Discrepancy in a rate of change question Upon solving the following rates of change question from a first year undergraduate course in calculus, I came across two possibilities for the ODE modelling the situation, and I can't understand why. The question is as follows:

An initially unpolluted lake of $10^9$ litres has a river flowing through it at $10^6$ litres per day. A factory is built which discharges $10^4$ litres per day of pollutant into the lake. Assume that the total volume of liquid in the lake remains at a constant $10^9$ litres.

If we let $x(t)$ be the amount of pollutant in the lake in litres at time $t$, then the amount of pollutant at time $t + \delta t$ should be $$x(t + \delta t) = x(t) + 10^4 \delta t - \lambda_1 10^6 \delta t $$ where $\lambda_1$ is the fraction of the $10^6 \delta t$ litres of water expelled from the lake that is pollutant. The fraction $\lambda_1$ should be the concentration of pollutant in the water after clean water and pollutant have been introduced over the time $\delta t$, so
\begin{align}
\lambda_1 = \frac{x(t) + 10^4 \delta t}{10^9 + 10^4 \delta t + 10^6 \delta t}
\end{align} Hence
$$
\frac{x(t + \delta t) - x(t)}{\delta t} = 10^4 - \frac{x(t) + 10^4 \delta t}{10^9 + 10^4 \delta t + 10^6 \delta t} 10^6 
$$ Taking the limit as $\delta t \to 0$ we obtain that $$\frac{dx}{dt} = 10^4 - 10^{-3} x$$ which is what I thought the solution was.
The problem is that if we let $c(t)$ be the amount of clean water in the lake in litres at time $t$, then by a similar argument we have that $$ \frac{dc}{dt} = 10^6 - 10^{-3}c$$ which does not respect the condition $\frac{dx}{dt} + \frac{dc}{dt} = 0$, which comes from $x + c = 10^9$. The solution at the back of the booklet gives the solution $$\frac{dx}{dt} = 10^4 - \frac{10^4 + 10^6 }{10^9} x$$ The only way I can think of arriving at this solution is to assume that $$ \frac{dx}{dt} = 10^4 - \lambda x$$ and $$ \frac{dc}{dt} = 10^6 - \lambda c$$ and solving for $\lambda$ using the condition that $x' + c' = 0$.
My question is: What is wrong with my solution and why does this discrepancy occur?

As pointed out in Lutz Lehmann's comment and Ryan G's answer, the rate at which fluid flows out of the system is $10^4 + 10^6$ litres per day, not $10^6$ litres per day.
Fixing this, we have that
$$x(t + \delta t ) = x(t) + (\text{added pollutant in the time } \delta t) - (\text{pollutant that has left the system in time } \delta t)$$ The pollutant added to the system in the time $\delta t$ is $10^4 \delta t$. The total amount of fluid that has exited the system in the time $\delta t$ is $(10^4 + 10^6) \delta t$, and so
$$ x(t + \delta t) = x(t) + 10^4 \delta t - \rho (10^4 + 10^6) \delta t \tag{1}$$ where $\rho$ is the concentration of pollutant in the fluid that has flown out of the system over the time $\delta t$. We could calculate $\rho$ in a couple of ways:

*

*We could assume that firstly, the pollutant and clean water enter the system so that the total amount of fluid in the system is $10^9 + (10^4 + 10^6) \delta t$. At this point $$\rho = \frac{x(t) + 10^4 \delta t}{10^9 + (10^4 + 10^6) \delta t}$$ and afterwards this excess fluid flows out of the system.


*The second way is to just take $\rho$ to be the concentration of pollutant at time $t$, i.e. $$\rho = \frac{x(t)}{10^9}$$
In either case, from $(1)$ we obtain that
$$ \frac{x(t+ \delta t) - x(t)}{\delta t} = 10^4 - \rho (10^4 + 10^6) $$ and so
\begin{align}
\lim_{\delta t \to 0} \frac{x(t+ \delta t) - x(t)}{\delta t} &= 10^4 - \left( \lim_{\delta t \to 0} \rho \right)(10^4 + 10^6) \\ &= 10^4 - \frac{10^4 + 10^6}{10^9} x(t)
\end{align} I.e. we have that
\begin{align}
\frac{dx}{dt} = 10^4 - \frac{10^4 + 10^6}{10^9} x
\end{align} which is the correct answer.
 A: 
If we let $x(t)$ be the amount of pollutant in the lake in litres at time $t$, then the amount of pollutant at time $t + \delta t$ should be $$x(t + \delta t) = x(t) + 10^4 \delta t - \lambda_1 10^6 \delta t $$ where $\lambda_1$ is the fraction of the $10^6 \delta t$ litres of water expelled from the lake that is pollutant. The fraction $\lambda_1$ should be the concentration of pollutant in the water after clean water and pollutant have been introduced over the time $\delta t$, so \begin{align}
\lambda_1 = \frac{x(t) + 10^4 \delta t}{10^9 + 10^4 \delta t + 10^6 \delta t}
\end{align}

Firstly, the rate of total outflow equals $(10^6+10^4)$ L/day, instead of just $10^6$ L/day.
Secondly, here's the correct model (which is consistent with the book's answer):
$$\frac{dx}{dt} = \text{rate of increase of pollutant}\\=\text{rate of pollutant inflow}-\text{rate of pollutant outflow}\\=\text{rate of pollutant inflow}\\-\text{pollutant concentration}\times\text{rate of total outflow}\\=10^4 - \frac x{10^9} ({10^6 + 10^4})\\=10^4 - \frac{10^6 + 10^4}{10^9} x.$$

Re: your addendum
Rewrite all instances of $\rho$ to $\rho(t)$ to remind self that it varies with time, and notice that $\displaystyle\lim_{\delta t \to 0} \rho=\displaystyle\lim_{\delta t \to 0} \rho(t)=\rho(t),$ by definition.
So, attempting to express $\rho$ in terms of both $t$ and $\delta t$ (as you did in the first bullet point) is unnecessary. That expression for $\rho$ is anyway incorrect (e.g., the total fluid is always $10^9$ litres), even though the error doesn't break the solution.
