A pharmacist mixed some 10%-saline solution
with some 15%-saline solution to obtain 100 mL of a 12%-saline solution. How much of the 10%-saline solution did the pharmacist use in the mixture?
A 60 mL
B 45 mL
C 40 mL
D 25 mL CAN SOME GIVE AN EQUATION TO HELP MY SOLVE THIS PROBLEM THIS IS THE ONLY WAY TO SOLVE THis problem
x ml of 10% --->0.1 x salt
(100-x ml) of 15% ----> 0.15(100-x) salt
total salt = .1x + 15 - .15x = 15-.05x salt
so
15-.05x = .12(100) = 12
3 = .05x
x = 60 ml
To solve this problem, we can set up a simple equation based on the principle of mixing two solutions with different concentrations.
Let's assume the pharmacist used x mL of the 10%-saline solution. Since they mixed it with the 15%-saline solution to obtain a total volume of 100 mL, the amount of the 15%-saline solution used would be 100 - x mL.
The equation for the overall concentration can be expressed as:
(Concentration of Solution 1 * Volume of Solution 1) + (Concentration of Solution 2 * Volume of Solution 2) = (Concentration of Mixture * Total Volume of Mixture)
Plugging in the given values:
(0.10 * x) + (0.15 * (100 - x)) = 0.12 * 100
Now, we can solve this equation to find the value of x, which will represent the amount (in mL) of the 10%-saline solution used in the mixture.
0.10x + 0.15(100 - x) = 0.12 * 100
0.10x + 15 - 0.15x = 12
-0.05x = 12 - 15
-0.05x = -3
x = -3 / -0.05
x = 60
Therefore, the pharmacist used 60 mL of the 10%-saline solution in the mixture.
Hence, the correct answer is option A: 60 mL.