SOLUTION: How much of a 10% saline solution should be mixed with a 20% saline solution to obtain 1000 milliters of a 12% saline solution? Use a system of equations and substitution to solve

Question 288736: How much of a 10% saline solution should be mixed with a 20% saline solution to obtain 1000 milliters of a 12% saline solution?
Use a system of equations and substitution to solve each problem.
Thanks
Found 3 solutions by richwmiller, texttutoring, checkley77:
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
.10x+.2y=1000*.12
x+y=1000
x=800 y=200

Answer by texttutoring(324)   (Show Source): You can put this solution on YOUR website!
Let V1 = the volume of the 10% solution.
Let V2= the volume of the 20% solution.
Working in Litres (1 Litre = 1000mL), and decimals instead of percents (i.e. 12%=0.12)
Eqn 1: 0.10*V1 + 0.20*V2 = 0.12*1.0
Eqn 2: V1 + V2 = 1.0 L

Isolate V2 in equation 2:

V2 = 1 - V1

Substitute this value for V2 into Eqn 1:
0.10*V1 + 0.20(1-V1) = 0.12
-0.10*V1 + 0.20 = 0.12
-0.10*V1 = -0.08
V1 = 0.8

Eqn 2 tells us that V1 + V2 must equal 1.00 Litres
Therefore V2=1-0.8
V2=0.2 L

So you would need 800 mL (0.8 L) of the 10% solution and 200mL (0.2L) of the 20% solution.

Hope this helps. If you have any questions, just let me know!
Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
A single equation is a much simplier method.
.20x+.10(1,000-x)=.12*1,000
.20x+100-.10x=120
.10x=120-100
.10x=20
x=20/.10
x=$200 invested @ 20%.
1,000-200=$800 invested @ 10%.
Proof:
.20*200+.10*800=.12*1,000
40+80=120
120=120
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