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# A certain car traveled twice as many miles from Town A to Town B as it

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A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  27 Jan 2011, 15:35
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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
[Reveal] Spoiler: OA
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  27 Jan 2011, 16:28
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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:
$$average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5$$.

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.
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  27 Jan 2011, 17:07
excellent explanation. Thank you.
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  27 Jan 2011, 23:02
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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.
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  29 Jan 2011, 09:10
good job. thanks for that. i like these kind of questions.
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  29 Jan 2011, 19:58
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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%?
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: Do till 740 :) Joined: 13 Jun 2011 Posts: 113 Concentration: Strategy, General Management GMAT 1: 460 Q35 V20 GPA: 3.6 WE: Consulting (Computer Software) Followers: 1 Kudos [?]: 8 [0], given: 19 Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink] 15 Jan 2012, 06:29 Karishma, Why is this not working ? 12*(2/3) + 18 *(1/3) /1 14?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5953 Location: Pune, India Followers: 1522 Kudos [?]: 8381 [1] , given: 192 Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink] 15 Jan 2012, 23:59 1 This post received KUDOS Expert's post 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 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 to Town B as it [#permalink]  16 Jan 2012, 09:41
Thanks karishma..
Guess to excel in gmat you need pay attention to every single word in the question before jumpin on answer it
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  17 Jan 2012, 04:24
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)
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  26 Dec 2012, 08:46
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.
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  19 Nov 2014, 07:17
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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  21 Nov 2014, 06:17
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

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Re: A certain car traveled twice as many miles from Town A to Town B as it [#permalink]  14 Jan 2015, 00:41
Let distance A to B = 2, then distance B to C = 1

Overall Average $$= \frac{Total distance}{Total consumption} = \frac{2+1}{\frac{2}{12} + \frac{1}{18}} = \frac{3}{4} * 18 = 13.5$$

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Re: A certain car traveled twice as many miles from Town A to Town B as it   [#permalink] 14 Jan 2015, 00:41
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