There is an elevator in a mining shaft that moves above and below - Brahmagupta’s Rules for Multiplication and Division of Integers

part 2 - Example 2 - Brahmagupta’s Rules for Multiplication and Division of Integers - Chapter 2 Class 7 - Operations with Integers (Ganita Prakash II) - Class 7 (Ganita Prakash 1, 2 & old NCERT)
part 3 - Example 2 - Brahmagupta’s Rules for Multiplication and Division of Integers - Chapter 2 Class 7 - Operations with Integers (Ganita Prakash II) - Class 7 (Ganita Prakash 1, 2 & old NCERT) part 4 - Example 2 - Brahmagupta’s Rules for Multiplication and Division of Integers - Chapter 2 Class 7 - Operations with Integers (Ganita Prakash II) - Class 7 (Ganita Prakash 1, 2 & old NCERT) part 5 - Example 2 - Brahmagupta’s Rules for Multiplication and Division of Integers - Chapter 2 Class 7 - Operations with Integers (Ganita Prakash II) - Class 7 (Ganita Prakash 1, 2 & old NCERT)

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Transcript

Example 2 (b) There is an elevator in a mining shaft that moves above and below the ground. The elevator’s positions above the ground are represented as positive integers and positions below the ground are represented as negative integers. (b) If it begins to descend from 15 m above the ground, what will be its position after 45 minutes?Given that Elevator descends into the shaft So, it is going down Now, it moves 3 meters per minute And, we need to find position after 45 minutes Now, Initial Position = 15 m Position after 1 minute = 15 – 3 × 1 Position after 45 minute = 15 – 3 × 45 = 15 – 135 = –135 + 15 = – (135 – 15) = – 120 m Thus, elevator will be 120 metres below the ground.

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