a person was in charge of ordering 26 pizzas for the office party. he ordered three types of pizza: cheese…
a person was in charge of ordering 26 pizzas for the office party. he ordered three types of pizza: cheese, pepperoni, and supreme. the cheese pizzas cost $6 each, the pepperoni pizzas cost $9 each, and the supreme pizzas cost $12 each. he spent exactly twice as much on the pepperoni pizzas as he did on the cheese pizzas. if the person spent a total of $222 on pizza, how many pizzas of each type did he buy?\nthe person ordered □ cheese pizzas, □ pepperoni pizzas, and □ supreme pizzas.
Answer
Explanation:
Step1: Define variables for each pizza type
Let $c$ = number of cheese pizzas, $p$ = number of pepperoni pizzas, $s$ = number of supreme pizzas.
Step2: Set up total pizza count equation
Total pizzas equal 26: $$c + p + s = 26$$
Step3: Set up cost relationship equation
Pepperoni cost = 2×cheese cost: $$9p = 2(6c) \implies 9p = 12c \implies 3p = 4c \implies p = \frac{4}{3}c$$
Step4: Set up total cost equation
Total spent is $222: $$6c + 9p + 12s = 222$$ Simplify by dividing by 3: $$2c + 3p + 4s = 74$$
Step5: Substitute $p$ into total pizza equation
$$c + \frac{4}{3}c + s = 26 \implies \frac{7}{3}c + s = 26 \implies s = 26 - \frac{7}{3}c$$
Step6: Substitute $p$ and $s$ into simplified cost equation
$$2c + 3\left(\frac{4}{3}c\right) + 4\left(26 - \frac{7}{3}c\right) = 74$$ Simplify term by term: $$2c + 4c + 104 - \frac{28}{3}c = 74$$ $$6c - \frac{28}{3}c = 74 - 104$$ $$\frac{18}{3}c - \frac{28}{3}c = -30$$ $$-\frac{10}{3}c = -30$$
Step7: Solve for $c$
Multiply both sides by $-\frac{3}{10}$: $$c = (-30) \times \left(-\frac{3}{10}\right) = 9$$
Step8: Solve for $p$ using $p = \frac{4}{3}c$
$$p = \frac{4}{3} \times 9 = 12$$
Step9: Solve for $s$ using total pizza equation
$$s = 26 - 9 - 12 = 5$$
Answer:
The person ordered 9 cheese pizzas, 12 pepperoni pizzas, and 5 supreme pizzas.