I have a restaurant with service time being exponentially distributed. Let's say the food-serving time has a mean of 30 minutes, and the checkout time has a mean of 10 minutes.
Would $\lambda$ in this problem $=\frac{1}{30+10}$ or $=\frac{1}{30} + \frac{1}{10}$?
Given two independent exponential random variables $X$ and $Y$ with distributions $f_X(t) = \mu e^{-\mu t}$ and $f_Y(t) = \lambda e^{-\lambda t}$, the distribution of the sum $X+Y$ is given by the convolution: \begin{align} f_{X+Y}(t) &= \int_0^t \mu e^{-\mu \tau}\lambda e^{-\lambda (t-\tau)}\;d\tau\\ &=\lambda\mu e^{-\lambda t}\int_0^t e^{-(\mu-\lambda)\tau}\;d\tau\\ &=\frac{\lambda\mu}{\lambda-\mu}\left[e^{-\mu t} - e^{-\lambda t}\right] \end{align}
The mean is given by \begin{align} \frac{\lambda\mu}{\lambda-\mu}\int_0^\infty t\left[e^{-\mu t} - e^{-\lambda t}\right]\;dt&= \frac{\lambda\mu}{\lambda - \mu}\left[\frac{1}{\lambda^2} - \frac{1}{\mu^2}\right]\\ &= \frac{\lambda+\mu}{\lambda\mu} \end{align} which in your case is $40$ minutes.