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Anton [14]
3 years ago
6

How many liters of a 10% alcohol solution must be mixed with 90 liters of a 60% solution to get a 30% solution?

Mathematics
1 answer:
Setler [38]3 years ago
4 0
Since 10% = 0.10 by moving the decimal two spots to the left, we can say that 0.10L is the amount of alcohol with 10% concentration in L liters. Similarly, 0.60L is 60% for L liters and 0.30L is the amount of liters with 30%. If the amount of 10% alcohol solution in liters is x and the 60% in y, we get 0.10x+0.60y=0.30L due to that the liters add up to get a 30% amount of alcohol (this equation represents the amount of alcohol). Next, x+y=L. Since we know that we have 90 liters of 0.60, y=90 and we have 
0.10x+54=0.30L 
x+90=L. Plugging x+90 for L, we get 0.10x+54=0.3(x+90)=0.3x+27. Subtracting 27 and 0.1x from both sides, we get 0.2x=27. Next, we can multiply both sides by 5 (since 1/0.2=5) to get x=135 liters
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