Direction Sense Test Questions and Answers

Answer: C

He is 5 km from home.

Walking 4 km north and then 8 km south leaves him 4 km south of his starting line, and he is 3 km west. The straight-line distance is the hypotenuse:

\[ d = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\ \text{km} \]

7 km adds the two offsets instead of combining them at right angles. 8 km is the last leg alone, and 6 km does not follow from the moves.

Answer: B

She will have to walk 7 km.

Take home as the origin, with north up.

1. 5 km south: she is 5 km south of home.
2. Facing south, a right turn faces west; 2 km west: 2 km west, 5 km south.
3. Facing west, a right turn faces north; 5 km north: back on the level of home, 2 km west of it.
4. Facing north, a left turn faces west; 5 km west: 7 km west of home.

She ends 7 km due west of home, so the straight distance back is 7 km.

17 is the total distance walked, not the straight distance home. 5 and 15 come from mistaking the direction of one of the turns.

Answer: C

He should walk keeping the polestar to his left.

The polestar marks north. Facing east, north is on the left hand, so walking east means keeping the polestar to the left.

Facing the polestar would take him north, walking away from it south, and keeping it on the right would take him west, away from the village.

The polestar (Polaris) stays almost fixed over the north pole of the sky, which is why it has long been used to find north at night.

Answer: C

Ashu is 500 m, South-East of the starting point.

In the morning the Sun is in the east, so he first walks 150 m east. A right turn from east faces south: 400 m south. A left turn from south faces east again: another 150 m east.

In total he is 300 m east and 400 m south of the start:

\[ \begin{aligned} \text{distance} &= \sqrt{300^2 + 400^2} \\ &= \sqrt{250000} = 500\ \text{m} \end{aligned} \]

700 m is the sum of the three legs, not the straight-line distance, and the direction is south-east because he ends up east and south of where he began.

Answer: C

The collinear points are A, R and M.

Take A as the origin. R is 3 m east, at (3, 0). S is 4 m south of R, at (3, −4). H is 5 m west of S, at (−2, −4). M is 4 m north of H, at (−2, 0).

A (0, 0), R (3, 0) and M (−2, 0) all lie on the same east-west line, so they are collinear. A, R, S form a right angle at R, and the other triples are not on one line either.

Answer: D

He is 13 m, South-East of the starting point.

He walks 20 m south and then 8 m back north, so he is 12 m south of the start. He then walks 5 m east. His displacement is 12 m south and 5 m east:

\[ \sqrt{12^{2} + 5^{2}} = \sqrt{169} = 13\ \text{m} \]

Being south and east of the start, he is to the south-east. 15 m South ignores the eastward walk, 7 m East ignores the net southward distance, and North-West points the wrong way.

Answer: A

He is now facing South-east.

Follow the turns from north-west:

1. 90° clockwise from north-west: north-east.
2. 180° anticlockwise from north-east: south-west.
3. Another 90° anticlockwise (the same direction) from south-west: south-east.

Overall he turned 90° clockwise and 270° anticlockwise, a net 180°, so he faces the opposite of north-west, which is south-east. South-west is where he faces after the second turn only.

Answer: D

He is 4 Km from the starting point.

Trace the route from the start O, taking north as up:

1. 3 Km west.
2. Facing west, a left turn faces south: 3 Km south.
3. Facing south, a right turn faces west: 1 Km west.
4. Facing west, a right turn faces north: 3 Km north.

The two 3 Km legs south and north cancel, so he ends level with the start. Only the westward legs remain:

\[ 3 + 1 = 4\ \text{Km} \]

7 Km comes from adding the westward and one north-south leg, and 5 Km from treating the result as a 3-4-5 triangle; both forget that the south and north legs cancel. 6 Km adds the two 3 Km legs.

Answer: B

Ravi is 5 km from the start.

Facing east, a right turn points him south, and another right turn points him west. So he goes 8 km east, 3 km south, then 12 km west. He ends \(12 - 8 = 4\) km west and 3 km south of the start.

\[ \begin{aligned} d &= \sqrt{4^{2} + 3^{2}} \\ &= \sqrt{25} = 5\ \text{km} \end{aligned} \]

7 km adds the two offsets instead of using Pythagoras, and 8 km or 6 km come from ignoring that the westward 12 km overshoots the start by 4 km.

Answer: B

The river finally flows east.

Trace each turn:

• It starts flowing east.
• A left turn makes it flow north.
• Going round a hillock in a semicircle reverses its direction, so it now flows south.
• A left turn at a right angle from south points it east.

North is the direction after the first turn only, and west would follow from a right turn at the end. South is the direction before the final turn.