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 \(\frac{2}{3}\). If x liters of water are then removed from the barrel, the amount of water in the barrel will decrease to \(\frac{2}{5}\) 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 \(\frac{2}{3}\)We 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 \(\frac{2}{5}\) 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