A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?
Answer & explanation
Answer: (d) 52
Work backwards from D’s 10 coins. Each giver handed over half his coins plus 2, so he had 2 × (coins given − 2): C had 16, B had 28 and A had 52.
- D now has 10 coins (the smallest two-digit number), all received from C.
- C gave half his coins + 2 = 10, so half = 8 and C had 16 — all received from B.
- B gave half + 2 = 16, so half = 14 and B had 28 — all received from A.
- A gave half + 2 = 28, so half = 26 and A had 52.
- Check: A (52) gives 26 + 2 = 28 to B; B gives 14 + 2 = 16 to C; C gives 8 + 2 = 10 to D.
Remember · For ‘gives half and k more’ chains, reverse each step: original = 2 × (amount given − k).
Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·