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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] New post 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!
[Reveal] Spoiler: OA
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Re: A certain car traveled twice as many miles from Town A [#permalink] New post 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.

Answer: B.

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] New post 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] New post 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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Re: A certain car traveled twice as many miles from Town A [#permalink] New post 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!
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Re: A certain car traveled twice as many miles from Town A [#permalink] New post 22 Feb 2012, 03:50
Expert's post
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.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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

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Re: A certain car traveled twice as many miles from Town A [#permalink] New post 22 Feb 2012, 04:30
Expert's post
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.
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Re: A certain car traveled twice as many miles from Town A [#permalink] New post 26 May 2013, 04:47
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Re: A certain car traveled twice as many miles from Town A [#permalink] New post 25 Jul 2014, 07:02
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Re: A certain car traveled twice as many miles from Town A   [#permalink] 25 Jul 2014, 07:02
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