A container has 3 liters of pure wine. 1 liter from the container is taken out and 2 liter water is added. The process is repeated several times. After 19 such operations, quantity of wine in the mixture is (A) 2/7 L (B) 3/7 L (C) 5/14 L (D) 5/19 L (E) 6/19L A container has 3 liters of pure lime juice. 1 liter from the container is taken out and 2 liter water is added. The process is repeated several times. After 19 such operations, quantity of lime juice in the mixture is (A) 2/7 L (B) 3/7 L (C) 5/14 L (D) 5/19 L (E) 6/19L

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A container has 3 liters of pure wine. 1 liter from the container is

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KarishmaB

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07 Feb 2013, 02:57

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y7214001

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 . :oops:

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.

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05 Jul 2013, 06:47

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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v1 = volume of liquid withdrawn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,

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16 Aug 2013, 11:56

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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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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

KarishmaB

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19 Aug 2013, 05:33

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emailmkarthik

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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29 May 2019, 09:46

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CFfinal= (100%*20!3!)/22!= (100%*2)/(7*22)
Amount= CFfinal * Vfinal
Vfinal= 22
Amount= 2/7

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08 Jul 2019, 23:24

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I still didn't get the logic to solve this. Could anyone please explain logically how to deduce a formula for this.

-Jyoti

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01 Jul 2020, 02:41

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Bunuel

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.

I just can't understand how you got to the 6/3L, it's unclear, and how you where did the 6 came from.

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03 Jul 2020, 20:44

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VeritasKarishma

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

Great explanation - very helpful!

Just a quick question; what if the same question asked to find the volume of water instead of milk after 19 operations? This would not apply right?

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04 Dec 2020, 05:56

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VeritasKarishma

y7214001

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 .

Hi VeritasKarishma

I'm stuck with the last bit here, it'll be great if you can help

I've come to the part where I have (2*3)/(21*22) after all cancellations which ultimately gives me 1/77. In your blog there was another sum where we started with a 50% concentration solution. So the final concentration was 50 * 9/10 * 9/10. In this case, since it's pure lemon juice, shouldn't it be 100% * 1/77 ? Where am I going wrong? And how do I get the answer after this?

KarishmaB

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07 Dec 2020, 02:35

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pearljiandani

VeritasKarishma

y7214001

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 .

Yes, the initial concentration is 100% which is the same as 1 (because 100% = 100/100). That is why I have used 1. I am using concentration in terms of fractions.

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14 May 2021, 03:57

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Quantity of wine left after

1 operation = 2/3 of 3 L
2 ops = 3/4 * 2/3 of 3 L
3 ops = 4/5 * 3/4 * 2/3 of 3 L

19 ops?

3 L * (20!/(21! except for the missing 2 in 2/3)) -->

3*2*(20!/21!) = 6/21 = 2/7 L

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16 May 2021, 13:39

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VeritasKarishma

Hi VeritasKarishma,

Can you please share first 4 iteration with formula:$C_i * V_i = C_f * V_f$.

I get how you got:$C_f = 1 * 2/4$

Similarly can you explain using this formula how you got 3/5, 4/6, 5/7?

This will help in clearly understanding the operation of the formula.

Thanks.

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17 May 2021, 00:25

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beeblebrox

VeritasKarishma

First check this: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

$C_f = C_i * (V_i/V_f)$

$C_2 = C_1 * (V_1/V_2)$

In the beg, we have 3 lt wine. 1 lt is removed and we have 2 lt wine.
Next, 2 lt water is added to give 4 lt of mixture. So we get 2/4.

$C_2 = 1 * (2/4)$

Next, 1 lt mixture is removed to get 3 lt mixture.
And then 2 lt water is added to get 5 lt mixture. So we get 3/5.

$C_3 = C_2 * (3/5)$
$C_3 = 1 * (2/4) * (3/5)$

Next, 1 lt mixture is removed to get 4 lt mixture.
And then 2 lt water is added to get 6 lt mixture. So we get 4/6.

$C_4 = C_3 * (4/6)$
$C_4 = 1 * (2/4) * (3/5) * (4/6) $

and so on...

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23 Jun 2021, 14:28

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VeritasPrepKarishma

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.

Karishma how did you arrive as in 1/77 ? Please explain

KarishmaB

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23 Jun 2021, 22:43

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Ved22

Karishma how did you arrive as in 1/77 ? Please explain

Note how the numerators and denominators get cancelled:

$C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * (\frac{6}{8}) * (\frac{7}{9}) .......* (\frac{19}{21}) * (\frac{20}{22})$

From the first two terms, the numerators are intact but the denominators (4 and 5) get cancelled by the numerators of next two terms (4 and 5).
The denominators of the next two terms (6 and 7) are cancelled off by the numerators of the further next two terms (6 and 7) and so on...

For the last two terms, their numerators will get cancelled by the previous denominators but their denominators will stay intact. So we will be left with

$C_f = 1 * (2 * 3 * \frac{1}{21} * \frac{1}{22})$

$C_f = 1 * \frac{1}{77}$

P.S - To tag me, please use "@VeritasKarishma" (without the quotes). I am less likely to miss it then.

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