Discussion
Three mixtures of volumes 3 litres, 4 litres and 5 litres, of milk and water, respectively, with \(\frac{4}{5}\) parts of milk, \(\frac{3}{4}\) parts of milk and \(\frac{5}{7}\) parts of milk, are mixed. What should be the quantity of pure milk in litres to be added to the mixture to have \(\frac{6}{7}\) part milk in the resulting mixture?
Answer: C
The pure milk to be added is 9.2 litres. Milk in the three mixtures is 3 x 4/5 = 2.4, 4 x 3/4 = 3 and 5 x 5/7 = 25/7 litres, so the total milk is about 8.971 litres in a 12-litre mixture. After adding x litres of pure milk the milk fraction must be 6/7, so (8.971 + x)/(12 + x) = 6/7, which gives 62.8 + 7x = 72 + 6x and x = 9.2. The values 8.2, 8.7 and 9.7 do not satisfy this equation. A check on the final mixture: the milk is 8.971 + 9.2 = 18.171 litres in a total of 21.2 litres, and 18.171 divided by 21.2 is 0.857, which is 6/7 as required.
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