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I think it's almost impossible within 2-3 minutes to figure out and build the pattern of 6/3, 6/4, 6/5,..... Just wonder how did you notice that 3L-1L=2=6/3L and why not 4/2L or 10/5L?

This is not a GMAT question, so don't worry about it.

Hi Bunuel,

How do you figure out which question is a GMAT type and which one is not?

Just curious

First of all the question has only three answer choices instead of five and also its difficulty level is higher than that of GMAT's.
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First operation: 3L-1L=2=6/3L of wine left, total 4L; #2: 6/3L-(6/3)/4=6/3-6/12=18/12=6/4L of wine left, total 5L; #3: 6/4L-(6/4)/5=6/4-6/20=24/20=6/5L, total 6L; #4: 6/5L-(6/5)/6=6/5-6/30=30/30=6/6L, total 7L; ....

At this point it's already possible to see the pattern: x=6/(n+2)

n=19 --> x=6/(19+2)=6/21=2/7L

Answer: A.

nice.. I just gave up after the first 2 iterations..

The question can be solved in under a minute if you understand the concept of concentration and volume.

Removal and addition happen 19 times so:

\(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\) All terms get canceled (4 in num with 4 in den, 5 in num with 5 in den etc) and you are left with \(C_f = \frac{1}{77}\) Since Volume now is 22 lt, Volume of wine = \(22*(\frac{1}{77}) = \frac{2}{7}\)

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

2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)

3. Amount of A = Concentration of A * Volume of mixture Amount = C*V ( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt) When you add water to this solution, the amount of milk does not change. So Initial Conc * Initial Volume = Final Conc * Final Volume \(C_i * V_i = C_f * V_f\) \(C_f = C_i * (V_i/V_f)\) In the question above, we find the final concentration of wine. Initial concentration \(C_i\) = 1 (because it is pure wine) When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt. So \(C_f = 1 * 2/4\) Since it is done 19 times, \(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\)

The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has \(V_f\) as 22, you get amount of wine = 1/77 * 22 = 2/7 lt

Karishma!!!!!

Great Explanation. Kudos +1
_________________

If you find my posts useful, Appreciate me with the kudos!! +1

The question can be solved in under a minute if you understand the concept of concentration and volume.

Removal and addition happen 19 times so:

\(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\) All terms get canceled (4 in num with 4 in den, 5 in num with 5 in den etc) and you are left with \(C_f = \frac{1}{77}\) Since Volume now is 22 lt, Volume of wine = \(22*(\frac{1}{77}) = \frac{2}{7}\)

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

2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)

3. Amount of A = Concentration of A * Volume of mixture Amount = C*V ( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt) When you add water to this solution, the amount of milk does not change. So Initial Conc * Initial Volume = Final Conc * Final Volume \(C_i * V_i = C_f * V_f\) \(C_f = C_i * (V_i/V_f)\) In the question above, we find the final concentration of wine. Initial concentration \(C_i\) = 1 (because it is pure wine) When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt. So \(C_f = 1 * 2/4\) Since it is done 19 times, \(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\)

The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has \(V_f\) as 22, you get amount of wine = 1/77 * 22 = 2/7 lt

If the operation is only done 19 times then where and why does "22" Lt pop up in the final volume of mixture I was following how the demoninators increased but dont understand the "22".

Also if 1 L of wine is removed every operation how is the concentration of the wine mixture go up since part of it is being removed...only thing that is increasing the total volume of the solution..

If the operation is only done 19 times then where and why does "22" Lt pop up in the final volume of mixture I was following how the demoninators increased but dont understand the "22".

Also if 1 L of wine is removed every operation how is the concentration of the wine mixture go up since part of it is being removed...only thing that is increasing the total volume of the solution..

Thanks a lot.

After the first step, the volume is 4 lt. After the second, it will be 5 lt. By the same logic, after the 19th step, it will be 19+3 = 22. or Initial volume is 3 lt and you add net 1 lt in every step. So after the 19th step you will have 3+19 = 22 lt

From a homogeneous mixture, if you remove some quantity of the mixture, the concentration of the elements stays the same. e.g., say you have a solution of 50% milk. If you take out some solution, what will be the concentration of milk in the leftover solution? It will still be 50%. The quantity of milk will reduce but not the concentration. Check out this post for more details: http://www.veritasprep.com/blog/2012/01 ... -mixtures/ _________________

The question can be solved in under a minute if you understand the concept of concentration and volume.

Removal and addition happen 19 times so:

\(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\) All terms get canceled (4 in num with 4 in den, 5 in num with 5 in den etc) and you are left with \(C_f = \frac{1}{77}\) Since Volume now is 22 lt, Volume of wine = \(22*(\frac{1}{77}) = \frac{2}{7}\)

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

2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)

3. Amount of A = Concentration of A * Volume of mixture Amount = C*V ( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt) When you add water to this solution, the amount of milk does not change. So Initial Conc * Initial Volume = Final Conc * Final Volume \(C_i * V_i = C_f * V_f\) \(C_f = C_i * (V_i/V_f)\) In the question above, we find the final concentration of wine. Initial concentration \(C_i\) = 1 (because it is pure wine) When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt. So \(C_f = 1 * 2/4\) Since it is done 19 times, \(C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})\)

The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has \(V_f\) as 22, you get amount of wine = 1/77 * 22 = 2/7 lt

Kudos +1 Karishma

Is there an fast way to compute the result of the multiplacation series like we have for \(Cf\)? I actually did the long way .

You need to observe here that other than first two numerators and last two denominators, all other terms will cancel out. First term's denominator will cancel out third term's numerator. Second term's denominator will cancel out fourth term's numerator. The last two denominators will have no numerators to cancel them out. The first two numerators have no denominators to cancel them out. Usually, in such expressions (where terms have a pattern), things simplify easily. You just need to observe the pattern.
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Re: A container has 3L of pure wine. 1L from the container is [#permalink]

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16 Aug 2013, 11:56

I made a guess on this question.

If we are left with 4L of mixture which has 2L of wine and 2L of water after 1st process, the ratio of wine is about 1/2. So after 19 successive processes, ratio must be significantly less than 1/2.

Option B is little less than 1/2 so can't be the answer and we are left with option A and C. At least this helped me narrowed down to two options in 15 sec.
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And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

Usually, in replacement questions, you remove n lts and put back n lts. So initial and final volume in each step is the same. That is why you get (V1/V2)^n.

In case V1 and V2 are different in subsequent steps, you use those volumes V1/V2 * V3/V4 *.....
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