SSC CPO Tier-1 27 June 2024 Shift 2 Question Paper with Solution

# Q1 of 197

Manish starts from Point A and drives 12 km towards East. He then takes a left turn, drives 5 km, turns left and drives 30 km. He then takes a left turn and drives 20 km. He takes a final left turn, drives 18 km and stops at Point Q. How far (shortest distance) and towards which direction should he now drive to reach Point A again? (All turns are 90° turns only.)

Options
A.

10 km towards South

B.

15 km towards North

C.

10 km towards North

D.

15 km towards South

Show Answer
Correct Answer
B 15 km towards North
Solution

The correct option is 15 km towards North.


Let us analyze the step-by-step movement of Manish starting from Point A on a standard 2D Cartesian coordinate plane where Point A is at the origin (0,0).


Step 1: Start at Point A
Initial coordinates: (0,0)


Step 2: Drives 12 km towards East
Moving East increases the x-coordinate by 12 km.
New position: (12,0)


Step 3: Takes a left turn and drives 5 km
Facing East, a left turn means driving North, which increases the y-coordinate by 5 km.
New position: (12,5)


Step 4: Takes a left turn and drives 30 km
Facing North, a left turn means driving West, which decreases the x-coordinate by 30 km.
New position: (12-30,5)=(-18,5)


Step 5: Takes a left turn and drives 20 km
Facing West, a left turn means driving South, which decreases the y-coordinate by 20 km.
New position: (-18,5-20)=(-18,-15)


Step 6: Takes a final left turn, drives 18 km and stops at Point Q
Facing South, a left turn means driving East, which increases the x-coordinate by 18 km.
Coordinates of Point Q: (-18+18,-15)=(0,-15)


Finding the distance and direction to reach Point A from Point Q:
Point A is at (0,0) and Point Q is at (0,-15).
The difference in x-coordinates is 0 km.
The difference in y-coordinates is:

0-(-15)=15 km


Since Point Q is directly below (South of) Point A by 15 km, Manish needs to drive 15 km towards North to reach Point A again.

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