Question 251525
This becomes a standard logic puzzle once you figure out what the number of toys are.

We know that we are looking for 5 numbers which add to 30, all different, all greater than 2. If we take the 4 smallest numbers, then 30 minus that is our largest possible number, 30-(3+4+5+6)=12. So we must choose from the numbers 3-12. We know we need a 5, and we need one number that is 3 times the other. This means we must either take 3,5,9 or 4,5,12. If we take 3,5,9 then we need two more numbers that sum to 13. the only choice is 6,7. If we take 4,5,12 then the remaining two numbers must add to 9, the only choices are 3,6.

So the numbers must be 3,5,6,7,9 or 3,4,5,6,12.

I can't see any way to eliminate either of these choices from the constraints, but every time I try to solve 3,4,5,6,12 it leads to a contradiction. The solution to 3,5,6,7,9 is:

Cher,yellow,train,6
Johnny,green,car,3
Jane	,red,ball,5
Sue,blue,spin,9
Marcia,orange,sled,7

Thanks for this, it was fun. I'd like to see how to eliminate the other set of numbers other than trial and error though. Where did you find it?