This solution explains how to calculate the profit percentage when a milkman mixes water into pure milk twice and sells the mixture at the cost price.
Let's assume the initial quantity of pure milk is 100 units.
Let the cost price of 1 unit of pure milk be $C.
Therefore, the initial Cost Price (CP) of the milk is:
$ \text{CP} = 100 \text{ units} \times C/\text{unit} = 100C $
The milkman first mixes 10% water in pure milk.
Water added = $ 10\% \times 100 \text{ units} = 10 \text{ units} $.
The new volume of the mixture after the first addition is:
$ \text{Volume}_1 = 100 \text{ units} (\text{milk}) + 10 \text{ units} (\text{water}) = 110 \text{ units} $
He then adds 10% more water based on the previous mixture's volume (110 units).
Additional water added = $ 10\% \times 110 \text{ units} = 11 \text{ units} $.
The final volume of the mixture is:
$ \text{Final Volume} = \text{Volume}_1 + \text{Additional Water} = 110 \text{ units} + 11 \text{ units} = 121 \text{ units} $
The milkman sells the final mixture (121 units) at the cost price. This implies the selling price per unit volume is the same as the cost price per unit volume of pure milk ($C$).
Total Selling Price (SP) = $ \text{Final Volume} \times C/\text{unit} $
$ \text{SP} = 121 \text{ units} \times C/\text{unit} = 121C $
Profit is the difference between the Selling Price and the Cost Price.
$ \text{Profit} = \text{SP} - \text{CP} $
$ \text{Profit} = 121C - 100C = 21C $
The profit percentage is calculated based on the initial Cost Price.
$ \text{Profit Percentage} = \left( \frac{\text{Profit}}{\text{CP}} \right) \times 100 $
$ \text{Profit Percentage} = \left( \frac{21C}{100C} \right) \times 100 $
$ \text{Profit Percentage} = \frac{21}{100} \times 100 = 21\% $
By adding water twice, the milkman increases the volume. When this larger volume is sold at the original cost price per unit, it results in a significant profit.
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