A 10- tank and a pail contained a total of 8346 ml of water. After 314 ml of water was poured from the pail into the tank, the tank had 12 times as much water as the pail. How much water was there in the tank at first?
Let x be the initial amount of water in the tank.
The amount of water in the pail initially is 8346ml - x.
After pouring 314 ml of water from the pail into the tank, the amount of water in the tank is x + 314 ml.
The amount of water in the pail after pouring into the tank is 8346ml - x - 314ml = 8032 ml - x.
It was given that the tank had 12 times as much water as the pail, so x + 314 = 12(8032 - x).
Expanding the equation gives x + 314 = 96384 - 12x.
Combining like terms gives 13x + 314 = 96384.
Subtracting 314 from both sides gives 13x = 96070.
Dividing both sides by 13 gives x = 7405 ml.
Therefore, the initial amount of water in the tank is 7405 ml. Answer: \boxed{7405}.
Let's denote the amount of water in the tank at first as T and the amount of water in the pail as P.
From the given information, we can set up the following equations:
1. T + P = 8346 (equation 1) - this represents the total amount of water in the tank and pail.
2. T + 314 = 12P (equation 2) - this represents the amount of water in the tank after pouring 314 ml from the pail.
To solve these equations, we can substitute equation 1 into equation 2:
(8346 - P) + 314 = 12P
Simplifying this equation, we get:
8346 + 314 - P = 12P
Combine like terms to get:
8660 = 13P
Divide both sides by 13 to solve for P:
P = 8660 / 13 = 667.69 (rounded to the nearest hundredth)
Now, we can substitute the value of P back into equation 1 to find the value of T:
T + 667.69 = 8346
T = 8346 - 667.69 = 7678.31 (rounded to the nearest hundredth)
Therefore, there was approximately 7678.31 ml of water in the tank at first.