A desalination process is able to remove 2/3 of salt in each cycle. If we have 1 liter of sea water that has 5% salt (i.e. 5 g in a liter), how much salt would have been removed from the liter of water after 3 cycles of the desalination process?  Choose the correct choice from the following options
A) 4.15	B) 4.21	C) 4.52	D) 4.81	E) 4.93
X is the initial quantity of salt in the water.   After a cycle of desalination the quantity of salt left would be X1=X-2/3X i.e. X1=1/3X   Same thing for X2, X3 respectively representing the quantity of salt left after 2, 3 cycles of desalination.   So we have:   X1=1/3 X   X2=1/3 X1   X3=1/3 X2   By multiplying both ends of the equations we get:   X1*X2*X3 = (1/3)^3 X*X1*X2   Simplification leaves us with:   X3 = (1/3)^3 X   X3 is the quantity of salt left after three cycles. So the cumulative quantity of salt (Q3) which has been removed after 3 cycles is: Q3= X-X3=X-(1/3)^3 X i.e. Q3 = X (1-(1/3)^3)   In this case, given X was: X=5 g   So we have Q3=5 * (1-1/27) = 5 * 26/27 = 4.814 g rounded Q3=4.81   Answer: 4.81
<result>D</result>
A desalination process is able to remove 2/3 of salt in each cycle. If we have 1 liter of sea water that has 5% salt (i.e. 5 g in a liter), how much salt would have been removed from the liter of water after 3 cycles of the desalination process?