Discussion

Answer: B

Only cube (ii) can be made from the net, so the answer is Only (ii) can correspond to the unfolded cube in Panel I.

Label the squares of the net: the centre square (dashed, with a vertical bar) is F2; above it is F1 (a shaded triangle in its lower-right corner); below it, in order, are F3 (a horizontal bar through its middle) and F4 (the lower half shaded); to the left and right of F2 are the left square (an L of gray with a white upper-right quarter) and the right square (a gray upper-right quarter). Folding F2 to the front puts F1 on top, F3 underneath, F4 at the back, and the left and right squares at the sides.

Cube (ii): its front face is white with a gray triangle in a corner, the pattern of F1; its top face is gray over the front half, the pattern of F4; and its right face has a gray quarter, the pattern of the net's right square. These three faces meet at one corner of the folded cube, and the orientations of the patterns agree with how the net folds, so (ii) is possible.

Cube (i): the right face, gray over its lower half and white above, can only be F4. The front face, gray with a white lower-left quarter, can only be the net's left square. The top face must meet both of those, which leaves F3, the barred square. But the bar on F3 runs parallel to the edge that F3 shares with F2, so on the cube it is perpendicular to the edge that F3 shares with the left square. In (i) the top bar is drawn parallel to the front edge, so (i) is impossible.