Question

# In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of $$2.5\%$$ per hour. Find the bacteria at the end of $$2$$ hours, if the count was initially $$5,06,000$$.

Solution

## We have, $$P_0$$ $$=$$ Original count of bacteria $$=506000$$, Rate of increase $$=R =2.5\%$$ per hour, Time $$=2$$ hours $$\therefore$$ Bacteria count after $$2$$ hours $$=$$ $$P$$ $$P = P_0(1 +\dfrac{R}{100})^T$$$$=506000\times {\left(1+\dfrac{2.5}{100}\right)}^{2}$$ $$=506000\times \dfrac{102.5}{100}\times \dfrac{102.5}{100}$$$$=531616.25=531616$$ (approx)Mathematics

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