Arnold needs a 25% solution of nitric acid. He has 20 milliliters (ml) of a 30% solution. How many ml of a 15% solution should he add to obtain the required 25% solution?

$10$ ml of 15% solution to be added to get 25% solution.
In $20$ml of %30%nitric acid solution, nitric acid content is $20 \cdot \left(0.3\right) = 6 m l$. Let $x$ ml of 15% solution is added. So nitric acid content in $\left(20 + x\right)$ ml solution is $6 + .15 x$, Now by condition, $6 + .15 x = 0.25 \left(20 + x\right) \mathmr{and} 6 + .15 x = 5 + .25 x \mathmr{and} .1 x = 1 \therefore x = 10 m l$[Ans]