A manufacturer has 600 litres of a 12% solution of acid. How many litres (x) of a 30% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%?

It is given that solution 1 contains 12% acid in 600 litres, and solution 2 contains 30% acid in x litres

=> Acid in solution 1 = $$\ \frac{12}{100}\times600$$ = 72 litres

=> Acid in solution 2 = $$\ \frac{30}{100}\times$$ x = 0.3x litres

In the final mixture, acid quantity = (72+0.3x)litres

It is given that the acid content in the resulting mixture will be more than 15% but less than 18%

The equation of acid content will be = 15% of (600+x) < 72+0.3x < 18% of (600+x)   { Final mixture will have a volume of 600+x litres}

=> 90+0.15x < 72+0.3x < 108 + 0.18x

So solving both sides

=> 18 < 0.15x  and 0.12x < 36

Therefore, x lies in 120 < x < 300