Question 1208622
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With each stroke, a pump removes 2/5 of the air in a container. After 3 strokes what is the fraction of the original amount 
of air in the container that is left?
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<pre>
Let M be the initial mass of the air in the container.


After first stroke,  the remaining mass of the air in the container is  {{{(1-2/5)*M}}} = {{{(3/5)M}}}.


Thus we see that the coefficient from the current air mass to remaining mass
after the first stroke is 3/5.


    +------------------------------------------------------+
    |  This coefficient remains the same at every stroke.  |
    +------------------------------------------------------+


After second stroke, the remaining mass of the air in the container is  {{{(3/5)*(3/5)*M}}} = {{{(3/5)^2*M}}}.


After third stroke,  the remaining mass of the air in the container is  {{{(3/5)*(3/5)^2*M}}} = {{{(3/5)^3*M}}} = {{{(27/125)*M}}}.


<U>ANSWER</U>.   After third stroke,     the remaining mass of the air in the container is  {{{(27/125)*M}}}.
          The remaining fraction of the original mass is  {{{27/125}}}.
</pre>

Solved.