To obtain 890 liters of a production line solution that contains 2 percent of chemical a a sample of a solution containing 12 percent of chemical a and another containing 1 percent of chemical a are mixed how many liters of the 12 percent sample are required

the question is of a mixture of solutions containing a certain chemical

finally the mixture should contain 2 % of chemical A with a volume of 890 L

solution A has a percentage of 12 %

solution B has a percentage of 1 %

x volume is taken from solution A

volume taken from A and B should be equal to 890 L

therefore if x volume is taken from A then (890 - x) volume should be taken from solution B

percentage multiplied by volume added gives the amount of chemical A

therefore we can write the following equation

12 % * x + 1 % * (890 - x) = 2 % x 890

0.12x + 8.9 - 0.01x  = 17.8

0.11x  = 8.9

x = 80.9 L

therefore 80.9 L of 12 % solution is required

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i think its 5049.

i hope this !