A pharmacist is to prepare 15 milliliters of special eye drops for a glaucoma patient. The eye drop solut...?

Question

(I) A pharmacist is to prepare 15 milliliters of special eye drops for aglaucoma patient. The eye-drop solution must have a 2% active ingredient, but the pharmacist only has 10% solution and 1% solution in stock. Can the pharmacist use the solutions she has in stock to fill the prescription? If she can, show the amount (volume in milliliters) of the 10% solution and the 1% solution that she would use to fill the prescription.
*************************************************************
(II) The same pharmacist receives a large number of prescriptions of special eye drops for glaucoma patients. The eye drop prescriptions vary in volume but each requires a 2% active ingredient. Help the pharmacist find a convenient way to determine the approximate amounts of the 10% solution and 1% solution needed for a given volume of eye drops.
(hint: since the volume of the final prescription varies, you are not looking for exact amount of 10% or 1% solutions as you did for question #1)

Answer 1

(I)
Step 1: Understand the problem. Let x = the number of mL of the 10% solution used. Since the total solution must be 15 mL, we can use (15 - x) for the other part. So (20 - x) = the number of mL of the 1% solution used. In this problem a chart or table is very helpful:
(A) - Number of mL of Solution
(B) - Percent Pure Active Ingredient
(C) - Number of mL of Pure Active Ingredient (or A times B)

~10% Solution~
(A) x
(B) 10%
(C) 0.10(x)

~1% Solution~
(A) 15 - x
(B) 1%
(C) 0.01(15 - x)

~Final 2% Solution~
(A) 15
(B) 2%
(C) 0.02(15)

Step 2: Write an equation from the enteries in (C):
0.10x + 0.01(15 - x) = 0.02(15) --- Multiply through parentheses
0.10x + 0.15 - 0.01x = 0.30 --- Combine like terms...
0.09x + 0.15 = 0.30 --- Subtract 0.15 from each side...
0.09x = 0.15 --- Divide both sides by 0.09...
x = 1_2/3 or approximately 1.67
If x = 1.67, then 15 - x = 15 - 1.67 = 13_1/3 or approximately 13.33. Thus, the pharmacist needs 1.67 mL of the 10% solution and 13.33 mL of the 1% solution to make a 2% solution with a volume of 15 mL.

ANSWER: Yes, the pharmacist can use the solutions that she has in stock to fill the perscription. She needs to use 1.67 milliliters of the 10% solution and 13.33 milliliters of the 1% solution to make a 2% solution with a volume of 15 milliliters.

************ ************ ************ ************

(II)
First we need to define some variables. We can let V = the volume of 2% solution needed. If we let x = mL of 10% solution, then V - x = mL of 1% solution. We set up our "chart" the same way as before, we just replace some things:

(A) - Number of mL of Solution
(B) - Percent Pure Active Ingredient
(C) - Number of mL of Pure Active Ingredient (or A times B)

~10% Solution~
(A) x
(B) 10%
(C) 0.10(x)

~1% Solution~
(A) V - x
(B) 1%
(C) 0.01(V - x)

~Final 2% Solution~
(A) V
(B) 2%
(C) 0.02(V)

Again, using our values in (C), we write an equation:
0.10(x) + 0.01(V - x) = 0.02(V)
0.10x + 0.01V - 0.01x = 0.02(V)
0.09x + 0.01V = 0.02V
0.09x = 0.01V
9x = V
Example: Let's see if our problem in part (1) works for this new formula. Remember that x = amount of 10% solution and V - x = amount of 1% solution.
For V = 15
9x = 15
x = 1.67 mL and 15 - x = 13.33 mL
Therefore, our formula seems to work!

ANSWER: A convenient way to determine the appropriate amounts of 10% solution and 1% solution for a given volume of eyedrops is by using the following formula:
9x = V; where V is volume, x is volume of 10% solution to use, and (V - x) is the volume of 1% solution to use.

Answer 2

(I) Dilute 3 mL of the 10% solution to 15 mL.

(II) A 10% stock solution will give 5 times its volume of 2% solution. If the required volume of 2% solution is more than 5 times the volume of the stock, the pharmacist will not have enough stock.

The 1% solution in these problems is completely useless.

Answer 3

1)
x = amount of 10% solution in mL
y = amount of 1% solution in mL

x + y = 15
.1x + .01y = 15(.02)

Solve it yourself.

2)
Introduce a third variable z, the volume to be produced.

x + y = z
.1x + .01y = .02z

Solve once by eliminating x and finding y in terms of z and then solve again by eliminating y and finding x in terms of z.

Answer 4

when I did someone's homework ... I got paid.