Bonnie volunteers to bring bags of candy to her child's class for the Halloween party this year. She buys one bag of candy A containing 120 pieces of candy, one bag of candy B containing 440 pieces of candy, and one bag of candy C containing
520 pieces of candy. She needs to use all the candy to create identical treat bags. How many treat bags can Bonnie make so
that each one has the same number and variety of candy? How many of each type of candy will be in each bag?

Bonnie can make 40 treat bags. Each treat bag has three pieces of Candy A, 11 pieces of Candy B, and 13 pieces of Candy C.

Step-by-step explanation:

Since there's no number of student's in her child's class, I did this:

I found the common factors of 120, 440, and 520. And found out that the GCF (greatest common factor) is 40.

120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.

440: 1, 2, 4, 5, 8, 10, 11, 20, 22, 40, 44, 55, 88, 110, 220, 440.

520: 1, 2, 4, 5, 8, 10, 13, 20, 26, 40, 52, 65, 104, 130, 260, 520.

After finding out that the GCF is 40, I took the numbers 120, 440, and 520, and divided them by 40.

120 divided by 40 = 3

440 divided by 40 = 11

520 divided by 40 = 13

Therefore Bonnie can make 40 treats bags with no candy pieces left behind.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

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step-by-step explanation:

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