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UPSC CSE CSAT 2023 · Paper II (CSAT) · Question 18 · Puzzles, arrangements & general mental ability

125 सर्वसम घन एक घनाकार खंड के रूप में व्यवस्थित किए गए हैं । कितने घन हर पार्श्व से अन्य घनों…

UPSC Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Question 18 — Puzzles, arrangements & general mental ability.

125 सर्वसम घन एक घनाकार खंड के रूप में व्यवस्थित किए गए हैं । कितने घन हर पार्श्व से अन्य घनों द्वारा घिरे हुए हैं ?

Read this question in English

125 identical cubes are arranged in the form of a cubical block. How many cubes are surrounded by other cubes from each side?

(a) 27 · (b) 25 · (c) 21 · (d) 18

उत्तर और व्याख्या · explanation in English

Answer: (a) 27

125 cubes make a 5 × 5 × 5 block. A cube surrounded on every side cannot lie on any outer face, so remove one layer from each side: the hidden core is 3 × 3 × 3 = 27 cubes.

  1. 125 = 5 × 5 × 5, so the block is 5 cubes long, wide and high.
  2. Every cube in the outer layer has at least one face exposed.
  3. Removing one layer from each side leaves (5 − 2) × (5 − 2) × (5 − 2) = 3 × 3 × 3 = 27 cubes, each touched by cubes on all sides.

Remember · Fully hidden cubes in an n × n × n block = (n − 2)³.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2023, Q18 — minimalistias.com/pyqs/csat/2023/q18

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