Ken baked, frosted, and decorated a rectangular cake for the last Math Club meeting. The cake was 3 inches high, 12 inches wide, and 16 inches long. He centered the cake on a piece of cardboard whose rectangular top surface had been covered with aluminum foil, as shown in the figure below.
1. Ken used a piece of cardboard large enough to allow the cardboard to extend 2 inches beyond the cake on all sides. What is the area, in square inches, of the aluminum foil that is exposed on the top surface of the cardboard?
2. At the Math Club meeting, Principal Gonzales cut the entire cake into pieces. Each piece is 2 inches wide, 2 inches long, and 3 inches high. What is the number of pieces Principal Gonzales cut the cake into?
3. The Math Club will pay Ken $5.00 for preparing the cake and will also pay him for the cost of the cake mix at $1.73, the frosting mix at $2.67, and the sales tax of 5% on these 2 items. What is the total amount the Math Club will pay Ken?
1. The correct answer is E.
Because the cardboard extends 2 inches beyond the cake on all sides, the cardboard forms a rectangle whose length and width are each 4 inches longer than the length and width of the rectangular cake.
The area of aluminum foil that is exposed equals the area of the foil covering the entire cardboard surface (or area of the cardboard) minus the area of the foil covered by the cake. So the area, in square inches, of aluminum foil exposed equals
20(16) - 16(12) = 320 - 192 = 128
2. The correct answer is D.
The length of the cut cake is (16 inches) ÷ (2 inches per piece) = 8 pieces. The width of the cut cake is (12 inches) ÷ (2 inches per piece) = 6 pieces. Because the cake has 6 rows each containing 8 pieces, the cake was cut into 6(8) = 48 pieces.
3. The correct answer is D.
Expenses for Ken’s cake:
Cake preparation - $5.00
Cake mix - $1.73
Frosting mix - $2.67
Tax on items purchased 0.05($1.73 + $2.67) = $0.22
The total amount the Math Club will pay Ken is the sum of the costs shown in the table,
$5.00 + $1.73 + $2.67 + $0.22 = $9.62