The number of bacteria in a stagment pond doubles every hour. A farmer sprays it every two hours and kils 3/4 of bacteria immediately. If there are 10,000 live bacteria left after the farmer sprays at noon, how many live bacteria are there at 11 p.m. that night?
To find the number of live bacteria at 11 p.m., we need to determine the number of bacteria after each two-hour spraying interval until that time.
Let's break down the information provided:
- At noon, there were 10,000 live bacteria remaining after spraying.
- The number of bacteria in the pond doubles every hour.
- The farmer sprays the pond every two hours, killing 3/4 of the bacteria immediately.
First, let's determine the number of bacteria before the noon spraying. We know that the number of bacteria doubles every hour, so if there were 10,000 after the noon spraying, then before the spraying, there were 10,000 / 3/4 = 13,333 bacteria.
Now, we need to consider the intervals between each spraying session until 11 p.m. Initially, there is a two-hour interval between noon and the first spraying at 2 p.m. During these two hours, the bacteria double in number, reaching 13,333 * 2 = 26,666 bacteria.
After the first spraying at 2 p.m., 3/4 of the bacteria are killed, leaving 26,666 * 1/4 = 6,666 bacteria. Now, there are 10 hours remaining until 11 p.m., which is equivalent to five two-hour spraying cycles.
During each two-hour cycle, the bacteria double in number, and the farmer sprays them again, killing 3/4 of the bacteria.
Let's calculate the number of bacteria after each cycle:
1st spraying (2 p.m.): 6,666 * 2 / 4 = 3,333 bacteria remaining
2nd spraying (4 p.m.): 3,333 * 2 / 4 = 1,666 bacteria remaining
3rd spraying (6 p.m.): 1,666 * 2 / 4 = 833 bacteria remaining
4th spraying (8 p.m.): 833 * 2 / 4 = 416 bacteria remaining
5th spraying (10 p.m.): 416 * 2 / 4 = 208 bacteria remaining
Finally, at 11 p.m., after the last spraying, there are 208 bacteria remaining in the pond.
Therefore, at 11 p.m., there are 208 live bacteria in the stagnant pond.