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

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.

Answer: A

In the morning sun rises in the East. Hence then any shadow falls in the West. Since Nikharia's shadow was exactly to the right of Nivedita. Hence Nikharia is facing towards North.

Answer: B

Correct answer: 13 km

The northward legs come to \( 2 + 3 = 5 \) km and the eastward to \( 10 + 2 = 12 \) km. By Pythagoras the distance is \[ \sqrt{5^2 + 12^2} = \sqrt{169} = \mathbf{13}\text{ km} \]

Answer: A

Correct answer: 32 m, South

Adding the legs: west 15, then south 20, then east 15, then south 12. The east and west cancel, leaving \( 20 + 12 = 32 \) metres. So Jayant is 32 m south of X.

Answer: D

Correct answer: None of these

Put X at the origin. Y lies east of X, Z lies south of X, and P lies south of Z - so P is directly south of X and Y is to its east. P therefore lies to the south-west of Y, which is not among the choices offered.