Algebra mixture problems are the problems which derive at a final solution by adding solution of two or more results of the same problem. A given algebra word problem consists of data which are arranged to solve part by part and then to derive the solution for end results.
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Amount 
Cost 
Value 
300 dollar alloy 
x 
300  300x 
100 dollar alloy  120 
150 
120 * 150 
Mixture  x + 120 
200 
200(x + 120) 
Let ounces of the 10 % of alcohol = x
PERCENT  AMOUNT OF SOLUTION  AMOUNT OF TARGET  
1ST CONTAINER  0.10  X  0.10X 
2ND CONTAINER  0.24  12  0.24(12) 
MIXTURE  0.15  X + 12  0.15(X + 12) 
=> 10 X + 0.24 (12) = 0.15 (X + 12)
=> 0.10 X + 2.88 = 0.15X + 1.80
=>10 X + 288 = 15 X + 180 (Multiply by 100)
=> 10 X  15 X = 180  288
=> 5X =  108
X = $\frac{108}{5}$ = 21.6
Therefore, 21.6 OUNCES @ 10 %.
The primary technique that we will develop here is to trace the target ingredient in the problem. The equation will state that the amount of alcohol in the first container and the amount of alcohol in the second will equal the amount of alcohol in the final mixture.
Table Showing Final Mixture

PERCENT  AMOUNT OF SOLUTION  AMOUNT OF TARGET 
1ST CONTAINER  0.50  10  5 
2ND CONTAINER  0.20  x 
0.20 x 
MIXTURE  0.30  x + 10  0.30 (x + 10) 
The amount of alcohol in the first container plus
the amount of alcohol in the second container equals the amount of
alcohol in the final mixture.
Now, we solve the equation
5 + 0.20x = 0.30(x + 10)
5 + 0.20x = 0.30x + 3
Subtract 0.20x from both the sides
5 + 0.20x  0.20x = 0.30x + 3  0.20x
5 = 0.10x + 3
5  3 = 0.10x + 3  3
$\frac{2}{0.10}$ = $\frac{0.10x}{0.10}$
X = 20.
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