Cube & Dice — PPSC previous year questions

Red is opposite black. Red is face down, so black is on top. The four side faces are blue, white, brown and the sixth colour. Blue touches both white and brown. So blue cannot be opposite either of them. Blue must be opposite the sixth colour. That leaves white and brown as the remaining pair of opposite faces.

> 3 cm cube cut into 1 cm cubes → 3 × 3 × 3 = 27 small cubes > Exactly one painted face = the centre cube of each face = 1 per face > 6 faces × 1 = 6 The six face-centre cubes have exactly one red side; corner cubes have three and edge cubes have two.

On a die every face is adjacent to the four faces that are not opposite it. Here 1 is opposite 2. So the four faces around 1 are 3, 4, 5 and 6, and the same four surround 2. So 2 is necessarily adjacent to 4, and statement (b) must be true. The others are not forced. Among 3, 4, 5 and 6 there are two opposite pairs. The data do not fix whether 5 is opposite 6 or opposite 3 or 4. So 5 and 6 may or may not be adjacent.

List which faces are seen touching each other in the four pictures. > 6 touches 3 and 2 (I), 2 and 4 (II), 5 and 4 (III) — that is four different faces, so 6 lies opposite 1 > The other four, 2, 3, 4 and 5, form a ring round the 6–1 axis > 2 touches 4 (II) and 3 (I), so 2 lies opposite 5 > The only pair left is 3 and 4

In the first view A touches K (top) and B (right); in the second view M touches H and K; in the third view P touches B and H. K is adjacent to A, B, M and H, so K is opposite P. H is adjacent to M, K, P and B, so H is opposite A, and B is then opposite M. > opposite pairs: K–P, H–A, B–M So the letter opposite A is H.

125 = 5³, so the cube is cut 5 × 5 × 5. The small cubes with no painted face form the inner cube of side (5 − 2) = 3, giving 3³ = 27 cubes.

Cube (i) shows 4 on top, 2 in front and 3 on the right; cube (ii) shows 5, 1 and 2. So 2 touches 4, 3, 5 and 1, which leaves 2 opposite 6. Matching the position of the shared face 2 in views (i) and (ii) fixes the other pairs as 1–4 and 3–5. > opposite pairs: 2–6, 1–4, 3–5 > cube (i): top 4, front 2, right 3 → bottom 1, back 6 > cube (iii) has 1 on top and 6 in front, i.e. cube (i) turned upside down about its left–right axis That turn leaves the right face where it was, so ? = 3.

Read the three visible faces of each position — top, left and right. I = 3, 4, 2; II = 4, 3, 5; III = 5, 1, 4; IV = 2, 4, 1. Faces that appear together in one position are adjacent. So list the adjacent pairs. > I: 3–4, 3–2, 4–2 II: 4–3, 4–5, 3–5 > III: 5–1, 5–4, 1–4 IV: 2–4, 2–1, 4–1 > 4 is adjacent to 3, 2, 5 and 1 → 4 is opposite 6 > 3 is adjacent to 4, 2 and 5 → 3 is opposite 1 or 6; 6 is taken by 4, so 3 is opposite 1 Cross-check: positions I and II share the faces 3 and 4. So the remaining faces, 2 and 5, are opposite each other. That leaves 3–1 and 4–6 as the other opposite pairs.

Each pair of opposite faces has its own colour. So a small cube shows three different colours only at a corner of the big cube. Three mutually adjacent faces meet there. A cube has 8 corners, however finely it is cut — here 4 × 4 × 4 = 64 pieces. So 8 cubes have three faces painted in different colours.

The 8 cm cube cut into 2 cm blocks gives 4 × 4 × 4 = 64 small cubes. A cube with exactly one painted face lies in the middle of a face, away from every edge: (4 − 2)² = 4 per face. Red covers only one pair of opposite faces. > one red face only = 2 × (4 − 2)² = 2 × 4 = 8 Had all six faces been painted, the single-face count would be 6 × 4 = 24 — the pair-of-faces condition must be read carefully.