Many enzyme reactions follow the rate law $-\dfrac{d[\text{S}]}{dt} = \dfrac{V_{\max}[\text{S}]}{K_m + [\text{S}]}$, where S is the substance the enzyme acts on. This rate is nearly zero order when $[\text{S}]$ is much larger than $K_m$, and nearly first order when $[\text{S}]$ is much smaller. For one enzyme at a fixed temperature, $V_{\max} = 1.00 \times 10^{-3}\ \text{mol L}^{-1}\ \text{min}^{-1}$ and $K_m = 5.00 \times 10^{-3}\ \text{mol L}^{-1}$. The amount of enzyme stays constant. A solution starts with $[\text{S}]_0 = 2.00 \times 10^{-2}\ \text{mol L}^{-1}$. How long does it take for $[\text{S}]$ to fall to $5.00 \times 10^{-3}\ \text{mol L}^{-1}$? Give the answer in $\text{min}$, to two decimal places. Take $\ln 2 = 0.6931$.