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

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05 Feb 2011, 10:15
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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07 Feb 2011, 17:00
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09 Feb 2011, 07:43
Bunuel,

Do you know some similar questions that are more GMAT-like?

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09 Feb 2011, 23:23
ye, some more questions like that will be great.
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12 Feb 2011, 12:54
What is the source for this question? is it a real gmat question? can we meet questions similar to this one?

thanks.
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12 Feb 2011, 12:59
144144 wrote:
What is the source for this question? is it a real gmat question? can we meet questions similar to this one?

thanks.

It was mentioned several times on the previous page that it's not a GMAT question!
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12 Feb 2011, 14:12
Bunuel - i am sry. i read the post before and i just forgot.
its very late here and im out of focus.

didnt mean to make u angry.

have a good day.
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01 Aug 2011, 01:45
Alternative explanation from Dabral, always enjoyed his problem solving approach.

http://www.gmatquantum.com/shared-posts ... stion.html
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Last edited by rahul on 08 Aug 2011, 00:31, edited 1 time in total.
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02 Aug 2011, 08:44
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

nice.. I just gave up after the first 2 iterations..
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16 Sep 2011, 21:04
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

Karishma!!!!!

Great Explanation. Kudos +1
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16 Sep 2011, 21:17
rahul wrote:
Alternative explanation from Dabral, always enjoyed his problem solving approach.

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

Rahul

Really great explanation at the above mentioned link.

Thanks
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24 Nov 2011, 23:24
too tough to figure out the pattern during exam pressure

I don't think I would be able to answer it during exam
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06 Feb 2013, 09:08
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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06 Feb 2013, 20:11
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 23 Dec 2012 Posts: 3 GMAT Date: 05-28-2013 GPA: 3.25 WE: Operations (Transportation) Followers: 0 Kudos [?]: 5 [0], given: 7 Re: Ratio and Proportion [#permalink] ### Show Tags 06 Feb 2013, 23:54 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 . Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7380 Location: Pune, India Followers: 2290 Kudos [?]: 15142 [0], given: 224 Re: Ratio and Proportion [#permalink] ### Show Tags 07 Feb 2013, 02:57 y7214001 wrote: 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 . It would have taken forever! $$C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * ....... (\frac{18}{20}) * (\frac{19}{21}) * (\frac{20}{22})$$ 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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05 Jul 2013, 06:47
Hi Bunuel and VeritasPrepKarishma,

Would it be possible to use this formula in this case?

New Concentration of wine= Old concentration of wine * (V1/V2)^n

n= number of iterations (it is 19 in this case)
v1 = volume of liquid withdrawn
v1 = initial volume of liquid

I noticed a similar formula being used here:
a-20-litre-mixture-of-milk-and-water-contains-milk-and-water-22212.html

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

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16 Aug 2013, 18:20
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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19 Aug 2013, 05:33
emailmkarthik wrote:
Hi Bunuel and VeritasPrepKarishma,

Would it be possible to use this formula in this case?

New Concentration of wine= Old concentration of wine * (V1/V2)^n

n= number of iterations (it is 19 in this case)
v1 = volume of liquid withdrawn
v1 = initial volume of liquid

I noticed a similar formula being used here:
a-20-litre-mixture-of-milk-and-water-contains-milk-and-water-22212.html

Regards,

Actually, it is a play on the same formula.

Cf = Ci * (V1/V2)*(V3/V4).....

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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Re: Ratio and Proportion   [#permalink] 19 Aug 2013, 05:33

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