College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.1 - Page 19: 135


3 drinks per hour

Work Step by Step

Let w be equal to 150 pounds. Use the formula $$BAC= \frac{600n}{w(0.6n+169)}$$ where w is weight in pounds, n is number of drinks consumed in a hour, and BAC is the blood alcohol content. Substitute 150 in for w. $$BAC= \frac{600n}{150(0.6n+169)}$$ The BAC has to be under 0.08 grams because that is the legal alcohol limit. Substitute numbers in for n, the number of drinks per hour, to try to find an acceptable amount of drinks able to be consumed without the BAC going over 0.08 grams. Let's try n=3. $$BAC= \frac{600(3)}{150(0.6(3)+169)}$$ $$BAC= \frac{1800}{150(1.8+169)}$$ $$BAC= \frac{1800}{150(170.8)}$$ $$BAC= \frac{1800}{25,620}$$ This gives an approximate BAC of 0.070. Now let's try n=4 and see if the BAC is over or under 0.08. $$BAC= \frac{600(4)}{150(0.6(4)+169)}$$ $$BAC= \frac{2400}{150(2.4+169)}$$ $$BAC= \frac{2400}{150(171.4)}$$ $$BAC= \frac{2400}{25,710}$$ This gives an approximate BAC of 0.093, which is over the legal limit. Because 0.093 is a higher BAC than the limit of 0.08, the number of drinks one may legally consume per hour is 3, with a BAC of 0.070. The answer is 3 drinks each hour.
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