5 litres are drawn from a cask full of wine and is then filled with water. This operation is performed once more. The ratio of the quantity of wine now left in cask to that of the water is 16:9. How much wine the cask hold originally? (In litres)
Calculating Original Cask Capacity
This problem involves calculating the original volume of wine in a cask after a process of repeated dilution with water.
Problem Analysis
- Initial state: A cask is full of wine.
- Operation: 5 litres are drawn, and the cask is filled with water.
- Repetitions: This operation is performed twice in total.
- Final State: The ratio of wine to water is 16:9.
- Objective: Determine the original total capacity of the cask.
Applying the Dilution Formula
Let $V$ be the original capacity of the cask in litres.
Let $x$ be the amount drawn and replaced, $x = 5$ litres.
Let $n$ be the number of times the operation is performed, $n = 2$.
The quantity of the original liquid (wine) remaining after $n$ operations is given by the formula:
$ \text{Remaining Wine} = V \times \left( 1 - \frac{x}{V} \right)^n $Substituting the known values ($x=5$, $n=2$):
$ \text{Remaining Wine} = V \times \left( 1 - \frac{5}{V} \right)^2 $Using the Final Ratio
The final ratio of wine to water is given as 16:9.
This means the proportion of wine in the cask after the operations is:
$ \text{Proportion of Wine} = \frac{\text{Quantity of Wine}}{\text{Quantity of Wine} + \text{Quantity of Water}} = \frac{16}{16 + 9} = \frac{16}{25} $Therefore, the actual quantity of wine remaining in the cask is:
$ \text{Remaining Wine} = V \times \frac{16}{25} $Solving for Cask Capacity (V)
Equating the two expressions for the remaining wine:
$ V \times \left( 1 - \frac{5}{V} \right)^2 = V \times \frac{16}{25} $Divide both sides by $V$ (assuming $V \neq 0$):
$ \left( 1 - \frac{5}{V} \right)^2 = \frac{16}{25} $Take the square root of both sides:
$ 1 - \frac{5}{V} = \sqrt{\frac{16}{25}} $ $ 1 - \frac{5}{V} = \frac{4}{5} $Rearrange the equation to solve for $\frac{5}{V}$:
$ \frac{5}{V} = 1 - \frac{4}{5} $ $ \frac{5}{V} = \frac{1}{5} $Cross-multiply to find $V$:
$ V = 5 \times 5 $ $ V = 25 $The original capacity of the cask was 25 litres.
Important Questions from Mixture Problems
- A bucket contains liquid A and B in the ratio 17:18. 140 litre of the mixture is taken out and filled with 140 litre of B. Now the ratio changes to 7:8. Find the quantity of liquid B initially. (In litres)
- A trader mixes 245 kg of rice at Rs. 155 per kg with 255 kg of rice of other variety at Rs. 155 per kg and sells the mixture at Rs.186 per kg. Find his profit percent.
- A trader mixes 255 kg of rice at Rs.170 per kg with 265 kg of rice of other variety at Rs. 170 per kg and sells the mixture at Rs.204 per kg. Find his profit percent.
- In what ratio must a grocer mix two varieties of pulses costing Rs.66 and Rs.88 per kg respectively so as to get a mixture worth Rs.76 per kg?
- A shopkeeper bought 10 kg wheat at ₹20 per kg and 15 kg wheat at ₹30 per kg and mixed them.
At what price per kg should he sell the mixed variety of wheat to earn 10% profit?
