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I am trying to do a mixing problem here and I don't know how to set up the problem....

Brine containing 1 lb/gal of salt is poured at 1 gal/min into a tank that initially contained 100 gallons of fresh water. The mixture is stirred constantly and is drained of at 1 gal/min.

What is the actual amount of salt in the tank at time , t ?

I was thinking

dA/dt = (concentration in )(rate of waterflow) - (concentration out )(rate of water exiting)

and have it in the formula

(0)(1)-(A(t)/100)(1)

dA/dt = -A/100 * 1

dA/dt = -A/ 0.01

integral of 1 / A da = integral of -1/ 0.01 dt

ln|A| = 1/ 0.01t + C

I don't how to go from here...am I on the right track ?

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The outline of the analysis is right, the details are not. The rate of intake of salt (in pounds per minute) is $1$. The rate of outflow of salt is $\frac{A}{100}\cdot 1$. Thus $$\frac{dA}{dt}=1-\frac{A}{100}.$$ We have the initial condition $A(0)=0$. Now solve the differential equation, by whatever technique you like. Note for example that the equation is separable. Or make the substitution $y=1-\frac{A}{100}$.

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