# a cask containing 20 liters of wine was emptied of one-fifth of its contents and then filled with water. if this is done 20 times, how many wine will remain in the cask?

## First term=20-1/5(20)=16

Second term= 12.8

r=4/5

An=A1 (r^n-1)

A20=16(4/5)^20-1

A20=0.23058 Liters

## To find out how much wine will remain in the cask after being emptied and refilled 20 times, let's break it down step by step:

1. Initially, the cask contains 20 liters of wine.

2. Each time the cask is emptied, one-fifth (1/5) of the wine is removed.

- 1/5 of 20 liters = 4 liters of wine removed.

3. After removing 4 liters of wine, the cask is filled with water.

4. The process is repeated 20 times.

Let's calculate the amount of wine remaining after each cycle:

1st cycle: 20 - 4 = 16 liters of wine remaining, and 4 liters of water

2nd cycle: 16 - 4 = 12 liters of wine remaining, and 8 liters of water (Previously 4 liters of water added in the previous step, and 4 liters more now)

3rd cycle: 12 - 4 = 8 liters of wine remaining, and 12 liters of water (Previously 8 liters of water added in the previous step, and 4 liters more now)

..........

Continuing this pattern for 20 cycles, we can calculate the wine remaining as follows:

4th cycle: 8 - 4 = 4 liters of wine remaining, and 16 liters of water

5th cycle: 4 - 4 = 0 liters of wine remaining, and 20 liters of water

After the 5th cycle, there is no more wine left in the cask. The cask will now be filled with 20 liters of water.

Therefore, after 20 cycles of emptying and refilling, there will be 0 liters of wine remaining in the cask.

## To find out how much wine will remain in the cask after 20 iterations of emptying and refilling, we need to calculate the amount of wine after each iteration.

Let's go through the process step by step:

1. Initially, the cask contains 20 liters of wine.

2. In the first iteration, one-fifth of the cask (20/5 = 4 liters) is emptied, leaving 16 liters of wine. Then, the cask is filled with water.

3. In the second iteration, the same process is repeated. One-fifth of the current contents (16/5 = 3.2 liters) is emptied, leaving 12.8 liters of wine. Water is added to fill the cask again.

4. This process continues for 20 iterations.

We can simplify the calculation by recognizing that each time, four-fifths (or 80%) of the wine remains in the cask. Therefore, the amount of wine after each iteration is given by multiplying the previous amount by 80%.

Starting with 20 liters of wine, the amount of wine after each iteration is:

1st iteration: 20 liters * 0.8 = 16 liters

2nd iteration: 16 liters * 0.8 = 12.8 liters

3rd iteration: 12.8 liters * 0.8 = 10.24 liters

...

20th iteration: x liters * 0.8 = ?

To find out how much wine remains after the 20th iteration, we need to continue multiplying the previous amount by 0.8:

20th iteration: x * 0.8 * 0.8 * 0.8 * ... (20 times)

To calculate this using an exponential equation, we can use the formula:

x * (0.8) ^ 20

Evaluating this expression, we get:

x ≈ 20 * (0.8) ^ 20

x ≈ 20 * 0.01152921

x ≈ 0.2305842 liters

Therefore, after 20 iterations, approximately 0.2305842 liters of wine will remain in the cask.

## Well, let's do some math. If one-fifth of the wine is poured out and then the cask is filled with water, that means that each time this process is repeated, the proportion of wine in the cask becomes four-fifths of what it was before. So, after 20 repetitions, the remaining wine can be calculated as:

20 liters * (4/5)^20 ≈ 0.31 liters

So, it looks like you'll have about 0.31 liters of wine left in the cask after 20 times. That's enough for a very tiny celebration!

## after each drain-fill cycle, there is 0.8 the amount of wine as before.

So, after 20 cycles, the wine remaining is

20(0.8^20) liters