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A container has 3L of pure wine. 1L from the container is [#permalink]

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03 Nov 2009, 23:04

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37% (07:06) correct
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A container has 3L of pure wine. 1L from the container is taken out and 2L water is added.The process is repeated several times. After 19 such operations, qty of wine in mixture is

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
_________________

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)

I understood everything until the step of *3/5. can u please explain?

bc every time we take out 1 liter of mixture and add 2 liters of water. how is *3/5*4/6****calculate that?

thanks a lot.

First time, \(C_f = 1 * 2/4\) Second time when you remove 1 litre from 4 litres, Initial Volume becomes 3 lts. When you add 2 lts of water back, final volume becomes 5 lts.

Now for second step, \(C_f = C_i * 3/5\) \(C_i\) for the second step is the final concentration obtained from step 1 i.e. \(C_i\) for second step = 1 * 2/4 So, \(C_f = 1 * 2/4 * 3/5\) and so on for every subsequent step... It doesn't matter whether we have 19 or 119 steps, we still get our answer in 2 steps.
_________________

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.
_________________

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.

I don't understand how the sequence could result in 2/7.

6/3 is supposed to be the ratio of 2

The first sequence begins by 3L-1L= 2L then 2L+2L=4L the process goes on until it becomes 21L because the process is like simply adding one over and over again.

I'm really confused please help me!

After the first operation there will be 2L of wine left. 2 can be expressed as a fraction and be written as 6/3. Then as you can see in the solution, for any operation the amount of wine can be expressed as 6/(# of operations+2) L.

Anyway: this is not a GMAT question, so don't worry about it. _________________

Q: A container has 3L of pure wine. 1L from the container is taken out and 2L water is added.The process is repeated several times. After 19 such operations, qty of wine in mixture is

1. 2/7 L 2. 3/7 L 3. 6/19L

Thats a too tough to calculated.

Got (2/7)L in A - After a lengthy calculation.
_________________

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.

Bunuel could you please explain the calculation for the 2nd op... Thanks

After 1st operation 2=6/3L of wine is left out of total 4L. When then for 2nd operation we remove 1L of mixture (or 1/4th of total mixture), we remove 2*1/4 of wine as well.

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.

1) in 1st operation 1L drawn off. so 2L wine left (6/3L). & added 2L water. so total=4L 2) how come it be 6/4L ?? not getting ?

Thanks. Cannot imagine that I will be able to solve smth like this one on the real test, besides I thought that quant is my strength.
_________________

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?

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.

Bunuel could you please explain the calculation for the 2nd op... Thanks

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.

Folks, have we seen any task like this, but easier? It can be useful to apply this concept on a gmat-like example.

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.

I don't understand how the sequence could result in 2/7.

6/3 is supposed to be the ratio of 2

The first sequence begins by 3L-1L= 2L then 2L+2L=4L the process goes on until it becomes 21L because the process is like simply adding one over and over again.

I'm really confused please help me!
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

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.

I don't understand how the sequence could result in 2/7.

6/3 is supposed to be the ratio of 2

The first sequence begins by 3L-1L= 2L then 2L+2L=4L the process goes on until it becomes 21L because the process is like simply adding one over and over again.

I'm really confused please help me!

After the first operation there will be 2L of wine left. 2 can be expressed as a fraction and be written as 6/3. Then as you can see in the solution, for any operation the amount of wine can be expressed as 6/(# of operations+2) L.

Anyway: this is not a GMAT question, so don't worry about it.

Thank you for taking the time anyway. It's good to understand solutions to difficult problems b/c it makes future problems easier to tackle.
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

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
_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

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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