rtaha2412 wrote: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

3x juice, 2x water
=> 5x = 10 liters
So, in the beginning: 6 liters of juice, 4 liters of water (Total 10 liter)

First time operation :
5 liters withdrawn => Leftover juice and water = 3 liters of juice, 2 liters of water (50% of each)
5 liters juice added => 8 liters of juice, 2 liters of water

Second time operation :
5 liters withdrawn => Leftover juice and water = 4 liters of juice, 1 liters of water (50% of each)
5 liters juice added => 9 liters of juice, 1 liter of water

Hence E

Problem states that the mixture was in a cranberry:water ratio of 3:2, meaning it was 60% cranberry and 40% water. It also says 5 L of mixture were removed and replaced with 5 L of pure juice, not once but twice, so:

Quantity of cranberry = 10(.6)(.25) + 5(.5) + 5 = 9

Mixture = 10

Quantity of H2O = 10 - 9 = 1

Ratio = 9:1, E

Juice Water

6 4


5 liters removed. Remaining left:

3 2

5 liters juice added

8 2

5 liters removed, Remaining Left:

4 1

5 liters juice added

9 1 (Final Answer)

rtaha2412 wrote: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