| September 09, 2009 10:19:24 |
#7 |
| paul.anuj |
Member Since: February 2008 Total Comments: 5 |
RE: there is 4 cube.They arrange in side by side.The black color is paint ed on the 4 cubes. 1st and 3rd cube is cube is cut by 64 pieces and 2nd 3rd by 27 pieces.Then1)how many pieces painted only one side?2)how many pieces painted only two side?how many
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No of cubes painted on one side for 1st & 3rd cube= 2[6(n-2)^2] =2[6(4-2)^2] = 48
No of cubes painted on one side for 2nd & 4th cube=2[6(n-2)^2] =2[6(3-2)^2] = 12
Total No of cubes painted on one side = 50.
No of cubes painted on two sides for 1st & 3rd cube= 2[12(n-2)] =2[12(4-2)] = 48
No of cubes painted on two sides for 2nd & 4th cube= 2[12(n-2)] =2[12(3-2)] = 24
Total No of cubes painted on one side = 48.
No of cubes painted on three sides for every cube will always be 8.
Hence Total No of cubes painted on three sides = 8*4= 24.
1st and 3rd cube has been cut into 64 pieses which is (n*n*n) hence n=4.
2nd and 3rd cube has been cut into 27 pieces. hence n=3.
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