Question - Bacterial Colony Doubling Problem
Solution:
Let the initial number of bacteria be $$ P_0 $$.Given that the colony doubles every 3 hours, the number of bacteria after 3 hours ($$ P_3 $$) is $$ 2P_0 $$.The number of bacteria after 6 hours ($$ P_6 $$) is $$ 2P_3 = 2 \times 2P_0 = 2^2P_0 $$.The number of bacteria after 9 hours ($$ P_9 $$) is $$ 2P_6 = 2 \times 2^2P_0 = 2^3P_0 $$.The number of bacteria after 12 hours ($$ P_{12} $$) is $$ 2P_9 = 2 \times 2^3P_0 = 2^4P_0 $$.Thus, after 12 hours, the number of bacteria is $$ P_{12} = 2^4P_0 $$.Since $$ P_0 = 10 $$:$$ P_{12} = 2^4 \times 10 $$$$ P_{12} = 16 \times 10 $$$$ P_{12} = 160 $$Therefore, the number of bacteria in the colony after 12 hours is 160. The answer is option C.
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