Dilution with 3 Different Percentages I'm trying to solve this problem.  


  How many grams of a solution at $25\%$ do you have to add to $100\ \mathrm{g}$ of a solution at $40\%$ to obtain a solution at $30\%$.


My Attempt

It's a bit more than a simple diluition. So far I've calculated the mass of the solute in $100\ \mathrm{g}$ at $40\%$: $(40\times 100\ \mathrm{g})/100 = 40\ \mathrm{g}$

Then the mass of the solution at $30\%$ so: $(40\ \mathrm{g}/30)\times 100 = 133\ \mathrm{g}$

$133\ \mathrm{g} - 100\ \mathrm{g} = 33\ \mathrm{g}$ (of solvent you need to add)

Now, how does that $25\%$ fit in all this? Can anybody point me to a useful formula or an example? Thanks!
OK, for what it's worth:
Let $x_f$ be the final solution mass and $x$ be the mass of the 25% solution. 
The dissolved mass can be defined by
$$0.3x_f~=~40+0.25x$$

The solution mass is:
$$x_f=100+x$$

Substitute $x_f$ into the first equation and solve. You should get the 25% solution mass to be added and then solve the easy solution mass equation to check your answer.