a mixture of alcohol and water is 35% acid. if the mixture contains a total of 40L, How many liters of pure acid are in the mixture? how many liters of pure water are in the mixture?
Are you asking about alcohol or acid?
both. I think.
To solve this problem, we need to set up a system of equations based on the given information.
Let's assume there are x liters of pure acid in the mixture and y liters of pure water. Since the mixture contains a total of 40L, we can express this as:
x + y = 40 -- Equation 1
We also know that the mixture is 35% acid. This means that the acid concentration is 35/100, which can be written as 0.35. Since the acid concentration is determined by the amount of pure acid in the mixture, we can express this as:
x / (x + y) = 0.35 -- Equation 2
Now we have a system of equations (Equation 1 and Equation 2) that we can solve simultaneously to find the values of x and y.
To solve this system of equations, we can use substitution or elimination method. In this case, let's use the substitution method.
Rearrange Equation 1 to solve for x:
x = 40 - y
Substitute this value of x into Equation 2:
(40 - y) / (40 - y + y) = 0.35
(40 - y) / 40 = 0.35
Now, we can solve for y:
40 - y = 0.35 * 40
40 - y = 14
-y = 14 - 40
-y = -26
y = 26
Now we have the value of y, which represents the number of liters of pure water in the mixture. Substitute this value back into Equation 1 to find x:
x + 26 = 40
x = 40 - 26
x = 14
Therefore, there are 14 liters of pure acid and 26 liters of pure water in the mixture.