A squirrel starts climbing up a tree at the speed of 6 metres a minute but after every 6 metres it slips down 4 metres in the next minute. It will be able to reach the top 120 metres high in

### Concept & Strategy This is a classic "cyclic progress" problem (often framed as a monkey climbing a pole). The crucial insight is that during the *last* cycle, the target is reached during the upward phase, ending the journey before any backward slip occurs. ### Step-by-Step Solution * **Given:** * Target height = 120 metres. * Climb phase = 6 metres in 1 minute. * Slip phase = 4 metres in 1 minute. * **Deduction:** Calculate the net progress of one complete cycle: $$\text{Cycle duration} = 1 + 1 = 2 \text{ minutes}$$ $$\text{Net gain per cycle} = 6 - 4 = 2 \text{ metres}$$ * **Calculation:** Subtract the final unabated climb (6m) from the total height to find the distance that must be covered via the slow, slipping cycles: $$\text{Cyclic distance} = 120 - 6 = 114 \text{ metres}$$ Calculate the number of complete cycles required to cover 114 metres: $$\text{Cycles} = \frac{114}{2} = 57 \text{ cycles}$$ Calculate the time taken for these 57 cycles: $$57 \times 2 = 114 \text{ minutes}$$ Finally, add the 1 minute for the last 6-metre climb to reach the 120m top: $$\text{Total Time} = 114 + 1 = 115 \text{ minutes}$$ ### Exam Strategy & Shortcut Use the standardized formula for alternating positive/negative work problems to find the number of pair cycles: $$\text{Cycles} = \frac{\text{Total Distance} - \text{Final Jump}}{\text{Net Gain}}$$ $\text{Cycles} = \frac{120 - 6}{2} = 57$. Total time is $57 \times 2 = 114$ mins, plus 1 minute for the final jump = 115 mins. ### Common Pitfall Dividing the total height (120) by the net cycle gain (2) directly to get 60 cycles, and thus concluding 120 minutes. This ignores the physical reality that once the top is reached at 115 minutes, the squirrel grabs the edge and won't slip down again. ### Final Answer **Therefore, the correct answer is 115 minutes.**