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Q.In a certain culture of bacteria, the rate of increase of bacteria is proportional to the number present. It is found that there are 10,000 bacteria at the end of 3 hours and 40,000 bacteria at the end of 5 hours. Determine the number of bacteria present in the beginning.

CBSECBSE Class XII Board 2022Subjective· 4mImportance★★★★★
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From N=N0ektN=N_0e^{kt}: dividing the 5-hour and 3-hour readings gives e2k=4e^{2k}=4, so ek=2e^{k}=2; then 10000=N0(ek)3=8N010000=N_0(e^{k})^3=8N_0, giving N0=1250N_0=1250.

Exponential growth: if dNdt=kN\dfrac{dN}{dt}=kN then N=N0ektN=N_0e^{kt}, where N0=N_0= initial number, k=k= growth constant, t=t= time, and N=N= number at time tt.

  1. The model is N=N0ektN=N_0e^{kt}. Given: at t=3t=3, N=10,000N=10{,}000; at t=5t=5, N=40,000N=40{,}000.
  2. Write both: 10000=N0e3k10000=N_0e^{3k} ... (i);   40000=N0e5k\;40000=N_0e^{5k} ... (ii). …

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