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DGCA CPL AIR NAVIGATION · ASKED IN REAL EXAMS

Critical Point & Point of No Return

34 questions on this topic, recalled by candidates after DGCA sittings between 2019 and 2026. Together they were asked 53 times; the most repeated come first. Pick a year to see only what was asked that year.

Asked in:
Q1Asked in 5 sittings:Oct 2021 · Attempt 2Aug 2023 · OLODENov 2023 · OLODE (30 Nov)Jun 2024 · RegularJun 2025 · Regular 02

CP 480 , what happens when tailwind changes to 30kts from 20 kts, what happens to CP ?

  1. A.less than 480
  2. B.more than 480
  3. C.does not change
Show answer

A. less than 480

DCP = D x H / (O+H). A stronger tailwind increases O and reduces H, which moves the CP closer: less than 480.

Q2Asked in 5 sittings:Oct 2023 · OLODEDec 2023 · RegularJul 2025 · OLODE 05May 2026 · OLODE 04Jul 2026 · OLODE 05

Which statement about the Critical Point (CP, equal-time point) and the Point of No Return (PNR) is correct?

  1. A.The CP depends on the fuel on board; the PNR depends only on the wind
  2. B.The CP depends only on distance and groundspeeds (not on fuel); the PNR depends on the safe endurance and the groundspeeds
  3. C.Both the CP and the PNR depend on the safe endurance
  4. D.Neither the CP nor the PNR is affected by the wind
Show answer

B. The CP depends only on distance and groundspeeds (not on fuel); the PNR depends on the safe endurance and the groundspeeds

CP distance from departure = D x H / (O + H): only the total distance and the groundspeeds out (O) and home (H) appear, so fuel never moves it. PNR distance = E x O x H / (O + H): it depends on the safe endurance E and the two groundspeeds, and not on the distance to the destination at all.

Q3Asked in 3 sittings:Jun 2024 · RegularJun 2025 · Regular 02Mar 2026 · Regular 01

An aircraft has a TAS of 300 kt and a safe endurance of 10 hours. The wind component is 50 kt head on the outbound leg. What is the distance to the Point of No Return?

  1. A.1458 NM
  2. B.1622 NM
  3. C.1544 NM
  4. D.1500 NM
Show answer

A. 1458 NM

O = 250, H = 350. PNR = E x O x H / (O + H) = 10 x 250 x 350 / 600 = 1458 NM (time 5.83 h).

The question was recalled without its options; these options were written to it.

Q4Asked in 3 sittings:Feb 2026 · OLODE 02May 2026 · OLODE 04Sep 2026 · Regular 03

Distance A to B 1200 NM. Normal TAS 400 kt, TAS with one engine inoperative 300 kt. Wind component 40 kt head on the way out. Find the distance from A to the CP for an engine failure, and the time to reach it.

  1. A.680 NM, 1 h 53 min
  2. B.660 NM, 1 h 50 min
  3. C.520 NM, 1 h 27 min
  4. D.680 NM, 2 h 37 min
Show answer

A. 680 NM, 1 h 53 min

The CP for an engine failure uses the REDUCED TAS, because that is the speed you will fly after the failure: O = 300 - 40 = 260, H = 300 + 40 = 340, CP = 1200 x 340 / 600 = 680 NM. You get there at the NORMAL groundspeed (400 - 40 = 360 kt): 680 / 360 = 1.89 h = 1 h 53 min.

Q5Asked in 2 sittings:Jan 2024 · Regular (Day 2)Jul 2026 · OLODE 05

P to Q 2000 NM, track 090. TAS: 4 engines 180 kt, 3 engines 150 kt. Wind 090/20 for the first 1000 NM, 150/20 for the rest. Fuel on board 1600 gal, reserve 200 gal; consumption 4 engines 100 gal/h, 3 engines 80 gal/h. Find the distance and time from P to the critical point (engine failure).

  1. A.1097 NM, 6 h 49 min
  2. B.1000 NM, 6 h 15 min
  3. C.903 NM, 5 h 39 min
  4. D.1278 NM, 7 h 53 min
Show answer

A. 1097 NM, 6 h 49 min

Use 3-engine groundspeeds for the CP: first half out 130 / home 170 kt, second half out 139 / home 159 kt. Equal times forward and back come at 1097 NM. Time to get there on 4 engines (GS 160 then 169 kt): 1000/160 + 97/169 = 6 h 49 min. With different winds on each part of the route, work it part by part - one formula for the whole route does not apply.

The question was recalled without its options; these options were written to it.

Q6Asked in 2 sittings:Jan 2025 · OLODE 01Apr 2025 · OLODE 03

Distance A to B 2346 NM, groundspeed out 365 kt, groundspeed back 480 kt. What is the time from A to the point of equal time (PET)?

  1. A.167 min
  2. B.219 min
  3. C.260 min
  4. D.197 min
Show answer

B. 219 min

PET distance = 2346 x 480 / (365 + 480) = 1333 NM; time = 1333 / 365 = 3.65 h = 219 min.

The question was recalled without its options; these options were written to it.

Q7Asked in 2 sittings:Mar 2024 · Regular 01May 2024 · OLODE 04

For a flight from station 'Z', aircraft cruises 235 kts, track outbound 090 (T), FOB 2500 lb, normal fuel Consumption 255 lb/hr and in case of one engine failure, reduced TAS and F/C are 180 kt TAS and 190 lb/hr, respectively. Average wind component + 30 kt, What shall be the radius of action if one engine fails at the time of returning back ?

  1. A.1122 nm
  2. B.1143 nm
  3. C.0985 nm
Show answer

A. 1122 nm

Endurance = 2500/255 = 9.8 h. Out GS = 235 + 30 = 265 kt; home GS with one engine = 180 - 30 = 150 kt at 190 lb/h. Fuel: 255X/265 + 190X/150 = 2500 gives X = 1,122 NM.

Q8Asked in 2 sittings:Nov 2023 · OLODE (30 Nov)May 2024 · OLODE 04

An aircraft with a TAS of 300 kt flies over water between two airfields 500 NM apart. The wind component is 60 kt head on the way out. What is the distance from the first airfield to the critical point?

  1. A.250 NM
  2. B.200 NM
  3. C.300 NM
  4. D.280 NM
Show answer

C. 300 NM

GS out O = 300 - 60 = 240 kt, GS home H = 300 + 60 = 360 kt. CP = D x H / (O + H) = 500 x 360 / 600 = 300 NM. A headwind out pushes the CP beyond the mid-point (into wind).

The question was recalled without its options; these options were written to it.

Q9Asked in 2 sittings:May 2024 · OLODE 04Feb 2026 · OLODE 02

Distance A to B 2368 NM, groundspeed out 365 kt, groundspeed home 480 kt, safe endurance 8 h 30 min. What is the time to the PNR?

  1. A.290 min
  2. B.209 min
  3. C.219 min
  4. D.190 min
Show answer

A. 290 min

Time to PNR = E x H / (O + H) = 8.5 x 480 / 845 = 4.83 h = 290 min (distance 290/60 x 365 = 1764 NM). Note 219 min is the PET time for similar figures - do not mix the two formulas.

The question was recalled without its options; these options were written to it.

Q10Asked in 2 sittings:Oct 2021 · Attempt 2Mar 2025 · Regular 01

DCP is independent of fuel endurance ?

  1. A.True
  2. B.False
  3. C.DCP is dependent on fuel endurance
Show answer

A. True

The distance to the critical point depends on distance and the wind-affected speeds only, not on fuel or endurance.

Q11Asked in 2 sittings:Sep 2023 · Regular 03Jan 2026 · OLODE 01

Distance 600 NM, groundspeed out 200 kt, groundspeed home 250 kt. The time from departure to the critical point is:

  1. A.100 min
  2. B.80 min
  3. C.90 min
  4. D.120 min
Show answer

A. 100 min

CP distance = D x H / (O + H) = 600 x 250 / 450 = 333 NM; time at the outbound GS = 333 / 200 = 1.67 h = 100 min.

Q12Asked in:Jun 2023 · Regular (ATPL)

When calculating the Point of Safe Return, which endurance is used?

  1. A.total endurance
  2. B.safe endurance (usable fuel excluding the reserve, divided by fuel flow)
  3. C.endurance to dry tanks plus 30 minutes
  4. D.half the total endurance
Show answer

B. safe endurance (usable fuel excluding the reserve, divided by fuel flow)

The PSR/PNR is worked on SAFE endurance - fuel that may be used without touching the reserve. Using total endurance would plan the reserve away. Papers give both figures to see whether you pick the right one.

The question was recalled without its options; these options were written to it.

Q13Asked in:Jun 2023 · Regular (ATPL)

Which of these does NOT affect the position of the point of equal time (PET / CP)?

  1. A.the wind component
  2. B.the TAS
  3. C.the fuel on board
  4. D.the distance between the aerodromes
Show answer

C. the fuel on board

PET = D x H / (O + H): distance and the two groundspeeds (TAS and wind) move it; fuel does not appear in it.

The question was recalled without its options; these options were written to it.

Q14Asked in:Jan 2019

You fly from C to D, a distance of 450 NM. The WC C - D is +30, and the WC D - Cis -40. TAS is 160 Kt and reduced TAS is 130 Kt. The Fuel Flow is 165 kg/hr, and the Safe endurance when overhead C is 4 hours. Calculate PNR for return to C. What is the distance from PNR to D?

  1. A.244 nm
  2. B.205 nm
  3. C.150 nm
Show answer

B. 205 nm

Return GS on reduced TAS = 130 - 40 = 90 kt; out GS 190 kt. PNR distance from C = 4 x 190 x 90 / 280 = 244 NM, so 450 - 244 = 205 NM from D.

Q15Asked in:Jan 2024 · Regular (Day 2)

Same flight (P to Q 2000 NM, 4 engines out at 100 gal/h, return on 3 engines at 80 gal/h, 1400 gal usable). Find the distance and time from P to the PNR.

  1. A.1278 NM, 7 h 53 min
  2. B.1097 NM, 6 h 49 min
  3. C.1400 NM, 8 h 36 min
  4. D.1150 NM, 7 h 11 min
Show answer

A. 1278 NM, 7 h 53 min

Fuel out (4 engines) + fuel back (3 engines) = 1400 gal: out 1000/160 x 100 + (x-1000)/169 x 100; back (x-1000)/159 x 80 + 1000/170 x 80. Solving gives x = 1278 NM, reached after 7 h 53 min.

As recalled, this question depended on a chart or table that is not reproduced here; the key is the recalled paper's answer.

The question was recalled without its options; these options were written to it.

Q16Asked in:Jul 2024 · OLODE 05

Distance 1020 NM, track 190, W/V 180/40, TAS 280 kt. What are the groundspeed and the flight time?

  1. A.240 kt, 4 h 15 min
  2. B.320 kt, 3 h 11 min
  3. C.280 kt, 3 h 39 min
  4. D.260 kt, 3 h 55 min
Show answer

A. 240 kt, 4 h 15 min

The wind is only 10 deg off the nose: headwind 40 cos 10 = 39 kt, drift about 1.4 deg, GS = 280 - 39 = 240 kt. Time = 1020 / 240 = 4.24 h = 4 h 15 min.

The question was recalled without its options; these options were written to it.

Q17Asked in:Jul 2025 · OLODE 05

Track 225(T), distance 477 NM, normal TAS 470 kt, TAS after an engine failure 180 kt, wind component 30 kt head on the outbound leg. Find the distance and time from departure to the critical point.

  1. A.278 NM / 38 min
  2. B.199 NM / 27 min
  3. C.239 NM / 33 min
  4. D.278 NM / 111 min
Show answer

A. 278 NM / 38 min

Engine-failure CP uses the reduced TAS: O = 180 - 30 = 150, H = 180 + 30 = 210, CP = 477 x 210 / 360 = 278 NM. Time to it at the normal groundspeed 470 - 30 = 440 kt: 278 / 440 = 0.63 h = 38 min.

Q18Asked in:Mar 2026 · Regular 01

Distance A to B is 360 NM. Wind component A to B is 15 kt head, B to A 15 kt tail. TAS 180 kt. What is the distance from the equal-time point (CP) to B?

  1. A.170 NM
  2. B.195 NM
  3. C.180 NM
  4. D.165 NM
Show answer

D. 165 NM

O = 165 kt, H = 195 kt. CP from A = 360 x 195 / 360 = 195 NM, so CP to B = 360 - 195 = 165 NM.

The question was recalled without its options; these options were written to it.

Q19Asked in:May 2025 · OLODE 04

An aircraft was over 'A' at 1435 flying direct to 'B'. Distance A to B 2900 NM, TAS 470 kt, mean wind component out +55 kt, back -75 kt, safe endurance 9 h 30 min. What is the distance from A to the Point of Safe Return (PSR)?

  1. A.2844 NM
  2. B.1611 NM
  3. C.1759 NM
  4. D.2141 NM
Show answer

D. 2141 NM

O = 470 + 55 = 525 kt, H = 470 - 75 = 395 kt. Time to PSR = E x H / (O + H) = 9.5 x 395 / 920 = 4.08 h; distance = 4.08 x 525 = 2141 NM.

The question was recalled without its options; these options were written to it.

Q20Asked in:Sep 2026 · Regular 03

The Critical Point (Equal Time Point) is the point along track where:

  1. A.fuel remaining equals fuel used
  2. B.the time to continue to destination equals the time to return to departure
  3. C.the headwind becomes a tailwind
  4. D.the aircraft can no longer return to departure
Show answer

B. the time to continue to destination equals the time to return to departure

At the CP the time to go on and the time to turn back are equal, so it tells the crew which way to go after an engine failure or medical emergency. The point beyond which you cannot return is the PNR.

The question was recalled without its options; these options were written to it.

Q21Asked in:Sep 2026 · Regular 03

The Point of No Return (Point of Safe Return) is:

  1. A.the point at which the time to continue equals the time to return
  2. B.the furthest point along track from which the aircraft can still return to the departure aerodrome with its reserves intact
  3. C.the point where half the fuel has been used
  4. D.the mid-point of the route corrected for wind
Show answer

B. the furthest point along track from which the aircraft can still return to the departure aerodrome with its reserves intact

Beyond the PNR there is not enough usable fuel to get back with reserves intact, so the decision to go on is final. The point where continuing and returning take equal time is the CP (equal-time point), a different point.

The question was recalled without its options; these options were written to it.

Q22Asked in:Sep 2023 · Regular 03

The PNR was planned at 580 NM with a fuel flow of 200 lb/h. In flight the actual fuel flow is 230 lb/h (TAS and wind as planned). Where is the new PNR?

  1. A.504 NM
  2. B.580 NM
  3. C.667 NM
  4. D.460 NM
Show answer

A. 504 NM

The PNR distance is proportional to the safe endurance, which is fuel / fuel flow. Same fuel, higher flow: 580 x 200 / 230 = 504 NM. More fuel flow always brings the PNR closer to departure.

The question was recalled without its options; these options were written to it.

Q23Asked in:Jan 2025 · OLODE 01

Usable fuel (excluding reserve) 4000 kg. Out: TAS 250 kt, 800 kg/h, 20 kt headwind. Return at a reduced TAS of 200 kt using 700 kg/h, with the 20 kt wind now behind. Find the time to the PNR.

  1. A.2 h 37 min
  2. B.2 h 23 min
  3. C.3 h 00 min
  4. D.2 h 10 min
Show answer

A. 2 h 37 min

With different fuel flows out and back, solve on FUEL, not time. GS out 250 - 20 = 230 kt, GS home 200 + 20 = 220 kt. x/230 x 800 + x/220 x 700 = 4000, so x (3.478 + 3.182) = 4000 and x = 601 NM. Time = 601 / 230 = 2.61 h = 2 h 37 min.

The question was recalled without its options; these options were written to it.

Q24Asked in:Dec 2024 · Regular 04

Same aircraft (fuel flow 473 lb/h, 2200 lb usable, track 047, W/V 200/30, TAS 240 kt normal / 190 kt with an engine failed). The time to the critical point is 35 min less than the time to the PNR. What is the stage distance?

  1. A.898 NM
  2. B.918 NM
  3. C.985 NM
  4. D.1020 NM
Show answer

B. 918 NM

Time to PNR = E x H / (O + H) = 4.65 x 213 / 479 = 124 min, so time to CP = 89 min; at GS 266 kt that is 395 NM. The engine-failure CP uses the reduced TAS (GS 216 out, 163 home): D = 395 x (216 + 163) / 163 = 918 NM.

The question was recalled without its options; these options were written to it.

Q25Asked in:Oct 2021 · Attempt 2

When actual FC is 220 but planned was 200, What happens to PNR

  1. A.20% less
  2. B.20% more
  3. C.10% less
Show answer

C. 10% less

Actual consumption is 10% higher, so endurance and the distance to the PNR fall by about 10%.

Q26Asked in:Oct 2021 · Attempt 2

Distance 2400 NM, TAS 300 kt. With a 50 kt headwind out (50 kt tail home) the CP was planned at 1400 NM from departure. In flight the wind is found to be a 50 kt TAILwind out instead. Where is the new CP?

  1. A.1000 NM
  2. B.1200 NM
  3. C.1400 NM
  4. D.1600 NM
Show answer

A. 1000 NM

O = 350, H = 250: CP = 2400 x 250 / 600 = 1000 NM. The CP always moves INTO the wind: headwind out puts it beyond the mid-point, tailwind out brings it back towards departure.

Q27Asked in:Oct 2021 · Attempt 2

Fuel on board 14,000 lb, reserve to be kept 2,000 lb. Twin-engine aircraft, fuel flow 1,000 lb/h per engine, TAS 180 kt, wind component 45 kt head out. Distance to the PNR?

  1. A.506 NM
  2. B.540 NM
  3. C.608 NM
  4. D.450 NM
Show answer

A. 506 NM

Safe endurance = 12,000 / 2,000 = 6 h. O = 135, H = 225. PNR = 6 x 135 x 225 / 360 = 506 NM (time 3 h 45 min).

Q28Asked in:May 2024 · OLODE 04

Given a distance (DCP) of 500 NM with a 20 kt tailwind, what will be the new DCP if the wind changes to a 30 kt tailwind?

  1. A.LESS than 500 Nm
  2. B.More than 500 Nm
  3. C.equal to 500 Nm
Show answer

A. LESS than 500 Nm

DCP = D x H / (O+H): a stronger tailwind lowers H and raises O, so the CP moves closer to the departure point.

Q29Asked in:May 2024 · OLODE 04

The fuel flow increases from 200 to 220 kg/h (other factors unchanged). The PNR distance:

  1. A.increases by about 10 %
  2. B.decreases by about 10 %
  3. C.does not change
  4. D.decreases by about 20 %
Show answer

B. decreases by about 10 %

PNR distance is proportional to safe endurance = fuel / fuel flow: 200/220 = 0.91, a decrease of 9 % - about 10 %.

The question was recalled without its options; these options were written to it.

Q30Asked in:May 2024 · OLODE 04

Dist. = 800 nm, TRK = 330°, W/V = 270°/30 kts, TAS = 240 kts DCP?

  1. A.524 Nm
  2. B.425 Nm
  3. C.480 Nm
Show answer

B. 425 Nm

Out GS = 223.6 kt (track 330), home GS = 253.6 kt (track 150). DCP = 800 x 253.6 / (223.6 + 253.6) = 425 NM.

Q31Asked in:Mar 2022

Track 200(T), W/V 047/30 kt, TAS 240 kt (190 kt if an engine fails). Fuel on board (usable) 2200 lb. The distance to the PNR is 550 NM. What is the fuel consumption?

  1. A.473 lb/h
  2. B.400 lb/h
  3. C.550 lb/h
  4. D.510 lb/h
Show answer

A. 473 lb/h

Wind 27 kt behind going out: GS out 266 kt, GS home 213 kt. Endurance E = PNR x (O + H) / (O x H) = 550 x 479 / 56,700 = 4.65 h. Fuel flow = 2200 / 4.65 = 473 lb/h.

The question was recalled without its options; these options were written to it.

Q32Asked in:Mar 2022

True track 190, W/V 150/40, TAS 280 kt, distance 1020 NM. Find the distance and time to the critical point.

  1. A.566 NM, 2 h 17 min
  2. B.510 NM, 2 h 03 min
  3. C.454 NM, 1 h 50 min
  4. D.566 NM, 1 h 50 min
Show answer

A. 566 NM, 2 h 17 min

Wind 40 deg off the nose: headwind 40 cos 40 = 31 kt. GS out about 248 kt, GS home (track 010, wind 40 deg behind) about 309 kt. CP = 1020 x 309 / (248 + 309) = 566 NM; time = 566 / 248 = 2.28 h = 2 h 17 min.

Q33Asked in:Oct 2023 · OLODE

The distance from departure to the PNR depends on:

  1. A.safe endurance, outbound and homebound groundspeeds - not on the distance to the destination
  2. B.the distance between the two aerodromes only
  3. C.the outbound groundspeed only
  4. D.the reserve fuel only
Show answer

A. safe endurance, outbound and homebound groundspeeds - not on the distance to the destination

PNR = E x O x H / (O + H). Move the destination further away and the PNR does not move; burn fuel faster (lower E) or meet a stronger wind and it moves back towards departure.

Q34Asked in:Aug 2026 · OLODE 06

A four-engine aircraft flies out at TAS 400 kt using 4000 kg/h into a 40 kt headwind. If it has to return on three engines, its TAS is 350 kt and fuel flow 3500 kg/h (40 kt tailwind home). Fuel available (excluding reserve) 24,000 kg. Distance to the PNR?

  1. A.1195 NM
  2. B.1290 NM
  3. C.1080 NM
  4. D.1350 NM
Show answer

A. 1195 NM

With different fuel flows, solve on fuel: x/360 x 4000 + x/390 x 3500 = 24000, so x (11.11 + 8.97) = 24000 and x = 1195 NM (time out 3 h 19 min).

These are candidates' recollections, not official DGCA papers: wording can differ from the paper as set. Every answer was checked against the DGCA CARs, AIP India, ICAO documents or worked calculation (independent check completed 17 Oct 2026). Spotted a mistake? Tell us and we will check it against the source.