Carcass wrote:
If 5 liters of water are added to a barrel when it is half full of water, the amount of water in the barrel will increase by
23. If x liters of water are then removed from the barrel, the amount of water in the barrel will decrease to
25 of the capacity of the barrel. What is the value of X?
Let v = the ORIGINAL volume of water in the tank (i.e., when the tank was HALF full)
If 5 liters of water are added to a barrel when it is half full of water, the amount of water in the barrel will increase by 23We can write: v + 5 = v + (2/3 of v)
In other words: v + 5 = v + 2v/3
Subtract v from both sides of the equation: 5 = 2v/3
Multiply both sides by 3 to get: 15 = 2v
Solve: v = 15/2 =
7.5Important: Since v = the volume of water when the tank was HALF full, we can conclude that of the total CAPACITY of the barrel =
15 liters
Also important: Since we added 5 liters of water to the tank, the tank NOW contains
12.5 liters of water
If x liters of water are then removed from the barrel, the amount of water in the barrel will decrease to 25 of the capacity of the barrel. We can write:
12.5 - x = 2/5 of
15In other words: 12.5 - x = 6
Solve: x = 6.5
Answer: 6.5
Cheers,
Brent