The time rate of charge $y'= \large\frac{dy}{dt}$

of $y(t) $ =the inflow of salt - outflow

The inflow $=10 \;lb/min$

(5 gal brive, each containing 2 lb)

$y(t) $ is total amount of salt.

tank is always contains 200 gal because 5 gal flow in and 5 gal flow out/ min

Thus 1 gal contains $\large\frac{y(t)}{200}$$lb$ of salt.

Hence 5 out flowing gal containing $\large\frac{5 y(t)}{200}=\frac{y(t)}{200}$

$\qquad=0.025\;y(t) \;lb$

This is outflow.

$y'= $ salt inflow-out flow rate

$y'=10-0.025\;y$

Initially $y(0)=40$

$y'= -0.025 (y-400)$

$\large\frac{dy}{y- 400}$$=-0.025\;at$

$y-400=ce^{-0.025 t}$

$y(0)^{-400}=40-400=-360=C$

$y(t)=400-360e^{-0.025 \;t}\;lb$

Hence c is the correct answer.