I have ten copies of each of the twelve pentominoes. Can I use all of them to completely tile twelve 5 x 10 rectangles?
It is easier to halve the size of the puzzle - using 5 sets of pentominoes to cover 6 of those rectangles. If that can be solved, you can simply use two copies of that as a solution.
I'll restrict things even further by assuming that each rectangle has no duplicate pieces. So each rectangle is covered using a single pentomino set with two pieces omitted. If the six omitted pairs of pieces form a single pentomino set, then the six rectangles are covered by exactly five sets.
So now I just have to look at ways of covering a single 5x10 rectangle with one set of pentominoes, and see which pentomino pairs can be omitted.
As there is no no-computers tag, I used a computer for this. It turns out that
you can omit almost any pair of pentominoes and still be able to cover the rectangle. The only pair for which this is not possible is the P and F pentomino pair.
So I split the pentominoes into six pairs but avoiding the (P,F) pair, and then tile rectangles while omitting each pair in turn.
Here is one such solution:
P.S.: If you omit I&L, I&P, Y&P, or N&P from a pentomino set then the 5x10 rectangle has a unique solution up to rotation/reflection.