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Re: A certain car traveled twice as many miles from Town A [#permalink]
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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]
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]
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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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Re: A certain car traveled twice as many miles from Town A [#permalink]
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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]
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
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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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.
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Re: A certain car traveled twice as many miles from Town A [#permalink]
plugin a number for distance
Let B to c=10
then as per question A to B=20
Total distance=30

Fuel used
B to C=10/18=8/9(Distance/Mileage per Gallon)
A to B=20/12=5/3

Total mileage=Total distance/Total fuel used=30/(5/3)+(8/9)=27/9=13.5
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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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



We can use the following formula:

Average = (total distance)/(total gallons)

We can let the distance from Town A to Town B = 2d and the distance from Town B to Town C = d. Thus, the total gallons of fuel consumed from Town A to Town B = 2d/12, and the total gallons of fuel consumed from Town B to Town C = d/18; thus:

Average = 3d/(2d/12 + d/18)

Average = 3d/(6d/36 + 2d/36)

Average = 3d/(8d/36) = (36 x 3d)/8d = (9 x 3)/2 = 13.5

Answer: B
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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Given: 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.

Asked: What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

Let us assume that distance between town A to town B = 2x miles
Distance between town B to town C = x miles

Town A to town B: -
Distance = 2x miles
Mileage = 12 miles/gallons
Fuel consumption = 2x/12 = x/6 gallons

Town B to Town C: -
Distance = x miles
Mileage = 18 miles/gallons
Fuel consumption = x/18 gallons

Town A to Town C: -
Distance = x + 2x = 3x miles
Fuel consumption = x/6 + x/18 = 4x/18 = 2x/9
Mileage = 3x/(2x/9) = 27/2 = 13.5 miles/gallons

IMO B
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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Re: A certain car traveled twice as many miles from Town A [#permalink]
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