1. The problem statement, all variables and given/known data A 500-gallon tank originally contains 350 gallons of pure water. A 2-pound per gallon brine solution is pumped into the tank at 10 gallons per minute and the well-stirred mixture is pumped out at one-half that rate. How much salt is in the tank at th emoment of overflow? Answer: 510 lb 2. Relevant equations dA/dt = [(rate in)(Concentration in)] - [(rate out)(concentration out)] 3. The attempt at a solution Here is what I did. dA/dt = (2lb/gal * 10 gal/min) - [A/(350+5t) * 5] dA/dt = 20 - A/(70+t) Rearrange: dA/dt + A/(70+t) = 20 Initial condition: A (0) = 0 eintgral[1/(70+t)] = 70 + t A(t) = [1/(70+t)] * [10* integral(t+70)dt + c A(t) = 5(t+70) + c/(t+70) A(0) = 0; c = -24,500 Thus; A(t) = 5(t+70) - 24500/(t+70) A (10)/500 = 400 - 24500/80 = 93.75 lb/gal This answer is incorrect, as the answer should be 510 lb. PLEASE HELP!! Thanks!!