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ser-zykov [4K]
3 years ago
14

Suppose there is a pile of​ quarters, dimes, and pennies with a total value of ​$1.06. ​How much of each coin can be present wit

hout being able to make change for a​ dollar? If there are multiple selections of coins that will​ work, choose the selection with the largest total number of coins.
Mathematics
1 answer:
SpyIntel [72]3 years ago
5 0
Hello,

I note (a,b,c) the result of a quarters, b dimes and c pennies:


2 solutions:
106=( 3, 3, 1)=( 1, 8, 1)



106=( 0, 0, 106) but : 100= 0*25+ 0*10+ 100
106=( 0, 1, 96) but : 100= 0*25+ 1*10+ 90
106=( 0, 2, 86) but : 100= 0*25+ 2*10+ 80
106=( 0, 3, 76) but : 100= 0*25+ 3*10+ 70
106=( 0, 4, 66) but : 100= 0*25+ 4*10+ 60
106=( 0, 5, 56) but : 100= 0*25+ 5*10+ 50
106=( 0, 6, 46) but : 100= 0*25+ 6*10+ 40
106=( 0, 7, 36) but : 100= 0*25+ 7*10+ 30
106=( 0, 8, 26) but : 100= 0*25+ 8*10+ 20
106=( 0, 9, 16) but : 100= 0*25+ 9*10+ 10
106=( 0, 10, 6) but : 100= 0*25+ 10*10+ 0
106=( 1, 0, 81) but : 100= 1*25+ 0*10+ 75
106=( 1, 1, 71) but : 100= 1*25+ 1*10+ 65
106=( 1, 2, 61) but : 100= 1*25+ 2*10+ 55
106=( 1, 3, 51) but : 100= 1*25+ 3*10+ 45
106=( 1, 4, 41) but : 100= 1*25+ 4*10+ 35
106=( 1, 5, 31) but : 100= 1*25+ 5*10+ 25
106=( 1, 6, 21) but : 100= 1*25+ 6*10+ 15
106=( 1, 7, 11) but : 100= 1*25+ 7*10+ 5
106=( 1, 8, 1) is good
106=( 2, 0, 56) but : 100= 2*25+ 0*10+ 50
106=( 2, 1, 46) but : 100= 2*25+ 1*10+ 40
106=( 2, 2, 36) but : 100= 2*25+ 2*10+ 30
106=( 2, 3, 26) but : 100= 2*25+ 3*10+ 20
106=( 2, 4, 16) but : 100= 2*25+ 4*10+ 10
106=( 2, 5, 6) but : 100= 2*25+ 5*10+ 0
106=( 3, 0, 31) but : 100= 3*25+ 0*10+ 25
106=( 3, 1, 21) but : 100= 3*25+ 1*10+ 15
106=( 3, 2, 11) but : 100= 3*25+ 2*10+ 5
106=( 3, 3, 1) is good
106=( 4, 0, 6) but : 100= 4*25+ 0*10+ 0






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Torrance is shopping for a school party. His donation to the party is snack bags and juice boxes. Snack bags come in packages of
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Torrance must purchase 5 packages of snack bags and 6 packages of juice boxes.

Step-by-step explanation:

Given that:

Snack bags come in packages of 12.

Juice bags come in packages of 10.

To find:

Fewest number of packages of each product so that there are same number of snack bags and juice boxes.

Solution:

Number of snack bags when 1 package is bought = 12

Number of snack bags when 2 package is bought = 24

Number of snack bags when 3 package is bought = 36

Number of snack bags when 4 package is bought = 48

Number of snack bags when 5 package is bought = 60

Number of snack bags when 6 package is bought = 72

Number of juice bags when 1 package is bought = 10

Number of juice bags when 2 package is bought = 20

Number of juice bags when 3 package is bought = 30

Number of juice bags when 4 package is bought = 40

Number of juice bags when 5 package is bought = 50

Number of juice bags when 6 package is bought = 60

Number of juice bags when 7 package is bought = 70

We can see that when 5 packages of snack bags are bought and 6 packages of juice bags are bought, 60 bags of each are bought.

This can be found by finding the LCM as well.

LCM of 10 and 12 is 60.

It means we need to buy 60 bags of each item.

And we can easily find the number of packages for each.

Number of packages of snack bags to be bought = \frac{60}{12} =5

Number of packages of juice bags to be bought = \frac{60}{10} = 6

Torrance must purchase 5 packages of snack bags and 6 packages of juice boxes.

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Answer:

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Guess and Check:

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