Gaurav walks 20 metres towards North. He then turns left and walks 40 ...
Final distance from original position = AE = AD + DE = 60 metres
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Gaurav walks 20 metres towards North. He then turns left and walks 40 ...
Solution:
To solve the given problem, we can draw a diagram as follows:
North
|
|
|
| A
| |
| |
| |
| |
| |
| |
| |
| |
|_______|__________ East
B
Where A and B represent the positions of Gaurav after he takes the turns.
Now, let's calculate the distance between Gaurav's final position and his original position:
- Gaurav walks 20 metres towards North. So, his position changes from O to A, where OA = 20 metres.
- Gaurav then turns left and walks 40 metres. So, his position changes from A to B, where AB = 40 metres and angle AOB = 90 degrees (since he turned left).
- Gaurav again turns left and walks 20 metres. So, his position changes from B to C, where BC = 20 metres and angle BOC = 90 degrees (since he turned left again).
- Finally, Gaurav moves 20 metres after turning to the right. So, his position changes from C to D, where CD = 20 metres and angle COD = 90 degrees (since he turned right).
Now, we need to find the distance between Gaurav's final position D and his original position O.
Using Pythagoras theorem, we can calculate OD as follows:
OD^2 = OA^2 + AB^2 + BC^2 + CD^2
OD^2 = 20^2 + 40^2 + 20^2 + 20^2
OD^2 = 1600
OD = 40 metres
Therefore, Gaurav is 40 metres away from his original position. Hence, option (c) is the correct answer.
Gaurav walks 20 metres towards North. He then turns left and walks 40 ...
Option c as first he moves 20 m north then 40 m left means towards westnow 20 m toward north and then again 20 m to west total distance from west( last position) to north 40+20=60