5 litres of wine is removed from a cask full of wine and is replaced with water. Five litres of this mixture is then removed and replaced with water. If the ratio of wine to water in the cask is now 12 : 13, how much wine did the cask hold?
Answers
Step-by-step explanation:
Initial ratio of wine and water = 1:0
Current ratio of wine and water = 12:13
Suppose that the initial quantity of wine = x litres
Then 5 litres of wine was removed and replaced with water. So the ratio of wine and water after first draw = (x-5):5
Then again 5 litres of mixture was removed and replaced with water. This time wine as well as water were removed from the cask.
So the quantity of wine removed = 5*(x-5)/x = (5x-25)/x
& the quantity of water removed = 5*5/x = 25/x
So the remaining wine in mixture = (x-5)-(5x-25)/x
= (x^2–5x-5x+25)/x
= (x^2–10x+25)/x
& the remaining water in mixture = 5-25/x +5
= (10x–25)/x
So the ratio of wine and water after second draw = (x^2–10x+25)/x : (10x–25)/x
(x^2–10x+25)/x : (10x–25)/x = 12:13
13*(x^2–10x+25) = 12*(10x–25)/x
13x^2–130x+325 = 120x-300
13x^2+325+300 = 120x+130x
13x^2+625 = 250x
13x^2-250x+625 = 0
Solve this using the Quadratic formula which provides the solution for ax2+bx+c=0.
After solving it we would get x = 16.277 or 2.954
As we know that the cask definitely contains more than 5 litres of mixture, 2.954 is not correct.
So the correct capacity of the cask = 16.277 litres
So the share of wine in mixture = 16.277*12/(12+13)
= 16.277*12/25
= 7.813 Litres
& the share of water in mixture = 16.277*13/(12+13)
= 16.277*13/25
= 8.464 Litres