Sixty-four cubes of sides 2 cm each are combined to form a cube of side 8 cm. If four of the smaller cubes along the diagonal of a surface are removed from the surface of the large cube, which one of the following statements about the surface area of this solid object is true ?
It is equal to the surface area of the large cube.
It is less than the surface area of the large cube.
It is more than the surface area of the large cube.
Insufficient data
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2023
The correct option is C) It is more than the surface area of the large cube.
A large cube of side 8 cm is formed from 64 smaller cubes of sides 2 cm (8/2 = 4, so it’s a 4x4x4 arrangement of small cubes).
The surface area of the large cube is 6 * (8 cm)^2 = 6 * 64 sq cm = 384 sq cm.
Four smaller cubes are removed along the diagonal of a surface (face) of the large cube. Consider one face as a 4×4 grid of the smaller cubes’ exposed faces. The diagonal consists of the 4 cubes at positions (0,0), (1,1), (2,2), (3,3) in this grid (assuming indexing from a corner 0,0).
These 4 small cubes correspond to positions in the 4x4x4 large cube grid. Let’s say the face is the top face (z=3, using 0-3 indexing). The cubes are (0,0,3), (1,1,3), (2,2,3), (3,3,3).
– Cubes (0,0,3) and (3,3,3) are corner cubes of the large cube assembly (3 faces originally exposed on the large cube’s surface).
– Cubes (1,1,3) and (2,2,3) are ‘face’ cubes of the large cube assembly (1 face originally exposed on the large cube’s surface).
When a small cube is removed from the surface of the large cube:
– Surface area is lost from the original surface of the large cube.
– New surface area is gained from the faces of the removed small cube that were previously internal.
– For a corner cube (like (0,0,3) or (3,3,3)): 3 faces (each 2×2=4 sq cm) were exposed on the large cube surface (total 12 sq cm). When removed, the 3 interior faces become exposed (total 12 sq cm). Net change in surface area = -12 + 12 = 0.
– For a face cube (like (1,1,3) or (2,2,3)): 1 face (4 sq cm) was exposed on the large cube surface. When removed, the 5 interior faces become exposed (total 5 * 4 = 20 sq cm). Net change in surface area = -4 + 20 = +16 sq cm.
The four removed cubes consist of 2 corner cubes and 2 face cubes of the large cube assembly.
Total change in surface area = (2 * Change from corner cube) + (2 * Change from face cube)
Total change = (2 * 0) + (2 * +16 sq cm) = 0 + 32 sq cm = +32 sq cm.
The surface area of the solid object after removing the cubes is the original surface area plus the net change.
New Surface Area = 384 sq cm + 32 sq cm = 416 sq cm.