A bottle is full of dettol. One-third of it is taken out and then an equal amount of water is poured into the bottle to fill it. This operation is done four times. Find the final ratio of dettol and water in the bottle

Options

A. `13:55`
B. `20:74`
C. `16:65`
D. `10:48`

Answer

To solve the problem, we will follow these steps:

### Step 1: Initial Setup
Let the total volume of the bottle be \( V \). Initially, the bottle is full of Dettol, so:
- Volume of Dettol = \( V \)
- Volume of Water = \( 0 \)

### Step 2: First Operation
In the first operation, one-third of the Dettol is taken out:
- Volume of Dettol taken out = \( \frac{1}{3}V \)
- Remaining Dettol = \( V - \frac{1}{3}V = \frac{2}{3}V \)

Now, we pour in one-third of the total volume as water to fill it back up:
- Volume of Water added = \( \frac{1}{3}V \)
- Total Volume of Water = \( \frac{1}{3}V \)

After the first operation:
- Volume of Dettol = \( \frac{2}{3}V \)
- Volume of Water = \( \frac{1}{3}V \)

### Step 3: Second Operation
In the second operation, we again take out one-third of the mixture (Dettol + Water). The total volume is still \( V \), so:
- Volume of mixture taken out = \( \frac{1}{3}V \)

The ratio of Dettol to Water after the first operation is:
- Dettol = \( \frac{2}{3}V \)
- Water = \( \frac{1}{3}V \)

Thus, the fraction of Dettol in the mixture is:
\[ \text{Fraction of Dettol} = \frac{\frac{2}{3}V}{V} = \frac{2}{3} \]
And the fraction of Water is:
\[ \text{Fraction of Water} = \frac{\frac{1}{3}V}{V} = \frac{1}{3} \]

Now, the volume of Dettol taken out in this operation:
\[ \text{Dettol taken out} = \frac{1}{3} \times \frac{2}{3}V = \frac{2}{9}V \]
And the volume of Water taken out:
\[ \text{Water taken out} = \frac{1}{3} \times \frac{1}{3}V = \frac{1}{9}V \]

Remaining volumes after the second operation:
- Remaining Dettol = \( \frac{2}{3}V - \frac{2}{9}V = \frac{6}{9}V - \frac{2}{9}V = \frac{4}{9}V \)
- Remaining Water = \( \frac{1}{3}V - \frac{1}{9}V = \frac{3}{9}V - \frac{1}{9}V = \frac{2}{9}V \)

Now we add \( \frac{1}{3}V \) of water:
- Total Water after adding = \( \frac{2}{9}V + \frac{1}{3}V = \frac{2}{9}V + \frac{3}{9}V = \frac{5}{9}V \)

After the second operation:
- Volume of Dettol = \( \frac{4}{9}V \)
- Volume of Water = \( \frac{5}{9}V \)

### Step 4: Third Operation
Repeat the same process:
- Total mixture = \( \frac{4}{9}V + \frac{5}{9}V = V \)
- Dettol taken out = \( \frac{1}{3} \times \frac{4}{9}V = \frac{4}{27}V \)
- Water taken out = \( \frac{1}{3} \times \frac{5}{9}V = \frac{5}{27}V \)

Remaining volumes:
- Remaining Dettol = \( \frac{4}{9}V - \frac{4}{27}V = \frac{12}{27}V - \frac{4}{27}V = \frac{8}{27}V \)
- Remaining Water = \( \frac{5}{9}V - \frac{5}{27}V = \frac{15}{27}V - \frac{5}{27}V = \frac{10}{27}V \)

Adding water:
- Total Water after adding = \( \frac{10}{27}V + \frac{1}{3}V = \frac{10}{27}V + \frac{9}{27}V = \frac{19}{27}V \)

After the third operation:
- Volume of Dettol = \( \frac{8}{27}V \)
- Volume of Water = \( \frac{19}{27}V \)

### Step 5: Fourth Operation
Repeat the same process:
- Total mixture = \( \frac{8}{27}V + \frac{19}{27}V = V \)
- Dettol taken out = \( \frac{1}{3} \times \frac{8}{27}V = \frac{8}{81}V \)
- Water taken out = \( \frac{1}{3} \times \frac{19}{27}V = \frac{19}{81}V \)

Remaining volumes:
- Remaining Dettol = \( \frac{8}{27}V - \frac{8}{81}V = \frac{24}{81}V - \frac{8}{81}V = \frac{16}{81}V \)
- Remaining Water = \( \frac{19}{27}V - \frac{19}{81}V = \frac{57}{81}V - \frac{19}{81}V = \frac{38}{81}V \)

Adding water:
- Total Water after adding = \( \frac{38}{81}V + \frac{1}{3}V = \frac{38}{81}V + \frac{27}{81}V = \frac{65}{81}V \)

### Final Volumes
After the fourth operation:
- Volume of Dettol = \( \frac{16}{81}V \)
- Volume of Water = \( \frac{65}{81}V \)

### Step 6: Final Ratio
Now, we find the final ratio of Dettol to Water:
\[ \text{Ratio of Dettol to Water} = \frac{\frac{16}{81}V}{\frac{65}{81}V} = \frac{16}{65} \]

Thus, the final ratio of Dettol to Water in the bottle is \( 16:65 \).

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