Question 27

From a container of milk, 5 litres of milk is replaced with 5 litres of water. This process is repeated again.Thus in two attempts the ratio of milk and water became 81 : 19. The initial amount of milk in the container was

Solution

Remaining milk = Initial concentration (1 - $$\frac{\textrm{quantity taken out}} {\textrm{Total amount}}$$)

Concentration of milk in the final mixture = $$\frac{81}{81 + 19} = \frac{81}{100}$$

Let initially, milk in the container = $$x$$ litres

=> $$81 = 100 (1 - \frac{5}{x})^2$$

=> $$\frac{81}{100} = (1 - \frac{5}{x})^2$$

=> $$(1 - \frac{5}{x})^2 = (\frac{9}{10})^2$$

=> $$1 - \frac{5}{x} = \frac{9}{10}$$

=> $$\frac{5}{x} = 1 - \frac{9}{10}$$

=> $$\frac{5}{x} = \frac{10 - 9}{10} = \frac{1}{10}$$

=> $$x = 5 \times 10 = 50$$ litres


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