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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?

rrauenza
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  • I had a hard time picking a tag, which makes me think this question might not be appropriate. Please let me know if it is not. – rrauenza Sep 23 '15 at 01:51
  • Well, I think I'm with you. I expect they meant to ask "how many balls did he put in green bags?". That's easy enough, you get 14. But as I see it he could have put 2 each in 7 green bags and 21 red bags, or 7 each in 2 green and 6 red, or 14 each in 1 green bag and 3 red red ones. Not horribly difficult, but for the fifth grade? – lulu Sep 23 '15 at 01:58
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    I looked through the standards you linked to. The first two seem unrelated but the third one works. I expect your son was meant to note that the green bag must hold $\frac 14$ of the balls, hence $14$. That leaves $42$ for the red ones and now he can solve the problem by enumerating the common factors of $14$ and $42$. No algebra needed, really. But...at a minimum the question should have indicated that there were multiple solutions. Or added a clause like "there were more than $25$ kids at the party!". Or tried a simpler question. – lulu Sep 23 '15 at 02:06

2 Answers2

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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).

Community
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lulu
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My fifth grader got this question too and I was very happy to see it pop up in Google after only typing in "Raul has" ...

Ours reads "how many does he put into each green bag"? I'm guessing it should have been something like "If he puts 1 ball into each bag how many green bags does he have?"

In any event 5th graders haven't studied ratios yet and based on how badly this question is written it makes you wonder how carefully this math program was put together ...

JEsse
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  • I haven't been impressed with Houghton Mifflin Harcourt's math texts even before common core. My other son is in middle school and their books are much better: http://cpm.org/ – rrauenza Oct 22 '15 at 14:19