cecconis wrote:
A company produces a cleaning solution that has 60% concentration of Chemical K. The current production batch contains 200L of an ingredient with a 30% concentration of Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How much pure chemical K must be added to the batch to attain the correct concentration of Chemical K?
A. 150L
B. 175L
C. 200L
D. 225L
E. 250L
We can create the equation:
(200 x 0.3 + 150 x 0.4 + K)/(350 + K) = 0.6
(60 + 60 + K)/(350 + K) = 3/5
5(120 + K) = 1050 + 3K
600 + 5K = 1050 + 3K
2K = 450
K = 225
Alternate Solution:
We have 200 liters of 30% solution and 150 liters of 40% solution, and we will add x liters of 100% solution, thus obtaining (200 + 150 + x) liters of a 60% solution. We can summarize this in the following equation:
200(0.30) + 150(0.40) + x(1.00) = (200 + 150 + x)(0.60)
60 + 60 + x = 120 + 90 + 0.6x
120 + x = 210 + 0.6x
0.4x = 90
x = 225 liters
Answer: D
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