Direction Sense Test Questions and Answers

Answer: B

He is 15 meter from his starting point.

Track his position, taking north as up:

• 10 m north: 10 m north of the start.
• Left turn, so west, 15 m: 10 m north, 15 m west.
• Right turn, so north again, 5 m: 15 m north, 15 m west.
• Right turn, so east, 15 m: 15 m north, 0 m west.

The last move cancels the westward walk, so he ends up directly north of the start, 10 + 5 = 15 m away.

35 m is the total of the north and east-west legs without allowing for direction, and 10 m counts only the first leg.

Answer: C

Shyam is 10 km. West from school.

He walks 5 km north. Facing north, a left turn points him west, and he goes 10 km. Facing west, another left turn points him south, and he goes 5 km.

The 5 km south cancels the 5 km north, so he ends up directly west of the school, 10 km away. East would follow from right turns, and north or south would ignore that the two 5 km legs cancel.

Drawing the route on rough paper with north at the top is the safest way to solve direction questions.

Answer: C

The person is 10m East of point A.

Trace the walk: 10 m north; right turn means east, 5 m; right again means south, 10 m, back to the starting latitude; left turn from south means east, 5 m more.

The north and south legs cancel, and the two east legs add to 10 m, so the person ends 10 m due east of A. 5 m East counts only one of the eastward legs, and the West options take a left turn as west.

Answer: B

The person ends up South-East of the start.

Starting facing east: 20 m east; turn left (north), 20 m; turn right (east again), 40 m; then turn right (south), 40 m.

East in total: \(20 + 40 = 60\) m. North-south: 20 m north, then 40 m south, leaving 20 m south. A point 60 m east and 20 m south of the start lies to the south-east. 'East' ignores the net southward movement, and the west options misread the turns.

Answer: A

C is to the south-west of D.

Place B at the centre. A is west of B, and C is south of A, so C is to the south-west of B. E is north of B, and D is east of E, so D is to the north-east of B.

In this grid D is up and to the right of C, so C lies down and to the left of D: south-west. North-west would need C above D, and 'south-south' is not a direction.

Answer: D

The answer is 15 km.

Put home at the origin, with east to the right and north up. A cycles 10 km south to (0, -10). Facing south, a right turn faces west: 5 km takes A to (-5, -10). Facing west, a right turn faces north: 10 km takes A to (-5, 0). Facing north, a left turn faces west: 10 km takes A to (-15, 0).

A is now on the same east-west line as home, 15 km to its west, so the straight ride home is 15 km due east. The distances 20, 25 and 27 km do not match the path.

Answer: A

He is 6 Km from the starting point.

Combine the movements along each direction separately:

\[ \begin{aligned} \text{East-West} &: 18 - 12 = 6\ \text{km East} \\ \text{North-South} &: 15 - 15 = 0 \end{aligned} \]

The southward and northward legs cancel, so he ends 6 km due east of the start. 33 km and 60 km add up distances travelled instead of the displacement, and 12 km is the length of the westward leg alone.

Answer: B

The shortest distance is 10 metre.

Put A at the origin. Then B = (0, 6), C = (3, 6) and D = (-5, 0). C is 8 m east of D and 6 m north of it.

\[ CD = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10\ \text{m} \]

9 and 11 come from adding sides instead of using Pythagoras, and 8 is only the east-west separation.

Answer: D

He is 60 metres from the start.

Trace the walk on axes, starting at (0, 0).

• 20 m north: (0, 20).
• Left turn, now facing west; 40 m: (-40, 20).
• Left turn, now facing south; 20 m: (-40, 0).
• Right turn, now facing west; 20 m: (-60, 0).

He ends 60 m due west of the start. 20 m is only the last leg, and 40 or 50 m would come from turning the wrong way at the last step.

Answer: D

Kunal moves from A 10 Km northwards upto B, then moves 6 Km southwards upto C, turns towards east and moves 3 km upto D.
Then AC = (AB-BC) = 4 Km

So Kunal distance from starting point A
\begin{aligned}
AD = \sqrt{AC^2 + CD^2} \\
= \sqrt{4^2+3^2} \\
= \sqrt{25} = 5
\end{aligned}

So AD is 5 Km also with reference to starting point Kunal's direction is North-East.