Mr. X starts walking towards south from point A. After walking 20 m., he reaches at point B. Now he turns left and walks 20 m and reaches at point C. Now he turns 45° anticlockwise, walks a distance 20√3 m and reaches at the point D. What is the shortest distance between point A and point D ?

Correct — option (3), 20√5 m.

Place A at the origin, with east as +x and north as +y.

Step 1 — A to B: 20 m south. B = (0, −20).
Step 2 — Facing south, a left turn faces east. B to C: 20 m east. C = (20, −20).
Step 3 — Facing east, a 45° anticlockwise turn faces north-east. C to D: 20√3 m north-east.

Step 4 — Join A and C: AC = √(20² + 20²) = √800 = 20√2 m.
Step 5 — From C, A lies to the north-west and D to the north-east. These directions are 90° apart, so angle ACD = 90°.

Step 6 — Pythagoras in triangle ACD:
AD² = (20√2)² + (20√3)²
AD² = 800 + 1200 = 2000
AD = √2000 = √(400 × 5) = 20√5 m (about 44.7 m)

Check with coordinates: D = (20 + 10√6, −20 + 10√6), and AD² = (20 + 10√6)² + (10√6 − 20)² = 1000 + 400√6 + 1000 − 400√6 = 2000.

The idea to remember: after a series of turns, look for a right angle between the line back to the start and the last leg.

Why the others are wrong

• (1)
  20√3 m — 20√3 m is the length of CD, the last leg of the walk, not the distance from A.

  A is 20√2 m from C, along a line at right angles to CD. So AD is the hypotenuse of a right triangle whose one leg is CD, and it must be longer than CD: AD² = 800 + 1200 = 2000.

• (2)
  40 m — 40 m would need AD² = 1600. The Pythagoras step gives that only if AC is taken as 20 m: 400 + 1200 = 1600.

  But A to C crosses the diagonal of a 20 m by 20 m corner, so AC = 20√2 m and AC² = 800. With the correct AC, AD² = 2000 and AD = 20√5 m.

• (4)
  40√3 m — 40√3 m is about 69.3 m. The two-leg route A → C → D is only 20√2 + 20√3 ≈ 28.3 + 34.6 = 62.9 m.

  By the triangle inequality, a straight line cannot be longer than a two-leg path between the same points, so 40√3 m is impossible.

In a direction-sense problem, every turn is measured from the direction the person is facing. A left turn is 90° anticlockwise and a right turn is 90° clockwise. A 45° turn from a cardinal direction lands midway between two cardinal directions: north-east, south-east, south-west or north-west.

The shortest distance between two points is the straight line joining them. When the relevant legs meet at a right angle, Pythagoras' theorem gives it directly.

When they do not, split each leg into an east–west part and a north–south part, add the parts, and apply Pythagoras to the totals.

Key facts

• A left turn is 90° anticlockwise and a right turn is 90° clockwise, measured from the direction the person faces.
• Facing east, a 45° anticlockwise turn points north-east; a 45° clockwise turn points south-east.
• In this walk, AC = 20√2 m towards the south-east, CD = 20√3 m towards the north-east, and angle ACD = 90°.
• AD² = (20√2)² + (20√3)² = 800 + 1200 = 2000, so AD = 20√5 m, about 44.7 m.
• Triangle inequality: any side of a triangle is shorter than the sum of the other two sides.

Common traps

• "Left" depends on the facing direction. Walking south, a left turn points east, not west.
• From east, anticlockwise turns towards north. A 45° anticlockwise turn from east is north-east, not south-east.
• Test the answer against the path: the straight-line distance cannot exceed AC + CD, about 62.9 m here.

UnlockIAS will link similar questions from RAS Pre 2021 and 2013 here once those papers are published on this site.

• practice — not a real PYQ

  Starting from point O, Ravi walks 10 m east to P. He turns 90° anticlockwise and walks 10 m to Q. He then turns 45° clockwise and walks 10√2 m to R. What is the shortest distance between O and R?

  • (a)20 m
  • (b)20√2 m
  • (c)30 m
  • (d)10√5 m
  Answer(2) — O = (0, 0); P = (10, 0); facing east, 90° anticlockwise faces north, so Q = (10, 10); facing north, 45° clockwise faces north-east, and 10√2 m north-east adds (10, 10), so R = (20, 20). OR = √(400 + 400) = 20√2 m.

  Option (1), 20 m, results from turning 45° anticlockwise instead (R = (0, 20)); options (3), 30 m, and (4), 10√5 m, do not come from the path as described.

• practice — not a real PYQ

  Meena walks 30 m north from point P, turns right and walks 40 m to point Q. How far is Q from P?

  • (a)10 m
  • (b)50 m
  • (c)70 m
  • (d)25 m
  Answer(2) — Facing north, a right turn faces east, so Q is 40 m east and 30 m north of P. PQ = √(30² + 40²) = √(900 + 1600) = √2500 = 50 m.

  Option (1), 10 m, subtracts the legs; option (3), 70 m, adds them, which is the walking distance, not the straight line; option (4), 25 m, has no basis in the path.