# 2008 iTest Problems/Problem 5

## Problem

Jerry recently returned from a trip to South America where he helped two old factories reduce pollution output by installing more modern scrubber equipment. Factory A previously filtered $80$% of pollutants and Factory $B$ previously filled $72$% of pollutants. After installing the new scrubber system, both factories now filter $99.5$% of pollutants.

Jerry explains the level of pollution reduction to Michael, "Factory $A$ is the much larger factory. It's four times as large as Factory $B$. Without any filters at all, it would pollute four times as much as Factory $B$. Even with the better pollution filtration system, Factory $A$ was polluting nearly three times as much as Factory B."

Assuming the factories are the same in every way except size and previous percentage of pollution filtered, find $a+b$ where $a/b$ is the ratio in lowest terms of volume of pollutants unfiltered from both factories $\textit{after}$ installation of the new scrubber system to the volume of pollutants unfiltered from both factories $\textit{before}$ installation of the new scrubber system.

## Solution

Let $x$ be the volume of the pollutants unfiltered in factory B. That means the volume of pollutants in factory A is $4x$.

The old filter filtered 80% of factory A’s pollutants, so the volume of pollutants of factory A with the old filter is $(0.2)4x = 0.8x$. Similarly, the old filter filtered 72% of factory A’s pollutants, so the volume of pollutants of factory A with the old filter is $(0.28)x = 0.28x$. The total amount of pollutants under the old filtration system is $1.08x$.

Using the same methodology, filtering 99.5% of the pollutants for both factories means that the volume of pollutants for both factories is $0.005(4x+x) = 0.025x$. The ratio of the total pollutants before the new filtration to the total pollutants after the new filtration is $\tfrac{1.08x}{0.025x} = \tfrac{1080}{25} = \tfrac{216}{5},$ so $a+b=\boxed{221}.$