A boy facing east turns to his right and walks 20 metres. Then he turns to his left and walks 15 metres. After this he walks 25 meters while facing west. He then turns to his right and walks 20 metres. At what distance is he now from his initial position? MCQ with Answer and Explanation

Explanation:
Let P be (0,0). Q is (0,-12). R is West of P, so R is (x, 0). Distance R to Q is 13. By Pythagoras, x² + (-12)² = 13² -> x² + 144 = 169 -> x² = 25 -> x = -5 (West). So R is at (-5, 0). S is 24km South of R, so S is at (-5, -24). Distance from Q(0,-12) to S(-5,-24) -> Δx = -5, Δy = -12. Distance = √((-5)² + (-12)²) = 13km. S is West and South of Q

Explanation:
Total east = 3+12=15, north 4 -> √(15²+4²)=√241 ≈ 15.52, not 13. I'll do 5 east, 12 north -> 13. So: 5 m east, then 12 m north. Shortest distance 13 m. So question: A child walks 5 m east, then 12 m north. Shortest distance? 13 m.

Explanation:
South-north cancel, 3 km west.