if eric has 32 fruit slices and 48 crackers , what is the greatest number of bags he can prepare with no snacks left over ?
Here's how I would solve it. I'm sure there's some algebraic way, but this works for me. 32 & 48 are both divisible by a few different numbers. What's the greatest common factor of the two? There's your answer.
4
divisible by 8 so 4 bags with 8 slices and 12 crackers
four is one of the common factors, but not the greatest
Are you assuming that each bag has to have the same number of fruit slices and crackers? The problem doesn't actually state that. Without that constraint, but assuming each bag must have at least one fruit slice or cracker, he can create 80 bags, each having either one fruit slice or one cracker.
i'm thinking that the best answer is 4 bags with 8 slices and 12 crackers
....wow ktklown, I would not have thought of that. That's definitely true. niya_baybiiee, is there anything stating that the bags have to have both snacks? because if not, ktklown is definitely correct with 80.
16 bags with 2 slices and 3 crackers would be better
probably not 80
Sharon, 4, 8, and 12 all have a common factor, therefore that is not the correct answer if we are assuming that both snacks need to be there.
Yep, ktklown, I agree with 16 bags.
"8 slices and 12 crackers" you know must be wrong bcause 8 and 12 themselves have a common factor
16 sounds right
That's what I was shooting for with having niya finding the GCF
i was thinking least number
Note that the constraint "each bag must have at least one kind of each snack" would give yet a different answer -- 32 bags, each of which has one fruit slice, and half of which have one cracker, the other half have two crackers
But most likely they want "each bag has the same contents", in which case you want 16 bags
Oh word problems, how easily you can go wrong. Sharon, the question states "greatest amount of bags," therefore you want the greatest factor, not the least common factor.
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