word problem

Question

a chemist need to mix a solution that is 12% iodine with one that is 50% iodine to obtain 600 liters of a mixture that is 36% iodine. Find the number of liters of 12% solution needed for the mixture. Round to the nearest tenth.

Answer 1

First set up the variables:

Let x = the amount of 12% iodine, and let y = the amount of 50% iodine.

 

So we know immediately that x + y = 600, since that's the total amount we need.

Since the 12% solution is 12% iodine, that means that there will be .12 x of actual iodine in that solution.

Similarly, there will be .50 y of actual iodine in the 50% solution.

We want the 600 ml that we get at the end to be a 36% solution, so there will have to be 36% of 600 , or 216, of actual iodine in that 600 ml. That means that .12 x + .50 y = 216.

So now we have 2 equations we can solve together:

x + y = 600
.12 x + .50 y = 216

Mulitply both sides of the first equation by 2:

x + y = 600
.24 x + y = 432

Subtract the first equation from the second:

.76 x = 168

Divide both sides by .76:

 x = 221.1

Since x is the amount of the 12% solution, that is the answer.

 

So you need 221.1 liters of the 12% solution.