Q55. 55. There are 7 places A, B, C, D, E, F and G in a city connected by various roads AB, AC, CD, DE, BF, EG and FG. A is 6 km south of B. A is 10 km west of C. D is 5 km east of E. C is 6 km north of D. F is 9 km west of B. F is 12 km north of G. A person travels from D to F through these roads. What is the distance covered by the person?

Let's map the locations based on the given information. We can use a coordinate system. Let A be at (0,0). * A: (0,0) * A is 6 km south of B => B is 6 km north of A. B: (0,6) * A is 10 km west of C => C is 10 km east of A.

C: (10,0) * C is 6 km north of D => D is 6 km south of C. D: (10,-6) * D is 5 km east of E => E is 5 km west of D. E: (5,-6) * F is 9 km west of B. F: (0-9, 6) = (-9,6) * F is 12 km north of G => G is 12 km south of F.

G: (-9, 6-12) = (-9,-6) Now, we need to find the distance from D(10,-6) to F(-9,6) using the given roads: AB, AC, CD, DE, BF, EG, FG. Lengths of relevant roads: CD = 6 km (from C(10,0) to D(10,-6)) AC = 10 km (from A(0,0) to C(10,0)) AB = 6 km (from A(0,0) to B(0,6)) BF = 9 km (from B(0,6) to F(-9,6)) DE = 5 km (from D(10,-6) to E(5,-6)) EG = 14 km (from E(5,-6) to G(-9,-6) is a horizontal distance: |5 - (-9)| = 14 km) FG = 12 km (from F(-9,6) to G(-9,-6)) Possible paths from D to F: Path 1: D → C → A → B → F Distance = CD + AC + AB + BF = 6 + 10 + 6 + 9 = 31 km.

Path 2: D → E → G → F Distance = DE + EG + FG = 5 + 14 + 12 = 31 km. Both valid paths give a distance of 31 km. This question requires careful mapping and pathfinding, a common type of logical reasoning problem.