SOLUTION: A chemist has 10% and 50% solutions of acid available. How many liters of each solution should be mixed to obtain 120 liters of 18% acid solution?
Question 1201269: A chemist has 10% and 50% solutions of acid available. How many liters of each solution should be mixed to obtain 120 liters of 18% acid solution?
x = amount of the 10% solution = first batch
120-x = amount of the 50% solution = second batch
0.10x = amount of pure acid from the first batch
0.50(120-x) = 60-0.5x = amount of pure acid from the second batch
We want 120 liters of 18% solution.
Therefore, we want 0.18*120 = 21.6 liters of pure acid.
The two pure acid subtotals 0.10x and 60-0.50x must add to the 21.6 figure.
0.10x + 60-0.50x = 21.6
-0.40x + 60 = 21.6
-0.40x = 21.6-60
-0.40x = -38.4
x = -38.4/(-0.40)
x = 96 liters of the 10% solution is needed.
120-x = 120-96 = 24 liters of the 50% solution is also needed.
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Method 2
The gap from 10% to 18% is 8
The gap from 18% to 50% is 32
The ratio 32:8 reduces to 4:1
We'll need 4 times as much of the 10% solution compared to the 50% solution.
Think of 4:1 as 4x:1x
Their parts add to 4x+1x = 5x
That total must be 120 liters
5x = 120
x = 120/5
x = 24
amount of 10% solution = 4x = 4*24 = 96 liters
amount of 50% solution = 1x = 1*24 = 24 liters
The ratio 96:24 reduces to 4:1.
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Check:
total = 96+24 = 120
10% of 96 = 0.10*96 = 9.6 liters of pure acid from the first batch
50% of 24 = 0.50*24 = 12 liters of pure acid from the second batch
9.6+12 = 21.6 liters of pure acid total
21.6/120 = 0.18 = 18% solution
We have confirmed the answers.
This type been asked and answered many thousands of times.
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(