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Having trouble with one quirk to this mixing tank problem:

All large mixing tank initially holds 300 gallons of water, with 50 lbs of salt already in it (initial value problem). A brine solution of 2 lb/gal is pumped in at 3 gallons per minute. The water is pumped out at the faster rate of 3.5 gal/min. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0.

I came up with

dA/dt + 7A/600 = 6. A(0) = 50.

Now I understand intuitively that the volume of the tank is decreasing so the 600 in the denominator is changing with time. The answer is

dA/dt + 7A/(600-t) = 6. A(0) = 50

My question is how do they know to subtract exactly t. The tank is losing 0.5 gal/min. Thanks in advance,

Lee.