### Q26: After getting around a turning point, what should a glider pilot be prepared for? (2,00 P.)... ^t30q26
- A) For weakening thermals due to the progressing time
- B) For a changed horizontal picture due to lower cloud bases
- C) For increased cloud dissipation due to the progressing time
- D) For a changed cloud picture due to the apparently changed position of the sun

**Correct: D)**

> **Explanation:** The correct answer is D because when a glider turns 90 or 180 degrees at a waypoint, the pilot's entire visual perspective of the sky shifts dramatically. The sun appears to have moved relative to the heading, and cumulus clouds that were behind or beside the aircraft now appear in different positions. This perceptual shift can make the sky look completely different. A is wrong because thermal weakening is a time-of-day issue, not a turning-point issue. B is wrong because cloud bases do not change from turning. C is wrong because cloud dissipation is unrelated to heading changes.

### Q27: According ICAO, what symbol indicates a group of unlighted obstacles? ^t30q27
![ICAO Obstacle Symbols](figures/t30_q27.svg)
- A) D
- B) C
- C) B
- D) A

**Correct: B)**

> **Explanation:** The correct answer is B (symbol C in the figure) because ICAO Annex 4 chart symbology uses distinct symbols to differentiate between single obstacles versus groups, and lighted versus unlighted. The symbol for a group of unlighted obstacles is specifically designated in the PFP-061 reference figure as C. A, C, and D represent other obstacle categories such as single obstacles, lighted groups, or other types. Knowing these symbols is critical for cross-country planning and obstacle avoidance.

### Q28: According ICAO, what symbol indicates a civil airport (not international airport) with paved runway? ^t30q28
![ICAO Airport Symbols](figures/t30_q28.svg)
- A) C
- B) A
- C) B
- D) D

**Correct: B)**

> **Explanation:** The correct answer is B (symbol A in the figure) because ICAO aeronautical chart symbology differentiates airports by civil versus military status, international versus domestic, and runway surface type. A civil domestic airport with a paved runway has a specific symbol shown as A in the PFP-062 annex. A, C, and D represent other aerodrome categories such as international airports, military airfields, or unpaved-runway airports. Glider pilots use these symbols when planning outlanding fields or alternate airports.

### Q29: According ICAO, what symbol indicates a general spot elevation? ^t30q29
![ICAO Spot Elevation Symbols](figures/t30_q29.svg)
- A) C
- B) B
- C) A
- D) D

**Correct: A)**

> **Explanation:** The correct answer is A (symbol C in the figure) because ICAO charts use specific symbols to differentiate between general spot elevations, surveyed elevation points, and obstruction heights. A general spot elevation marks a notable terrain high point for situational awareness and is depicted according to ICAO Annex 4 standards. B, C, and D represent other elevation-related symbols such as maximum elevation figures or obstruction markers. Familiarity with these symbols is essential for terrain clearance planning.

### Q30: What distance can be covered during a glide in a glider plane with glide ratio 1/30 from a height of 1500 m? (Neglect wind and thermal effects)... ^t30q30
- A) 45 NM
- B) 30 km
- C) 45 km
- D) 81 NM

**Correct: C)**

> **Explanation:** The correct answer is C because glide distance equals glide ratio multiplied by height: 30 x 1,500 m = 45,000 m = 45 km. The glide ratio of 1:30 means the glider covers 30 metres horizontally for every 1 metre of height lost. A is wrong because 45 NM equals approximately 83 km, which would require a glide ratio of about 1:55. B is wrong because 30 km would correspond to a glide ratio of only 1:20. D is wrong because 81 NM (150 km) would require a glide ratio of 1:100. Always verify that units are consistent — mixing nautical miles and metres is a common exam trap.

### Q31: Why can wing loading be increased when soaring conditions are good? ^t30q31
- A) Because the stall speed diminishes.
- B) Because the glider achieves a better glide ratio at high speed even though the minimum speed rises.
- C) Because the glider can fly more slowly and achieves a better glide ratio.
- D) Because the glider has a better climb rate even though it must fly more slowly.

**Correct: B)**

> **Explanation:** The correct answer is B because in strong thermal conditions, the glider benefits from flying faster between thermals (MacCready theory). Adding water ballast increases wing loading, which shifts the speed polar to the right — improving the glide ratio at high cruising speeds while accepting a higher stall and minimum sink speed. A is wrong because increasing wing loading raises the stall speed. C is wrong because higher wing loading means the glider must fly faster, not slower. D is wrong because a heavier glider has a worse climb rate in thermals due to its higher minimum sink speed.

### Q32: The tail wheel of a glider was not removed before departure. What will be the consequence? ^t30q32
- A) Better manoeuvrability at departure.
- B) The centre of gravity shifts forward.
- C) No consequence. The wheel represents only a tiny fraction of the total weight of the glider and has no effect on the centre of gravity.
- D) The centre of gravity will be further aft and possibly too far aft, which is dangerous.

**Correct: D)**

> **Explanation:** The correct answer is D because the tail wheel is mounted at the extreme rear of the fuselage, far aft of the nominal C.G. Even though its absolute mass is small, its large moment arm produces a significant moment that shifts the C.G. aftward — potentially beyond the aft limit, making the aircraft pitch-unstable and difficult to control. A is wrong because the tail wheel does not improve manoeuvrability. B is wrong because the tail wheel is aft of the C.G., so its presence shifts the C.G. backward, not forward. C is wrong because the long arm amplifies the effect of even a small mass.

### Q33: The pilot exceeds the maximum cockpit payload by 10 kg. What has to be done? ^t30q33
- A) Trim aft.
- B) Trim forward.
- C) Reduce the payload.
- D) Compensate by reducing the water ballast slightly.

**Correct: C)**

> **Explanation:** The correct answer is C because the maximum seat load is a certification limit that cannot be circumvented. Exceeding it may place the C.G. outside the forward limit and subjects the structure to loads beyond what was tested. The only remedy is to reduce the payload until the limits are respected. A and B are wrong because trimming changes the aerodynamic forces on the elevator but does not alter the aircraft's mass or C.G. position. D is wrong because reducing water ballast changes total mass but does not address the specific seat load limitation.

### Q34: What propels a pure glider forward? ^t30q34
- A) Ascending air currents.
- B) Drag directed forward.
- C) The component of gravity acting in the direction of the flight path.
- D) A tailwind.

**Correct: C)**

> **Explanation:** The correct answer is C because in steady gliding flight, the weight vector can be resolved into two components: one perpendicular to the flight path (balanced by lift) and one along the flight path. This along-path component of gravity provides the forward-driving force that balances drag and maintains airspeed. A is wrong because ascending air can reduce the descent rate but does not propel the glider forward through the air. B is wrong because drag always opposes the direction of motion. D is wrong because a tailwind affects groundspeed but does not propel the aircraft through the airmass.

### Q35: The current mass of an aircraft is 610 kg and the centre of gravity (C.G.) position is at 80.0. You remove a 10 kg item of baggage located at a moment arm of 150. Which is the new centre of gravity? ^t30q35
- A) 75.0
- B) 81.166
- C) 70.0
- D) 78.833

**Correct: D)**

> **Explanation:** The correct answer is D. The calculation proceeds as follows: Initial moment = 610 x 80.0 = 48,800. Removed moment = 10 x 150 = 1,500. New total moment = 48,800 - 1,500 = 47,300. New mass = 610 - 10 = 600 kg. New C.G. = 47,300 / 600 = 78.833. Since the baggage was located aft of the current C.G. (arm 150 > 80), removing it shifts the C.G. forward — consistent with the result (78.833 < 80.0). A (75.0) and C (70.0) are too far forward. B (81.166) incorrectly shows a rearward shift.

### Q36: The empty mass of the Discus B is 245 kg. You are planning to carry 184 kg of water ballast. What is the maximum load at the pilot's seat? ^t30q36
> **Extract from the Discus B Flight Manual — Loading table with water ballast**
> ![[figures/t30_q36.png]]
> Max. permitted all-up weight including water ballast : **525 kg**
> Lever arm of water ballast : **203 mm aft of datum (BE)**

> *Table of water ballast loads at various empty weights and seat loads:*

| Empty mass (kg) | Seat load 70 kg | 80 kg | 90 kg | 100 kg | 110 kg |
|---|---|---|---|---|---|
| 220 | 184 | 184 | 184 | 184 | 184 |
| 225 | 184 | 184 | 184 | 184 | 184 |
| 230 | 184 | 184 | 184 | 184 | 184 |
| 235 | 184 | 184 | 184 | 184 | 180 |
| 240 | 184 | 184 | 184 | 184 | 175 |
| 245 | 184 | 184 | 184 | 180 | 170 |
| 250 | 184 | 184 | 184 | 175 | 165 |

> *Water ballast in both wing tanks (kg). For empty mass 245 kg and ballast 184 kg: the maximum seat load is **90 kg** (column 90 kg → value 184, but column 100 kg → 180 and column 110 kg → 170; with ballast=184 required, read the 245 kg row and find the seat load corresponding to ballast=184, i.e. max 90 kg permitted according to the table).*
- A) 100 kg
- B) 110 kg
- C) 90 kg
- D) 80 kg

**Correct: C)**

> **Explanation:** The correct answer is C (90 kg). Reading the Discus B loading table at the row for empty mass 245 kg: with a seat load of 90 kg the permitted water ballast is 184 kg (matching our requirement), but at 100 kg seat load only 180 kg of ballast is permitted, and at 110 kg only 170 kg. Since we need the full 184 kg of ballast, the maximum seat load that still allows this is 90 kg. A (100 kg) and B (110 kg) would require reducing the water ballast below 184 kg. D (80 kg) is unnecessarily restrictive — the table shows 184 kg is still permitted at 90 kg.

### Q37: What important principle must be observed when making an off-field landing on sloping terrain? ^t30q37
- A) Only land with airbrakes fully extended.
- B) Land facing uphill with an approach speed slightly above normal.
- C) Always land into wind regardless of the slope.
- D) The landing flare must be initiated at a greater height than usual.

**Correct: B)**

> **Explanation:** The correct answer is B because landing uphill uses the slope to decelerate the glider — gravity assists braking, dramatically shortening the ground roll. A slightly higher approach speed provides a safety margin against wind shear and turbulence near unfamiliar terrain. A is wrong because full airbrakes may not always be appropriate on short or steep fields. C is wrong because on significant slopes, landing uphill takes priority over landing into wind. D is wrong because the flare height should be adapted to the terrain, but this is not the primary principle.

### Q38: You must land in heavy rain. What must you pay particular attention to? ^t30q38
- A) The approach speed is lower than usual because rain slows the aircraft.
- B) The landing is performed as in dry conditions.
- C) Due to poor visibility, the approach angle must be shallower than usual.
- D) A higher approach speed must be used.

**Correct: D)**

> **Explanation:** The correct answer is D because heavy rain on the wing surface degrades the aerodynamic profile through increased roughness, potentially raising the stall speed. A higher approach speed provides an adequate safety margin. A is wrong because rain does not lower the safe approach speed — if anything, the stall speed increases. B is wrong because rain significantly changes conditions (reduced visibility, wet surfaces, degraded aerodynamics). C is wrong because a shallower approach reduces obstacle clearance margins and extends the final approach in poor visibility.

### Q39: You are taking off from a grass runway that has become waterlogged after several days of rain. What should you expect? ^t30q39
- A) The takeoff distance is likely to be longer.
- B) The glider is wet and has reduced performance.
- C) The wet grass offers less resistance, which is why the takeoff distance will be shorter.
- D) The glider may skid sideways (aquaplaning).

**Correct: A)**

> **Explanation:** The correct answer is A because a waterlogged grass runway creates greater rolling resistance due to soft ground deformation and water drag on the wheels, slowing acceleration and increasing the takeoff distance. B is wrong because while a wet glider has slightly degraded performance, the primary issue is the runway condition. C is wrong because wet, soft grass increases resistance rather than reducing it. D is wrong because aquaplaning occurs on hard surfaces with standing water, not on soft grass — and the question asks about takeoff distance, not directional control.

### Q40: Which of these statements is correct at a speed of 170 km/h, taking into account the following speed polar? ^t30q40
> **ASK 21 Speed Polar:**
> ![[figures/t30_q40.png]]
> *Two curves: G=470 kp (light mass, min sink rate ~0.657 m/s at ~75 km/h) and G=570 kp (heavy mass, min sink rate ~0.724 m/s). The best glide ratio is read from the tangent from the origin. At 170 km/h, the sink rate is higher for G=570 kp than for G=470 kp.*
- A) Regardless of the mass of the ASK21, the sink rate stays constant.
- B) As the mass of the ASK21 rises, the sink rate increases.
- C) As the mass of the ASK21 increases, the sink rate increases.
- D) As the mass of the ASK21 decreases, the glide angle improves.

**Correct: C)**

> **Explanation:** The correct answer is C because at 170 km/h, reading both polar curves, the heavier configuration (570 kp) shows a higher sink rate than the lighter one (470 kp). A heavier glider requires more lift to maintain flight, producing greater induced drag and therefore a higher sink rate at any given speed. A is wrong because the two curves clearly show different sink rates at 170 km/h. B and C state the same thing — sink rate increases with mass — which is correct. D is wrong because at high speeds the glide angle is not necessarily better at lower mass.

### Q41: Which is the speed at the minimum sink rate in still air for a mass of 450 kg? ^t30q41
> **Speed Polar (AIRSPEED):**
> ![[figures/t30_q41.png]]
> *Two curves: 450 kg and 580 kg. The minimum sink rate (top of the curve) for 450 kg is at approximately 75 km/h. The 580 kg curve is shifted to the right (higher speeds) and downward (greater sink rate).*
- A) 75 km/h
- B) 95 km/h
- C) 50 km/h
- D) 140 km/h

**Correct: A)**

> **Explanation:** The correct answer is A because the minimum sink rate speed corresponds to the highest point on the speed polar curve — where the sink rate is smallest. For 450 kg, this peak occurs at approximately 75 km/h. This speed maximises flight endurance in still air and is optimal for centring thermals. B (95 km/h) would be closer to the best-glide speed or the minimum-sink speed at higher mass. C (50 km/h) is below the stall speed. D (140 km/h) is far into the high-speed range where sink rate is much greater.

### Q42: From what altitude on the route between Murten (approx. N46°56'/E007°07') and Neuchâtel aerodrome (approx. N46°57'/E006°52') are you required to request permission to cross the PAYERNE TMA? ^t30q42
- A) 950 m AMSL (3100 ft).
- B) 3050 m AMSL (FL 100).
- C) 700 m AMSL (2300 ft).
- D) At any altitude since the lower limit of the TMA is represented by the ground surface (GND).

**Correct: C)**

> **Explanation:** The correct answer is C because on the route between Murten and Neuchatel, the relevant sector of the PAYERNE TMA has a lower limit at 700 m AMSL (2300 ft). Below this altitude, flight can proceed in uncontrolled airspace without clearance. Above 700 m AMSL, ATC authorisation is required. A (950 m) does not match the published boundary. B (FL 100) is far too high — that is the upper limit of some TMAs, not the lower limit here. D is wrong because the TMA does not extend to the ground in this sector.

### Q43: In which airspace class are you flying at 1400 m AMSL (QNH 1013 hPa) over Birrfeld aerodrome (47°25'36"N/007°14'02"E), and what are the visibility and cloud distance minima in that airspace? ^t30q43
- A) Airspace class E, horizontal visibility 5 km, horizontal cloud distance 1.5 km, vertical 300 m.
- B) Airspace class D, horizontal visibility 5 km, horizontal cloud distance 1.5 km, vertical 300 m.
- C) Airspace class G, horizontal visibility 1.5 km, clear of cloud with permanent ground contact.
- D) Airspace class C, horizontal visibility 5 km, horizontal cloud distance 1.5 km, vertical 300 m.

**Correct: A)**

> **Explanation:** The correct answer is A because at 1400 m AMSL over Birrfeld, you are in Class E airspace. VFR minima in Class E require 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance. B is wrong because Class D applies within specific CTRs or TMAs, not over Birrfeld at this altitude. C is wrong because Class G applies below a certain altitude and has reduced minima. D is wrong because Class C begins at a higher altitude in this area (typically FL 130 in Switzerland).

### Q44: The route shown below towards SCHWYZ (dotted line) is planned for 20 June 2015 (summer time) between 1515–1545 LT at 6500 ft AMSL. Which of the following statements is correct? ^t30q44
> **DABS — Daily Airspace Bulletin Switzerland (extract)**
> ![[figures/t30_q44.png]]

| Firing-Nr D-/R-Area NOTAM-Nr | Validity UTC | Lower Limit AMSL or FL | Upper Limit AMSL or FL | Location | Center Point | Covering Radius | Activity / Remarks |
|---|---|---|---|---|---|---|---|
| B0685/14 | 0000–2359 | 900m / 3000ft | FL 130 | SION TMA SECT 1 | 461610N 0072940E | 4.7 KM / 2.5 NM | TMA SECT 1 ACT HX ONLY |
| W0912/15 | 1145–1300 | GND | FL 120 | MORGARTEN | 470507N 0083758E | 10.0 KM / 5.4 NM | R-AREA ACT. ENTRY PROHIBITED. FOR INFO CTC ZURICH INFO 124.7 |
| W0957/15 | 1400–1700 | 2150m / 7000ft | FL 120 | HINWIL | 471721N 0084859E | 7.0 KM / 3.8 NM | TEMPO R-AREA ACTIVE. ENTRY PROHIBITED. CTC 118.975 |
| W0960/15 | 0800–1700 | GND | 1200m / 4050ft | 1.7 KM SE CERNIER | 470352N 0065442E | 1.5 KM / 0.8 NM | D-AREA ACT |
- A) It is not possible to fly the planned route that day.
- B) You can ignore the DABS as it only applies to commercial aviation.
- C) You can pass through all relevant danger and restricted areas below 1000 ft AGL or above 12,000 ft AMSL.
- D) The route can be flown without coordination between 1500 and 1600 LT.

**Correct: D)**

> **Explanation:** The correct answer is D. On 20 June 2015 (CEST = UTC+2), the planned time of 1515-1545 LT corresponds to 1315-1345 UTC. Zone W0912/15 (MORGARTEN) was active 1145-1300 UTC and has already expired. Zone W0957/15 (HINWIL) activates at 1400 UTC (1600 LT) — it is not yet active. The route can therefore be flown without coordination between 1500 and 1600 LT. A is wrong because the route is flyable during the given time window. B is wrong because the DABS applies to all airspace users including gliders. C is wrong because it incorrectly suggests blanket altitude-based exemptions.

### Q45: According to the ICAO aeronautical chart at 1:500,000, at what altitude over Schwyz (approx. 47°01' N, 8°39' E) must you request permission to enter Class C airspace? ^t30q45
- A) FL 90
- B) 4500 ft
- C) FL 130
- D) FL 195

**Correct: C)**

> **Explanation:** The correct answer is C because over Schwyz, the Swiss ICAO 1:500,000 chart shows Class C airspace beginning at FL 130. Below FL 130, the airspace is Class E. Entering Class C requires ATC clearance regardless of flight rules. A (FL 90) is below the actual boundary. B (4500 ft) is far too low and in uncontrolled airspace. D (FL 195) is the upper limit of Swiss controlled airspace, not the lower limit of Class C over Schwyz.

### Q46: Until what time is La Côte aerodrome (LSGP) open in the evening? ^t30q46
> **AD INFO 1 — LA CÔTE / LSGP**
> ![[figures/t30_q46.png]]

| Data | Value |
|--------|--------|
| ICAO | LSGP |
| Elevation | 1352 ft (412 m) |
| ARP | 46°24'23"N / 006°15'28"E |
| Runway | 04 / 22 — true/mag: 041°/040° and 221°/220° |
| Dimensions | 560 x 30 m — GRASS |
| LDG distance available | 490 m |
| TKOF distance available | 490 m |
| SFC strength | 0.25 MPa |
| Status | Private — Airfield, **PPR** |
| Location | 25 km NE Geneva |
| Hours MON–FRI | 0700–1200 LT / 1400–**ECT –30 min** |
| Hours SAT/SUN | 0800–1200 LT / 1400–**ECT –30 min** |
| ECT reference | → VFG RAC 1-1 |

> *ECT = End of Civil Twilight. The aerodrome closes 30 minutes before end of civil twilight.*
- A) Until half an hour before the start of civil twilight.
- B) Until half an hour before sunset.
- C) Until half an hour before the end of civil twilight.
- D) Until the end of civil twilight.

**Correct: C)**

> **Explanation:** The correct answer is C because the AD INFO sheet for LSGP shows afternoon hours as "1400-ECT -30 min," meaning the aerodrome closes 30 minutes before the end of civil twilight. A is wrong because it references the start of civil twilight, not the end. B is wrong because sunset occurs earlier than the end of civil twilight. D is wrong because the aerodrome closes 30 minutes before ECT, not at ECT itself.

### Q47: On which frequency do you receive information about winch launches at Gruyères aerodrome (LSGT) at weekends? ^t30q47
> **Visual Approach Chart — GRUYÈRES / LSGT**
> ![[figures/t30_q47.png]]
> AD **124.675** — PPR — ELEV 2257 ft (688 m)

> *Key chart data (altitudes in ft, magnetic headings):*

| Data | Value |
|--------|--------|
| ICAO | LSGT |
| AD Frequency | **124.675 MHz** |
| Elevation | 2257 ft (688 m) |
| Status | PPR |
| Minimum AD overfly altitude (MNM ALT) | **4000 ft** |
| Glider ARR/DEP sector W (GLD ARR/DEP W) | **MAX 3100 ft** |
| Glider ARR/DEP sector E (GLD ARR/DEP E) | **MAX 3600 ft** |
| HEL ARR/DEP | 3000 ft |
| Preferred ARR sectors | WEST and EAST |
| CTN (cross-country traffic) | 3000 ft |
| MNM AD overfly | 4000 ft |
| Class C airspace above | FL 100 / 119.175 GENEVA DELTA |
| Winch launches | Intensive SAT/SUN (CTN: Intense winch launching SAT/SUN) |
| Nearby VOR/DME | SPR R076, 113.9 MHz |

> *Noise-sensitive areas (yellow) around Bulle/Broc. Avoid overflying the field during PJE (parachute dropping). Contact RTF 5 min before ETA.*
- A) 113.9
- B) 124.675
- C) 119.175
- D) 110.85

**Correct: B)**

> **Explanation:** The correct answer is B (124.675 MHz) because this is the aerodrome frequency shown on the Visual Approach Chart for LSGT Gruyeres. Local traffic information, including intensive winch launching activity on weekends, is broadcast on this frequency. A (113.9) is the VOR/DME SPR navigation frequency. C (119.175) is the Geneva Delta sector frequency for Class C airspace above. D (110.85) is not shown on this chart and does not relate to LSGT operations.

### Q48: What distance do you cover in 90 minutes at a ground speed of 90 km/h? ^t30q48
- A) 90 km
- B) 135 km
- C) 100 km
- D) 120 km

**Correct: B)**

> **Explanation:** The correct answer is B because distance = speed x time. Ground speed = 90 km/h, time = 90 minutes = 1.5 hours. Distance = 90 x 1.5 = 135 km. A (90 km) results from incorrectly using 1 hour instead of 1.5 hours. C (100 km) and D (120 km) do not correspond to any correct calculation. Remember to convert minutes to hours before multiplying: 90 minutes = 1.5 hours, not 0.9 hours.

### Q49: At an altitude of 6000 m, the airspeed indicator shows 160 km/h (IAS). The true airspeed (TAS)… ^t30q49
- A) is lower than the IAS.
- B) is also 160 km/h.
- C) can be higher or lower than the IAS depending on atmospheric pressure and temperature.
- D) is higher than the IAS.

**Correct: D)**

> **Explanation:** The correct answer is D because the airspeed indicator measures dynamic pressure, which depends on air density. At 6000 m, air density is significantly lower than at sea level. For the pitot tube to register the same dynamic pressure (same IAS), the aircraft must be moving faster through the thinner air. TAS increases by approximately 2% per 300 m of altitude gain, so at 6000 m, TAS is roughly 40% higher than IAS. A is wrong because TAS is always higher than IAS at altitude. B is wrong because they only equal each other at sea level in ISA conditions. C is wrong because at any altitude above sea level, TAS is always higher than IAS.

### Q50: You are flying in wave lift at 6000 m altitude. Which is the maximum speed you may fly? ^t30q50
- A) In the low-density air, at a higher speed than usual.
- B) Below the red V_NE mark on the airspeed indicator, according to the speed-altitude table displayed on the instrument panel.
- C) At the same speed as at sea level since V_NE is an absolute value.
- D) Maximum within the green arc.

**Correct: B)**

> **Explanation:** The correct answer is B because at high altitude the true airspeed corresponding to a given IAS is much higher, and it is the TAS that determines aerodynamic loads on the structure. Glider flight manuals provide a speed-altitude table (or V_NE reduction curve) displayed in the cockpit, giving the corrected maximum IAS at each altitude. At 6000 m, the allowable IAS is lower than the sea-level V_NE mark. A is wrong because you must fly slower (lower IAS), not faster. C is wrong because V_NE as indicated must be reduced with altitude. D is wrong because the green arc alone does not account for altitude corrections.