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Weighted Average and Mixture Problems on the GMAT

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Re: Weighted Average and Mixture Problems on the GMAT  [#permalink]

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New post 21 Nov 2018, 07:41
1
Bunuel wrote:
Zap the Weighted Average Brutes

BY KARISHMA, VERITAS PREP


I visit the GMAT Club forum regularly and discuss some ideas, some methodologies there. The weighted averages method I discussed in my previous two posts is one of my most highly appreciated inputs on the forum. People love how easily they can solve some of the most difficult questions by just drawing a scale or using a ratio. If you are not a Quant jock, I am sure you feel a chill run down your spine every time you see a mixtures problem. But guess what, they are really simple if you just use the same weighted average concepts we discussed in the previous two posts. Let’s look at a mixtures question in detail:

Question: Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

(A) 3 gallons
(B) 4 gallons
(C) 5 gallons
(D) 10 gallons
(E) 12 gallons

Solution:

Let’s try and imagine what is taking place here:

Image

Imagine two tumblers, one containing Mixture A and the other containing Mixture B. The mixtures from the two tumblers are poured into a new big empty tumbler. The first thing that we can conclude from this step is that V1 + V2 gallons = 15 gallons because the total volume when you combined the mixtures turned out to be 15 gallons (in the big tumbler).

Can you say anything about the relative value of V1 and V2? Can you say which one will be greater? We discussed in the previous post that when we find the weighted average of two quantities, the average is closer to the one which has higher weight because that quantity ‘pulls’ the average toward itself. So if the new mixture contains 30% alcohol, which of the two mixtures – A (20% alcohol) or B (50% alcohol), will have higher volume? Obviously A because it pulled the average toward itself. The average, 30% alcohol, is closer to 20% than to 50%. Now, let’s calculate exactly how much of the 15 gallons was mixture A and how much was mixture B.

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A1)}\)

\(\frac{w_1}{w_2}= \frac{(50 – 30)}{(30 – 20)} = \frac{2}{1}\)

Ratio of volume of mixture A : volume of mixture B = 2:1.

So out of a total of 15 gallons, 10 gallons will be mixture A and 5 gallons will be mixture B. (Check out the post on Ratios if you are not sure how we arrived at this.)

Let’s try the same thing using the scale method:

Image

V1 : V2 = 2:1

So out of a total of 15 gallons, 10 gallons will be mixture A and 5 gallons will be mixture B. It wasn’t very bad, was it?

This question is discussed HERE.

And now, let’s try a quick example that looks ominous. People end up making tons of equations with ‘x’ and ‘y’ to solve this one but we can do it in a moment!

Question: There are 2 bars of copper-nickel alloy. One bar has 2 parts of copper to 5 parts of nickel. The other has 3 parts of copper to 5 parts of nickel. If both bars are melted together to get a 20 kg bar with the final copper to nickel ratio of 5:11. What was the weight of the first bar?

(A) 1 kg
(B) 4 kg
(C) 6 kg
(D) 14 kg
(E) 16 kg

Solution:

This question might look different from the question above but it is actually very similar to it. Focus on any one of the two elements, say Copper (or Nickel if you fancy it more. Either way, you will get the same answer). Forget about the other one. If you notice, in the question above there are two elements in each mixture too (obviously, that is why they are called mixtures!) – Alcohol and Water. 20% Alcohol means that the rest 80% is Water. We just work with the concentration of Alcohol because it is given explicitly to us. We could as well say that mixture A has 80% water and mixture B has 50% water and the combined mixture has 70% water. Still the ratio obtained will be 2:1. But, still, somehow, giving the ratio of both Copper and Nickel in this question throws people off.

Let me quote one of my favorite GMAT instructors here: “And this is one of many reasons why McDonald’s, the greatest marketing organization on earth, is a lot like the GMAT, the test that can help you get into the greatest marketing MBA programs on earth. They know how to make those tiny tweaks that make massive differences, at least in perception to the end consumer.” (Check out the complete(-ly) delicious article that discusses Shamrock Shakes here.)

Anyway, let’s get back to this question. I am keen to show you that it is, in fact, just like the previous question.

First bar is 2/7th copper (2 parts copper and 5 parts nickel to get a total of 7 parts). Second bar is 3/8th copper and the combined alloy is 5/16th copper.

Now, let’s calculate the ratio of weights of the two bars.

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\)

\(\frac{w_1}{w_2} = \frac{(\frac{3}{8} – \frac{5}{16})}{(\frac{5}{16} – \frac{2}{7})} = \frac{7}{3}\)

So weight of first bar: weight of second bar = 7:3

Out of a total of 20 kgs, the weight of the first bar was 14 kg and the weight of the second bar was 6 kg.

Or we can use the scale method too.

Image

So the ratio of their weights is 1:3/7 or 7:3

Out of a total of 20 kgs, the weight of the first bar was 14 kg and the weight of the second bar was 6 kg.

This question is discussed HERE.

Hope these questions don’t look very intimidating now. Use this method for your mixture problems; they won’t feel like problems any more!

Attachment:
Ques3.jpg

Attachment:
Ques2.jpg

Attachment:
Ques42.jpg
Please explain the last part, I am getting first bar as 14kg and second bar as 6 kg. W1*(1/16)=w2*(3/7*16)

Sent from my ONE A2003 using GMAT Club Forum mobile app
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Re: Weighted Average and Mixture Problems on the GMAT  [#permalink]

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New post 21 Nov 2018, 08:18
Unnathi wrote:
Bunuel wrote:
Zap the Weighted Average Brutes

BY KARISHMA, VERITAS PREP


I visit the GMAT Club forum regularly and discuss some ideas, some methodologies there. The weighted averages method I discussed in my previous two posts is one of my most highly appreciated inputs on the forum. People love how easily they can solve some of the most difficult questions by just drawing a scale or using a ratio. If you are not a Quant jock, I am sure you feel a chill run down your spine every time you see a mixtures problem. But guess what, they are really simple if you just use the same weighted average concepts we discussed in the previous two posts. Let’s look at a mixtures question in detail:

Question: Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

(A) 3 gallons
(B) 4 gallons
(C) 5 gallons
(D) 10 gallons
(E) 12 gallons

Solution:

Let’s try and imagine what is taking place here:

Image

Imagine two tumblers, one containing Mixture A and the other containing Mixture B. The mixtures from the two tumblers are poured into a new big empty tumbler. The first thing that we can conclude from this step is that V1 + V2 gallons = 15 gallons because the total volume when you combined the mixtures turned out to be 15 gallons (in the big tumbler).

Can you say anything about the relative value of V1 and V2? Can you say which one will be greater? We discussed in the previous post that when we find the weighted average of two quantities, the average is closer to the one which has higher weight because that quantity ‘pulls’ the average toward itself. So if the new mixture contains 30% alcohol, which of the two mixtures – A (20% alcohol) or B (50% alcohol), will have higher volume? Obviously A because it pulled the average toward itself. The average, 30% alcohol, is closer to 20% than to 50%. Now, let’s calculate exactly how much of the 15 gallons was mixture A and how much was mixture B.

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A1)}\)

\(\frac{w_1}{w_2}= \frac{(50 – 30)}{(30 – 20)} = \frac{2}{1}\)

Ratio of volume of mixture A : volume of mixture B = 2:1.

So out of a total of 15 gallons, 10 gallons will be mixture A and 5 gallons will be mixture B. (Check out the post on Ratios if you are not sure how we arrived at this.)

Let’s try the same thing using the scale method:

Image

V1 : V2 = 2:1

So out of a total of 15 gallons, 10 gallons will be mixture A and 5 gallons will be mixture B. It wasn’t very bad, was it?

This question is discussed HERE.

And now, let’s try a quick example that looks ominous. People end up making tons of equations with ‘x’ and ‘y’ to solve this one but we can do it in a moment!

Question: There are 2 bars of copper-nickel alloy. One bar has 2 parts of copper to 5 parts of nickel. The other has 3 parts of copper to 5 parts of nickel. If both bars are melted together to get a 20 kg bar with the final copper to nickel ratio of 5:11. What was the weight of the first bar?

(A) 1 kg
(B) 4 kg
(C) 6 kg
(D) 14 kg
(E) 16 kg

Solution:

This question might look different from the question above but it is actually very similar to it. Focus on any one of the two elements, say Copper (or Nickel if you fancy it more. Either way, you will get the same answer). Forget about the other one. If you notice, in the question above there are two elements in each mixture too (obviously, that is why they are called mixtures!) – Alcohol and Water. 20% Alcohol means that the rest 80% is Water. We just work with the concentration of Alcohol because it is given explicitly to us. We could as well say that mixture A has 80% water and mixture B has 50% water and the combined mixture has 70% water. Still the ratio obtained will be 2:1. But, still, somehow, giving the ratio of both Copper and Nickel in this question throws people off.

Let me quote one of my favorite GMAT instructors here: “And this is one of many reasons why McDonald’s, the greatest marketing organization on earth, is a lot like the GMAT, the test that can help you get into the greatest marketing MBA programs on earth. They know how to make those tiny tweaks that make massive differences, at least in perception to the end consumer.” (Check out the complete(-ly) delicious article that discusses Shamrock Shakes here.)

Anyway, let’s get back to this question. I am keen to show you that it is, in fact, just like the previous question.

First bar is 2/7th copper (2 parts copper and 5 parts nickel to get a total of 7 parts). Second bar is 3/8th copper and the combined alloy is 5/16th copper.

Now, let’s calculate the ratio of weights of the two bars.

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\)

\(\frac{w_1}{w_2} = \frac{(\frac{3}{8} – \frac{5}{16})}{(\frac{5}{16} – \frac{2}{7})} = \frac{7}{3}\)

So weight of first bar: weight of second bar = 7:3

Out of a total of 20 kgs, the weight of the first bar was 14 kg and the weight of the second bar was 6 kg.

Or we can use the scale method too.

Image

So the ratio of their weights is 1:3/7 or 7:3

Out of a total of 20 kgs, the weight of the first bar was 14 kg and the weight of the second bar was 6 kg.

This question is discussed HERE.

Hope these questions don’t look very intimidating now. Use this method for your mixture problems; they won’t feel like problems any more!

Attachment:
Ques3.jpg

Attachment:
Ques2.jpg

Attachment:
Ques42.jpg
Please explain the last part, I am getting first bar as 14kg and second bar as 6 kg. W1*(1/16)=w2*(
3/7*16)

Sent from my ONE A2003 using GMAT Club Forum mobile app


You are getting the right answer as Karishma got. So, what else you want to clarify? Please explain.
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Re: Weighted Average and Mixture Problems on the GMAT  [#permalink]

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New post 18 Apr 2019, 23:39
Hi,

I am having difficulty understanding this re-arrangement of the formula:

So, Average Age, \(A_{avg} = \frac{(A_1*w_1 + A_2*w_2)}{(w_1 + w_2)}\)

Now if we re-arrange this formula, we get, \(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\)


Bunuel wrote:

Weighted Average and Mixture Problems on the GMAT




Heavily Weighted Weighted Averages

BY KARISHMA, VERITAS PREP


Today, I will delve into one of the most important topics (ubiquitous application) that are tested on GMAT. It is also one of the topics that will appear time and again during MBA e.g. in Corporate finance, you might be taught how to find ‘Weighted Average Cost of Capital’. So it will be highly beneficial if you have a feel for weighted average concepts.

The first question is – What is Weighted Average? Let me explain with an example.

A boy’s age is 17 years and a girl’s age is 20 years. What is their average age?

Simple enough, isn’t it? \(Average \ age = \frac{(17 + 20)}{2} = 18.5\)

It is the number that lies in the middle of 17 and 20. (Another method of arriving at this number would be to find the difference between them, 3, and divide it into 2 equal parts, 1.5 each. Now add 1.5 to the smaller number, 17, to get the average age of 18.5 years. Or subtract 1.5 from the greater number, 20, to get the average age of 18.5 years. But I digress. I will take averages later since it is just a special case of weighted averages.)

Now let me change the question a little.

There are 10 boys and 20 girls in a group. Average age of boys is 17 years and average age of girls is 20 years. What is the average age of the group?

Many people will be able to arrive at the following:

\(Average \ Age = \frac{(17*10 + 20*20)}{(10 + 20)} = 19\) years

Average age will be total number of years in the age of everyone in the group divided by total number of people in the group. Since the average age of boys is 17, so total number of years in the 10 boys’ ages is 17*10. Since the average age of girls is 20, the total number of years in the 20 girls’ ages is 20*20. The total number of boys and girls is 10 + 20. Hence you use the expression given above to find the average age. I hope we are good up till now.

To establish a general formula, let me restate this question using variables and then we will just plug in the variables in place of the actual numbers above (Yes, it is opposite of what you would normally do when you have the formula and you plug in numbers. Our aim here is to deduce a generic formula from a specific example because the calculation above is intuitive to many of you but the formula is a little intimidating.)

There are w1 boys and w2 girls in a group. Average age of boys is A1 years and average age of girls is A2 years. What is the average age of the group?

\(Average \ Age = \frac{(A_1*w_1 + A_2*w_2)}{(w_1 + w_2)}\)

This is weighted average. Here we are not finding the average age of 1 boy and 1 girl. Instead we are finding the average age of 10 boys and 20 girls so their average age will not be 18.5 years. Boys have been given less weightage in the calculation of average because there are only 10 boys as compared to 20 girls. So the average has been found after accounting for the weightage (or ‘importance’ in regular English) given to boys and girls depending on how many boys and how many girls there are. Notice that the weighted average is 19 years which is closer to the average age of girls than to the average age of boys. This is because there are more girls so they ‘pull’ the average towards their own age i.e. 20 years.

Now that you know what weighted average is and also that you always knew the weighted average formula intuitively, let’s move on to making things easier for you (Tougher, you say? Actually, once people know the scale method that I am going to discuss right now (It has been discussed in our Statistics and Problem Solving book too), they just love it!)

So, Average Age, \(A_{avg} = \frac{(A_1*w_1 + A_2*w_2)}{(w_1 + w_2)}\)

Now if we re-arrange this formula, we get, \(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\)

So we have got the ratio of weights w1 and w2 (the number of boys and the number of girls). How does it help us? Knowing this ratio, we can directly get the answer. Another example will make this clear.

John pays 30% tax and Ingrid pays 40% tax. Their combined tax rate is 37%. If John’s gross salary is $54000, what is Ingrid’s gross salary?

Here, we have the tax rate of John and Ingrid and their average tax rate. A1 = 30%, A2 = 40% and Aavg = 37%. The weights are their gross salaries – $54,000 for John and w2 for Ingrid. From here on, there are two ways to find the answer. Either plug in the values in the formula above or use the scale method. We will take a look at both.

1. Plug in the formula

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}= \frac{(40 – 37)}{(37 – 30)} = \frac{3}{7}\)

Since A1 is John’s tax rate and A2 is Ingrid’s tax rate, w1 is John’s salary and w2 is Ingrid’s salary

\(\frac{w_1}{w_2} = \frac{John’s \ Salary}{Ingrid’s \ Salary} = \frac{3}{7} = \frac{54,000}{Ingrid’s \ Salary}\)

So Ingrid’s Salary = $126,000

It should be obvious that either John or Ingrid could be A1 (and the other would be A2). For ease, it a good idea to denote the larger number as A2 and the smaller as A1 (even if you do the other way around, you will still get the same answer)

2. Scale Method

On the number line, put the smaller number on the left side and the greater number on the right side (since it is intuitive that way). Put the average in the middle.

Image

The distance between 30 and 37 is 7 and the distance between 37 and 40 is 3 so \(w_1:w_2 = 3:7\) (As seen by the formula, the ratio is flipped).

Since \(w_1 = 54,000\), \(w_2\) will be 126,000

So Ingrid’s salary is $126,000.

This method is especially useful when you have the average and need to find the ratio of weights. Check out next week’s post for some 700 level examples of weighted average.

Attachment:
Ques21.jpg
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Weighted Average and Mixture Problems on the GMAT  [#permalink]

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New post 18 Apr 2019, 23:57
2
Quote:
divyajeswani"]Hi,

I am having difficulty understanding this re-arrangement of the formula:

So, Average Age, \(A_{avg} = \frac{(A_1*w_1 + A_2*w_2)}{(w_1 + w_2)}\)

Now if we re-arrange this formula, we get, \(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\)


divyajeswani


\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}\)

\(w_1(A_{avg} - A_1) = w_2(A_2 - A_{avg})\)

\(w_1*A_{avg} - w_1*A_1 = w_2*A_2 - w_2*A_{avg}\)

\(w_1*A_{avg} + w_2*A_{avg} = w_2*A_2 + w_1*A_1\)

\(A _{avg} (w_1 + w_2) = w_1*A_1 + w_2*A_2\)

\(A_{avg} = \frac{w_1*A_1 + w_2*A_2}{w_1 + w_2}\)


Hope this helps.

All the best.
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Re: Weighted Average and Mixture Problems on the GMAT  [#permalink]

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New post 30 Nov 2019, 02:40
Breaking down the replacement question.

I was wondering if we can simply identify that the alcohol concentration so that we can simply use compounding formula when concentration is reducing..
Say

Alcohol final concentration after n replacements =Ai(1-10/100)(1-10/100)..and so on

Say we have a solution of 50% alcohol and water and there are two replacements of 20% of the solution
We can simply relate this as --starting amount of alcohol = 50grams

After two replacements ,
Final amount of alcohol = 50(0.9)(0.9)
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Re: Weighted Average and Mixture Problems on the GMAT  [#permalink]

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New post 01 Dec 2019, 23:56
prabsahi wrote:
Breaking down the replacement question.

I was wondering if we can simply identify that the alcohol concentration so that we can simply use compounding formula when concentration is reducing..
Say

Alcohol final concentration after n replacements =Ai(1-10/100)(1-10/100)..and so on

Say we have a solution of 50% alcohol and water and there are two replacements of 20% of the solution
We can simply relate this as --starting amount of alcohol = 50grams

After two replacements ,
Final amount of alcohol = 50(0.9)(0.9)


Absolutely! It is discussed in detail on our blog here: https://www.veritasprep.com/blog/2012/0 ... -mixtures/

Final concentration = Initial Concentration * (Initial Volume/Final Volume)^n
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Re: Weighted Average and Mixture Problems on the GMAT   [#permalink] 01 Dec 2019, 23:56

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