Weighted Average and Mixture Problems on the GMAT

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

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.

But the actual question is something else: What percentage of original alcohol was replaced?

We need to find the percentage of original alcohol that was replaced, not the percentage of original solution! Now, here is the interesting thing: Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced. So the answer will still be 80%. I will explain this point in detail since it is extremely important while dealing with the really treacherous replacement questions.

Let’s say we have 100 liters of 50% alcohol solution (so alcohol = 50 liters and water = 50 liters). When we remove 80% of the solution, we remove 80 liters of the solution. In the solution we remove, we will still have 50% alcohol i.e. we will have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, we will have 10 liters alcohol and 10 liters water. So amount of alcohol removed is 40/50 = 80%.

This question is discussed HERE.

Important Points to Remember:

1. When a fraction of a homogenous solution is removed, the percentage of either part does not change. If milk:water = 1:1 in initial solution, it remains 1:1 in the leftover solution.

2. When you add one component to a solution, the amount of the other component does not change. In milk and water solution, if you add water, amount of milk is still the same (not percentage but amount). If milk:water = 1:1 in 10 liters of solution, it means, milk = 5 liters and water = 5 liters. Now, if you add 2 liters of water, amount of water = 7 liters but amount of milk is still 5 liters. The percentage of milk has changed but the amount of milk is still the same.

3. Amount of A = Concentration of A * Volume of the mixture

\(Amount = C*V\)

In a 10 liter mixture of milk and water, if milk is 50%, amount of milk = 50%*10 = 5 liter

When you add water to this solution, the amount of milk does not change (as discussed in point 2 above). The concentration of milk changes of course since the solution is diluted.

Amount of milk before addition = Amount of milk after addition

So Initial Concentration of milk * Initial Volume of solution = Final Concentration of milk * Final Volume of solution

Ci * Vi = Cf * Vf

Or

Cf = Ci * (Vi/Vf)

Remember, this is the relation between the initial and final concentration of milk since the amount of milk remains the same. The amount of water does not remain the same since more water is added. Hence, this relation does not hold for water.

Go through these points repeatedly till you are very comfortable with them!

Example 2: 10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

(A) 3%
(B) 20%
(C) 25%
(D) 36%
(E) 40%

Solution: In each step, we are replacing the solution with water. Every time we remove p% of the solution, the amount of alcohol goes down but the concentration of alcohol in the mixture remains the same (point 1 above). When we add water, the amount of alcohol remains the same.

Let’s try and perform the steps to see what happens:

Step 1: 10% of a 50% alcohol solution is removed – In the leftover solution, concentration of alcohol remains the same i.e. 50%. If initial volume of the solution was 10 liters, new volume is 9 liters.

Step 2: Water is added to the solution to replace the 10% shortfall – the concentration of alcohol changes now (but the amount of alcohol is still the same). Also, the volume of the solution is 10 liters again. In this new solution,

The concentration of alcohol after this step Cf1 = (50%)*(9/10) (using point 3)

Step 3: 10% of the solution with concentration of alcohol = Cf1 is removed – In the leftover solution, concentration of alcohol is still Cf1. The volume of the solution reduces to 9 liters again.

Step 4: Water is added to the solution to replace the 10% shortfall – The concentration of alcohol changes now. Also, the volume of the solution is 10 liters again. In this new solution,

The concentration of alcohol after this step Cf2 = Cf1*(9/10) = (50%)*(9/10) *(9/10)

Step 5: 10% of the solution with concentration of alcohol = Cf2 is removed – In the leftover solution, concentration of alcohol is still Cf2. The volume of the solution reduces to 9 liters again.

Step 6: Water is added to the solution to replace the 10% shortfall – The concentration of alcohol changes now. Also, the volume of the solution is 10 liters again. In this new solution,

The concentration of alcohol after this final step Cf3 = Cf2*(9/10) = 0.5*(9/10) *(9/10) *(9/10)

The concentration changes only when water is added. Each time water is added, the concentration becomes (9/10)th of the previous concentration.

Final concentration of alcohol = \((50\%) *(\frac{9}{10}) *(\frac{9}{10}) *(\frac{9}{10}) = 36.45\%\)

Answer (D) This question is discussed HERE.

Now try the following question to see if the theory makes sense to you:

Example 2: 20% of a 40% alcohol solution is removed and replaced with water. From the resulting solution, again 20% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

Solution:

Concentration of alcohol in the final solution = \((40\%) * (\frac{8}{10}) * (\frac{8}{10}) * (\frac{8}{10}) = 20.48\%\)

I will leave you here with a complicated question. See if you can arrive at the answer on your own. If not, let me know!

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

(A) 2/7 L
(B) 3/7 L
(C) 5/14 L
(D) 5/19 L
(E) 6/19L

This question can be solved in under a minute if you understand the concept of concentration and volume. Take your time and see if you can do it on your own!

This question is discussed HERE.
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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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Let me start here discussion with a question from our Arithmetic book. I love this question because it is very crafty (much like actual GMAT questions, I assure you!) It looks like a calculation intensive question and makes you spend 3-4 minutes (scribbling furiously) but is actually pretty straight forward when understood from the ‘weighted average’ perspective. We looked at an easier version of this question in the last post.

Question: John and Ingrid pay 30% and 40% tax annually, respectively. If John makes $56000 and Ingrid makes $72000, what is their combined tax rate?

A. 32%
B. 34.4%
C. 35%
D. 35.6%
E. 36.4%

Solution: If we do not use weighted averages concept, this question would involve a tricky calculation. Something on the lines of:

Total Tax = \((\frac{30}{100})*56000 + (\frac{40}{100})*72000\)

Tax Rate = \(\frac{Total \ Tax}{(56000 + 72000)}\)

But we know better! The big numbers – 56000 and 72000 are just a smokescreen. I could have as well given you $86380 and $111060 as their salaries; I would have still obtained the same average tax rate! What is important is not the actual values of the salaries but the relation between the values i.e. the ratio of their salaries. Let me show you.

We need to find their average tax rate. Since their salaries are different, the average tax rate is not (30 + 40)/ 2. We need to find the ‘weighted average of their tax rates’. In the last post, we discussed

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

The ratio of their salaries \(\frac{w_1}{w_2} = \frac{56000}{72000} = \frac{7}{9}\)

\(\frac{7}{9} = \frac{(40 – T_{avg})}{(T_{avg} – 30)}\)

\(T_{avg} = 35.6%\)

Imagine that! No long calculations! In the last post, when we wanted to find the average age of boys and girls – 10 boys with an average age of 17 yrs and 20 girls with an average age of 20 yrs, all we needed was the relative weights (relative number of people) in the two groups i.e. 1:2. It didn’t matter whether there were 10 boys and 20 girls or 100 boys and 200 girls. It’s exactly the same concept here. It doesn’t matter what the actual salaries are. We just need to find the ratio of the salaries.

Also notice that the two tax rates are 30% and 40%. The average tax rate is 35.6% i.e. closer to 40% than to 30%. Doesn’t it make sense? Since the salary of Ingrid is $72,000, that is, more than salary of John, her tax rate of 40% ‘pulls’ the average toward itself. In other words, Ingrid’s tax rate has more ‘weight’ than John’s. Hence the average shifts from 35% to 35.6% i.e. toward Ingrid’s tax rate of 40%.

This question is discussed HERE.

Let’s now look at PS question no. 148 from the Official Guide which is a beautiful example of the use of weighted averages.

Question: If a, b and c are positive numbers such that [a/(a+b)]*20 + [b/(a+b)]*40 = c and if a < b, which of the following could be the value of c?

(A) 20
(B) 24
(C) 30
(D) 36
(E) 40

Solution: Let me tell you, it isn’t an easy question (and the explanation given in the OG makes my head spin).

First of all, notice that the question says: ‘could be the value of c’ not ‘is the value of c’ which means there isn’t a unique value of c. ‘c’ could take multiple values and one of those is given in the options. Secondly, we are given that a < b. Now how does that figure in our scheme of things? It is not an equation so we certainly cannot use it to solve for c. If you look closely, you will notice that the given equation is

\(\frac{(20*a + 40*b)}{(a + b)}= c\)

Does it remind you of something? It should, considering that we are doing weighted averages right now! Isn’t it very similar to the weighted average formula we saw in the last post?

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

So basically, c is just the weighted average of 20 and 40 with a and b as weights. Since a < b, weightage given to 20 is less than the weightage given to 40 which implies that the average will be pulled closer to 40 than to 20. So the average will most certainly be greater than 30, which is right in the middle of 40 and 20, but will be less than 40. There is only one such number, 36, in the options. ‘c’ can take the value ‘36’ and hence, (D) will be the answer. Elementary, isn’t it? Not really! If you do not consider it from the weighted average perspective, this question can torture you for hours.

These are just a couple of many applications of weighted average. Next week, we will review Mixtures, another topic in which weighted averages are a lifesaver
_________________

DS Mixture Problems
PS Mixture Problems

Mixture Tips and Tricks

Mixture Problems Made Easy
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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

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

Really appreciate the work involved behind the topics posted! Quite useful! Thank you!

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
= \(0.4 * 94 + 0.3 * 88 + 0.2 * 98 + 0.1 * 85 = 37.6 + 26.4 + 19.6 + 8.5 = 92.1\)

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 = \(22 * \frac{1}{77} = \frac{2}{7} L\)

Its indeed very useful and thanks Karishma and Bunuel for creating and sharing such useful material.

Thanks for sharing this . It certainly makes Weighted average questions look easy.

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.

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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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
_________________

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