Question 102401
With mixture problems, you always should try to figure out how much "pure stuff" you have to begin with and how much you will need at the end.

The chemist has 60 grams of solution that is 70% acid, to .7*60 = 42g of pure acid. Water is considered pure water with no acid.

The final solution will be 60 grams of the original solution + x grams of pure water. It will contain 42 grams of acid (all of which came from the original solution).

We want this final solution to be 40% pure acid. We can represent this as {{{ .4(60 + x) = 42}}}. That simplifies to  {{{24 + .4x = 42}}}. Multiply by 10 to remove the decimal points, so we now have: {{{240 + 4x = 420}}}

Subtracting 240 from both sides, {{{4x = 180}}}. Dividing both sides by 4, we have {{{x = 45}}}. So you need to add 45 g of pure water to the original solution.

ALWAYS check your answer! 60g + 45g = 105g of solution in total. It contains 42 grams of pure acid. So the soluton is {{{ 42/105 = .4}}}. Check.