mile We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing). Case 2. Car moves A to B @ 12mph for 2 hrs B to C @ 18 mph for 1 hr We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here! I think that the issue was over complicated above: {average rate}={total distance}/{total time}; {average salary}={total salary}/{total # of employees}; {average miles per gallon}={total miles}/{total gallons}; ... Keep it simple. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2134 Kudos [?]: 13648 [0], given: 222 Re: A certain car traveled twice as many miles from Town A [#permalink] ### Show Tags 22 Feb 2012, 04:30 docabuzar wrote: Many Thanks. Just to make myself clearer, we can also say that: Case 1. Car moves A to B @ 12mph and covers 2 miles B to C @ 18 mph and covers 1 mile We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing). Case 2. Car moves A to B @ 12mph for 2 hrs B to C @ 18 mph for 1 hr We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here! You are right. Case 1 doesn't work but case 2 does. What is the unit of the denominator in average speed? Average speed = miles/hour. So you can use weighted averages when the weights are time (i.e. in hours). That is the reason your case 2 works. (By the way, you yourself said speed is miles/hour and time is in hours. Why do you think the units don't match?) When you want to find weighted average speed, distance cannot be the weights, it has to be time. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
  
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  26 May 2013, 04:47
  Bumping for review and further discussion.
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  25 Jul 2014, 07:02
  Hello from the GMAT Club BumpBot!
  
  Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
  
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  15 Feb 2015, 14:43
  Hello Experts!
  I am quiet fascinated with The Bunuel approach... of using algebraic expression.
  Kindly assess my approach which I will try next time onward to got a shortcut way
  step 1) took LCM of 12 and 18.. came as 36. just multiplied by 10...(to make easy calculation)
  step 2) 360 distance between B to C... do 360/18 hence 20 gallons used
  step 3) twice distance.. hence 360*2= 720... do as above.. 720/12= 60 gallons used
  step 4) total gallons.. 20+60= 80 gallons
  step ) total miles= 360+720= 1080 miles
  hence.. average of whole journey = 1080/80 which comes to 13.5
  Thanks
  Celestial
  
  Kudos if it makes sense...
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  06 May 2016, 13:10
  Hello from the GMAT Club BumpBot!
  
  Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
  
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  12 Aug 2016, 16:13
  Hi!
  Kindly share your views on my above approach plz..
  experts...
  thanks
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  Re: A certain car traveled twice as many miles from Town A [#permalink]
  
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  14 Aug 2016, 23:57
  Celestial09 wrote:
  Hello Experts!
  I am quiet fascinated with The Bunuel approach... of using algebraic expression.
  Kindly assess my approach which I will try next time onward to got a shortcut way
  step 1) took LCM of 12 and 18.. came as 36. just multiplied by 10...(to make easy calculation)
  step 2) 360 distance between B to C... do 360/18 hence 20 gallons used
  step 3) twice distance.. hence 360*2= 720... do as above.. 720/12= 60 gallons used
  step 4) total gallons.. 20+60= 80 gallons
  step ) total miles= 360+720= 1080 miles
  hence.. average of whole journey = 1080/80 which comes to 13.5
  Thanks
  Celestial
  
  Kudos if it makes sense...
  
  While you approach is not wrong, you have made lots of double and tedious calculations which is quite time consuming. You can do it without multiplying by 10
  
  You got LCM 36. so we will considered total miles traveled under 18 Miles/gallon to be 36. Hence it took 2 gallons
  Now, the distance traveled under 12M/ Gallon will be 72 and total gallons used will be 6.
  
  therefore no. of gallons used traveling both route = 2+6 -= 8 and
  total miles traveled is 36+72 = 108.
  
  average = 108/8= 13.5 M/Gallon
  
  its much more easier than you steps, which include unnecessary multiplication and double division. Its quite time consuming in GMAT.
  Re: A certain car traveled twice as many miles from Town A   [#permalink] 14 Aug 2016, 23:57
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