A radioactive element X converts into another stable element Y. Half-life of  X is 2h. Initially only X is present. After time t, the ratio of atoms of  X and Y is found to be 1/4, then t in hours is 

Let N₀ be the no. of atoms of X at time =0. Then at t=4 hrs (i.e, two half-lives)

$\begin{array}{l}{\mathrm{N}}_{\mathrm{x}}=\frac{{\mathrm{N}}_0}{4}\text{ and }{\mathrm{N}}_{\mathrm{y}}=\frac{3{\mathrm{N}}_0}{4}\Rightarrow \frac{{\mathrm{N}}_{\mathrm{x}}}{{\mathrm{N}}_{\mathrm{y}}}=\frac{1}{3}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{and},\text{ at t=6 h:\hspace{0.17em}\hspace{0.17em}}{\mathrm{N}}_{\mathrm{x}}=\frac{{\mathrm{N}}_0}{8}\mathrm{and} {\mathrm{N}}_{\mathrm{y}}=\frac{7{\mathrm{N}}_0}{8}\\ \Rightarrow \hspace{0.17em}\frac{{\mathrm{N}}_{\mathrm{x}}}{{\mathrm{N}}_{\mathrm{y}}}=\frac{1}{7}\\ \mathrm{The}\text{ given ratio }\frac{\text{1}}{\text{4}}\mathrm{lies}\text{ between }\frac{\text{1}}{\text{3}}\text{ and }\frac{\text{1}}{\text{7}}\text{.\hspace{0.17em}}\\ \mathrm{Therefore},\mathrm{t}\text{ lies between 4h and 6 h.}\end{array}$