At some instant, a radioactive sample $S_1$ having an activity $5\mu Ci$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10\mu Ci$. The half lives of $S_1$ and $S_2$ are

For sample 1 we can write,

$N_1={N_o}_1{e}^{-{\lambda }_{1}t}$

$N_2={N_o}_2{e}^{-{\lambda }_{2}t}$

Given that at a particular time t $N_1=2N_2$

Also given that

For sample 1, 

$A_1=-\frac{d{N}_1}{dt}={N_o}_1{\lambda }_1{e}^{-{\lambda }_{1}t}=5\mu C_i$

$A_2=-\frac{d{N}_2}{dt}={N_o}_2{\lambda }_2{e}^{-{\lambda }_{2}t}=10\mu C_i$

Solvong above equations we get,

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{4}$

$\frac{{halflife}_1}{{halflife}_2}=\frac{ln2}{{\lambda }_1}\times \frac{{\lambda }_2}{ln2}=4$

$T_{S_1}=4T_{S_2}$
$\Rightarrow T_{S_1}=20$ years and $T_{S_2}=5$ years is the possible answer.