Explanation

(a) We know,
$N = N_{0}\left( \frac{1}{2} \right)^{n}$
Where $N_{0}$ is original number of atoms, $n$ is number of half-lives
$\ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{t}{T_{1/2}} = \frac{180}{60} = 3$
$\therefore\ \ \ \ \ \ \ \ \frac{N}{N_{0}} = \left( \frac{1}{2} \right)^{n} = \ \left( \frac{1}{2} \right)^{3} = \frac{1}{8}$
$\therefore\ \ \ \ \ \ \ \ N\ = \ \frac{N_{0}}{8}\ = \ 0.125\ N_{0}\ = \ 12.5\ \%\ N$
Amount decayed$= \ 100 - 12.5\ = 87.5\ \%\ N$