1. ## Mixture math problem!

Here is the problem:

Jane is asked to make a 37 milliliter solution of 12% HCL. However, only 8% and 15% HCL is available in the lab. How many milliliters of each solution should be use?

I started out by setting up the variables like this, but then I don't know what to do afterwards. Here is how I set up the variables.

Let x=number of ml of 8% HCl
y=number of ml of 15% HCL

Any help on how to solve this would be great appreciated!

2. 8%*x+15%*y=12%*(x+y) and x+y=37

solve:x(ml)=111/7,y(ml)=148/7

over

Originally Posted by AceComical
Here is the problem:

Jane is asked to make a 37 milliliter solution of 12% HCL. However, only 8% and 15% HCL is available in the lab. How many milliliters of each solution should be use?

I started out by setting up the variables like this, but then I don't know what to do afterwards. Here is how I set up the variables.

Let x=number of ml of 8% HCl
y=number of ml of 15% HCL

Any help on how to solve this would be great appreciated!

3. How did you get solve:x(ml)=111/7,y(ml)=148/7?

4. Hello, AceComical!

Jane is asked to make a 37 milliliter solution of 12% HCL.
However, only 8% and 15% HCL is available in the lab.
How many milliliters of each solution should be use?

I started out by setting up the variables like this,
but then I don't know what to do afterwards.
Here is how I set up the variables.

Let $x$ = number of ml of 8% HCl
Let $y$ = number of ml of 15% HCL . . . . Good!

We know that the total solution is 37 ml: . $x + y \:=\:37\;\;{\color{blue}[1]}$

Next, we have to "talk" our way through the problem . . .

We have $x$ ml of 8% acid.
. . It contains:. $0.08x$ ml of acid.

We have $y$ ml of 15% acid.
. . It contains: . $0.15y$ ml of acid.

Together, they contain: . ${\color{blue}0.08x + 0.15y}$ ml of acid.

But we are told that the final mixture is 37 ml which is 12% acid.
. . It contains: . $0.12(37) \:=\:{\color{blue}4.4}$ ml of acid.

We just described the final amount of acid in two ways.

There is our equation: . $0.08x + 0.15y \:=\:4.4\;\;{\color{blue}[2]}$

We have: . $\begin{array}{cccc}x + y &=& 37 & {\color{blue}[1]} \\ 0.08x + 0.15y &=& 4.4 & {\color{blue}[2]} \end{array}$

$\begin{array}{ccccc}\text{Multiply {\color{blue}[1]} by 15:} & 15x + 15y &=& 555 \\ \text{Multiply {\color{blue}[2]} by 100:} & 8x + 15y &=& 440 \end{array}$

Subtract: . $7x \:=\:115 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{115}{7}}$

Substitute into [1]: . $\frac{115}{7} + y \:=\:37 \quad\Rightarrow\quad\boxed{ y \:=\:\frac{144}{7}}$

5. Awesome, thanks for the help wood and Soroban!