1. For a first-order reaction, the rate constant is related to half-life by $k=\frac{\ln 2}{t_{1/2}}$

2. The molecule undergoes two independent first-order reactions simultaneously, so the overall rate constant is the sum: $k_{\text{total}}=k_1+k_2$

3. Compute the individual rate constants: $k_1=\frac{\ln 2}{12}$ $k_2=\frac{\ln 2}{3}$

Thus, $k_{\text{total}}=\ln 2\left(\frac{1}{12}+\frac{1}{3}\right)$ $=\ln 2\left(\frac{1+4}{12}\right)=\ln 2\cdot \frac{5}{12}$

4. If both reactions are occurring, the reactant disappears with effective first-order constant $k_{\text{total}}$. For 50% consumption, $\frac{N}{N_0}=\frac{1}{2}$ For a first-order process, $t=\frac{\ln 2}{k_{\text{total}}}$

Substitute: $t=\frac{\ln 2}{(\ln 2)(5/12)}=\frac{12}{5}=2.4\text{ min}$

5. Nearest integer: $2.4 \approx 2$

Therefore, the required time is $\boxed{2\text{ min}}$