A climber tries to reach the top of an object which is 130 m high. He can climb one metre per minute. However, as soon as he completes 5 metres, he slips down by one metre. The climber will reach the top in :
2 hours 10 minutes
2 hours 45 minutes
2 hours 42 minutes
2 hours 40 minutes
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2023
The climber makes a net progress of 4m every 5 minutes of climbing until the point where the final climb to the top begins. Let’s find out how many 4m segments are needed to get close to 130m. The climber reaches heights that are multiples of 4m after each full cycle of climbing 5m and slipping 1m. The last segment climbed to reach the top will be less than or equal to 5m, and no slip will occur upon reaching 130m.
Consider the height reached after N full cycles (ending with a slip): H = 4N meters, taking 5N minutes. The final climb starts from H and goes up to 130m. This final climb must be <= 5m (from H). So, 130 - H <= 5, which means H >= 125. We need the smallest multiple of 4 that is greater than or equal to 125. 4 * 31 = 124 (not >= 125). 4 * 32 = 128. So, N = 32 cycles are completed.
After 32 cycles, the height reached is 128 m (after the 32nd slip). Time taken for 32 cycles = 32 * 5 minutes = 160 minutes.
From 128 m, the climber needs to climb 130 m – 128 m = 2 meters. This climb takes 2 minutes (at 1 m/min). Since 2m is less than 5m, no slip occurs upon reaching 130m.
Total time = Time for 32 cycles + Time for final climb = 160 minutes + 2 minutes = 162 minutes.
162 minutes = 2 hours and 42 minutes (120 minutes + 42 minutes).