Bunuel wrote:

Official Solution:

A 10 liter mixture of cranberry juice and water contains juice and water in the ratio of 3 : 2. 5 liters of the mixture are removed and replaced with pure juice and the operation is repeated once more. At the end of the two removals and replacements, what is the ratio of juice to water in the resulting mixture?

A. 5 : 3

B. 6 : 4

C. 8 : 2

D. 17 : 3

E. 9 : 1

The initial mixture contains 6 liters of juice and 4 of water. 50% of the mixture is removed each time (5 liters is 50% of 10 liters of initial mixture). This means we have to remove 50% of juice as well as 50% of water. After the first removal of 5 liters it has 3 liters of juice and 2 of water. Adding 5 liters of juice gives us a mixture of 8 liters of juice and 2 of water. Removing 5 liters of the mixture, we have 4 liters of juice and 1 of water. Adding 5 liters of juice again gives us a 9 : 1 ratio of juice to water.

Answer: E

Hi Bunuel,

great solution..Could u please tell me why doesnt this mixture replacement formula work here ?Its working fine with other problem..m i missing something ?

Suppose a container contains x of liquid from which y units are taken out and replaced by water.

After n operations, the quantity of pure liquid = \(A = Q( 1-\frac{q}{Q} )^n\)

Heres how m trying to solve =

For juice=

\(A= 6(1 - \frac{5}{10}) ^2 + 10\)

\(=\frac{23}{2}\)

For water=

\(A2= 4(1-\frac{5}{10})^2\)

\(=1\)

\(\frac{A1}{A2}=\frac{11.5}{1}\)

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