Track / Stage 5 / 109 of 400 #709 of 1964

Problem 709

AIME late

Algebra Difficulty 5.2 Find the answer

Impurities constitute $20 \%$ of the total volume of the solution. What is the smallest number of filters through which the solution must be passed so that the final impurity content does not exceed $0.01 \%$ , if each filter absorbs $80 \%$ of the impurities? (It is known that $\lg 2 \approx 0.30$ .)

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution.

Impurities constitute $\frac{1}{5}$ of the solution. After the first filtration, $\left(\frac{1}{5}\right)^{2}$ of the impurities will remain, and after the $k$ -th filtration, $-\left(\frac{1}{5}\right)^{k+1}$ . According to the condition, $\left(\frac{1}{5}\right)^{k+1} \leq 10^{-4} ;-(k+1) \lg 5 \leq-4$ , from which $k \geq 4.7$ .

Answer: 5 filters.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.