Question: I have encountered many mixing problem like this but I just can't answer this one.
Here is the problem,

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> A mass of inert material containing $15 \text{ lb}$ of salt in its pores is agitated with $10 \text{ gal}$ of water initially fresh. The salt dissolves at a rate which varies jointly as the number of pounds of undissolved salt and the difference between the concentration of the solution and that of a saturated solution ($3 \text{ lb/gal}$). If $9 \text{ lb}$ are dissolved in $10 \text{ min}$, when will $90\%$ be dissolved?
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Here's my solution:  

Let $P$ be the amount of dissolved salt, $15-P$ be the undissolved. The concentration of the solution $= \frac P {10} \text{ lb/gal}$. Then,
$$\frac{\mathrm dP}{\mathrm dt} = (15-P)\left(3-\frac P {10}\right)\text;$$
$$\frac{\mathrm dP}{\mathrm dt} = (15-P)(30-P)/(10)\text.$$
Since this is separable, I used partial fraction then integrate both side and got an answer of around $11 \text{ min}$, which is wrong according to the textbook. The answer key provided an answer of $30.5 \text{ min}$. I don't really know whats wrong in my equation above.

Answer: 
$\frac{dP}{dt} = k (15-P) (3- \frac{P}{10})$ where $k$ is a constant multiplier.

$dt = \frac{10}{k(15-P)(30 - P)} dP$

$ \displaystyle \int dt = 10 = \frac{1}{k}\int\_0^9\frac{10}{k(15-P)(30 - P)} dP = \frac{1}{k} \frac{2}{3} \ln \frac{7}{4}$ (time taken is $10$ mins).

So, $k = \frac{1}{15} \ln \frac{7}{4} \approx 0.0373$

Now for $90\%$ of salt = $13.5$ lb. For it to dissolve, time taken will be

$ \displaystyle \int dt = t = \frac{1}{k}\int\_0^{13.5}\frac{10}{k(15-P)(30 - P)} dP = \frac{1.1365}{k}$

So, $t = \frac{1.1365}{0.0373} \approx 30.5$ minutes.
The setup is not quite right. The wording of the problem is a little poor, but ultimately it means that the rate of change of the amount of salt is *proportional* to the remaining salt and the difference between the concentrations. As such, our equations should read

$$\frac{dP}{dt}=k(15-P)\left(3-\frac{P}{10}\right)$$

Comparing to what you had, we have a proportionality constant and no $dP/dt$ term on the right. The general solution (using separation of variables) is

$$P=30-\frac{15}{1+Ce^{-kt/2}}$$

(can you do this?) There are two conditions we need to satisfy here: namely

$$P(0)=0,\;\;\;\;P(10)=9$$

(Note that we are using $t$ here measured in minutes). The first condition gives us the constant $C$ as

$$0=P(0)=30-\frac{15}{1+C}\implies C=-1/2\implies P=30\left(1-\frac{1}{2-e^{-kt/2}}\right)$$

The second condition gives us $k$, which is

$$9=P(10)=30\left(1-\frac{1}{2-e^{-5k}}\right)\implies k=-\frac{1}{5}\ln\left(\frac{4}{7}\right)$$

Our final equation is then

$$P=30\left(1-\frac{1}{2-\left(4/7\right)^{t/10}}\right)$$

The question asks at what time $t$ is $P$ at 90% of the total amount of salt, so for what $t$ is $P(t)=13.5$. We have

$$13.5=30\left(1-\frac{1}{2-(4/7)^{t/10}}\right)\implies t=\frac{10\ln(11/2)}{\ln(7/4)}\approx 30.46$$

Hope this helps!