CSAT (Aptitude)·Revision Notes

Cube and Dice — Revision Notes

Updated 19 Oct 2026

⚡ 30-Second Revision

Identify the task first. Cube: 6 faces, 12 edges, 8 vertices. Each face has 4 neighbours and 1 opposite. Fully painted n³ subdivision: 8; 12(n − 2); 6(n − 2)²; (n − 2)³ for 3, 2, 1 and 0 painted faces, with n ≥ 2.

2-Minute Revision

Use a reference orientation for folding and rolls. A corner view contains three mutually adjacent faces; four distinct known neighbours identify the remaining opposite. A–B–C–D with E above B and F below gives A–C, B–D and E–F opposite. For n = 4, the painted categories are 8 + 24 + 24 + 8 = 64. Exactly two excludes corners; at least two includes them. Uniform selection is an extra assumption for probability.

5-Minute Revision

Reconstruct the six positions for the stated net without looking at the answer. Then derive the four painted categories for n = 5: 8, 36, 54 and 27, totaling 125. Explain why 27/125 is the unpainted probability under uniform selection. Finally roll a labelled cube right and state which faces remain fixed. Check the derivation, not just the final numbers.

Vyyuha Quick Recall

Map, classify, count, check: map faces; separate paint categories; count each cube once; check the total.