A pharmacist has 40% and 60% iodine solutions on hand. How many liters of each iodine solution will be required to produce 6 liters of 50% iodine mixture?
If there are x liters of 40%, then the rest (6-x) is 60%
so, add up the amounts of iodine...
.40x + .60(6-x) = .50*6
To solve this problem, we can use a mixture equation:
Let x be the amount (in liters) of the 40% iodine solution needed.
Therefore, the amount of the 60% iodine solution needed would be (6 - x) liters.
According to the mixture equation, for the resulting solution to have a 50% iodine concentration, the total amount of iodine from the 40% solution plus the total amount of iodine from the 60% solution should be equal to 50% of the total solution volume.
Now let's calculate the amount of iodine from each solution:
Iodine from the 40% solution = 0.4x (40% = 0.4 as a decimal)
Iodine from the 60% solution = 0.6(6 - x) (60% = 0.6 as a decimal)
Putting it all together, the equation becomes:
0.4x + 0.6(6 - x) = 0.5(6)
Now we can solve for x:
0.4x + 3.6 - 0.6x = 3
Combining like terms:
-0.2x + 3.6 = 3
Subtracting 3.6 from both sides:
-0.2x = -0.6
Dividing both sides by -0.2:
x = -0.6 / -0.2
x = 3
Therefore, you would need 3 liters of the 40% iodine solution and 6 - 3 = 3 liters of the 60% iodine solution to produce 6 liters of a 50% iodine mixture.