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8 litres are drawn from a cask filled with wine and is then filled wit 10 Dec 2020, 06:30
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8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres
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A
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8 litres are drawn from a cask filled with wine and is then filled wit 29 Dec 2020, 12:27
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Let a cask be x Litres ----Total solution

If i take out 8 Litres out of x Litres, wine left out will be x(1-8/x)

For example- If you take out 10 Litres out of 100 Litres of pure milk solution, the remaining milk solution will be 100*(1-10/100) = 90 ]

This process is carried out three more times i.e. total four times

So, wine left = x(1-8/x)^4

Wine/Total solution= 16/81

{x(1-8/x)^4 } / x = 16/81

x= 24 Litres

Option A
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8 litres are drawn from a cask filled with wine and is then filled wit 29 Dec 2020, 17:37
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Bunuel
8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres
Show more

Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)

\( (1 - \frac{8}{x})^4 = 16/81\)

\(1 - \frac{8}{x} = 2/3\)

\(\frac{8}{x} = \frac{1}{3}\)

\(x = 24\)

Ans: A

A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;

Assuming the total volume was 50L and we took away 10L of the solution each time:

Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...

As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.
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8 litres are drawn from a cask filled with wine and is then filled wit 10 Dec 2020, 07:17
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Hi Bunuel,

I am not sure whether the question is correct.

My Understanding -

1. 8 litres of wine is removed in every operation and then the 8 litres is replaced with water.
2. The total quantity would remain the same.
3. The operation is performed 4 times , therefore the total quantity of wine removed is 8 * 4 = 32 litres

Let X be the total quantity and therefore the ratio is (X-32)/X = 16/81

Solving for X, we get x = 39.8 , not matching with any of the answer choices.

Pls help, where I am going wrong.
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8 litres are drawn from a cask filled with wine and is then filled wit 08 Jan 2021, 21:36
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Bunuel
8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres

Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)

\( (1 - \frac{8}{x})^4 = 16/81\)

\(1 - \frac{8}{x} = 2/3\)

\(\frac{8}{x} = \frac{1}{3}\)

\(x = 24\)

Ans: A

A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;

Assuming the total volume was 50L and we took away 10L of the solution each time:

Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...

As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.
Show more

Is this method bullet proof for all types of questions like these irrespective of how many times we remove and replace?

What if we replaced it with different amounts of water each time (less or more)? Can you show some examples?
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8 litres are drawn from a cask filled with wine and is then filled wit 15 Jan 2021, 21:20
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Hi CEdward thanks for the question and sorry for late reply,

This method works only if we fill the container back up to the original amount. Assuming that 8 liters was 10% of the volume of the container, every time we are effectively taking away 10% of the water AND 10% of the wine that is currently in the container (we fill back to full with water afterward). If we take out 8 liters after refilling, again we are taking out 10% wine and 10% water. If we take another 8 liters out again without refilling, the percentage taken out would be greater than 10% since we took some of the liquid out. You would just need to know the question is a lot more complicated without the refilling part!
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8 litres are drawn from a cask filled with wine and is then filled wit 15 Jan 2021, 23:37
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Bunuel
8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres

Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)

\( (1 - \frac{8}{x})^4 = 16/81\)

\(1 - \frac{8}{x} = 2/3\)

\(\frac{8}{x} = \frac{1}{3}\)

\(x = 24\)

Ans: A

A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;

Assuming the total volume was 50L and we took away 10L of the solution each time:

Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...

As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.
Show more



Hi,

I don't understand how you came up with such ratio ? And how is that ratio constant amongst all 4 cycles ?

Can anyone please explain ?

Thanks
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8 litres are drawn from a cask filled with wine and is then filled wit 16 Feb 2021, 08:58
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Bunuel
8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres

Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)

\( (1 - \frac{8}{x})^4 = 16/81\)

\(1 - \frac{8}{x} = 2/3\)

\(\frac{8}{x} = \frac{1}{3}\)

\(x = 24\)

Ans: A

A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;

Assuming the total volume was 50L and we took away 10L of the solution each time:

Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...

As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.



Hi,

I don't understand how you came up with such ratio ? And how is that ratio constant amongst all 4 cycles ?

Can anyone please explain ?

Thanks
Show more

Same doubt here. should the second act of removing 8 litres, remove some of wine and some of water as well, since the first round mixed 8 litres of water with the left over wine?

Bunuel chetan2u Please clarify!
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8 litres are drawn from a cask filled with wine and is then filled wit 16 Feb 2021, 10:30
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funkyakki
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8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres

Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)

\( (1 - \frac{8}{x})^4 = 16/81\)

\(1 - \frac{8}{x} = 2/3\)

\(\frac{8}{x} = \frac{1}{3}\)

\(x = 24\)

Ans: A

A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;

Assuming the total volume was 50L and we took away 10L of the solution each time:

Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...

As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.



Hi,

I don't understand how you came up with such ratio ? And how is that ratio constant amongst all 4 cycles ?

Can anyone please explain ?

Thanks

Same doubt here. should the second act of removing 8 litres, remove some of wine and some of water as well, since the first round mixed 8 litres of water with the left over wine?

Bunuel chetan2u Please clarify!
Show more


Hi

The ratio 4/5 has nothing to do with the water or wine, but the amount of total that is being taken out.

Let us talk of the example where the mix contains 50 liter and we are removing 10 liters. When you remove 10 liter out of 50, it means you have taken out 10/50 or 1/5 of total ( water and wine).

What is left ? 1-(1/5) or 4/5 of total and total is (w1+water 1) => 4/5 (w1+water 1)
Next time the quantity of w1 will reduce as we are left with only 4/5*w1. Add 10 liter water and water becomes 4/5(water1)+10

Again when you take out 10, the ratio left is 4/5. It has nothing to do with what is inside that mixture, but due to the fact that we have removed 10 out of 59 again.

Now we will be left with 4/5*4/5(w1)
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8 litres are drawn from a cask filled with wine and is then filled wit 11 Mar 2021, 12:04
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Hi CEdward thanks for the question and sorry for late reply,

This method works only if we fill the container back up to the original amount. Assuming that 8 liters was 10% of the volume of the container, every time we are effectively taking away 10% of the water AND 10% of the wine that is currently in the container (we fill back to full with water afterward). If we take out 8 liters after refilling, again we are taking out 10% wine and 10% water. If we take another 8 liters out again without refilling, the percentage taken out would be greater than 10% since we took some of the liquid out. You would just need to know the question is a lot more complicated without the refilling part!
Show more

TBH I am really stuck on these questions. I guess this one kind of makes sense. How does it differ from this question?

https://gmatclub.com/forum/from-a-solut ... l#p2741264

In the link above, one does the same thing almost except the final concentration is divided by the initial concentration, but the poster did not multiply by x (the unknown volume).

Bunuel can you help on this one? Been stuck on this for hours. Conceptually I understand what happens when we remove, say, X amount from the original solution. Because the solution is homogenous, the relative proportions remain the same in both the remaining and removed solutions. Where I get lost is trying to figure out what the initial volume is. I just get confused with all of the formulas.
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8 litres are drawn from a cask filled with wine and is then filled wit 12 Mar 2021, 08:30
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Bunuel
8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16:81. How much wine did the cask hold originally?

A. 24 litres

B. 44 litres

C. 45 litres

D. 49 litres

D. 50 litres

Set the total original volume of wine as x litres. Each time we take away 8 litres of the solution, we lose \(\frac{8}{x}\) of the wine (and water too!), and we are left with \(1 - \frac{8}{x}\) of the wine. This ratio does not change as the total volume is always filled back to x. If we perform this 4 times in total we multiply x by this ratio 4 times, so we have the equation:
\(\frac{x * (1 - \frac{8}{x})^4}{x} = \frac{16}{81} \)

\( (1 - \frac{8}{x})^4 = 16/81\)

\(1 - \frac{8}{x} = 2/3\)

\(\frac{8}{x} = \frac{1}{3}\)

\(x = 24\)

Ans: A

A couple of numbers can help us see we are multiplying the volume of wine by a fixed ratio each step too;

Assuming the total volume was 50L and we took away 10L of the solution each time:

Wine vs Water
50 0
40 10, taking away 10L would be taking away 8 wine 2 water.
32 18, taking away 10L would be taking away 6.4 wine and 3.6 water.
25.6 ...

As we can see every time we take away 10L the wine portion is multiplied by a fixed ratio of 4/5.
Show more


Hey, can you explain, how did you arrive at 40, 10; 32,18 etc? how is the 4/5 ratio determined?
how do you ascertain the initial mixture this way?
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