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