KCET · Maths · Sequences and Series
A population grows at the rate of \(10 \%\) of the population per year. How long does it take for the population to double?
- A \(20 \log 2 \mathrm{yr}\)
- B \(10 \log 2 \mathrm{yr}\)
- C \(5 \log 2 \mathrm{yr}\)
- D None of the Above
Answer & Solution
Correct Answer
(D) None of the Above
Step-by-step Solution
Detailed explanation
Given, \(r=10 \%\)
Let population is \(P\).
According to the question
\[
\begin{aligned}
2 \mathrm{P} &=\mathrm{P}\left(1+\frac{10}{100}\right)^{\mathrm{n}} \\
\Rightarrow \quad 2 &=\left(1+\frac{1}{10}\right)^{\mathrm{n}}
\end{aligned}
\]
\[
\begin{aligned}
\Rightarrow \quad \log 2 &=\mathrm{n}(\log 11-\log 10) \\
\Rightarrow \quad \mathrm{n} &=\frac{\log 2}{1.0414-1}=\frac{\log 2}{0.0414} \\
&=24.15 \log 2
\end{aligned}
\]
Hence, none of the given option is correct.
Let population is \(P\).
According to the question
\[
\begin{aligned}
2 \mathrm{P} &=\mathrm{P}\left(1+\frac{10}{100}\right)^{\mathrm{n}} \\
\Rightarrow \quad 2 &=\left(1+\frac{1}{10}\right)^{\mathrm{n}}
\end{aligned}
\]
\[
\begin{aligned}
\Rightarrow \quad \log 2 &=\mathrm{n}(\log 11-\log 10) \\
\Rightarrow \quad \mathrm{n} &=\frac{\log 2}{1.0414-1}=\frac{\log 2}{0.0414} \\
&=24.15 \log 2
\end{aligned}
\]
Hence, none of the given option is correct.
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