In the mixture of $\mathrm{P}-32$ and $\mathrm{P}-33$ initially $10 \%$ decay came from $\mathrm{P}-33$.
Hence initially $90 \%$ of the mixture is P-32 and $10 \%$ of the mixture is $\mathrm{P}-33$.
Let after time ' $t$ ' the mixture is left into $10 \%$ of $\mathrm{P}-3$ and $90 \%$ of $\mathrm{P}-33$.
Half life of both P-32 and P-33 are given as $14.3$ days and $25.3$ days respectively.
Let ' $x$ ' be total mass undecayed initially and ' $y$ ' be total mass undecayed finally.
Let initial number of $\mathrm{P}-32$ nuclides $=0.9 x$
Final number of P-32 nuclides $=0.1 y$
Similarly, initial number of $\mathrm{P}-33$ nuclides $=0.1 x$
Final number of $\mathrm{P}-33$ nuclides $=0.9 y$
For isotope P-32
$$
N=N_0 e^{-t / \pi / 2} \text { or } 0.1 y=0.9 \times e^{-t / \pi / 2}
$$
For isotope P-33
$$
N=N_0 e^{-t / \pi / 2} \text { or } 0.9 y=0.1 \times e^{-t / 25.3} \text {. }
$$
On dividing, we get
$$
\frac{1}{9}=9 \frac{e^{-t / 14.3}}{e^{-t / 25.3}} \text { or } \frac{1}{81}=e^{\left(\frac{t}{14.3}+\frac{t}{25.3}\right)}
$$
$$
\frac{1}{81}=e^{-t\left[\frac{11}{14.3 \times 25.3}\right]} \text { or } 81=e^t\left[\frac{11}{14.3 \times 25.3}\right]
$$
Taking $\log , \log _e 81=t\left(\frac{11}{14.3 \times 25.3}\right) \log _e e$
or $\log _e 81=t\left[\frac{11}{14.3 \times 25.3}\right]$ time $t=208.5$ days.