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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 66 cases of juice and 11 case of bottled water for a total of $135 \$ 135 . The science department bought 44 cases of juice and 22 cases of bottled water for a total of $110 \$ 110 . How much did a case of juice cost?\newlineChoose 11 answer:\newline(A) $12.50 \$ 12.50 \newline(B) $15.00 \$ 15.00 \newline(C) $20.00 \$ 20.00 \newline(D) $25.00 \$ 25.00

Full solution

Q. For a high school dinner function for teachers and students, the math department bought 66 cases of juice and 11 case of bottled water for a total of $135 \$ 135 . The science department bought 44 cases of juice and 22 cases of bottled water for a total of $110 \$ 110 . How much did a case of juice cost?\newlineChoose 11 answer:\newline(A) $12.50 \$ 12.50 \newline(B) $15.00 \$ 15.00 \newline(C) $20.00 \$ 20.00 \newline(D) $25.00 \$ 25.00
  1. Define Variables: Let's call the cost of a case of juice JJ and the cost of a case of bottled water WW. We have two equations based on the problem.\newline11) 6J+1W=1356J + 1W = 135\newline22) 4J+2W=1104J + 2W = 110
  2. Modify Second Equation: First, we can multiply the second equation by 12\frac{1}{2} to make it easier to eliminate WW.\newlineSo, 2J+W=552J + W = 55
  3. Eliminate W: Now, we subtract the modified second equation from the first equation to eliminate W.\newline(6J+1W)(2J+W)=13555(6J + 1W) - (2J + W) = 135 - 55\newlineThis simplifies to 4J=804J = 80
  4. Solve for J: Solving for J, we divide both sides by 44.J=804J = \frac{80}{4}J=20J = 20

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