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weighted average problems

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weighted average problems [#permalink] New post 25 Apr 2010, 08:52
Can somebody please share the quick process of solving weighted average problems ?
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Re: weighted average problems [#permalink] New post 25 Apr 2010, 09:21
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Here's a sample problem I just made up:

1/5 of the 20 students in a class scored 80 on a test, and the remaining students scored 100. What was the average test score for the class?

Formula (version 1)
(fractional weighting)(score) + (fractional weighting)(score) = weighted average score
(1/5)(80) + (4/5)(100) = 16 + 80 = 96

Formula (version 2)
[(subtotal # of students)(score) + (subtotal # of students)(score)]/[total # of students] = weighted average score
[(4)(80) + (16)(100)] /20 = [320+1600]/20 = 1920/20 = 96

{Note that the two formulas are equivalent, as fractional weighting = subtotal # of students/total # of students, e.g. 1/5 = 4/20}

Portion of the difference
The weighted average will always be betweent the two values/scores/amounts/etc. Here, it must be between 80 and 100, but where in between?

The difference is 20 = 100-80.
The 1/5 of students who score low pull the average score down from 100 by 1/5 of 20: 100-4 = 96
The 4/5 of students who score high pull the average score up from 80 by 4/5 of 20: 80 + 16 = 96
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Re: weighted average problems [#permalink] New post 11 Dec 2013, 08:48
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Re: weighted average problems [#permalink] New post 01 Aug 2015, 02:41
I was reading weighted average topics on Statistics chapter in mgmat word strategy book and couldn't understand this:

Quote:
"You need to make these differentials cancel out, so you should multiply both differentials by differ­ ent numbers so that the positive will cancel out with the negative."

can anyone explain why we have to cancel out or make the equation equals to zero?
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weighted average problems [#permalink] New post 01 Aug 2015, 03:55
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appleid wrote:
I was reading weighted average topics on Statistics chapter in mgmat word strategy book and couldn't understand this:

Quote:
"You need to make these differentials cancel out, so you should multiply both differentials by differ­ ent numbers so that the positive will cancel out with the negative."

can anyone explain why we have to cancel out or make the equation equals to zero?


I will use the same example to explain it.

The lean beef is 10% fat while the super lean meat is 3% fat. Both of these are combined in a certain ration to give you a 8% 'averaged' beef mixture. You need to calculate the ratio of the quantites of 10% vs 3%.

2 ways of looking at it:

Method 1: you have been given 2 entities , 10% and 3% that give you a 8% mixture. What this means is that the differentials (or differences from 8%) add up : (10-8)*x for 10% , (3-8)*y for 3% and (8-8) for the mixture itself.

Now as the differentials for x and y need to equate to differential for the combined (8%) : 2x-5y = 0



Method 2: Consider this scenario, let x and y be the quantities of 10% and 3% beef added to obtain a mixture of 8% beef

Thus you can set up the equation: (10%*x+3%y) = 8%*(x+y) --->\(0.1x+0.03y = 0.08x+0.08y\) --->\(0.02x = 0.05y\) ---> \(2x = 5y\)

So, you get the same answer by both the methods. Pick whichever explanation is better for you.
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Re: weighted average problems [#permalink] New post 02 Aug 2015, 19:31
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I will use the same example to explain it.

The lean beef is 10% fat while the super lean meat is 3% fat. Both of these are combined in a certain ration to give you a 8% 'averaged' beef mixture. You need to calculate the ratio of the quantites of 10% vs 3%.

2 ways of looking at it:

Method 1: you have been given 2 entities , 10% and 3% that give you a 8% mixture. What this means is that the differentials (or differences from 8%) add up : (10-8)*x for 10% , (3-8)*y for 3% and (8-8) for the mixture itself.

Now as the differentials for x and y need to equate to differential for the combined (8%) : 2x-5y = 0



Method 2: Consider this scenario, let x and y be the quantities of 10% and 3% beef added to obtain a mixture of 8% beef

Thus you can set up the equation: (10%*x+3%y) = 8%*(x+y) --->0.1x+0.03y=0.08x+0.08y --->0.02x=0.05y ---> 2x=5y

So, you get the same answer by both the methods. Pick whichever explanation is better for you.




Another, more conceptual way of looking at weighted averages (using Engr2012's example above) is this:
Mixing the 10% and 3% beef in equal amounts would yield a mixture that is 6.5%. Thus, you would need to use more of the 10% to create a mixture of 8%. Now, the difference between the 10% and the 3% is 7; and, the final 8% mixture is 5/7 of the way from the 3% to the 10%. Thinking of this in a ratio, then you would mix the 10% and the 3% in a 5:2 ratio to create the 8% mixture.

This method uses ratios to calculate the mixture as opposed to algebraic equations (differentials) and, to me, seems more straightforward. Let me know if you would like more information on this.
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Re: weighted average problems   [#permalink] 02 Aug 2015, 19:31
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