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?

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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?

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Answer 1 · 2015-11-07T22:34:31

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

Explanation:

If we denote:

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

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

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

$0.2 x + 0.12 y = 0.15 \cdot 40$ $0.2x+0.12y=0.15*40$ , 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.

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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?

1 Answer

Explanation:

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