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For a high school dinner function for teachers and students, the math department bought 6 cases of juice and 1 case of bottled water for a total of
$135. The science department bought 4 cases of juice and 2 cases of bottled water for a total of
$110. How much did a case of juice cost?
Choose 1 answer:
(A)
$12.50
(B)
$15.00
(c)
$20.00
(D)
$25.00

For a high school dinner function for teachers and students, the math department bought $6$ cases of juice and $1$ case of bottled water for a total of $\$ 135$ . The science department bought $4$ cases of juice and $2$ cases of bottled water for a total of $\$ 110$ . How much did a case of juice cost? $\newline$ Choose $1$ answer: $\newline$ (A) $\$ 12.50$ $\newline$ (B) $\$ 15.00$ $\newline$ (C) $\$ 20.00$ $\newline$ (D) $\$ 25.00$

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Counting principle

Full solution

Q. For a high school dinner function for teachers and students, the math department bought

6

cases of juice and

1

case of bottled water for a total of

\$ 135

. The science department bought

4

cases of juice and

2

cases of bottled water for a total of

\$ 110

. How much did a case of juice cost?

\newline

Choose

1

answer:

\newline

(A)

\$ 12.50

\newline

(B)

\$ 15.00

\newline

(C)

\$ 20.00

\newline

(D)

\$ 25.00

Set up equations: Let $J$ be the cost of a case of juice and $W$ be the cost of a case of bottled water. We can set up two equations based on the information given: $\newline$ For the math department: $6J + W = \$(135)$ $\newline$ For the science department: $4J + 2W = \$(110)$
Multiply first equation: We can multiply the first equation by $2$ to help eliminate $W$ : $\newline$ $2(6J + W) = 2(\$135)$ $\newline$ $12J + 2W = (\$270)$
Subtract equations: Now we subtract the second equation from the modified first equation to eliminate $W$ : $\newline$ $(12J + 2W) - (4J + 2W) = (\$270) - (\$110)$ $\newline$ $8J = (\$160)$
Solve for J: Divide both sides by $8$ to solve for J: $\newline$ $J = \frac{\$160}{8}$ $\newline$ $J = \$20$

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