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Manager  Joined: 19 Aug 2009
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A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Question Stats: 32% (02:31) correct 68% (02:26) wrong based on 481 sessions

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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

A. 2/7 L
B. 3/7 L
C. 6/19L

Originally posted by virtualanimosity on 03 Nov 2009, 22:04.
Last edited by Bunuel on 08 Jul 2019, 22:32, edited 2 times in total.
Edited the question and added the OA
Veritas Prep GMAT Instructor V
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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59
48
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
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Karishma
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Let's go step by step:

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

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SVP  Joined: 29 Aug 2007
Posts: 1712
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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virtualanimosity wrote:
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.
Intern  Joined: 25 Oct 2009
Posts: 16
Schools: Wharton, HBS, Stanford
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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1
Initial q-ty: 3L

1 operation: +2L-1L=+1L
19 operations: +19L

The final q-ty of mixture (or denominator) is 3L+19L=22L. How it can be transformed to X/7 or X/19? What have I missed?

P.s. Why only 3 answer choices?
Intern  Joined: 26 Aug 2010
Posts: 17
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 65785
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Kronax wrote:
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.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Bunuel wrote:
Let's go step by step:

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

Bunuel could you please explain the calculation for the 2nd op... Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 65785
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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2
tingle15 wrote:
Bunuel wrote:
Let's go step by step:

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

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.

Hope it's clear.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Bunuel wrote:
Let's go step by step:

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

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.

Math Expert V
Joined: 02 Sep 2009
Posts: 65785
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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mariyea wrote:
Bunuel wrote:
Let's go step by step:

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

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.

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.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Bunuel wrote:
mariyea wrote:
Bunuel wrote:
Let's go step by step:

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

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.

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. Senior Manager  Joined: 08 Nov 2010
Posts: 258
Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Karishma, Thanks a lot. +1.

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.
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Joined: 16 Oct 2010
Posts: 10784
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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144144 wrote:
Karishma, Thanks a lot. +1.

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.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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To solve this question was an exacting task.

Iteration-> Wine(in litres)##########Total Solution(in litres)
0th-> 3l##########3l = (0+3)l
1st-> 2l##########4l = (1+3)l
2nd-> 2-2/4 = 6/4=3/2##########5=(2+3)
3rd-> 3/2(1-1/5)=3/2*4/5##########6=(3+3)
4th-> (3/2)*(4/5)(1-1/6)=(3/2)*(4/5)*(5/6)##########7=(4+3)
5th-> (3/2)*(4/5)*(5/6)(1-1/7)=(3/2)*(4/5)*(5/6)*(6/7)##########8=(5+3)
6th-> (3/2)*(4/5)*(5/6)*(6/7)*(1-1/8)=(3/2)*(4/5)*(5/6)*(6/7)*(7/8)##########9=(6+3)
.
.
.
19th->(3/2)*(4/1)*(1/21)##########22litres=(19+3)

Ans: 2/7 litres of wine in 22 litres of solution.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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Alternative explanation from Dabral, always enjoyed his problem solving approach.

http://www.gmatquantum.com/shared-posts ... stion.html

Originally posted by rahul on 01 Aug 2011, 00:45.
Last edited by rahul on 07 Aug 2011, 23:31, edited 1 time in total.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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1
Wow, it took me quite a while to figure this out.

Basically, just examine how the wine content works out

at n=0 w=3L
n=1 w=2
n=2 w=1.5
n=3 w=1.2
n=4 w=1.0

thank figure out w at n=1 = 3 x (2/3)
w at n=2, w = 3 x (2/3) x (3/4)
at n=3, w = 3 x (2/3) x (3/4) * (4/5) and so forth

therefore at n=x, w = 3 x (2 / (x+2))
n=19 w = 3 x (2/21), w = 6 / 21 = 2 / 7

Took me 5 minutes, which is embarrassing....
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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VeritasPrepKarishma wrote:
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..

Thanks a lot.
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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pharm wrote:

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/
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Re: A container has 3L of pure wine. 1L from the container is taken out an  [#permalink]

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VeritasPrepKarishma wrote:
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 .  Re: A container has 3L of pure wine. 1L from the container is taken out an   [#permalink] 06 Feb 2013, 22:54

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