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UPSC CSE CSAT 2020 · Paper II (CSAT) · Question 80 · Arithmetic: percentage, ratio, averages, time & work

एक मेंढक फिसलन वाली दीवारों वाले 4·5 m गहरे एक सूखे कुएँ से बाहर आने की कोशिश करता है। प्रत्येक…

UPSC Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Question 80 — Arithmetic: percentage, ratio, averages, time & work.

एक मेंढक फिसलन वाली दीवारों वाले 4·5 m गहरे एक सूखे कुएँ से बाहर आने की कोशिश करता है। प्रत्येक बार मेंढक 30 cm फुदकता है और 15 cm नीचे सरक जाता है। कुएँ से बाहर आने के लिए मेंढक के लिए जरूरी फुदकन की संख्या क्या है?

Read this question in English

A frog tries to come out of a dried well 4.5 m deep with slippery walls. Every time the frog jumps 30 cm, slides down 15 cm. What is the number of jumps required for the frog to come out of the well?

(a) 28 · (b) 29 · (c) 30 · (d) 31

उत्तर और व्याख्या · explanation in English

Answer: (b) 29

Each jump-and-slide gains a net 15 cm, but the final jump does not slide back. The frog needs to be within 30 cm of the top, i.e. at 420 cm, before its last jump; 28 jumps with slides get it there, and the 29th jump takes it out.

  1. Well depth = 450 cm; net gain per jump-and-slide = 30 − 15 = 15 cm.
  2. The last jump of 30 cm must start from at least 450 − 30 = 420 cm.
  3. 420 ÷ 15 = 28 jumps (with slides) to reach 420 cm.
  4. The 29th jump lifts it from 420 to 450 cm and out — 29 jumps.
  5. Check: after 27 jumps it is at 405 cm; the 28th reaches only 435 cm and slides back to 420 cm.

Remember · Climb-and-slip problems: subtract the last full jump from the height, count net gains to reach that point, then add one jump.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2020, Q80 — minimalistias.com/pyqs/csat/2020/q80

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