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25 June, 09:38

Tim has $20 to buy snacks for 12 people in an office. Each person will get one snack. Tim is buying bags of pretzels that cost $1.50 per bag and bags of crackers that cost $2.00 per bag.

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Answers (2)
  1. 25 June, 09:53
    0
    Step-by-step explanation:

    Given:

    Total amount = $20

    Cost:

    1 bag of pretzel = $1.50

    1 bag of crackers = $2.00

    Let the number of bag of pretzel be x and number of bag of crackers be y.

    Total number of people = 12

    Total number of people = x + y

    1. x + y = 12

    Total cost = $20

    Total cost = 1.5x + 2y

    2. 1.5x + 2y = 20

    Solving equarion 1 and 2 simultaneously,

    x = 12 - y into equation 2,

    1.5 (12 - y) + 2y = 20

    0.5y = 20 - 18

    y = 4

    x = 8

    Therefore, the number of bag of pretzel be 8 while the number of bag of crackers be 4.
  2. 25 June, 10:19
    0
    4 bags of crackers and 8 bags of pretzels.

    Step-by-step explanation:

    Let number of crackers=c

    Let number of pretzels=p

    Tim is buying snacks for 12 people and each person gets only one snacks. Therefore:

    c+p=12. (Equation 1)

    Bags of pretzels costs $1.50 per bag.

    Bags of crackers that cost $2.00 per bag.

    Tim has $20 to buy snacks.

    1.50p+2c=20. (Equation 2)

    We then solve the two equations to find out how many bag of each type of snacks can be bought.

    c+p=12.

    1.50p+2c=20

    From Equation (1), c=12-p

    Substitute c=12-p into equation (2).

    1.50p+2 (12-p) = 20

    Opening the bracket

    1.50p+24-2p=20

    Next, I collect like terms

    1.50p-2p=20-24

    -0.5p=-4

    Divide both sides by - 0.5 to solve for p.

    p=8.

    Recall from Equation 1.

    c=12-p = 12-8=4

    Therefore, Tim can get 4 bags of crackers and 8 bags of pretzels.
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