SOLUTION: How many quarts of a 20% alcohol solution and how many quarts of pure alcohol should be mixed together to obtain 100 quarts that contain 30% alcohol? Can you please help me! I a

Question 455868: How many quarts of a 20% alcohol solution and how many quarts of pure alcohol should be mixed together to obtain 100 quarts that contain 30% alcohol?
Can you please help me! I am completely lost.
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
t = quarts of 20%
p = quarts of pure
-----------
t + p = 100 (total liquid)
0.20t + 1.00p = 0.30*100 (total alcohol)
2t + 10p = 300
t = 100 - p (from 1st eqn)
2(100-p) + 10p = 300
200 - 2p + 10p = 300
8p = 100
p = 12.5 quarts of pure
t = 87.5 quarts of 20%

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many quarts of a 20% alcohol solution and how many quarts of pure alcohol should be mixed together to obtain 100 quarts that contain 30% alcohol?
-------------------
Equation:
Quantity Equation:: x + y = 100 quarts
Alcohol Equation:: 0.20x+y = 0.30*100
---
Multiply thru 1st by 20
Multiply thru 2nd by 100
------
20x + 20y = 20*100
20x +100y = 30*100
-----------------------
Subtract and solve for "y":
80y = 10*100
y = 100/8 = 12.5 quarts (amt of pure alcohol needed)
---
Solve for "x":
x + 12.5 = 100
x = 87.5 quarts (amt of 20% alcohol needed)
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Cheers,
Stan H.
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