a chemist mixed a 15% glucose solution with a 35% glucose solution. This mixture produed 35 liters of 19% glucose solution. How many liters of each solution did the chemist use in the misture?
Let x = the volume of the weak solution
Let y = the volume of the strong solution
x + y = 35 (total volume)
0.15 x + 0.35 y = (0.19)*35 = 6.65
(amount of glucose in mixture)
Solve those two simultaneous equations.
To solve this problem, we can use the concept of mixture problems.
Let's assume that the chemist used x liters of the 15% glucose solution and y liters of the 35% glucose solution.
Since 35 liters of a 19% glucose solution was produced, the total volume of the mixture is 35 liters.
We can set up two equations based on the amount of glucose in each solution and the total volume of the mixture.
Equation 1: The amount of glucose in the 15% glucose solution mixed with x liters is 0.15x.
Equation 2: The amount of glucose in the 35% glucose solution mixed with y liters is 0.35y.
Equation 3: The amount of glucose in the final 19% glucose solution with a total volume of 35 liters is 0.19 * 35.
Now, we can create an equation by combining the amount of glucose in the two solutions:
0.15x + 0.35y = 0.19 * 35
Simplifying the equation, we have:
0.15x + 0.35y = 6.65
Since we want to find the values of x and y, we have a system of equations:
Equation 1: 0.15x + 0.35y = 6.65
Equation 2: x + y = 35
To solve this system of equations, we can use a method like substitution or elimination. Let's solve it using the substitution method.
Rearrange Equation 2 to express x in terms of y:
x = 35 - y
Substitute x in Equation 1 with this expression:
0.15(35 - y) + 0.35y = 6.65
Expand and solve for y:
5.25 - 0.15y + 0.35y = 6.65
0.2y = 6.65 - 5.25
0.2y = 1.4
y = 1.4 / 0.2
y = 7
Now, we can calculate the value of x by substituting y = 7 into Equation 2:
x = 35 - 7
x = 28
Therefore, the chemist used 28 liters of the 15% glucose solution and 7 liters of the 35% glucose solution in the mixture.