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# ounces of #20%# solution and #25# ounces of #12%# solution

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.