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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:



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:



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.



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!

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Hi Bunuel,

Can we get pdf of these somehow? I prefer offline reading and the problem is printing these distorts the equations so i dont know how to do that?

Please help with some suggestion!

Thanks
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Bunuel


Hi Bunuel,

Can we get pdf of these somehow? I prefer offline reading and the problem is printing these distorts the equations so i dont know how to do that?

Please help with some suggestion!

Thanks

You can click on print view at th top of each topic and then print as PDF (if you use chrome or safari ). This will give you a PDF of the entire topic.

Hope this helps.
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Bunuel


Hi Bunuel,

Can we get pdf of these somehow? I prefer offline reading and the problem is printing these distorts the equations so i dont know how to do that?

Please help with some suggestion!

Thanks

You can click on print view at th top of each topic and then print as PDF (if you use chrome or safari ). This will give you a PDF of the entire topic.

Hope this helps.



Hi,

Thanks for the reply. I tried using that but its distorting the divisions signs as it shows "/" by "(fraction {a}{b} "... Its not easy if you understand. I am using chrome!
Any other alternative?

Thanks
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Really appreciate the work involved behind the topics posted! Quite useful! Thank you!
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Hello Karishma,

Could you please explain what should be weight is this problem and why?I thought weight is 94,88,98 and 85 but its wrong.

In a translation class, 40% of the student's grade comes from exams, 30% from written assignments, 20% from conversational practice, and 10% from interpretation. If a student's grades are 94 for exams, 88 for written assignments, 98 for conversational practice, and 85 for interpretation, what is the student's overall grade in the course?


Many thanks in advance
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vaidhaichaturvedi
Hello Karishma,

Could you please explain what should be weight is this problem and why?I thought weight is 94,88,98 and 85 but its wrong.

In a translation class, 40% of the student's grade comes from exams, 30% from written assignments, 20% from conversational practice, and 10% from interpretation. If a student's grades are 94 for exams, 88 for written assignments, 98 for conversational practice, and 85 for interpretation, what is the student's overall grade in the course?


Many thanks in advance

Total grade = 40% of exams + 30% of assignment + 20% of Conversation + 10 % of interpretation
= \(0.4 * 94 + 0.3 * 88 + 0.2 * 98 + 0.1 * 85 = 37.6 + 26.4 + 19.6 + 8.5 = 92.1\)
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Quote:
Question 1: A container has 3 liters of pure lime juice. 1 liter from the container is taken out and 2 liter water is added. The process is repeated several times. After 19 such operations, quantity of lime juice in the mixture is

Initial concentration = 1 , Volume 3 L

Conc. after 1st Iteration, Volume = 3 L --> remove 1 L and 2 L -->\(\frac{v1}{v2} = \frac{(3-1)}{(3-1+2)} = \frac{2}{4}\)
Conc. after 2nd Iteration Volume = 4 L --> remove 1 L and 2 L --> \(\frac{v1}{v2} = \frac{(4-1)}{(4-1+2)} = \frac{3}{5}\)
Conc. after 3nd Iteration Volume = 5 L --> remove 1 L and 2 L --> \(\frac{v1}{v2} = \frac{(5-1)}{(5-1+2)} = \frac{4}{6}\)
...
...
Conc. after nth Iteration Volume = (n+2) L --> remove 1 L and 2 L --> \(\frac{v1}{v2} = \frac{((n+2)-1)}{((n+2)-1+2 )} = \frac{n+1}{n+3}\)
Conc. after 18th Iteration Volume = 20 L --> remove 1 L and 2 L --> \(\frac{v1}{v2} = \frac{(20-1)}{(20-1+2)} = \frac{19}{21}\)
Conc. after 19th Iteration Volume = 21 L --> remove 1 L and 2 L --> \(\frac{v1}{v2} = \frac{(21-1)}{(21-1+2)} = \frac{20}{22}\)


Final conc = \(1* \frac{2}{4} * \frac{3}{5} * \frac{4}{6} * \frac{19}{21} * \frac{20}{22}\) = \(\frac{1}{77}\).

After 19th iteration total volume of solution = 22L

Total volume of juice = \(22 * \frac{1}{77} = \frac{2}{7} L\)
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Its indeed very useful and thanks Karishma and Bunuel for creating and sharing such useful material.
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Thanks for sharing this . It certainly makes Weighted average questions look easy.
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Removal/Replacement in Mixtures
BY KARISHMA, VERITAS PREP

Here we are going to take removal/replacement in mixtures. For those of you who were looking forward to some more tricky probability questions, I will make up for your disappointment next week. Meanwhile, rest assured, replacement is a very interesting, not to mention useful, concept in GMAT. So brace yourself to learn some new things today.

First of all, many “replacement” questions are nothing but the plain old mixture questions, the type we discussed in this post, with an extra step. So don’t flip out the moment you read the word “replace.” Let me show you what I mean:

Example 1: If a portion of a 50% alcohol solution (in water) is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

Solution: What this question says is that a solution of 25% alcohol (say solution 1) is mixed with another solution of 50% alcohol (say solution 2) to give us a 30% alcohol solution. We don’t know in what quantities they were mixed. Can we find out the ratio in which these two solutions were mixed? If you are not sure, check out this post.

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)} = \frac{(50 – 30)}{(30 – 25)} = \frac{4}{1}\)

So the two solutions were mixed in the ratio 4:1. Mind you,

(volume of 25% alcohol solution) : (volume of 50% alcohol solution) = 4:1


Hi allpowerful VeritasPrepKarishma , :)

As per below mentioned concept (marked in green) about which one has more volume V1 or V2. Why in the above mentioned question A2 = 50 and not 25 ?
25 pulled volume to itself as it is closer to 30. should not it be like this:

\(A1 = 50\)
\(A2 = 25\)
\(Aavg = 30\)

\(\frac{25-30}{30-50}\) = \(\frac{-1}{-4}\) ?? please explain :)


"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."

many thanks! :)
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Bunuel
Removal/Replacement in Mixtures
BY KARISHMA, VERITAS PREP

Here we are going to take removal/replacement in mixtures. For those of you who were looking forward to some more tricky probability questions, I will make up for your disappointment next week. Meanwhile, rest assured, replacement is a very interesting, not to mention useful, concept in GMAT. So brace yourself to learn some new things today.

First of all, many “replacement” questions are nothing but the plain old mixture questions, the type we discussed in this post, with an extra step. So don’t flip out the moment you read the word “replace.” Let me show you what I mean:

Example 1: If a portion of a 50% alcohol solution (in water) is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

Solution: What this question says is that a solution of 25% alcohol (say solution 1) is mixed with another solution of 50% alcohol (say solution 2) to give us a 30% alcohol solution. We don’t know in what quantities they were mixed. Can we find out the ratio in which these two solutions were mixed? If you are not sure, check out this post.

\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)} = \frac{(50 – 30)}{(30 – 25)} = \frac{4}{1}\)

So the two solutions were mixed in the ratio 4:1. Mind you,

(volume of 25% alcohol solution) : (volume of 50% alcohol solution) = 4:1


Hi allpowerful VeritasPrepKarishma , :)

As per below mentioned concept (marked in green) about which one has more volume V1 or V2. Why in the above mentioned question A2 = 50 and not 25 ?
25 pulled volume to itself as it is closer to 30. should not it be like this:

\(A1 = 50\)
\(A2 = 25\)
\(Aavg = 30\)

\(\frac{25-30}{30-50}\) = \(\frac{-1}{-4}\) ?? please explain :)


"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."

many thanks! :)
Hi dave, let me try to explain this one as I understood it. It doesn't really matter which you choose to be A1 or A2 because A1 goes with w1 and A2 with w2 so if you decide A1 is 50 you will get the same answer because the ratio will become w2/w1 = (25-30/30-50) = 1/4. They said you should set A2 as the greatest number to ease things but A2 doesn't have to be the closest to the average. As long as you keep them linked with their "counterparts" you can switch them they way you like. I hope this helped. ;)
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Bunuel

Example 1: If a portion of a 50% alcohol solution (in water) is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

Solution: What this question says is that a solution of 25% alcohol (say solution 1) is mixed with another solution of 50% alcohol (say solution 2) to give us a 30% alcohol solution. We don’t know in what quantities they were mixed. Can we find out the ratio in which these two solutions were mixed? If you are not sure, check out this post.

w1w2=(A2–Aavg)(Aavg–A1)=(50–30)(30–25)=41w1w2=(A2–Aavg)(Aavg–A1)=(50–30)(30–25)=41

So the two solutions were mixed in the ratio 4:1. Mind you,

(volume of 25% alcohol solution) : (volume of 50% alcohol solution) = 4:1

Out of 5 parts of total solution obtained, 50% solution was 1 part while 4 parts was 25% solution. So what part of the 50% solution was removed and replaced by the 25% solution? Would you agree it is (4/5)th? We can say that 80% of the 50% solution was replaced by the 25% solution.


In this question my ratio comes out to be 1:4 rather then 4:1. In many other questions too i'm having trouble identifying which value must be selected for a1 or a2. Can anyone please help me out. I'll be grateful.

mikemcgarry
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VeritasPrepKarishma
Bunuel

Example 1: If a portion of a 50% alcohol solution (in water) is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

Solution: What this question says is that a solution of 25% alcohol (say solution 1) is mixed with another solution of 50% alcohol (say solution 2) to give us a 30% alcohol solution. We don’t know in what quantities they were mixed. Can we find out the ratio in which these two solutions were mixed? If you are not sure, check out this post.

w1w2=(A2–Aavg)(Aavg–A1)=(50–30)(30–25)=41w1w2=(A2–Aavg)(Aavg–A1)=(50–30)(30–25)=41

So the two solutions were mixed in the ratio 4:1. Mind you,

(volume of 25% alcohol solution) : (volume of 50% alcohol solution) = 4:1

Out of 5 parts of total solution obtained, 50% solution was 1 part while 4 parts was 25% solution. So what part of the 50% solution was removed and replaced by the 25% solution? Would you agree it is (4/5)th? We can say that 80% of the 50% solution was replaced by the 25% solution.


In this question my ratio comes out to be 1:4 rather then 4:1. In many other questions too i'm having trouble identifying which value must be selected for a1 or a2. Can anyone please help me out. I'll be grateful.

mikemcgarry

This question is discussed here: https://gmatclub.com/forum/if-a-portion ... 00271.html You can find there all sorts of different solutions. In case if the questions still remains, please post there.
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VeritasPrepKarishma
Bunuel

Example 1: If a portion of a 50% alcohol solution (in water) is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

Solution: What this question says is that a solution of 25% alcohol (say solution 1) is mixed with another solution of 50% alcohol (say solution 2) to give us a 30% alcohol solution. We don’t know in what quantities they were mixed. Can we find out the ratio in which these two solutions were mixed? If you are not sure, check out this post.

w1w2=(A2–Aavg)(Aavg–A1)=(50–30)(30–25)=41w1w2=(A2–Aavg)(Aavg–A1)=(50–30)(30–25)=41

So the two solutions were mixed in the ratio 4:1. Mind you,

(volume of 25% alcohol solution) : (volume of 50% alcohol solution) = 4:1

Out of 5 parts of total solution obtained, 50% solution was 1 part while 4 parts was 25% solution. So what part of the 50% solution was removed and replaced by the 25% solution? Would you agree it is (4/5)th? We can say that 80% of the 50% solution was replaced by the 25% solution.


In this question my ratio comes out to be 1:4 rather then 4:1. In many other questions too i'm having trouble identifying which value must be selected for a1 or a2. Can anyone please help me out. I'll be grateful.

mikemcgarry

Yes, I have seen this issue quite a few times before and hence, I have written posts and made videos on it:

Weighted Avg and Mixtures Basics:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg
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PayalGmat2018
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Hello Bunuel,
thanks for sharing this topic, it was really helpful in summarizing the complete topic. But I have a question :
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.

w1/w2=(A2–Aavg)/(Aavg–A1)

w1/w2=(3/8–5/16)/(5/16–2/7)=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.

__________________________________________________
Isn't this be total copper weight in the final bar instead of Copper - Nickel weight of first bar as was asked in the question? Since you calculated the copper ratio from the bar 1 and bar 2. Please elaborate.

Thanks in advance..
Payal
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