Bunuel wrote:
virupaksh2010 wrote:
Eight litres are drawn off from a vessel full of water and substituted by pure milk. Again eight litres of the mixture are drawn off and substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity of the vessel.
A. 21 liters
B. 22 liters
C. 20 liters
D. 14 liters
E. 28 liters
Let the capacity of the vessel be \(x\).
After the first removal there would be \(x-8\) liters of water left in the vessel. Note that the share of the water would be \(\frac{x-8}{x}\);
After the second removal, the removed mixture of 8 liters will contain \(8*\frac{x-8}{x}\) liters of water, so there will be \((x-8)-8*\frac{x-8}{x}=\frac{(x-8)^2}{x}\) liters of water left.
As the ratio of water to milk after that is \(\frac{9}{40}\), then the ratio of water to the capacity of the vessel would be \(\frac{9}{40+9}=\frac{9}{49}\).
So \(\frac{\frac{(x-8)^2}{x}}{x}=\frac{9}{49}\) --> \(\frac{(x-8)^2}{x^2}=\frac{9}{49}\) --> \(\frac{x-8}{x}=\frac{3}{7}\) --> \(x=14\).
Answer: D.
Bunuel, I understood your solution and I tried to work with milk too. what am I doing wrong?
1st pass: concentration of milk in solution: \(\frac{8}{x}\)
Removed: 8 litres of mixture --> 8 X \(\frac{8}{x}\)
Milk left: x - (8 X \(\frac{8}{x}\)) + 8 = \(\frac{x^2+8x-64}{x}\)\(\frac{Milk}{Capacity}\) = \(\frac{x^2+8x-64}{x^2}\) = \(\frac{40}{49}\)
What am I doing wrong?
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update: bolded working should be
Milk left: 8 - (8 X \(\frac{8}{x}\)) + 8 = \(\frac{16x-64}{x}\)but that would lead to a very ugly quadratic equation: 10\(x^2\) - 196x + 784 = 0 --> which gives the answer x =14
so my question is, is there anything in the question that prompted you to use water instead?