Weighted Averages

[Joey]

I don't understand how statement 1 is sufficient - (i understand why 2 is sufficient). Can you explain how knowing that this is a weighted average problems helps in arriving at a solution?

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs $3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs $5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. What is the cost per gallon of the fuel mixture currently in the vehicle’s tank?

1) Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles.

2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank

[RR]

i Vehicle burns 8 gallons for 200 miles. Which means that per gallon, it gives 200/8 = 25 miles
Fuel X gives 20 miles per gallon and fuel Y gives 40 miles per gallon. The ratio in which they should be taken so that the resultant mixture gives 25 miles
20 X + 40 Y
-------------- = 25
X + Y
Simplify by grouping X and Y together
X : Y = 3 : 1
Since I know the cost of each fuel type and the ratio in which they need to be taken, I can find out the cost per gallon of the mixture. SUFFICIENT

ii. The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank
The number of miles that can be covered per dollar of pure fuel X is 20/3
The number of miles that can be covered per dollar of pure fuel Y is 40/5.
I am not sure how you found this to be sufficient. Please explain.
I would guess INSUFFICIENT

[RonPurewal]

you could dive headlong into algebra, but you don't want to do that unless it's an absolute necessity.

here's a higher-level way of thinking about it.

first, you must understand the holy canon of weighted averages. among the most important facts concerning such averages is the following:
in taking a weighted average of 2 quantities, knowing the RATIO of the quantities is equivalent to knowing the VALUE OF THE AVERAGE itself.

example: let's say everyone in a class scores either 80 or 90 on an exam; then that's a weighted average of 80's and 90's (the two quantities in question).
* if you know the average score for the class, then you can determine the ratio of 80's to 90's. (note that you cannot determine the actual numbers of those scores without additional information, such as the total # of students in the class.)
* if you know the ratio of 80's to 90's (or even more specific information, such as the actual numbers thereof), then you can calculate the class average for the exam.

--

REPHRASE THE QUESTION:
the overall cost per gallon is a weighted average of $3/gallon (for fuel x) and $5/gallon (for fuel y).
therefore, the question can be rephrased as follows:
what's the ratio of gallons of fuel x to gallons of fuel y in the tank?


STATEMENT 1:
this is equivalent to giving the miles per gallon for the fuel mixture: 200/8 = 25 miles per gallon.
fuel x gets 20 miles/gallon, and fuel y gets 40 miles/gallon.

* conceptually: you can just realize that, the larger the ratio of fuel y to fuel x, the better the miles per gallon will be. therefore, each specific number of miles/gallon will correspond to a fixed ratio of fuel y to fuel x; any higher ratio would give more miles per gallon, and any lower ratio would give fewer miles per gallon.
therefore, this information is sufficient.

* algebraically / using templates: you can think of this as a weighted average of 20 miles/gallon and 40 miles/gallon. same concept applies as above. therefore, sufficient.


STATEMENT 2:
this is equivalent to giving the miles per dollar for the fuel mixture: 7 1/7 miles per dollar.
fuel x gets 20/3 = 6 2/3 miles/dollar, and fuel y gets 40/5 = 8 miles/dollar.

* conceptually: you can just realize that, the larger the ratio of fuel y to fuel x, the better the miles per dollar will be. therefore, each specific number of miles/dollar will correspond to a fixed ratio of fuel y to fuel x; any higher ratio would give more miles per dollar, and any lower ratio would give fewer miles per dollar.
therefore, this information is sufficient.

* algebraically / using templates: you can think of this as a weighted average of 6 2/3 miles/dollar and 8 miles/dollar. same concept applies as above. therefore, sufficient.

ans (d)

[700+]

first, you must understand the holy canon of weighted averages. among the most important facts concerning such averages is the following:
in taking a weighted average of 2 quantities, knowing the RATIO of the quantities is equivalent to knowing the VALUE OF THE AVERAGE itself.

example: let's say everyone in a class scores either 80 or 90 on an exam; then that's a weighted average of 80's and 90's (the two quantities in question).
* if you know the average score for the class, then you can determine the ratio of 80's to 90's. (note that you cannot determine the actual numbers of those scores without additional information, such as the total # of students in the class.)
* if you know the ratio of 80's to 90's (or even more specific information, such as the actual numbers thereof), then you can calculate the class average for the exam.

I didn't clearly understand the above example. Could you please explain it again.

[RonPurewal]

I didn't clearly understand the above example. Could you please explain it again.

this request is impossible to fulfull unless you specify what you didn't understand.

in any case, search the archives at this link for "weighted average" -- there is about half an hour on this topic at the end of one of the lectures:
http://www.manhattangmat.com/thursdays-with-ron.cfm

[jp.jprasanna]

REPHRASE THE QUESTION:
the overall cost per gallon is a weighted average of $3/gallon (for fuel x) and $5/gallon (for fuel y).
therefore, the question can be rephrased as follows:
what's the ratio of gallons of fuel x to gallons of fuel y in the tank?

Hi ron - One doubt here....

What is the cost per gallon of the fuel mixture currently in the vehicle’s tank?

Is basically (cost of fuel x + cost of fuel y) / (quantity of fuel x and quantity of fuel y)

So from statement 1 and 2 if i get x : y 3:1 then sol would be 3+5 / 3+1 = 8/4 i.e 2/1

Is this correct? Im a little confused.

[tim]

you have incorrectly calculated the cost of fuel. cost is unit price times gallons, not just unit price. instead of 3+5 you should have had 3*3+5*1 in your numerator..

[jnovo]

I'm having problems with Ron Purewal's explanation for Statement 2. Conceptually, it is correct to say that the 25 mpg will show the fixed ratio of fuel X to fuel Y, 3:1. Visually, this is achieved by: 1. calculating the difference between the two fuel types' different miles per gallon: X = 20, Y = 40, Y - X = 20; 2. Finding the difference between the mixture's mpg and either fuel's mpg, let's say X: Mix = 25 mpg, X = 20 mpg, Mix-X = 5; 3. And finally, stacking the differences as a fraction to find the proportion of the OTHER fuel mixture, in this case Y: (Mix - X)/(Y - X) = 5/20 = 1/4. This means Y makes up 1/4 of the total mixture, thus X makes up 3/4.

HOWEVER

One would assume that since this method works for MPG it should also work for MPD (Miles per dollar). BUT it DOES NOT.

X = 20/3 MPD, Y = 40/5 = 8 MPD, MIX = 50/7 MPD

Conceptually we would assume the ratio to be the same as when we are using MPG. But, using the same methodology, the proportion of Y comes out as 5/14.

(This reminds me of the Challenge problem on May 4, 2009, in which a wrong answer is achieved if one tries to weight two ratios using the numerator instead of the denominator, miles instead of gallons)

MY QUESTION:

Why are the proportions of the two fuels not equal when using MPG and MPD?

[tim]

it seems that you've already verified that the proportions are not equal. recognizing that is true and knowing that you have to be careful about this sort of thing is what's most important to get out of this problem. as for WHY they don't end up being equal, that's too theoretical to worry about for the GMAT; just remember that these quirks pop up from time to time..