29. 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?
Answer: B
Seven cans are required to paint the wall.
To determine the minimum number of cans needed, we first calculate the total area of the wall and then subtract the area of the blue circle. The resulting area painted white will dictate how many cans are needed, given each can covers 70 ft².
A) 6
Choosing 6 cans would mean covering a total area of 420 ft² (6 cans x 70 ft²/can). However, the area of the wall is 400 ft² (40 ft x 10 ft), and the area of the circle is 78.5 ft² (π x 5²). After subtracting the blue area, 400 ft² - 78.5 ft² = 321.5 ft² of white area remains, which requires more than 6 cans.
B) 7
Selecting 7 cans provides a coverage of 490 ft² (7 cans x 70 ft²/can). This is sufficient to cover the white area of 321.5 ft² remaining after accounting for the blue circle. Thus, 7 cans is the minimum required to paint the wall correctly.
C) 8
Opting for 8 cans would yield a coverage of 560 ft² (8 cans x 70 ft²/can), which exceeds the necessary coverage of 321.5 ft². While technically enough, it is not the minimum number of cans required, making this option incorrect.
D) 9
Choosing 9 cans results in 630 ft² (9 cans x 70 ft²/can) of coverage. Similar to option C, while this amount is more than adequate to cover the white area, it is not the least number of cans needed, rendering this option incorrect.
E) 10
Selecting 10 cans would cover 700 ft² (10 cans x 70 ft²/can), which is excessive for the area that needs painting. This option is incorrect as it does not reflect the minimum requirement.
Conclusion
In conclusion, 7 cans are necessary to adequately cover the white area of the wall after accounting for the blue circle. Other options either undercut the requirement or exceed the necessary amount, confirming that 7 is the optimal and minimum choice for this painting task.