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12 May 2016, 01:48
Kudos
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B
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Difficulty:
35%
(medium)
Question Stats:
74% (02:21) correct
26%
(02:14)
wrong
based on 121
sessions
History
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The solid brick shown is made of small bricks of side 1. When the large brick is disassembled into its component small bricks, the total surface area of all the small bricks is how much greater than the surface area of the large brick?
A. 32
B. 40
C. 60
D. 72
E. 80
Attachment:
2016-05-12_1244.png [ 1.52 KiB | Viewed 20192 times ]
12 May 2016, 04:08
Kudos
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From the figure: The large brick is made up of 12 cubes.
Length of the large brick = 3
Width of large brick = 2
Height of large brick = 2
Total Surface area of the large brick = Total Surface area of a Cuboid = 2(3 * 2 + 2 * 2 + 3 * 2) = 2 * 16 = 32
Total Surface area of all 12 smaller bricks = 12 * (6 * 1) = 72
TSA of all small bricks - TSA of large brick = 72 - 32 = 40
Answer: B
Length of the large brick = 3
Width of large brick = 2
Height of large brick = 2
Total Surface area of the large brick = Total Surface area of a Cuboid = 2(3 * 2 + 2 * 2 + 3 * 2) = 2 * 16 = 32
Total Surface area of all 12 smaller bricks = 12 * (6 * 1) = 72
TSA of all small bricks - TSA of large brick = 72 - 32 = 40
Answer: B
12 May 2016, 06:54
Kudos
Bookmarks
Surface area of 12 single bricks
12*6*1^2 (Surface area of a cube is 6L^2)= 72
Surface area of large brick 2(3*2+3*2+2*2)= 32 (Surface area of rectangular solid is 2(LB+BL+HB) )
Difference will be 72-32= 40
12*6*1^2 (Surface area of a cube is 6L^2)= 72
Surface area of large brick 2(3*2+3*2+2*2)= 32 (Surface area of rectangular solid is 2(LB+BL+HB) )
Difference will be 72-32= 40
