David is going to install tiles on the floor of two rooms. One of the floors is 18 feet long by 12 feet wide. The other floor is
24 feet long by 16 feet wide. His options at the tiles store are below. One tile costs $5 and covers 1 square feet of area. Four tiles costs $17 and covers 4 square feet of area. What is the minimum cost of installing the tiles on the two floors of David's two rooms?
Dimension of one of the floors of one room that David wants to install tiles is 18feet long by 12 feet wide Then Area of the above room = 18 * 12 square feet = 216 square feet Dimension of the floor of the other room that David wants to install tiles is 24 feet long and 16 feet wide Then Area of the other room = 24 * 16 square feet = 384 square feet Then The total square feet of the rooms that David wants to install tiles = 216 + 384 = 600 square feet Cost of the tile that covers 1 square feet = $5 Cost of the 4 tiles that cover 4 square feet = $17 Then Area that can be covered with 4 square feet of tiles = 600/4 square feet = 150 square feet Minimum cost of covering the two rooms that David wants to install tiles = 150 * 17 dollars = 2550 dollars So the minimum cost of installing the tiles on the two floors of David's two rooms is $2550. I hope the procedure is simple enough for you to understand.
