In Need of Help For College Algrbra Class

[sweetchildkisses]

Hi I am new to freemathhelp.com. My teacher I suggested I suggest this site out. I am having problems with this problem. Can anyone direct me how to get started?

A lab technician wishes to mix a 63% saline solution and a 11% saline solution to prepare 575 ml of a 32% saline solution. How much of each solution should be used?

Answer Choices
a. 342.79 mL of the 11% solution and 232.21 mL of the 63% solution.
b. 232.21 mL of the 11% solution and 342.79 mL of the 63% solution.
c. 287.5 mL of the 11% solution and 287.5 mL of the 63% solution.
d. 332.79 mL of the 11% solution and 242.21 mL of the 63% solution.

I am just confused, I think I may be using the wrong formula from my book. Any help would be appreciated!

 

[galactus]

Let x=the amount of the 63% solution. Therefore, the remaining amount must be 575-x for the 11%.

We want 575 ml of 32% solution.

\(\displaystyle .63x+.11(575-x)=.32(575)\)

solve for x.

 

[sweetchildkisses]

I am getting b, does that seem right?

 

[galactus]

Yep. Plug it in and see.