A radio isotope “A” undergoes simultaneous decay to yield “B” and “C” nuclei as 

Assuming that neither B nor C was present in the beginning. After how many hours will be the amount “C” be just double to amount of “A” remaining?

 $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[B\right]=2t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[C\right]$
Let after time “t”,  $\left[C\right]=2\left[A\right]$
                          $B=x\%$

C is formed at twice the rate of formation of B, the rate constant for decay of A is  
$$\begin{gathered} K=K_1+K_2 \\ =\frac{0.693}{9}+\frac{0.693}{4.5} \\ =\frac{0.693}{3}=\frac{\ln 2}{3} \\ \left[C\right]=2\left[A\right] \\ 2x=2(100-3x) \\ 4x=100 \\ x=25 \\ Kt=\ln \frac{a}{a-x} \\ \frac{\ln 2}{3}\times t=\ln \frac{100}{100-3x} \\ \ln 2^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\ln \frac{100}{100-3x} \\ {2}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{100}{100-3x} \\ {2}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{100}{25}=4 \\ {2}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=2^2 \\ t\text{/}3=2 \\ t = 6 \end{gathered}$$