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21 Mar 2022, 02:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
62% (02:33) correct
38%
(02:33)
wrong
based on 295
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In a certain population of microbes in a petri dish, the rate of increase in the population is 9 percent per hour. The time it takes for a given population to double can be estimated using the formula 70/t hours, where t is the rate of increase per hour expressed as a percent. If the number of microbes in the dish is 60,000 at the start of the first hour, what will be the approximate number of microbes after 48 hours?
A. 120,000
B. 420,000
C. 720,000
D. 1,900,000
E. 3,800,000
Are You Up For the Challenge: 700 Level Questions
A. 120,000
B. 420,000
C. 720,000
D. 1,900,000
E. 3,800,000
Are You Up For the Challenge: 700 Level Questions
21 Mar 2022, 18:22
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Bunuel
Breaking Down the Info:
\(\frac{70}{t} = \frac{70}{9}\) is the amount of time it takes for the number of microbes to double. We can divide 48 by this to find the number of times we double:
\(48/\frac{70}{9} = \frac{432}{70} \approx 6\). Thus we double more than 6 times.
\(2^6 = 64\), so we expect the answer to be more than \(60,000 * 64 = 3,840,000\).
Answer: E
14 Apr 2025, 00:06
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Bunuel
In a period of 70/9 hours, the number of microbes doubles i.e. gets multiplied by 2. In 48 hours, how many such periods are there? There are
\(\frac{48}{70/9} = 6\) (approximately) such periods.
In 6 periods, the number of microbes will become 60k * 2 * 2 * 2 * 2 * 2 * 2 = 60k * 64
This is somewhat more than 3600k and hence the answer is (E)
