2 jugs are filled with water. If you remove 1 liter of water from the first jug, then the two jugs contain the same amount of water. If you remove 2 liters from the second jug, then the second jug would contain half as much as the first jug. How many liters of water are in the largest jug?
Let the jugs hold x and y liters.
x-1 = y
y-2 = x/2
The jugs hold 6 and 5 liters.
To solve this problem, let's assume the capacity of the first jug is x liters and the capacity of the second jug is y liters.
According to the given information:
1) If 1 liter of water is removed from the first jug, the two jugs will contain the same amount of water. This implies that the first jug originally had x liters, and after removing 1 liter, it has (x - 1) liters. The second jug also has (x - 1) liters.
2) If 2 liters of water are removed from the second jug, it will contain half as much as the first jug. This means that the second jug originally had y liters, and after removing 2 liters, it has (y - 2) liters. The first jug still has (x - 1) liters.
From the above information, we can set up the following equations:
Equation 1: (x - 1) = (x - 1)
Equation 2: (y - 2) = (x - 1)/2
We can solve these equations simultaneously to find the values of x and y.
From Equation 1:
x - 1 = x - 1
0 = 0
From Equation 2:
y - 2 = (x - 1)/2
2y - 4 = x - 1
2y = x + 3
Since we have two variables, we need another equation to solve for their values. Unfortunately, the problem doesn't provide any more information or equations. Therefore, we cannot determine the exact values of x and y, which means we cannot determine the exact number of liters in the largest jug.