Let V_{t} = V = The volume in the tank at time t.

Let K_{x} = Some constant of integration.

V_{t+Δt} = V_{t} + 1*Δt - (Δt/100)*V_{t}.

dv/dt = (V_{t} - V_{t+Δt}) / Δt = 1 - V/100.

dv = (1 - V/100) dt

100/(100-V) dv = dt

Integrate each side:

-100*ln(100-V)=t + K_{1}

exponentiate each side:

(100-V)^{-100} = K_{2} * e^{t}

100-V = e^{-t/100} * K_{3}

V = 100 - K_{3}*e^{-t/100}

We know V=0 when t=0, thus K_{3}=100

V = 100*(1-e^{-t/100})

Substitute 50 for V and it is not difficult to solve for t=100*ln(2).

**Tristan Simbulan** sent in the following solution you may find better:

Let v be the volume at time t and let dv/dt the volume rate of change at time t.

Volume rate of change = rate in - rate out at time t. rate in = 1 gal. per second, rate out = v/100 * 1 gal. per sec.

Therefore dv/dt = 1 - v/100 * 1 ====> dv = 1/100 * (100 - v ) dt ======> dv/(100 - v ) = dt/100 - ln ( 100 - v ) = t/100 + C.

At t = 0 and v = 0, C = - ln 100. v = 100 * { 1 - e^(-t/100) } , let v = 50 .

Find value of t: e^(t/100) = 2, t = 100 * ln 2

Michael Shackleford, A.S.A.