 Mathematics
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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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.