8 litres of wine is drawn from a cask full of wine and is then filled

Although this question can be solved in the conventional method using expressions, a faster approach in this question would be to observe the numbers and pick options accordingly.

The withdrawal and replacement operations are being carried out four times. Each time, 8 litres are withdrawn and replaced. At the end of four operations, the ratio of wine to water is 16:65.

This means that the total volume left in the cask is a multiple of 81 (16+65). We know 81 = \(3^4\).
So, let’s pick up those options which are multiples of 3.

Let’s start with option A. Out of 18 litres, if 8 litres is removed, the concentration of wine left after 4 operations = \(\frac{5}{9} * \frac{5}{9} * \frac{5}{9} * \frac{5}{9}\). This will definitely not give us \(\frac{16}{65}\). Option A can be eliminated.

Let’s try option B. Out of 24 litres, if 8 litres is removed, the concentration of wine left after 4 operations = \(\frac{2}{3} * \frac{2}{3} * \frac{2}{3} * \frac{2}{3}\) = \(\frac{16}{81}\). Clearly, this is,

\(\frac{Wine}{Total}\) = \(\frac{16}{81}\) i.e. \(\frac{Wine}{Wine + Water}\) = \(\frac{16}{16 + 65}\). Therefore, the ratio of wine to water is 16:65, which is what is mentioned in the question.

So, the correct answer option is B.

As you can see, this approach where we try out options based on a certain logic worked really well and did not take too much time as well. So, in such questions, instead of always relying on an expression to help you solve the question, try using logic and the answer options instead.

Hope this helps!

Dear Aravind,

Can you please explain how did you get the ratio of 5/9 and 2/3 while working on options A & B?

When we have a situation where, from a solution containing 'xi' and 'yi' liters of two components X and Y, 'z' liters is drawn out and replaced by 'z' liters of Y and this process is repeated 'n' times, the remaining quantity of X(xr) is given by the following formula which is explained in the attachment:

xr = xi{(xi+yi-z)/(xi+yi)}^n

This is a generalized formula which can be applied to solve several variations of this problem (e.g. when both components are present in the original solution which I have come across in other similar problems)

In this particular case, 'yi' (water) is zero since the cask is filled entirely with wine. To simplify matters, let's say the volume of the cask is T so xi=xi+yi=T. So the formula becomes:

xr=T{(T-z)/T}^n=T{(T-8)/T}^4

The fraction of wine that is present in the final solution after the process is done 4 times is T{(T-8)/T}^4/T={(T-8)/T}^4 which is equal to 16/(65+16)=16/81=(2/3)^4....> (T-8)/T=2/3...> T=24

\(Cfinal=Cinitial(Vinitial/Vfinal)^{(repeated)}\)
Cinitial of wine = 100% = 1 ("a cask full of wine")
Cfinal of wine = 16/(65+16) = 16/81 ("ratio wine to water")
Vinitial in the cask = t-8 (liquid removed)
Vfinal in the cask = t-8+8 (liquid removed and water added)

find initial volume: \(Cfinal=Cinitial(Vinitial/Vfinal)^4=… 16/81=1(t-8/t)^4… 2/3=(t-8/t)… t=24\)

Answer (B)

Hi MartyMurray, would you be able to take a stab at this?

Frankly I find the answers quite lacking and require an amount of prior understanding to understand which I do not have.... Also, is there a formula/method to address such questions?

I find it quite hard to accept that a constant removal and addition can be equated to a proportional reduction at every step..­

 

­Let's think about what's going on in this scenario.

Each time the process occurs, 8 liters of liquid is removed and replaced with 8 liters that contains no wine.

Of course, in each case, 8 liters is going to represent the same fraction of the liquid that's in the cask.

For example, if the amount of liquid in the cask is 32 liters, 8 liters will be \(\frac{1}{4}\) of the liquid each time.

So, let's see what happens if we remove the same fraction of liquid from the cask multiple times.

Let's say that the number of liters in the cask is \(x\).

We remove 8 liters of wine.

So, when we remove 8 liters of wine, the fraction of the total wine that we have removed is \(\frac{8}{x}\).

Then, the number of liters of wine left in the cask is \(x - 8\).

The fraction of original wine that will be left will be \(\frac{x - 8}{x}\).

Then, we replace what we have removed with water, and there's still \(x - 8\) liters of wine in the cask, along with 8 ilters of water.

Simple enough.

Now, you might think that the next time, we'd remove a different fraction of the wine because, the next time we remove 8 liters of liquid, there's less wine in the cask. However, notice the following.

When we remove 8 liters, we're still removing \(\frac{8}{x}\) of the liquid in the cask, which presumably is a homogenous mix of wine and water.

So, when we remove 8 liters, we remove \(\frac{8}{x}\) of the wine and \(\frac{8}{x}\) of the water.

So, with each removal and replacement, we remove the same fraction of the remaining wine, which is \(\frac{8}{x}\), and replace it with water.

Now, the fraction of the wine that remains when we take out \(\frac{8}{x}\) of the wine is \(\frac{x - 8}{x}\) of the wine. So, after the process occurs 1 time, we have \(\frac{x - 8}{x}\) of the original amount.

After the process occurs 2 times, we have \((\frac{x - 8}{x})^2\) of the wine left.

After the process occurs 3 times, we have \((\frac{x - 8}{x})^3\) of the wine left.

After the process occurs 3 times, we have  \((\frac{x - 8}{x})^4\) of the wine left.

Thus, since the ratio of wine to water is 16:65, and thus the ratio of wine to total is \(\frac{16}{16 + 65}\) = \(\frac{16}{81}\),  \((\frac{x - 8}{x})^4 = \frac{16}{81}\).

Thus, \(\frac{x - 8}{x} =\) the fourth root of \(\frac{16}{81} = \frac{2}{3}\).

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

\(3(x - 8) = 2x\)

\(3x - 24 = 2x\)

\(x = 24\)

Correct answer: B­

hey MartyMurray, thank you for your reply.

I couldn't get past understanding how removal of 8 litres of wine equates to 8/x.

I do understand that amount of wine after the first removal is x-8, if x was the original amount of wine. Not sure why it has a denominator of x.

 

­If the volume of a container is reduced by x%, then every element in the container is reduced by the same x%.

Example:
A 100-liter container filled to capacity with a mixture of water and alcohol has 10% of its contents removed.

Case 1: The mixture is composed of 90 liters of water and 10 liters of alcohol
In this case, the volume of water decreases by 9 liters (10%) and the volume of alcohol by 1 liter (also 10%), with the result that the total decrease in volume \(= \frac{9+1}{100}\) = 10%

Case 2: The mixture is composed of 80 liters of water and 20 liters of alcohol
In this case, the volume of water decreases by 8 liters (10%) and the volume of alcohol by 2 liters (also 10%), with the result that the total decrease in volume \(= \frac{8+2}{100} =\) 10%

Case 3: The mixture is composed of 70 liters of water and 30 liters of alcohol
In this case, the volume of water decreases by 7 liters (10%) and the volume of alcohol by 3 liters (also 10%), with the result that the total decrease in volume \(= \frac{7+3}{100}\) = 10%

The composition of the mixture is irrelevant: whatever the composition, each element in the container will be reduced by the same percentage (10% in the examples above).

This principle will hold true no matter how many times the process is implemented.
For each x% reduction in the total volume, the volume of each element will be reduced by the same x%.
In the problem at hand, we keep removing 8 liters of a 24-liter cask -- a reduction of 1/3.
Implication:
With each removal, the volume of every element in the cask -- including the wine -- is reduced by the same fraction (1/3).
 ­

 

Responding to a pm:

From the theory of removal and replacement, we know
\(C_f = C_i * (\frac{V_i}{V_f})^n\)

Initially, we had 100% wine so concentration
\(C_i = 1\)
\(C_f = \frac{16}{(16+65)} = \frac{16}{81}\) (Final Concentration)

Since the entire removal and replacement process is done a total of 4 times, 
\(\frac{16}{81} = 1 * (\frac{V_i}{V_f})^n\)
\(\frac{V_i}{V_f} = \frac{2}{3}\)

So ratio of volume after removing wine and ­and after putting back water was 2:3. This means 1/3rd of wine was removed and replaced with water. Since 1/3rd was 8 ltrs, total wine volume must have been 24 ltrs. 

Answer (B)

I have discussed removal and replacement in detail in my content which you can check out for free tomorrow through 'Super Sundays' (details in my signature below and in this video:
https://youtu.be/gN_vlDpUflo )
 ­

­I've edited my explanation. See whether you understand now.

Hey MartyMurray, thanks this is really helpful and I can't thank you enough.

Would you be able to help me understand how you got from the first pour to the subsequent pours? i.e from ((x-8)/x) to ((x-8)/x)^2 and so on?

­

Bunuel , Can we have links of similar questions for practice ? I think this topic is very frequently tested in gmat.

Posted from my mobile device

­Try searching in Mixture Problems:

https://gmatclub.com/forum/search.php?s ... tag_id=114

Hey MartyMurray,

Would you be able to help explain how you got from (x-8)/x to the subsequent ((x-8)/x)^2 and so on?

I understand the first part on removal.

­