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Free Practice Questions for DHMİ ATCO Selection
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Sample DHMİ ATCO Selection Practice Questions
Try these sample questions to review concepts for the DHMİ ATCO Selection exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 30+ question experience with AI tutoring.
1A target moves 12 grid units east and then 5 units north. What is its straight-line distance from its starting point?
A.7 units
B.13 units
C.17 units
D.60 units
Explanation: The two movements form perpendicular sides of a right triangle. The displacement is √(12² + 5²) = √169 = 13 units, rather than the total distance travelled.
2Using bearings measured clockwise from north, a marker is on bearing 048° from a station. What is the bearing of the station from the marker?
A.132°
B.228°
C.318°
D.048°
Explanation: Opposite bearings differ by 180°. Adding 180° to 048° gives 228°.
3A pointer initially faces 075°. It rotates anticlockwise to 290°. Through how many degrees does it rotate?
A.145°
B.215°
C.115°
D.135°
Explanation: The pointer rotates 75° to north, then another 70° to 290°. Its total anticlockwise rotation is 145°.
4On a north-up grid, A is at (20 east, 35 north) and B at (20 east, 15 north). Where is A relative to B?
A.20 units south
B.20 units east
C.20 units north
D.50 units north
Explanation: The east coordinates are identical, so there is no east-west displacement. Subtracting the north coordinates gives 35 − 15 = 20 units north.
5A vehicle faces south. Another vehicle is at its 10 o'clock position, with 12 o'clock straight ahead. In which compass direction is the other vehicle?
A.Southwest
B.Northwest
C.Northeast
D.Southeast
Explanation: Ten o'clock is 60° to the left of the nose. From south, turning left points toward the southeast, on bearing 120°.
6Two targets are 150 NM apart and move directly toward one another at groundspeeds of 420 kt and 480 kt. With constant speeds, when will they meet?
A.10 minutes
B.12 minutes
C.15 minutes
D.20 minutes
Explanation: The closing speed is 420 + 480 = 900 NM/hour, or 15 NM/minute. Dividing 150 NM by 15 NM/minute gives 10 minutes.
7A target travels 70 NM at a constant groundspeed of 420 kt. How long does this take?
A.8 minutes
B.10 minutes
C.12 minutes
D.15 minutes
Explanation: A groundspeed of 420 kt is 7 NM per minute. The time is 70 ÷ 7 = 10 minutes.
8A lead target travels at 360 kt and a following target at 450 kt on the same straight track. Initial spacing is 30 NM. In this arithmetic exercise, how long until spacing decreases to a specified 5 NM buffer?
A.13 min 20 sec
B.20 min
C.16 min 40 sec
D.25 min
Explanation: The follower gains 90 NM/hour, or 1.5 NM/minute. Reducing spacing by 25 NM takes 25 ÷ 1.5 = 16⅔ minutes; the supplied buffer is an exercise assumption, not an operational clearance.
9A target travels 42 NM at 280 kt. If its groundspeed were instead 210 kt for the entire leg, how much later would it arrive?
A.1.5 minutes
B.2 minutes
C.4.5 minutes
D.3 minutes
Explanation: The first time is 42 ÷ 280 × 60 = 9 minutes. The second is 42 ÷ 210 × 60 = 12 minutes, a difference of 3 minutes.
10In a simplified exercise, a target must climb 3,000 ft while travelling 15 NM at 180 kt. What constant climb rate is required?
A.400 ft/min
B.800 ft/min
C.600 ft/min
D.1,000 ft/min
Explanation: At 180 kt the target covers 3 NM/minute, taking 5 minutes to travel 15 NM. Gaining 3,000 ft in 5 minutes requires 600 ft/min; this calculation alone does not establish safe traffic separation.
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