A sample contains equal numbers of nuclei of two independent radioactive isotopes, X (half-life 10 h) and Y (half-life 30 h), each decaying to a…

Activity of each isotope is $\lambda N=\lambda N_0e^{-\lambda t}$, and with equal initial numbers $N_0$ the initial total activity is $N_0(\lambda_X+\lambda_Y)$. Take $\lambda_X=\ln2/10$ and $\lambda_Y=\ln2/30=\lambda_X/3$. The condition for one quarter of the initial activity is $\dfrac{\lambda_Xe^{-\lambda_Xt}+\lambda_Ye^{-\lambda_Yt}}{\lambda_X+\lambda_Y}=\dfrac14$. Dividing by $\lambda_X$ and writing $u=e^{-\lambda_Yt}$ so that $e^{-\lambda_Xt}=u^3$, we get $u^3+u/3=\tfrac14\cdot\tfrac43=\tfrac13$, i.e. $3u^3+u-1=0$. Solving numerically gives $u=0.5366$. Then $t=-\ln u/\lambda_Y=0.6226/0.023105=26.95\ \mathrm{h}$. Neither the shorter nor the longer half-life alone gives this answer, because the faster isotope dominates early and the slower one dominates later, which is why the combined curve is not a single exponential.