A wall is 40 ft long and 10 ft high. A circle of radius 5 ft in the center is painted blue, the rest is painted white. One can covers 70 ft². What is the minimum number of cans required?
Seven cans are required to paint the wall.
To calculate the minimum number of cans needed, we first determine the total area of the wall and then subtract the area occupied by the blue circle. Finally, we divide the remaining area by the coverage of one can of paint.
Choosing 6 cans would imply covering an area of 420 ft² (6 cans x 70 ft²/can). However, the total area to be painted (400 ft²) minus the blue circle area (approximately 78.54 ft²) leaves us with around 321.46 ft². Since 6 cans cannot cover this area, this choice is insufficient.
Seven cans will cover 490 ft² (7 cans x 70 ft²/can), which is sufficient for the remaining area of approximately 321.46 ft² after accounting for the blue circle. This option correctly meets the requirement for the total area needing coverage.
While 8 cans would provide coverage for 560 ft² (8 cans x 70 ft²/can), it exceeds the necessary coverage. Although this choice ensures full coverage, it is not the minimum number required, thereby making it an inefficient option.
Nine cans would cover 630 ft² (9 cans x 70 ft²/can), which again exceeds the area needed. This option is unnecessarily high, thus not aligning with the question's requirement for the minimum number of cans.
Choosing 10 cans would cover 700 ft² (10 cans x 70 ft²/can), which is excessive for the area needing paint. Like the previous options, this does not represent the minimum required, as it wastes resources.
In summary, to effectively cover the remaining area of the wall after painting the blue circle, 7 cans of paint are required. This choice ensures sufficient coverage without excess, aligning with the calculated area of 321.46 ft² to be painted white. Thus, 7 is the minimum number of cans necessary to complete the project efficiently.
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