Explanation

(b) $N = N_{0}e^{- \lambda t}$
$\therefore 0.9N_{0} = N_{0}e^{- \lambda \times 5} \Rightarrow 5\lambda = \log_{e}\frac{1}{0.9}$ …(i)
and $xN_{0} = N_{0}e^{- \lambda \times 20} \Rightarrow 20\lambda = \log_{e}\left( \frac{1}{x} \right)$ …(ii)
Dividing (i) by (ii), we get
$\frac{1}{4} = \frac{\log_{e}{(1/0.9)}}{\log_{e}{(1/x)}} = \frac{\log_{10}{(1/0.9)}}{\log_{10}{(1/x)}} = \frac{\log_{10}{0.9}}{\log_{10}x}$
$\Rightarrow \log_{10}x = 4\log_{10}{0.9 \Rightarrow}x = 0.658 = 65.8\%$