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UPSC Prelims 2025 CSAT, question 65

Topic: Logical reasoning

Q65

There are 7 places A,B,C,D,E,FA,B,C,D,E,F and GG in a city connected by various roads AB,AC,CD,DE,BF,EGAB,AC,CD,DE,BF,EG and FGFG. AA is 6 km south of BB. AA is 10 km west of CC. DD is 5 km east of EE. CC is 6 km north of DD. FF is 9 km west of BB. FF is 12 km north of GG. A person travels from DD to FF through these roads. What is the distance covered by the person ?

  1. (a) 20 km
  2. (b) 25 km
  3. (c) 31 kmCorrect answer
  4. (d) 37 km

Solution.

Answer: (c) 31 km

  1. Setting A=(0,0)A = (0, 0) with East as +x+x and North as +y+y: B=(0,6)B = (0, 6) (AB=6AB = 6 km), C=(10,0)C = (10, 0) (AC=10AC = 10 km), D=(10,−6)D = (10, -6) (CD=6CD = 6 km), E=(5,−6)E = (5, -6) (DE=5DE = 5 km), F=(−9,6)F = (-9, 6) (BF=9BF = 9 km), and G=(−9,−6)G = (-9, -6) (FG=12FG = 12 km, and horizontal road EGEG connects (5,−6)(5, -6) to (−9,−6)(-9, -6) with length 1414 km).
  2. Along the upper route D→C→A→B→FD \to C \to A \to B \to F, the distance is CD+AC+AB+BF=6+10+6+9=31CD + AC + AB + BF = 6 + 10 + 6 + 9 = 31 km.
  3. Along the lower route D→E→G→FD \to E \to G \to F, the distance is DE+EG+FG=5+14+12=31DE + EG + FG = 5 + 14 + 12 = 31 km; thus the distance covered is 3131 km.