An elevator descends into a mine shaft at the rate of $6\text{ m/min}$. If the descent starts from $10\text{ m}$ above the ground level, how long will it take to reach a depth of $-350\text{ m}$?

Step-by-step solution

1. Determine the initial position of the elevator relative to ground level: $+10\text{ m}$.
2. Determine the final position of the elevator: $-350\text{ m}$.
3. Calculate the total distance to be descended: Total distance = Initial position - Final position = $10\text{ m}-(-350\text{ m})=10\text{ m}+350\text{ m}=360\text{ m}$.
4. Determine the rate of descent: $6\text{ m/min}$.
5. Calculate the time taken by dividing the total distance by the rate of descent: $\text{Time}=\frac{360\text{ m}}{6\text{ m/min}}=60\text{ minutes}$.

Concept tested: Subtraction and division of integers in context of height and depth

Published 20 August 2026 · How this solution was prepared and checked

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