Let x(t) denote the amount of salt (in kg) in the tank at time t. The tank starts with 18 kg of salt, so x (0) = 18.
The solution is drained from the tank at a rate of 90 L/min, so that the amount of salt in the tank changes according to the differential equation
dx(t)/dt = - (x(t) kg)/(9000 L) × (90 L/min) = -1/100 x(t) kg/min
or, more succintly,
x' = -1/100 x
This equation is separable as
dx/x = -1/100 dt
Integrating both sides gives
∫ dx/x = -1/100 ∫ dt
ln|x| = -1/100 t + C
x = exp(-1/100 t + C )
x = C exp(-t/100)
(a) Using the initial condition x (0) = 18, we find
18 = C exp(0) ==> C = 18
so that
x(t) = 18 exp(-t/100)
(b) After 20 minutes, we have
x (20) = 18 exp(-20/100) = 18 exp(-1/5) ≈ 14.74
so the tank contains approximately 14.74 kg of salt after this time.