There are three pillars X, Y and Z of different heights. Three spiders A, B and C start to climb on these pillars simultaneously. In one chance, A climbs on X by 6 cm but slips down 1 cm. B climbs on Y by 7 cm but slips down 3 cm. C climbs on Z by 6.5 cm but slips down 2 cm. If each of them requires 40 chances to reach the top of the pillars, what is the height of the shortest pillar?
Answer & explanation
Answer: (b) 163 cm
A spider does not slip back in the chance in which it reaches the top, so height = 39 net gains + one full climb. That gives X = 201 cm, Y = 163 cm and Z = 182 cm; the shortest pillar is 163 cm.
- Net gain per chance: A = 6 − 1 = 5 cm, B = 7 − 3 = 4 cm, C = 6.5 − 2 = 4.5 cm.
- In the 40th chance each spider reaches the top and does not slip, so height = 39 × (net gain) + (full climb).
- X = 39 × 5 + 6 = 201 cm; Y = 39 × 4 + 7 = 163 cm; Z = 39 × 4.5 + 6.5 = 182 cm.
- Shortest pillar = 163 cm (Y).
- Check for Y: after 38 chances B is at 152 cm and reaches only 159 cm in the 39th climb, so it truly needs the 40th.
Remember · Climb-and-slip problems: the last climb has no slip. Height = (n − 1) × net gain + one full climb.
Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·