A boy plays with a ball and he drops it from a height of 1.5 m. Every time the ball hits the ground, it bounces back to attain a height 4/5th of the previous height. The ball does not bounce further if the previous height is less than 50 cm. What is the number of times the ball hits the ground before the ball stops bouncing?
Answer & explanation
Answer: (b) 5
Why not the tempting option · UPSC's key is 5. A very literal reading of 'previous height' lets the ball rise once more, to 49.15 cm, and touch the ground a sixth time; but that final contact is the ball coming to rest, not a hit before it stops bouncing, so 5 remains the better count. In the exam, stop at the first height that breaks the threshold and do not add the final landing.
Each bounce reaches 4/5 of the previous height: 150 → 120 → 96 → 76.8 → 61.44 cm. The next rise would be only 49.15 cm, below the 50 cm limit, so the ball does not bounce again; by then it has hit the ground 5 times.
- Heights in cm: dropped from 150, the ball rises to 4/5 of each previous height — 120, 96, 76.8, 61.44.
- Every rise is preceded by a hit: hits after falling from 150, 120, 96, 76.8 and 61.44 cm — that is 5 hits.
- The rise after the 5th hit would be 4/5 × 61.44 = 49.152 cm, below 50 cm, so the ball does not bounce further.
- Number of times the ball hits the ground before it stops bouncing = 5.
Remember · List the geometric sequence of heights, stop at the first term that breaks the threshold, and count the hits up to that point; do not add an extra landing after the ball has stopped bouncing.
Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). ·