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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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weighted average problems [#permalink] New post 06 Aug 2015, 09:20
Thanks both of you for guiding through concept and examples. Yeah, the cancel out and ratio part is very handy so I am trying hard to get the main concept behind differentials. I need a bit clarification (if anyone can graphically show, that's great) on the relationship with weighted average and differentials.

Quote:
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.


What does it mean that differentials add up from both ends of weighted average?
If I can see it in this way:
3% ...............( going to add +5 >> )...............8%...............(<< going to add -2)...............10%
How the +5 and -2 add up to 0?

Last edited by appleid on 06 Aug 2015, 09:53, edited 1 time in total.
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Re: weighted average problems [#permalink] New post 06 Aug 2015, 09:46
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appleid wrote:
Thanks both of you for guiding through concept and examples. Yeah, the cancel out and ratio part is very handy so I am trying hard to get the main concept behind differentials. I need a bit clarification (if anyone can graphically show, that's great) on the relationship with weighted average and differentials.

Quote:
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.


What does it mean that differentials add up from both ends of weighted average?
If I can see it in this way:
3% ...............( going to add +5 >> )...............8%...............(<< going to add -2)...............10%
How the +5 and -2 add up to 0?


appleid -

The piece you are missing on the "differentials" is HOW MUCH of each differential. That is 5x = 2y where x is the amount of the 10% solution and y is the amount of the 3% solution.

Does this help?
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weighted average problems [#permalink] New post 06 Aug 2015, 09:51
appleid wrote:
Thanks both of you for guiding through concept and examples. Yeah, the cancel out and ratio part is very handy so I am trying hard to get the main concept behind differentials. I need a bit clarification (if anyone can graphically show, that's great) on the relationship with weighted average and differentials.

Quote:
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.


What does it mean that differentials add up from both ends of weighted average?
If I can see it in this way:
3% ...............( going to add +5 >> )...............8%...............(<< going to add -2)...............10%
How the +5 and -2 add up to 0?


You are missing one very important point that I had mentioned above.

+5 and -2 are fine but these value need to be multiplied by the respective quantitires (x and y resp.) to bring the net effect = 0

i.e. 5x -2y = 0 ---> 5x = 2y. This is what I had mentioned before. The differentials are of no use without multiplying them with the respective mixture quantities.

Makes sense?
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Re: weighted average problems [#permalink] New post 06 Aug 2015, 10:46
Engr2012 wrote:
appleid wrote:
Thanks both of you for guiding through concept and examples. Yeah, the cancel out and ratio part is very handy so I am trying hard to get the main concept behind differentials. I need a bit clarification (if anyone can graphically show, that's great) on the relationship with weighted average and differentials.

Quote:
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.


What does it mean that differentials add up from both ends of weighted average?
If I can see it in this way:
3% ...............( going to add +5 >> )...............8%...............(<< going to add -2)...............10%
How the +5 and -2 add up to 0?


You are missing one very important point that I had mentioned above.

+5 and -2 are fine but these value need to be multiplied by the respective quantitires (x and y resp.) to bring the net effect = 0

i.e. 5x -2y = 0 ---> 5x = 2y. This is what I had mentioned before. The differentials are of no use without multiplying them with the respective mixture quantities.

Makes sense?


Oh, I was thinking of quantities altogether but showed only +5 and -2 for adding up to 0. The main question is:

Why these differentials with respective quantities add up to 0? or why it bring to the net effect = 0?

Any theory/concept/visuals that'd help?
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weighted average problems [#permalink] New post 06 Aug 2015, 10:49
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

1.
differential for x + differential for y = differential for combined (x+y)
(10% - 8%) for x + (3%-8%) for y = (8%-8%) for combined x+y
(2)x + (-5)y = 0 (x+y)
2x - 5y = 0 # why the net = 0? because the difference from 8% to 8% is 0?


2.
(10%*x+3%y) = 8%*(x+y)
0.1x+0.03y=0.08x+0.08y
10x + 3y = 8x + 8y
10x - 8x = 8y - 3y
(10-8)x = (8-3)y
(10-8)x - (8-3)y = 0 #I think this is where I understood why the add up becomes 0. But I can't just understand same thing in method 1
2x - 5y = 0

or, from the fifth line,
(10-8)x = (8-3)y
2x = 5y
x/y = 5/2
x to y Ratio will be 5:2

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking
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weighted average problems [#permalink] New post 06 Aug 2015, 10:59
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

1.
differential for x + differential for y = differential for combined (x+y)
(10% - 8%) for x + (3%-8%) for y = (8%-8%) for combined x+y
(2)x + (-5)y = 0 (x+y)
2x - 5y = 0 # why the net = 0? because the difference from 8% to 8% is 0? Yes. This is correct.


2.
(10%*x+3%y) = 8%*(x+y)
0.1x+0.03y=0.08x+0.08y
10x + 3y = 8x + 8y
10x - 8x = 8y - 3y
(10-8)x = (8-3)y
(10-8)x - (8-3)y = 0 #I think this is where I understood why the add up becomes 0. But I can't just understand same thing in method 1 Method 1 is just another "fancy way" of quoting the same thing.
2x - 5y = 0

or, from the fifth line,
(10-8)x = (8-3)y
2x = 5y
x/y = 5/2
x to y Ratio will be 5:2

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking


You do not have to resort to #3 method to see that as 3% needs to go more towards 8% than 10%, there will be more of 3 and less of 10 (to compensate for more 'distance' to travel for 3%). Yes, you can just take the reciprocals of the "differentials" 2/7 and 5/7 to come up with 5/2 for x:y as x>y .
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Re: weighted average problems [#permalink] New post 06 Aug 2015, 11:02
Expert's post
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

1.
differential for x + differential for y = differential for combined (x+y)
(10% - 8%) for x + (3%-8%) for y = (8%-8%) for combined x+y
(2)x + (-5)y = 0 (x+y)
2x - 5y = 0 # why the net = 0? because the difference from 8% to 8% is 0?


2.
(10%*x+3%y) = 8%*(x+y)
0.1x+0.03y=0.08x+0.08y
10x + 3y = 8x + 8y
10x - 8x = 8y - 3y
(10-8)x = (8-3)y
(10-8)x - (8-3)y = 0 #I think this is where I understood why the add up becomes 0. But I can't just understand same thing in method 1
2x - 5y = 0

or, from the fifth line,
(10-8)x = (8-3)y
2x = 5y
x/y = 5/2
x to y Ratio will be 5:2

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking



appleid -

For #1, think of it as you did in #2. If 2x - 5y = 0, then manipulating the equation gives us 2x = 5y (by subtracting 5y from each side). This gives the same algebraic set up as #2.

For #3, you are 99% of the way there. Fractions and ratios go together. Think of an example - you have 3 apples and 2 pears in your basket - how many pieces of fruit do you have? Of course, the answer is five and, thus, the fractions would be 3/5 of your fruit is apples and 2/5 are pears while the ratio of apples to pears is 3:2.

Now in #3 above, you came up with 2/7 and -5/7. Forget the negative sign becasue we are talking about distance rather than "quantity", so the fractions are 2/7 and 5/7. Re-engineering this into ratios gives us a ratio of x:y as 2:5.

I understand your question, as I too am more visual, but I thought an explanation would be quicker. If you would like, I draw this out visually and upload the document. Let me know.
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Re: weighted average problems [#permalink] New post 06 Aug 2015, 11:03
appleid wrote:
Oh, I was thinking of quantities altogether but showed only +5 and -2 for adding up to 0. The main question is:

Why these differentials with respective quantities add up to 0? or why it bring to the net effect = 0?

Any theory/concept/visuals that'd help?


The best method (without getting confused) is method 2 mentioned in your post above.
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Re: weighted average problems [#permalink] New post 12 Aug 2015, 23:26
Engr2012 wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:
3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking


You do not have to resort to #3 method to see that as 3% needs to go more towards 8% than 10%, there will be more of 3 and less of 10 (to compensate for more 'distance' to travel for 3%).Yes, you can just take the reciprocals of the "differentials" 2/7 and 5/7 to come up with 5/2 for x:y as x>y .


Shouldn't there will be more of 10% (x) to weight toward 8%?
Would you mind explaining the reciprocal process for me?
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Re: weighted average problems [#permalink] New post 12 Aug 2015, 23:30
VeritasPrepDennis wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

2.
(10%*x+3%y) = 8%*(x+y)
(10-8)x = (8-3)y
2x = 5y
x/y = 5/2
x to y Ratio will be 5:2

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking



appleid -

For #1, think of it as you did in #2. If 2x - 5y = 0, then manipulating the equation gives us 2x = 5y (by subtracting 5y from each side). This gives the same algebraic set up as #2.

For #3, you are 99% of the way there. Fractions and ratios go together. Think of an example - you have 3 apples and 2 pears in your basket - how many pieces of fruit do you have? Of course, the answer is five and, thus, the fractions would be 3/5 of your fruit is apples and 2/5 are pears while the ratio of apples to pears is 3:2.

Now in #3 above, you came up with 2/7 and -5/7. Forget the negative sign becasue we are talking about distance rather than "quantity", so the fractions are 2/7 and 5/7. Re-engineering this into ratios gives us aratio of x:y as 2:5.

I understand your question, as I too am more visual, but I thought an explanation would be quicker. If you would like, I draw this out visually and upload the document. Let me know.


I am 100% clear about method 1 and 2 so focusing on ration thinking with WA. The situation we have now:
2/7 distance from x
5/7 distance from y

I understand the ratio of your simple example, its easy for me. Whereas, when it comes to finding ratio from reciprocal multiplication of x and y, the ratio I am getting is

2x/7 = 5y/7
2x/7 * 7/5y = 5y/7 * 7/2x
2x/5y = 5y/2x #stuck!

how would I think of this in a numerical ratio of x to y??

another way to think about it
2/7 distance from x (like 3/5 of apples)
5/7 distance from y (like 2/5 of pears)

ratio of x to y is 2:5 (like ratio of apples to pears is 3:2)

This is certainly not right according to my method 2, there's more of x (10%) than y. So the ratio should be x:y is 5:2.

It would be helpful if you please explain "Re-engineering" process, that'd be enough for now, visuals can be the last resort.

Thanks a lot. I completely understood the first two method by both of your help and now trying to get into ration thinking from weighted average. Excuse me if I am pushing it too much. I hope it will help me to connect WA, Fractions, Ratios better in long run. :)
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Re: weighted average problems [#permalink] New post 13 Aug 2015, 05:58
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appleid wrote:
Engr2012 wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:
3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking


You do not have to resort to #3 method to see that as 3% needs to go more towards 8% than 10%, there will be more of 3 and less of 10 (to compensate for more 'distance' to travel for 3%).Yes, you can just take the reciprocals of the "differentials" 2/7 and 5/7 to come up with 5/2 for x:y as x>y .


Shouldn't there will be more of 10% (x) to weight toward 8%?
Would you mind explaining the reciprocal process for me?


Yes, you are correct that a greater 'x' will push the number closer to 10.
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Re: weighted average problems [#permalink] New post 13 Aug 2015, 06:21
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appleid wrote:
VeritasPrepDennis wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

2.
(10%*x+3%y) = 8%*(x+y)
(10-8)x = (8-3)y
2x = 5y
x/y = 5/2
x to y Ratio will be 5:2

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
So, I know there's more quantity of x than y because the WA is closer to x, considering 6.5% average; but how would I think of this in a numerical ratio of x to y??.......should I think of reciprocal multiplication?? #I am stuck in this ratio thinking



appleid -

For #1, think of it as you did in #2. If 2x - 5y = 0, then manipulating the equation gives us 2x = 5y (by subtracting 5y from each side). This gives the same algebraic set up as #2.

For #3, you are 99% of the way there. Fractions and ratios go together. Think of an example - you have 3 apples and 2 pears in your basket - how many pieces of fruit do you have? Of course, the answer is five and, thus, the fractions would be 3/5 of your fruit is apples and 2/5 are pears while the ratio of apples to pears is 3:2.

Now in #3 above, you came up with 2/7 and -5/7. Forget the negative sign becasue we are talking about distance rather than "quantity", so the fractions are 2/7 and 5/7. Re-engineering this into ratios gives us aratio of x:y as 2:5.

I understand your question, as I too am more visual, but I thought an explanation would be quicker. If you would like, I draw this out visually and upload the document. Let me know.


I am 100% clear about method 1 and 2 so focusing on ration thinking with WA. The situation we have now:
2/7 distance from x
5/7 distance from y

I understand the ratio of your simple example, its easy for me. Whereas, when it comes to finding ratio from reciprocal multiplication of x and y, the ratio I am getting is

2x/7 = 5y/7
2x/7 * 7/5y = 5y/7 * 7/2x
2x/5y = 5y/2x #stuck!

how would I think of this in a numerical ratio of x to y??

another way to think about it
2/7 distance from x (like 3/5 of apples)
5/7 distance from y (like 2/5 of pears)

ratio of x to y is 2:5 (like ratio of apples to pears is 3:2)

This is certainly not right according to my method 2, there's more of x (10%) than y. So the ratio should be x:y is 5:2.

It would be helpful if you please explain "Re-engineering" process, that'd be enough for now, visuals can be the last resort.

Thanks a lot. I completely understood the first two method by both of your help and now trying to get into ration thinking from weighted average. Excuse me if I am pushing it too much. I hope it will help me to connect WA, Fractions, Ratios better in long run. :)




appleid -

You have got this 95% correct - but have unnecessarily complicated it at the end.

You are perfect up to this point : 2x/7 = 5y/7 and then make a minor algebra mistake.

If 2x/7 = 5y/7, you can multiply both sides by 7 and get 2x = 5y

If 2x = 5y, then y=2/5x and x=5/2y. Looking at it this way, you can see that you need MORE of x (the 10% solution) to make the 8% solution
Also, you can clearly see how the ratio of 5:2 now is represented by the fractions (ie, that you will use 2.5 times more x than y, or only 40% the amount of x for y)


In the end, if you get one method stick with it, although being flexible helps on the exam. Let me know if this makes sense.
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weighted average problems [#permalink] New post 14 Aug 2015, 11:06
VeritasPrepDennis wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

I am 100% clear about method 1 and 2 so focusing on ration thinking with WA. The situation we have now:
2/7 distance from x
5/7 distance from y

I understand the ratio of your simple example, its easy for me. Whereas, when it comes to finding ratio from reciprocal multiplication of x and y, the ratio I am getting is
2x/7= 5y/7
2x/7 * 7/5y = 5y/7 * 7/2x
2x/5y = 5y/2x #stuck!

how would I think of this in a numerical ratio of x to y??

another way to think about it
2/7 distance from x (like 3/5 of apples)
5/7 distance from y (like 2/5 of pears)

ratio of x to y is 2:5 (like ratio of apples to pears is 3:2)

This is certainly not right according to my method 2, there's more of x (10%) than y. So the ratio should be x:y is 5:2.

It would be helpful if you please explain "Re-engineering" process, that'd be enough for now, visuals can be the last resort.

Thanks a lot. I completely understood the first two method by both of your help and now trying to get into ration thinking from weighted average. Excuse me if I am pushing it too much. I hope it will help me to connect WA, Fractions, Ratios better in long run. :)


appleid -

You have got this 95% correct - but have unnecessarily complicated it at the end.

You are perfect up to this point : 2x/7 = 5y/7 and then make a minor algebra mistake.

If 2x/7 = 5y/7, you can multiply both sides by 7 and get 2x = 5y

If 2x = 5y, then y=2/5x and x=5/2y. Looking at it this way, you can see that you need MORE of x (the 10% solution) to make the 8% solution
Also, you can clearly see how the ratio of 5:2 now is represented by the fractions (ie, that you will use 2.5 times more x than y, or only 40% the amount of x for y)


In the end, if you get one method stick with it, although being flexible helps on the exam. Let me know if this makes sense.


Its all clear now, I was thinking of taking reciprocals and somewhat lost myself in working through fractions. Thanks both of you. I am compiling the steps in method 3:

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
As the differentials are distance from WA, you can omit negative sign for calculation and make them equal *

The fractions are:
2/7 distance from x = \(\frac{2}{7}x\)
5/7 distance from y = \(\frac{5}{7}y\)

As both distances come and meet at WA 8%* so,
\(\frac{2}{7}x = \frac{5}{7}y\)
\(2x = 5y\) #Multiply by 7 (Not taking reciprocals)
\(x = \frac{5y}{2}\)
\(\frac{x}{y} = \frac{5}{2}\)
So the ratio from the fraction is \(x:y = 5:2\)

Hope its perfect now. Please confirm me if everything is alright. :)

Another question (let it be last one..haha) to VeritasPrepDennis,
I tried to apply the same method on the example you gave me on 3 apple (a) and 2 pears(p):
\(\frac{3}{5}a = \frac{2}{5}p\)
\(3a = 2p\)
\(a = \frac{2p}{3}\)
\(\frac{a}{p} = \frac{2}{3}\)
So the ratio from the fraction is \(a:p = 2:3\)

Where am I getting it wrong? Is it because it's a simple fraction-ratio math compared to weighted average math?
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Re: weighted average problems [#permalink] New post 14 Aug 2015, 11:39
Expert's post
appleid wrote:
VeritasPrepDennis wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:

I am 100% clear about method 1 and 2 so focusing on ration thinking with WA. The situation we have now:
2/7 distance from x
5/7 distance from y

I understand the ratio of your simple example, its easy for me. Whereas, when it comes to finding ratio from reciprocal multiplication of x and y, the ratio I am getting is
2x/7= 5y/7
2x/7 * 7/5y = 5y/7 * 7/2x
2x/5y = 5y/2x #stuck!

how would I think of this in a numerical ratio of x to y??

another way to think about it
2/7 distance from x (like 3/5 of apples)
5/7 distance from y (like 2/5 of pears)

ratio of x to y is 2:5 (like ratio of apples to pears is 3:2)

This is certainly not right according to my method 2, there's more of x (10%) than y. So the ratio should be x:y is 5:2.

It would be helpful if you please explain "Re-engineering" process, that'd be enough for now, visuals can be the last resort.

Thanks a lot. I completely understood the first two method by both of your help and now trying to get into ration thinking from weighted average. Excuse me if I am pushing it too much. I hope it will help me to connect WA, Fractions, Ratios better in long run. :)


appleid -

You have got this 95% correct - but have unnecessarily complicated it at the end.

You are perfect up to this point : 2x/7 = 5y/7 and then make a minor algebra mistake.

If 2x/7 = 5y/7, you can multiply both sides by 7 and get 2x = 5y

If 2x = 5y, then y=2/5x and x=5/2y. Looking at it this way, you can see that you need MORE of x (the 10% solution) to make the 8% solution
Also, you can clearly see how the ratio of 5:2 now is represented by the fractions (ie, that you will use 2.5 times more x than y, or only 40% the amount of x for y)


In the end, if you get one method stick with it, although being flexible helps on the exam. Let me know if this makes sense.


Its all clear now, I was thinking of taking reciprocals and somewhat lost myself in working through fractions. Thanks both of you. I am compiling the steps in method 3:

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
As the differentials are distance from WA, you can omit negative sign for calculation and make them equal *

The fractions are:
2/7 distance from x = \(\frac{2}{7}x\)
5/7 distance from y = \(\frac{5}{7}y\)

As both distances come and meet at WA 8%* so,
\(\frac{2}{7}x = \frac{5}{7}y\)
\(2x = 5y\) #Multiply by 7 (Not taking reciprocals)
\(x = \frac{5y}{2}\)
\(\frac{x}{y} = \frac{5}{2}\)
So the ratio from the fraction is \(x:y = 5:2\)

Hope its perfect now. Please confirm me if everything is alright. :)

Another question (let it be last one..haha) to VeritasPrepDennis,
I tried to apply the same method on the example you gave me on 3 apple (a) and 2 pears(p):
\(\frac{3}{5}a = \frac{2}{5}p\)
\(3a = 2p\)
\(a = \frac{2p}{3}\)
\(\frac{a}{p} = \frac{2}{3}\)
So the ratio from the fraction is \(a:p = 2:3\)

Where am I getting it wrong? Is it because it's a simple fraction-ratio math compared to weighted average math?



appleid -

Yes, the "error" in the apples to pears analogy is that in setting up the equality it would actually be a/p : 3/2, or a/p = 3/2 ==> in multiplying both sides by p, we'd get a = 3/2p.

This is not a great example though.

Here is a better problem that ilustrates the concept.

Say you have unlimited quantities of 15% HCL (hydro-chloric acid) and 50% HCL. How many gallons of the 15% HCL will you need to use to make 7 gallons of a 30% HCL solution?
A) 2
B) 3
C) 4
D) 5
E) 5 2/3
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Re: weighted average problems   [#permalink] 14 Aug 2015, 11:39

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