Question

A paper shown in Panel I is folded along the dashed lines (- - -) to construct a cube. The shaded regions shown in Panel I appear on the outer surface of the cube. Referring to cubes shown in Panel II, which one of the options is correct?

• A. Only (i) can correspond to the unfolded cube in Panel I.
• B. Only (ii) can correspond to the unfolded cube in Panel I.
• C. Both (i) and (ii) can correspond to the unfolded cube in Panel I.
• D. Neither (i) nor (ii) can correspond to the unfolded cube in Panel I.

Hint:

Trace which squares of the net become adjacent faces on the cube, and check whether the rotational order of the shaded pattern around the diamond corner is preserved without any mirror flip.

Answer 1

The Correct Option is A

Step 1: Identify the faces and their neighbours in the flat net.
Panel I is a 6 square net joined along dashed fold lines. One square carries a large diamond shape (touching the midpoint of all four of its edges), and the squares directly adjoining it in the net carry shaded triangles formed by a diagonal cut, along with a fully shaded square and a small corner triangle further along the strip.

Step 2: Use the rule that folding never creates a mirror image.
When a flat net is physically folded along its crease lines into a cube, each face keeps a fixed rotational relationship with its neighbours, the same relationship you would get by folding a real sheet of paper. It is impossible for a genuine fold to reverse the left-right handedness of the pattern; only an illegal "flip over" of a face (which cannot happen during folding) would reverse it. So a correct 3D sketch of the folded cube must show the diamond face together with its shaded neighbours in the same rotational order (clockwise or counter clockwise) as they appear in the flat net, never mirrored.

Step 3: Track the diamond face and its shaded neighbour around a shared corner.
In the net, the square directly above the diamond square is split by a diagonal into a shaded triangle and an unshaded triangle, and this pair of faces shares one specific corner of the cube once folded. Walking around that shared corner in the net (diamond corner, then the shaded triangle's corner, in a fixed rotational direction) fixes exactly how those two faces, plus the third face meeting at that same corner, must appear together on the finished cube.

Step 4: Compare with cube (i).
In cube (i), the diamond sits on the top face exactly as in the net (touching all four edges of that face), and going around the corner shared with the top face, the neighbouring shaded triangle appears on the correct adjoining face in the same rotational order as in Panel I, with the larger shaded triangle on the top-back edge and a thin shaded sliver on the adjoining side face, matching how the two half-triangle faces in the net actually meet at that corner when folded.

Step 5: Compare with cube (ii).
In cube (ii), the diamond face itself is shown correctly, but the neighbouring shaded triangle has swapped to the opposite adjoining face (the large triangle now appears on the front-left face instead of matching the net's rotational order), which is the arrangement you would get only by mirroring the net rather than physically folding it. Since a real paper fold cannot produce a mirror image, cube (ii) cannot be a valid folding of Panel I.

Step 6: Conclude.
Only cube (i) preserves the exact rotational (non mirrored) relationship between the diamond face and its shaded neighbours that Panel I requires, so only (i) can correspond to the unfolded cube in Panel I.

\[ \boxed{\text{Only (i) is correct, Option (A)}} \]