In a particular dilution technique, 10% of the solution is removed and

We can start with 100 grams of pure alcohol and see how many times it takes the alcohol to drop below 65 grams where each time we remove 10% of the solution.

After the first time, we have 100 x 0.9 = 90 grams of alcohol left.

After the second time, we have 90 x 0.9 = 81 grams of alcohol left.

After the third time, we have 81 x 0.9 = 72.9 grams of alcohol left.

After the fourth time, we have 72 x 0.9 = 65.61 grams of alcohol left.

Without further calculation, we can see that after the fifth time, the amount of alcohol will drop below 65 grams (since the the fourth time, that amount is just above 65 grams).

Answer: C

Let the total volume of the solution be 100l
The initial concentration of alcohol is 100%.
Thus, the volume of alcohol is 100l.

After first dilution, the volume of the alcohol removed = 10% of 100l = 100*10/100 = 10l
Thus, the volume of alcohol left = 100 - 10 = 90l or 0.9*100

After the second dilution, volume of the alcohol left = [(100-10)/100]*90l = 0.9*90 = 81l

Similarly, after the third dilution, the volume of alcohol left = 0.9*81 = 72.9l.

Thus, after nth dilution, the volume of alcohol left = (0.9)^n*100

Since the percentage, if alcohol should be below 65% or 65*100/100 = 65l.

(0.9)^n*100 < 65l
(0.9)^n < 0.65

When n = 5, (0.9)^5 = 0.5905

Thus, after 5 dilutions, the concentration of alcohol will be below 65%.

Thus, the correct answer is C.