If one third of the air in the tank is removed by each stroke of an air pump, what fractional part of the total air is removed in 6 strokes?
If one third of the air is removed by each stroke, then two thirds of the air remains in the tank after each stroke.
After the first stroke, two thirds of the air remains.
After the second stroke, two thirds of what remains (two thirds of two thirds) is removed, leaving (2/3)*(2/3) = 4/9 of the original air remaining.
After the third stroke, two thirds of what remains (two thirds of four ninths) is removed, leaving (2/3)*(4/9) = 8/27.
After the fourth stroke, two thirds of what remains (two thirds of eight twenty-sevenths) is removed, leaving (2/3)*(8/27) = 16/81.
After the fifth stroke, two thirds of what remains (two thirds of sixteen eighty-firsts) is removed, leaving (2/3)*(16/81) = 32/243.
After the sixth stroke, two thirds of what remains (two thirds of thirty-two two hundred forty-thirds) is removed, leaving (2/3)*(32/243) = 64/729.
Therefore, after 6 strokes, 64/729 of the total air in the tank is removed.