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?

1. A. 20 km
2. B. 25 km
3. C. 31 km
4. D. 37 km

Answer: C

Explanation

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.