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

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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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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Mar 2012, 05:28, edited 1 time in total.
Edited the question and added the OA

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27 Jan 2011, 09:35
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17871 [28], given: 235 Math Expert Joined: 02 Sep 2009 Posts: 42341 Kudos [?]: 133163 [16], given: 12416 Re: Ratio and Proportion [#permalink] ### Show Tags 05 Nov 2009, 01:50 16 This post received KUDOS Expert's post 5 This post was BOOKMARKED 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 Answer: A. _________________ Kudos [?]: 133163 [16], given: 12416 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7747 Kudos [?]: 17871 [4], given: 235 Location: Pune, India Re: Ratio and Proportion [#permalink] ### Show Tags 28 Jan 2011, 12:55 4 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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19 Sep 2010, 01:55
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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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26 Jan 2011, 16:59
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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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04 Nov 2009, 11:32
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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.
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26 Sep 2010, 04:44
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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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02 Aug 2011, 02:54
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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 [#permalink]

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16 Aug 2013, 18:20
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The trick is to identify there should be a pattern as it is not possible to carry out all the calculations.

1. Initially wine was 3L.
2. After first operation wine was 2L
3. After second operation wine was 1.5L
4. After third operation wine was 1.2L

Now we can see the pattern (2) is 2/3 of (1), (3) is 3/4 of (2), (4) is 4/5 of (3) and so on

So in 3 operations wine left is 3 * 2/3 * 3/4 * 4/5 , after cancelling out of numbers we have 3* 2/5 = 1.2 L

So in 19 operations after cancelling out of numbers 3* 2/21 = 2/7 L of wine left
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07 Oct 2013, 02:56
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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

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 ?

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04 Nov 2009, 16:18
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?
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05 Nov 2009, 02:38
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.
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19 Sep 2010, 00:32
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?

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25 Sep 2010, 12:08
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

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29 Sep 2010, 12:49
Bunuel wrote:
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.

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

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26 Jan 2011, 16:46
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.

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26 Jan 2011, 17:36
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.
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28 Jan 2011, 11:25
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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05 Feb 2011, 06:15
Bunuel wrote:
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.

Hi Bunuel,

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

Just curious
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Re: Ratio and Proportion   [#permalink] 05 Feb 2011, 06:15

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