In order for $40 + 50e^{.25t}$ to be a solution, we need to be able to rewrite it into the form $40 + C_1 e^{25t}$ so that it is true for all $t$.
$$40 + 50e^{.25t}$$
$$40 + 50e^{25t-24.75t}$$
$$40 + (50e^{-24.75t}) e^{25t} $$
But we see that $50e^{-24.75t}$ is not a constant, as it contains a $t$. Therefore it couldn't possibly be a solution.
If by some chance the $.25$ should have been a $25$, then it should be straightforward to see what the answer would be.
Let's separate and integrate:
$$\frac{dA}{dt} = 25(A-40)$$ $$\frac{dA}{(A-40)} = 25 dt$$ $$\int \frac{dA}{(A-40)} = \int 25 dt$$ $$\ln\left|A-40\right| = 25t+C_0 $$ $$A = 40 + e^{25t+C_0}$$ $$A = 40 + C_1 e^{25t}$$
In order for $40 + 50e^{.25t}$ to be a solution, we need to be able to rewrite it into the form $40 + C_1 e^{25t}$ so that it is true for all $t$.
$$40 + 50e^{.25t}$$ $$40 + 50e^{25t-24.75t}$$ $$40 + (50e^{-24.75t}) e^{25t} $$ But we see that $50e^{-24.75t}$ is not a constant, as it contains a $t$. Therefore it couldn't possibly be a solution.
If by some chance the $.25$ should have been a $25$, then it should be straightforward to see what the answer would be.