The bacteria in a culture grows at 5% per hour of what it had been at the beginning of the hour.If the original count of bacteria in the culture at 11 a.m of one day was 18,00,000, find the count of bacteria at 1 p.m. on that day. *
Answers
Answer:
he number of the bactera at the onset=10000.
We know that Finalcount=initialcount±((initialcount)×
100
time×rate
).
∴ In the first hour, the number increases by 10%,so total number of count becomes
10000+(10000×
100
10
)=11000.
In the second hour, the number decreases by 10%
So the rate is negative.
∴ In the second hour, the number of bacteria is
11000−(11000×
100
10
)=9900.
In the third hour the number of bacteria increases by 10%
∴ In the third hour, the number of bacteria is
9900+(9900×
100
10
)=10890.
So, at the end of 3hours the number of bacteria is 10890.
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The count of bacteria at 1 p.m. on that day would be 19,84,500.
To solve the problem, we need to determine the number of bacteria at 1 p.m., given that the growth rate of the bacteria is 5% per hour.
Let's start by finding the number of bacteria at 12 p.m. (noon), one hour after the initial count at 11 a.m.
Number of bacteria at 12 p.m. = 18,00,000 + (5/100) x 18,00,000 = 18,00,000 x 1.05 = 18,90,000
Now, we can use the same formula to find the number of bacteria at 1 p.m., two hours after the initial count at 11 a.m.
Number of bacteria at 1 p.m. = 18,90,000 + (5/100) x 18,90,000 = 18,90,000 x 1.05 = 19,84,500
Therefore, the count of bacteria at 1 p.m. on that day would be 19,84,500.
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