Question

Duk Son wants to mix 20% saline solution with a 12% saline solution to make 40 ounces of a solution that is 15% saline. how many ounces of the 12% saline solution will he need?

Answer 1

He will need `15` ounces of `20%` solution and `25` ounces of `12%` solution

Explanation:

If we denote:

`x`- amount of `20%` solution (in ounces)
`y`- amount of `12%` solution (in ounces)

Then the conditions of the task lead to the following equations:

`x+y=40`, because the total amount of both solutions is `40` ounces.

`0.2x+0.12y=0.15*40`, this comes from the calculation of amount of saline in all 3 solutions (the two we mx and the result).

Now we can write the system of equations to solve:

`{(x+y=40),(0.2x+0.12y=0.15*40):}`

I started calculations with multiplying second equation by `100` to make the coefficients integer:

`{(x+y=40),(20x+12y=600):}`

Now we can multiply first equation by `(-12)` to make `y` coefficients opposite numbers:

`{(-12x-12y=-480),(20x+12y=600):}`

After adding both equations we get:

`8x=120` which leads to `x=15`

Now we can calculate `y` from the firs equation:

`15+y=40`

`y=25`

So the solution of the system of equations is: `{(x=15),(y=25):}`

The answer is: We need `15` ounces of `20%` solution and `25` ounces of `12%` solution.