A milkman purchased 10 litres of milk at rupees 7 per litre and forms ...
Problem Statement
A milkman purchased 10 litres of milk at rupees 7 per litre and forms a mixture by adding freely available water which constitutes 16.66 % of mixture later on he replaced the mixture by some freely available water and thus the ratio of milk is to water is 2:1 he then sold the new mixture at cost price of milk and replaced amount of mixture at twice the cost of milk then. what is the profit percentage? Explain in details.
Solution
Step 1: Initial Mixture
Initial Cost price of 10 litres of milk = 10 * 7 = 70
Let the amount of water added be x litres
Then, the amount of mixture = 10 + x litres
16.66% of the mixture is water. Hence,
x = (16.66/100) * (10 + x)
100x = 166.6 + 16.66x
83.34x = 166.6
x = 2 litres
Therefore, amount of water added = 2 litres and amount of milk = 10 litres
Hence, the initial mixture contains 10 litres of milk and 2 litres of water.
Step 2: Final Mixture
Let the amount of mixture replaced be y litres
Amount of milk in the mixture replaced = (2/3) * y litres
Amount of water in the mixture replaced = (1/3) * y litres
Amount of milk left in the mixture = 10 - (2/3) * y litres
Amount of water left in the mixture = 2 - (1/3) * y litres
Let the amount of water added be z litres
Then, the amount of mixture = 10 - (2/3) * y + z litres
Also, the ratio of milk to water in the final mixture is 2:1.
Therefore,
(10 - (2/3) * y)/(2 - (1/3) * y + z) = 2/1
10 - (2/3) * y = 4 - (2/3) * y + 2z
2z = (2/3) * y + 6
z = (1/3) * y + 3
Step 3: Calculation of Profit Percentage
The final mixture is sold at cost price of milk. Therefore, selling price of 1 litre of mixture = selling price of 1 litre of milk = Rs. 7
Cost price of 1 litre of initial mixture = 70/12 = Rs. 5.83
Cost price of 1 litre of final mixture = (10 - (2/3) * y + z)/(10 - (2/3) * y + z + y