A laboratory has 22 beakers of liquid in a fume hood, and 8 of the beakers have copper ions in them. Adding three drops of a solution will turn the liquid in a beaker blue if there are copper ions present. If the beakers are tested one by one for copper ions and 45 drops are used before all 8 beakers with copper ions are found, how many beakers without copper ions were tested?
There were 45 / 3 = <<45/3=15>>15 beakers tested to find all the beakers with copper ions.
Thus, 15 - 8 = <<15-8=7>>7 beakers without copper ions were tested.
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