InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
An elevator begins from 20 m above the ground. It descends into a mine shaft at the rate of 6 m per minute. What will beits position after 15 minutes ? |
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Answer» Since the elevator is going down, so the distance covered by it will be represented by a negative integer. Change in position of the elevator in one minute = – 6 m. Change in position of the elevator in 15 minutes = 15 × (-6) = – 90 m. So, the final position of the elevator – 20 + (-90) = – 70 m. The elevator is at 70 m below the ground level. |
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| 2. |
An elevator descends into a mine shaft at the rate of 6 meters per minute. If thedecent starts from 10 mabove the ground level, how long will it take to reach -350 m. |
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Answer» Total length to be descend = -350 – 10 = -360 m Rate of the speed of the elevator 6m per minute Total time to descend = 360/6 = 60minutes = 1 hour |
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| 3. |
Calculate :(+ 63) ÷ (- 9) |
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Answer» (+ 63) ÷ (- 9) We know, a ÷ (- b) = – (a ÷ b) = 63 ÷ (-9) = – (63 ÷ 9) = – 7 |
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