Weighted Average

[asth678]

My question is about the application of a technique which Ron taught- the shortcut approach to wt averages where you use the number line.I was able to apply that technique successfully to-Percents and Weighted Averages problem in the MGMAT FDP book-"Kris-P cereal is 10% sugar by weight, whereas healthier but less delicious
Bran-D cereal is 2% sugar by weight. To make a delicious and
healthy mixture that Is 4% sugar, what should be the ratio of Kris-P
cereal to Bran-D cereal, by weight?"

Can I also apply the same technique to Percent Change and Weighted Averages problems given in the MGMAT FDP book- question-A company sells only pens and pencils. The revenue from pen sales in 2007
was up 5% from 2006, but the revenue from pencil sales declined 13% over
the same period. If overall revenue was down 1% from 2006 to 2007, what
was the ratio of pencil revenues to pen revenues in 2oo6?.

If I can apply the same technique are there any tips-cause I couldn't solve this problem using the shortcut number line approach.

[tim]

Yep, same approach is good. First, let’s go through this one algebraically: Old pens = x, new pens = 1.05x. Old pencils = y, new pencils = .87y. New revenue = 99% of old revenue so (1.05x + .87y) = .99(x + y).

1.05x + .87y = .99x + .99y
.06x = .12y
6x = 12y
x = 2y

We sold twice as many pens as pencils. Now let’s try the number line approach: Place the pens at +5 on the number line, the pencils at -13, and the balance point at -1. Remember the distance from the balance point to an endpoint gets multiplied by the weight of that endpoint, and this number should be the same on both sides. 6 (5 minus -1) times the number of pens = 12 (-1 minus -13) times the number of pencils. Again, we see that we sold twice as many pens as pencils. This technique doesn’t work with all conceivable situations though (look up the problem about the election and the two parties for a good example), so you need to consider this technique carefully. And if you aren’t 100% sure you can use the number line method, don’t use it..

[Eddie Gutia]

Hi Ron,

Below is the "election and two parties" problem (Page 120, FDP strategies). This problem sounds exactly similar to "pen and pencil" problem. However, why does the number line approach doesn't work here?

Last year, all registered voters in Kummannia voted either for the revolutionary party or for the status quo party. This year, the number of revolutionary voters increased 10, while the number of status Quo voters increased 5%. No other votes were cast. If the number of total voters increased 8%, what fraction of voters voted revolutionary this year?

My analysis: The ratio of revolutionary voters to Status quo voters according to the number line technique should be 8-5:10-8. But this is not a correct answer. Why does the number line approach not work although the problem is exactly similar to "pen and pencil" problem?

[RonPurewal]

.

[Eddie Gutia]

Thanks Ron for the response. However, if you look at Tim's response to the original question, he solved the percentage CHANGE problem using the number line technique. Are you saying Tim's number line approach for percentage change problem worked by fluke for that particular problem?

[PetriF258]

Hi Guthey

Just my unprofessional opinion. The difference between the two questions is in the "required".

Question 1: what was the ratio of pencil revenues to pen revenues in 2oo6? PRIOR YEAR

versus

No other votes were cast. If the number of total voters increased 8%, what fraction of voters voted revolutionary this year? THIS YEAR

By using the number line method we could get the ratio of the prior year, but we need the ratio for the current year. Therefore it would not work...

As Tim showed, the number line method is almost just a shortcut of the algebra method, so using the algebra / number line method, you could get the ratio of the prior year, but you need extra calcs to get the current year ratio.

Bunuel also explains it on the GMAT club forum.

Regards

[Eddie Gutia]

Thanks PetriF258 for the reply. I appreciate it. Can you please share the link for the GMAT club forum post you are referring to?

[PetriF258]

No problem. Are we allowed to post links? I didn't notice anything about it in the rules. So please instructors, if it is not allowed please delete the link. It is a link to gmatclub. Thanks

http://gmatclub.com/forum/last-year-all ... 88126.html

[Eddie Gutia]

Excellent, excellent, Thank you..

[RonPurewal]

in general you shouldn't link to other forums, but i think it makes sense to make an exception here.

[RonPurewal]

also—
more generally, the same number-line shortcut should work for these percent change problems, too.

the reason is, essentially, that "percent changes" represent actual quantities themselves.
to see why this is true, ignore the original populations and just look at the "transactions" that take place.
e.g., in the pencil problem, we can think of the situation as an exchange, in which i give away a certain number of pencils (with each loss worth "-13") and, in exchange, acquire some number of pens ("+5").

so, shortcut should still work.

[RonPurewal]

in fact, the shortcut DOES work for this problem:

Last year, all registered voters in Kummannia voted either for the revolutionary party or for the status quo party. This year, the number of revolutionary voters increased 10, while the number of status Quo voters increased 5%. No other votes were cast. If the number of total voters increased 8%, what fraction of voters voted revolutionary this year?

My analysis: The ratio of revolutionary voters to Status quo voters according to the number line technique should be 8-5:10-8. But this is not a correct answer. Why does the number line approach not work although the problem is exactly similar to "pen and pencil" problem?

that ratio is 3:2.
the percent change is closer to %10 than to %5, so there should be more R voters.

thus R : SQ = 3:2.

this will totally work.

algebraically:
original quantities = 3x and 2x, respectively (R and SQ)
quantities after the changes = 1.10(3x) = 3.30x and 1.05(2x) = 2.10x, respectively
so, new total = 5.40x, which is exactly %8 more than the original total (5x)

you can also pick numbers to show the same thing.

[RonPurewal]

ok, i see what you did: in applying the number-line shortcut, you switched the numbers in the ratio.
i.e., you used R:SQ = 2:3, rather than 3:2.

so, the method you're currently using to assign these coefficients is not working.

fortunately, there's a very easy way to do that: just use common sense to figure out which way the ratio goes.

in any weighted average, one of the two items will have a stronger influence than the other. i.e., the weighted average will be closer to one of the two original values than to the other one.
whichever one that is gets the bigger coefficient/"weight".

in this example, since 8 is closer to 10 than to 5, there must be more R voters.

[Eddie Gutia]

Gracia Muchas en castellano

[RonPurewal]

you're welcome.