Direction and Distance Test Practice Questions — State PSC Exam — Page 2

Question 21Question →

A maritime patrol boat departs from its home harbor to secure coastal waters. It first sails $24\text{ km}$ strictly towards the West to investigate a radar blip. Finding nothing, it turns North and travels $7\text{ km}$ to a reported location. It then changes course, sailing $45\text{ km}$ East to follow a distress signal. Finally, it turns South and travels $27\text{ km}$ to successfully intercept the target. Calculate the exact straight-line distance from the home harbor to the interception point.

Show answer & explanation

Answer: 29

Answer

29

By resolving the boat's path into net vertical and horizontal components, we find its final position is

21\text{ km}

East and

20\text{ km}

South of its starting point. Using the Pythagorean theorem (

a^2 + b^2 = c^2

), the shortest straight-line distance is the hypotenuse:

\sqrt{21^2 + 20^2} = 29\text{ km}

.

Step-by-Step Solution

1

Calculate the net horizontal (East-West) displacement.

21\text{ km}

East

The boat initially travels

24\text{ km}

West, then later travels

45\text{ km}

East. The net horizontal movement is

45 - 24 = 21\text{ km}

in the East direction.

2

Calculate the net vertical (North-South) displacement.

20\text{ km}

South

The boat travels

7\text{ km}

North, and later travels

27\text{ km}

South. The net vertical movement is

27 - 7 = 20\text{ km}

in the South direction.

3

Apply the Pythagorean theorem to find the shortest straight-line distance.

\sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29\text{ km}

The net East and South displacements form the two perpendicular legs of a right-angled triangle relative to the starting point. The straight-line distance is the hypotenuse.

Key Concept

Vector displacement across cardinal directions and Pythagorean theorem calculation.

Question 22Question →

One evening, just before sunset, a security guard begins his patrol in an open plaza. He starts walking straight such that his shadow falls exactly to his right. He walks $20\text{ m}$ straight, then turns $135^\circ$ clockwise and walks $15\sqrt{2}\text{ m}$ . Finally, he turns $135^\circ$ counter-clockwise and walks another $15\text{ m}$ straight.

What is the shortest distance between his final position and his starting point, and in which direction is his final position relative to the starting point?

5\text{ m}

, North

35\text{ m}

, North-East

25\text{ m}

, North-East

25\text{ m}

, South-West

Show answer & explanation

Answer: $25\text{ m}$ , North-East

Answer

The shortest distance is

25\text{ m}

and the direction is North-East.

At sunset, the sun is in the West, casting shadows toward the East. Since the guard's shadow is exactly to his right, his right side faces East, meaning he begins by facing North. Walking 20 m North takes him to coordinate (0, 20). He then turns 135 degrees clockwise, which points him South-East. Moving 15√2 m in this direction shifts him 15 m East and 15 m South, placing him at (15, 5). Next, he turns 135 degrees counter-clockwise from South-East, returning him to a North-facing direction. He walks 15 m North, reaching the final coordinate of (15, 20). The shortest distance from the start (0, 0) is found using the Pythagorean theorem: √(15² + 20²) = √(225 + 400) = √625 = 25 m. Because the final position is 15 m East and 20 m North of the origin, the direction is North-East.

Step-by-Step Solution

1

Determine the initial facing direction based on the sun and shadow.

The guard is facing North.

Since it is evening (sunset), the sun is in the West and casts shadows to the East. If the shadow is to the guard's right, his right side points East, which means he must be facing North.

2

Calculate the position after the first movement.

Current position is

(0, 20)

.

Walking

20\text{ m}

straight while facing North changes the Y-coordinate by

+20

.

3

Apply the first turn and calculate the second position.

Current position is

(15, 5)

.

Turning

135^\circ

clockwise from North means the guard now faces South-East. Walking

15\sqrt{2}\text{ m}

South-East yields a horizontal displacement of

+15\text{ m}

(East) and vertical displacement of

-15\text{ m}

(South). New coordinates:

(0+15, 20-15) = (15, 5)

.

4

Apply the final turn and calculate the final position.

Final position is

(15, 20)

.

Turning

135^\circ

counter-clockwise from South-East means subtracting

135^\circ

, placing the guard back to facing North. Walking

15\text{ m}

straight (North) changes the Y-coordinate by

+15

. New coordinates:

(15, 5+15) = (15, 20)

.

5

Calculate the shortest distance and direction from the origin

(0,0)

.

Distance is

25\text{ m}

, Direction is North-East.

Applying the Pythagorean theorem:

\sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m}

. Since both X and Y coordinates are positive, the direction relative to the start is North-East.

Key Concept

Vector displacement combining angular turns, Cartesian coordinates, and shadow-based directional framing.

Question 23Question →

Four athletes are training on a large flat ground. Match each athlete's movement sequence with their final direction relative to their respective starting points.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Athlete P matches with East, Athlete Q matches with North-East, Athlete R matches with North-West, and Athlete S matches with South-West.

By placing each athlete's starting point at an origin

(0,0)

on a Cartesian plane, their movements can be translated into coordinate updates. P ends at

(40, 0)

which is East. Q ends at

(20, 20)

which is North-East. R ends at

(-60, 40)

which is North-West. S ends at

(-20, -50)

which is South-West.

Step-by-Step Solution

1

Calculate final position of Athlete P.

Starts at

(0,0)

. Moves to

(0, 30)

after going North, then to

(40, 30)

after running East, and finally down to

(40, 0)

after running South.

Tracking coordinates simplifies finding the final direction. The point

(40, 0)

lies exactly on the positive x-axis, meaning the direction is East.

2

Calculate final position of Athlete Q.

Starts at

(0,0)

. Moves to

(0, -50)

after going South. A left turn from South faces East, moving to

(20, -50)

. Another left turn faces North, moving

70\text{ m}

to

(20, 20)

.

The final point

(20, 20)

has positive x (East) and positive y (North) components, placing it in the North-East direction from the origin.

3

Calculate final position of Athlete R.

Starts at

(0,0)

. Moves to

(-40, 0)

after going West. A right turn from West faces North, moving to

(-40, 40)

. A left turn faces West, moving to

(-60, 40)

.

The final point

(-60, 40)

has a negative x (West) and positive y (North) component, placing it in the North-West direction.

4

Calculate final position of Athlete S.

Starts at

(0,0)

. Moves to

(60, 0)

after going East. A right turn faces South, moving to

(60, -50)

. Another right turn faces West, moving

80\text{ m}

to

(-20, -50)

.

The final point

(-20, -50)

has a negative x (West) and negative y (South) component, placing it in the South-West direction.

Key Concept

Tracking sequential movements using cardinal directions and angular turns to determine net displacement and final relative direction.

Estimated Time:2m 0s

Question 24Question →

A geologist leaves her base camp and drives a rover $30\text{ km}$ straight North. From there, she turns $135^{\circ}$ to her right and travels $10\sqrt{2}\text{ km}$ in a straight line. She then turns $45^{\circ}$ to her left and travels $20\text{ km}$ . Finally, she takes a $90^{\circ}$ turn to her right and drives another $20\text{ km}$ . What is her final position relative to the base camp?

30\text{ km}

West

10\text{ km}

West

30\text{ km}

East

30\text{ km}

South

Show answer & explanation

Answer: $30\text{ km}$ East

Answer

30\text{ km}

East

By tracing the movements on a Cartesian plane where North is the positive y-axis and East is the positive x-axis, the sequence unfolds as follows: the initial position is

(0, 30)

after the first drive. A

135^{\circ}

right turn points the rover South-East, and moving

10\sqrt{2}\text{ km}

adds a displacement of

(10, -10)

, resulting in position

(10, 20)

. A

45^{\circ}

left turn points the rover East, adding

(20, 0)

to reach

(30, 20)

. The final

90^{\circ}

right turn points the rover South, adding

(0, -20)

to land precisely at

(30, 0)

. The point

(30, 0)

is exactly

30\text{ km}

East of the starting base camp at

(0, 0)

.

Step-by-Step Solution

1

Map the first movement from the base camp.

Starting at origin

(0, 0)

, moving

30\text{ km}

North reaches coordinate

(0, 30)

while facing North.

Establishing the initial position and orientation on a Cartesian plane (where North is +y and East is +x) is necessary for calculating subsequent turns.

2

Calculate the second movement after a

135^{\circ}

right turn.

Turning

135^{\circ}

right from North changes the heading to South-East. Traveling

10\sqrt{2}\text{ km}

South-East yields a displacement of

+10

in the x-direction and

-10

in the y-direction. The new position is

(10, 20)

.

A

135^{\circ}

clockwise turn from North equates to a South-East direction. Using trigonometry,

10\sqrt{2} \times \cos(45^{\circ}) = 10

and

10\sqrt{2} \times \sin(45^{\circ}) = 10

for the respective axis displacements.

3

Calculate the third movement after a

45^{\circ}

left turn.

Turning

45^{\circ}

left (counter-clockwise) from South-East aligns the heading to East. Traveling

20\text{ km}

East shifts the position by

+20

on the x-axis, bringing the coordinates to

(30, 20)

.

Adjusting the South-East heading by

45^{\circ}

back towards the counter-clockwise direction points exactly East.

4

Calculate the final movement after a

90^{\circ}

right turn.

Turning

90^{\circ}

right (clockwise) from East changes the heading to South. Traveling

20\text{ km}

South shifts the position by

-20

on the y-axis, resulting in a final coordinate of

(30, 0)

.

A

90^{\circ}

right turn from East is South. Moving

20\text{ km}

South precisely negates the remaining

+20\text{ km}

vertical displacement.

5

Determine the final relative position.

The final coordinate is

(30, 0)

while the base camp is at

(0, 0)

. This represents a displacement of

30\text{ km}

directly East.

Comparing the final position coordinates directly to the origin determines the exact distance and direction.

Key Concept

Vector displacement combining angular turns and cardinal directions

Question 25Question →

A wildlife biologist tracking a radio-collared leopard starts at a research outpost and drives $10\text{ km}$ straight East. She then turns $45^{\circ}$ to her left and drives $10\sqrt{2}\text{ km}$ along a dirt trail. Next, she turns $90^{\circ}$ to her right and travels $4\sqrt{2}\text{ km}$ through the brush. Finally, she turns $135^{\circ}$ to her left and drives $1\text{ km}$ straight North to locate the leopard. What is the shortest straight-line distance, in kilometers, between the research outpost and the leopard's final location?

Show answer & explanation

Answer: 25

Answer

25

The correct answer is found by tracking the Cartesian coordinates after each vector movement. By breaking diagonal distances into their horizontal and vertical components, the final position is located exactly at

(24, 7)

. Using the distance formula

\sqrt{24^2 + 7^2}

, the shortest direct distance from the origin is

25\text{ km}

.

Step-by-Step Solution

1

Establish a coordinate system with the starting point at

(0,0)

and translate the first movement of

10\text{ km}

East.

Current position is

(10, 0)

.

East corresponds to the positive x-axis direction.

2

Determine the new heading after turning

45^{\circ}

left from East, and calculate the displacement for

10\sqrt{2}\text{ km}

.

New heading is North-East. Displacement is

(10, 10)

, making the current position

(20, 10)

.

A

45^{\circ}

left turn from East (

0^{\circ}

) results in

45^{\circ}

(NE). The vector components are

d \times \cos(45^{\circ})

and

d \times \sin(45^{\circ})

.

3

Determine the new heading after turning

90^{\circ}

right from North-East, and calculate the displacement for

4\sqrt{2}\text{ km}

.

New heading is South-East. Displacement is

(4, -4)

, making the current position

(24, 6)

.

A

90^{\circ}

right turn from NE (

45^{\circ}

) results in SE (

-45^{\circ}

). The vector components are

d \times \cos(-45^{\circ})

and

d \times \sin(-45^{\circ})

.

4

Determine the new heading after a

135^{\circ}

left turn from South-East, and calculate the displacement for

1\text{ km}

.

New heading is North. Displacement is

(0, 1)

, making the final position

(24, 7)

.

A

135^{\circ}

left turn (positive rotation) from SE (

-45^{\circ}

) yields

-45^{\circ} + 135^{\circ} = 90^{\circ}

(North).

5

Apply the Pythagorean theorem to find the straight-line distance from the origin

(0,0)

to the final coordinates

(24, 7)

.

The shortest distance is

25\text{ km}

.

The straight-line distance between two points on a Cartesian plane is given by

\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

.

Key Concept

Vector displacement, angular turns mapping to cardinal/ordinal directions, and the Pythagorean theorem.

Question 26Question →

Four marine exploration vessels (Alpha, Beta, Gamma, and Delta) are deployed for oceanographic surveying. Each vessel begins facing a specific cardinal direction and executes a sequence of angular maneuvers as described below:

- Vessel Alpha: Starts facing North, rotates $135^{\circ}$ clockwise, moves forward, and then rotates $45^{\circ}$ anti-clockwise.
- Vessel Beta: Starts facing East, rotates $90^{\circ}$ clockwise, moves forward, and then rotates $135^{\circ}$ clockwise.
- Vessel Gamma: Starts facing South, rotates $45^{\circ}$ anti-clockwise, moves forward, rotates $180^{\circ}$ clockwise, and finally rotates $90^{\circ}$ anti-clockwise.
- Vessel Delta: Starts facing West, rotates $135^{\circ}$ anti-clockwise, moves forward, rotates $90^{\circ}$ clockwise, and finally rotates $45^{\circ}$ anti-clockwise.

Match each vessel with its final facing direction.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Vessel Alpha matches with East; Vessel Beta matches with North-West; Vessel Gamma matches with South-West; Vessel Delta matches with South.

Each vessel's final direction is calculated by applying its sequence of clockwise and anti-clockwise rotations mathematically or visually, starting from its initial cardinal direction.

Step-by-Step Solution

1

Calculate the final direction for Vessel Alpha by summing its angular turns from North.

North (0°) + 135° CW - 45° ACW = 90° (East).

Clockwise (CW) turns add to the angle and anti-clockwise (ACW) turns subtract, assuming North is 0° and measuring clockwise.

2

Calculate the final direction for Vessel Beta by summing its angular turns from East.

East (90°) + 90° CW + 135° CW = 315° (North-West).

Tracking successive CW rotations from 90° gives the final orientation on the 360° compass.

3

Calculate the final direction for Vessel Gamma by summing its angular turns from South.

South (180°) - 45° ACW + 180° CW - 90° ACW = 225° (South-West).

Alternating CW and ACW turns are added and subtracted respectively.

4

Calculate the final direction for Vessel Delta by summing its angular turns from West.

West (270°) - 135° ACW + 90° CW - 45° ACW = 180° (South).

Tracking all ACW and CW maneuvers determines the exact final coordinate.

Key Concept

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