This is a homework problem my son brought home -- 5th grade.
Raul has 56 bouncy balls. He puts three times as many balls into red gift bags as he puts into green gift bags. If he puts the same number of balls into each bag, how many balls does he put into each bag?
It is supposed to reinforce these common core standards (per the homework sheet):
https://grade5commoncoremath.wikispaces.hcpss.org/5.NBT.1 https://grade5commoncoremath.wikispaces.hcpss.org/5.NBT.5 https://grade5commoncoremath.wikispaces.hcpss.org/5.NBT.6
It is from Houghton Mifflin Harcourt's California texts.
I went ahead and applied algebra (which isn't at his grade level of course) and treated it as simultaneous linear equations. I of course got multiple integer answers: Wolfram Solution
I suspect this is just a poorly written question for a 5th grader, but I don't see the specific error.
What is the author trying to teach/reinforce here?
To solve this without (much) algebra:
Note that green bags must hold $\frac 14$ of the balls (as we divided them, by color, in the ratio 3:1). Thus there are $14$ balls in the green bags, hence $42$ in the red bags.
Now we ask: how many might be in each bag? Well, whatever that number is it must divide both $14$ and $42$, hence it is a common factor of those two numbers. But we can list those common factors: $\{1,2,7,14\}$. Each of these gives us a solution!
I. 1 ball in each of 14 green and 42 red bags.
II. 2 balls in each of 7 green and 21 red bags.
III. 7 balls in each of 2 green and 6 red bags.
IV. 14 balls in each of 1 green and 3 red bags. (Unless you interpret the problem as requiring plural green bags.)
Unless we are told more information about the size of the party we really can't choose amongst these (though at a guess, we are after case II....I is too many and III and IV are too few).
To be clear: I would not assign this to a fifth grader unless I mentioned that there might be several acceptable answers (and probably not even then).