Quarter Wit, Quarter Wisdom: Removal/Replacement in Mixtures

Quarter Wit, Quarter WisdomToday, as requested by Pratap, 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.

w1/w2 = (A2 – Aavg)/(Aavg – A1) =  (50 – 30)/(30 – 25) = 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%

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) = .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%) *(9/10) *(9/10) *(9/10) = 36.45%

Answer (D)

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%) * (8/10) * (8/10) * (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!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

17 Responses

  1. JSmith says:

    A little help with the last one i’m not sure where to start
    Thanks

  2. Karishma says:

    Question: 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

    We have 3 liters of pure lime juice i.e. 100% lime juice. We are removing the solution and adding water to it. So we should work with the lime juice.

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

    Step 1: 1lt lime juice is removed – In the leftover solution, concentration of lime juice remains the same i.e. 100%. New volume of the solution is 2 liters.

    Step 2: 2 lt water is added to the solution. The concentration of lime juice changes now (but the amount of lime juice is still the same as in step 1 above). Also, the volume of the solution is 4 liters now. In this new solution,

    The concentration of lime juice after this step Cf1 = (100%)*(2/4) (using point 3 given in the post above)

    Step 3: 1lt solution is removed – In the leftover solution, concentration of lime juice remains the same i.e. Cf1. New volume of the solution is 3 liters.

    Step 4: 2 lt water is added to the solution. The concentration of lime juice changes now (but the amount of lime juice is still the same as in step 3 above). Also, the volume of the solution is 5 liters now. In this new solution,

    The concentration of lime juice after this step Cf2 = (Cf1)*(3/5)

    Step 5: 1lt solution is removed – In the leftover solution, concentration of lime juice remains the same i.e. Cf2. New volume of the solution is 4 liters.

    Step 6: 2 lt water is added to the solution. The concentration of lime juice changes now (but the amount of lime juice is still the same as in step 5 above). Also, the volume of the solution is 6 liters now. In this new solution,

    The concentration of lime juice after this step Cf3 = (Cf2)*(4/6)

    Step 7: 1lt solution is removed – In the leftover solution, concentration of lime juice remains the same i.e. Cf3. New volume of the solution is 5 liters.

    Step 8: 2 lt water is added to the solution. The concentration of lime juice changes now (but the amount of lime juice is still the same as in step 7 above). Also, the volume of the solution is 7 liters now. In this new solution,

    The concentration of lime juice after this step Cf4 = (Cf3)*(5/7)

    So, as of now, Cf4 = (100%)*(2/4)*(3/5)*(4/6)*(5/7)

    Do you see a pattern here?

    Can I say the final concentration after 19 steps will be

    Cf = (100%)*(2/4)*(3/5)*(4/6)*(5/7)*…*(17/19)*(18/20)*(19/21)*(20/22)

    Did you notice another thing? All terms get canceled (4 in the numerator in 4/6 with 4 in the denominator in 2/4, 5 in the numerator in 5/7 with 5 in the denominator in 3/5 etc) and you are left with

    Cf = 1*(2*3)/(21*22) (the first two numerators and the last two denominators)
    (You can write down all the terms to see how you are left with just these if you want)

    Cf = 1/77
    The final volume will be 22 lts (the last denominator)

    Hence the volume of lime juice = (1/77) * (22) = 2/7 lts

    Mind you, it is a very tricky question but you can do it very quickly once these concepts are clear in your head.

    • j says:

      hi,
      thnx for te good article.

      in the example above i dont understnad how come you dont recalculate the amount of juice every time, and just assume it stays the same and go on adding two litters.
      i mean: i would think that starting with 3/3 jiuce->2/4-> and now i would take off one litter, while knowing that the ratio stays the same it would be 1.5/3, next adding two litters would give 1.5/5, same method next stage would be 1.2/6 and so on…

      it seems you just left the amount of the solution in the nominator each step while adding amount of pure water to the denominator, i cant figure out your logic here,
      although it seems very elegant and im sure you are right, i guess i lack something here, thnx for your help!
      j:)

      • Karishma says:

        In every odd numbered step, volume of the solution goes down and hence the amount of juice goes down too. Notice step 1 where we had 3 lt juice which became 2 lt.
        In even numbered steps, the quantity of the juice does not change. Note step 2 where you just add 2 lt water. The concentration of the solution changes and so does the volume of the solution but the amount of juice remains the same since only water is added in this step.
        This is where we say amount of juice before this step is equal to amount of juice after this step so Ci*Vi = Cf*Vf
        Note that Vi is 2 lt, not 3 lt. (so we have taken the reduction is quantity of juice in step 1 into account)

  3. Pratap says:

    Thanks for remembering and posting the removal/replacement concepts Karishma. This is very helpful..

  4. Karishma says:

    No problem!

  5. Parvez says:

    Hello Karishma, did you cover sets? I have questions pertaining to sets especially with some type of problem; is it possible for you to answer them if I email you those?

    Thanks :)

  6. Karishma says:

    Hey Parvez,

    I haven’t discussed sets yet. I will do it in a couple of weeks. Send me the questions and I will discuss those in the post as well.

    kbansal@veritasprep.com

    Best

  7. Shoki says:

    10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand. how much of the mixture is to be removed and replaced with pure sand?

    Karishma, would you use the same approach for this question?

    • Karishma says:

      You could use either approach. As I said in the beg, of this post, “many replacement questions are nothing but the plain old mixture questions.”

      Approach 1:
      We can use the approach discussed in Example 1 to solve this question.
      A portion of 0% clay mixture (say mixture 1) is mixed with 70% clay mixture (say mixture 2) to give us a 50% clay mixture.

      w1/w2 = (A2 – Aavg)/(Aavg – A1) = (70 – 50)/(50 – 0) = 2/5

      So the two mixtures were mixed in the ratio 2:5. Mind you,

      (0% clay i.e. pure sand) : (70% clay i.e. the original mixture) = 2:5

      Out of 7 parts of total mixture obtained, pure sand was 2 parts. So we can say that (2/7)th of the mixture was replaced. What is (2/7)th of 10?
      (2/7)*10 = 20/7 kg was replaced by pure sand.

      or we can use the approach of Example 2:

      Since we add sand, we will work with clay.
      Step 1: x kg of mixture was removed out of 10 kg. The concentration of clay remains the same. The weight now is (10-x) kg
      Step 2: x kg of sand is added to the mixture to get a total of 10 kg again. Now the concentration of clay changes. The final concentration of clay is 50%.
      .5 = (.7)*(10-x)/10
      x = 20/7
      So 20/7 kg of the mixture was replaced by pure sand.

      • Manish says:

        Shoki,

        1 kg of mixture contains 7/10 kgs of Clay, we have to remove 2 kgs of Clay (to make it 5 kgs = 50%), so 2 kgs of Clay is present in ‘x’ kgs of mixture.

        7/10 – 1
        2 – x

        7x/10 = 2 => x =20/7

  8. Ram says:

    Wonderful, crystal clear explanation, making things easier. Hats off Karishma!

  9. Kevin says:

    Hi Karishma,

    Thank you for all the great information. Your blog has been very helpful.

    Regarding your lime juice questions, If we see the pattern, as you noted, we don’t have to scribble all the way to n/22. Would the GMAT ever give us a problem where we remove 1 and add 3, “n” number of times? I recall in a previous post you stated that GMAT problems don’t require convoluted calculations and if you have a conceptual understanding, you’ll be able to solve the problem quickly rather than using, as you say, “brute force.” However, if we were given a 100% solution and we remove 1 and add 3 and repeat this process 19 times, I don’t see how we could solve the problem other than applying time consuming multiplication: 100% x 2/5 x 4/7 x 6/9 x 8/11 x 10/13 x 12/15 x 14/17…. etc.

    On a separate note, Brian Galvin wrote about the “tale of ratio land and value land,” in a different blog. Seeing as how you are the ratio guru, I was hoping I could run something by you.

    He stated that if we’re asked to make proportional changes to a ratio, then we can determine the resulting ratio. He then gives the following example:

    “Say that there are a set of balloons of 3 different colors, and we know that the ratio of blue, green, and pink balloons is 3:4:5. If the GMAT testmakers asks us to double the number of blue balloons and halve the number of green balloons, then we should simply double the “3” to “6”, and halve the “4” to “2”….. and the resulting ratio would be 3:2:5.”

    However, wouldn’t the resulting ratio be 6:2:5?

    Your thoughts are very much appreciated and again thank you for the wealth of information.

    • Karishma says:

      Hey Kevin,

      We are glad you are enjoying our posts.

      As for your questions:

      I think its highly unlikely you would get more than 2-3 iterations. If the question does involve a large number of iterations, you would certainly not be required to write down all the factors. In GMAT, most of your time will be spent figuring out the question and developing a strategy – your next move. The actual calculations will take a very small fraction of your time. The numbers given in the questions will almost always fall beautifully in place.

      As for the ratio problem, the 3 is a typo. The article says clearly – double the “3″ to “6″. The ratio was meant to be 6:2:5.

  10. Mumbi says:

    Hi Karishma,
    Now i don’t understand in the example given above why we multiplied 8/10 three times.The question said the process was repeated only once.As in shouldn’t we have stopped here?
    The concentration of alcohol after this step Cf2 = Cf1*(9/10) = (50%)*(9/10) *(9/10)

    Thanks for the continuous good work

    • Karishma says:

      I don’t know if you are mixing two questions here but example 2 states that the replacement process took place 3 times. That is why we multiply by (8/10) three times.

  11. Srini says:

    Hi Karishma,

    I have some problem understanding the replacement concepts.

    Can you see these two questions-

    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?

    Example 2: If a portion of a 50% alcohol solution (in water) is MIXED with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

    I got a bit confused by your explanation of Example 1 at the top of this page as that explanation hold true for example 2(the one I gave here) as well.

    Can you please explain the difference in solving these two examples?

    Many thanks,
    Srini

Leave a Reply

Spam protection by WP Captcha-Free