MATH TEST

Question 1

What would have a faster Mach speed?
  250KIAS at FL400 OAT -49 degrees C.
  250KIAS at 15K on a standard day.
  300KIAS at 10K OAT 6 degrees C.
  300KIAS at 5K on a standard day.

Answer 1

250KIAS at FL400 OAT -49 degrees C.Correct
The coldest temperature is going to produce the lowest local speed of sound and thus the highest IMN. 250 at FL400 = 450KTAS, 300 at 10K = 350KTAS, 250 at 15K = 325KTAS, 300 at 5K = 325KTAS

Question 2

You are at FL350 going Mach .84, 112nm from XYZ VOR. You are cleared to cross 12nm prior to XYZ VOR at 10000 feet and 250KIAS. Indicated airspeed in the descent is 300KIAS. When do you start your descent?
  85nm.
  92nm.
  88nm.
  95nm.

Answer 2

92nm.Correct
25000ft of altitude to lose / 1000 x 3 =75nm. Reduce speed by 50kts at a rate of 10kts / 1 NM = 5nm. So far we have 80nm. We must be at 10000 feet by 12nm prior to the VOR, so we’ll add the 12 to the 80nm for a total of 92nm from the VOR.

Question 3

MDA for the LOC 27 is 450’AGL, How far from the runway is the VDP?

1. 4NM.
2. 5NM.
3. 8NM.
4. 0NM.

Answer 3

1.5NM.
Correct
450 ft/ 300ft/nm = 1.5nm (300 ft/nm is roughly a 3 degree glide path).
Visual Descent Point (VDP): A defined point on the final approach course of a non-precision straight-in approach from which a normal decent from the MDA to the runway touch-down zone may commence, provided the runway or approach lights are clearly visible to pilot.
To calculate a 3 degree decent angle from the VDP t the runway, divide the groundspeed by 2, then multiple the result by 10 (100kts (GS) 2 x 10 = 500 fpm.
EEPP p. 77.

Question 4

The bearing pointer moves from 5 degrees in front of the wing to 5 degrees behind the wing in 8 minutes, you are doing 360kts. How far from the station are you?
  280NM.
  240NM.
  288NM.
  265NM.

Answer 4

288NM.
Correct
360 NM/hr / 60 min/hr = 6 NM/min. 6NM / min x 8 min = 48 NM. 48NM / 10 degrees = 4.8 degrees / NM x 60 = 288nm. At 60nm, 1 degree = 1nm. At 120nm 1 degree = 2nm. At 180nm 1 degree = 3nm, at 240nm 1 degree = 4nm. So the answer makes sense. At 288nm 1 degree = 4.8nm

Question 5

What is the higher TAS?
  350KTAS at 10000ft.
  260KCAS at FL300 and -30deg C.
  350KTAS at FL200 and -20deg C.
  250KCAS at FL370 and -40deg C.

Answer 5

250KCAS at FL370 and -40deg C.
Correct
A quick ROT to use is KTAS = KIAS + ALT/200. So for 250 + 37000/200 = 250 + 185 = 435KTAS. 260 + 30000/200 = 410KTAS.

Question 6

142 kts IAS on ILS with 20 kt headwind. What should your VSI be?
  250.
  600.
  500.
  300.

Answer 6

600.
Correct
(120/2= 60 add a zero = 600)

Question 7

You’re holding in an aircraft that burns 3,000 lbs an hour on each of it’s three engines. You have 7500 lbs. of fuel on board. How long until fuel exhaustion?
  30 mins.
  40 mins.
  20 mins.
  50 mins.
Correct

Answer 7

50 mins.

Question 8

Flying direct to the station on the 180 degree radial, at 37000 feet, and 70 DME from the VOR. Approach tells you to cross 15DME on the 360 Radial at 10000 feet. What is your descent rate?
317 FT/NM (No airspeed given to convert it to FPM) or approx. 3 degree gradient.
340 FT/NM (No airspeed given to convert it to FPM) or approx. 3 degree gradient.
320 FT/NM (No airspeed given to convert it to FPM) or approx. 6 degree gradient.
Correct

Answer 8

317 FT/NM (No airspeed given to convert it to FPM) or approx. 3 degree gradient.
You need to lose 27000 feet and have 85NM to do it. 27000 ft/85NM = 317ft/NM
If you are given an airspeed in NM/min you can multiply it by 317ft/NM to get FPM.

Question 9

You are on a visual approach at Denver and are at 7000’ MSL, the TDZE is 5200’. When would you descend for a 3 degree glidepath?
  6NM.
  5NM.
  10NM.
  3NM.

Answer 9

6NM.
Correct
1800ft / 300ft/nm = 6NM.

Question 10

Two airplanes are traveling in the same direction 50nm apart. The one in front is going .76 IMN, and the one in the rear is going .86IMN. When will the one in the rear overtake the one in the front?
  35 Minutes.
  45 Minutes.
  40 Minutes.
  50minutes.

Answer 10

50minutes.
Correct
7.6NM/min and 8.6NM/min = 1NM/min advantage x 50 NM = 50 minutes.

Question 11

One dot off on the localizer at 6nm equals what offset?
.04nm or approximately 600ft.
.05nm or approximately 300ft.
.08nm or approximately 800ft.

Answer 11

.05nm or approximately 300ft.

VOR 2 degrees per dot with full scale deflection equal to 10 degrees
ILS localizer 0.5 degrees per dot with full scale deflection 2.5 degrees
ILS glidepath 0.14 degrees per dot with full scale deflection 0.7 degrees
So the old 60-to-1 rule comes into play here again. At 6mn there are 10 degrees/NM. So 1 degree is = to 0.1NM, and 0.5 degrees is = to 0.05NM which is about 300 feet.
ROT: 50’/NM per dot deviation (so 6NM = 300 ft.)
ROT: 24’/NM per dot deviation off glidepath.

Question 12

Winds are 340/28. RWY 1 in use. What is your crosswind?
  16.
  9.
  18.
  14.

Answer 12

14.
Correct
Rule of six should be applied here 1/2 of wind=14.

Question 13

MDA for the LOC 27 is 450’AGL, Time from FAF to MAP is 3:00. How long after the FAF will your VDP occur?

2: 45.
2: 30.
2: 15.
2: 10.

Answer 13

2:15.
Correct
Take 10% of the MDA (AGL) and subtract it from the total time. That is your time on the approach until reaching the VDP 450 x 10% = 45 sec. 3:00 – 0:45 = 2:15

Question 14

If you are on an ILS approach at 180KTS, what is your rate of descent?
  600 FPM.
  900 FPM.
  350 FPM.
  400FPM.

Answer 14

900 FPM.
Correct
Take ½ your Ground Speed and add a zero. 90+0 = 900 FPM.

Question 15

Fuel burn in holding is 3000lb/hr. per engine. You have 5000lbs of fuel, how long can you hold?
  35min.
  50min.
  44min.
  45min.

Answer 15

50min.
Correct
5000lbs / 6000lbs/hr. = 5/6 hr. or 50min.

Question 16

You are established on the 20DME arc to the north and you are going to intercept the 120R inbound to the VOR. Your flying 300KTAS, what is your lead point?
  128 radial.
  125 radial.
  139 radial.
  129 radial.

Answer 16

129 radial.
Correct
Turn radius = TAS/60min/hr – 2, so 300 / 60 – 2 = 3NM
Lead point = 60/DME x Turn radius (NM) = 60 / 20NM x 3NM = 9 degrees
Lead Radial = Radial +/- Lead radial. In this case we are approaching R120 from the south (north heading) so we would start a left hand turn at R129 to intercept the R120 inbound.

Question 17

On a 3 degree glide slope 140kts GS at 700 FPM rate of descent, the head winds increase, what adjustment must be made?
Continue at the current rate.
Increase descent rate.
Reduce rate of descent.

Answer 17

Reduce rate of descent.
Correct
140KGS = 700FPM, 20kt headwind reduces Kts GS to 120. 600FPM.

Question 18

You are at FL370, 0.86IMN. You need to cross the outer marker at 1600 feet. When do you start your descent? Your descent profile assumes idle power and .86IMN transitioning to 300KIAS, then 250KIAS at 10000ft. Assume the aircraft slows down at 10KTS/NM.
120NM FROM THE OUTER MARKER (115 from the Airfield).
105NM FROM THE OUTER MARKER (118 from the Airfield).
110NM FROM THE OUTER MARKER (115 from the Airfield).

Answer 18

110NM FROM THE OUTER MARKER (115 from the Airfield).
Correct
Round 1600 to 2000 37000-2000/1000 x 3 = 105NM
300KIAS-250KIAS = 50KTS / 10 KTS/NM = 5NM
Descent + Speed reduction = 110NM
Outer marker approx. 5nm from runway. So add 5NM if req.

Question 19

You have to descend 12,000’ in 20 miles, your speed is .7 Mach. What should your VSI be?
4100 FPM.
4200 FPM.
3800 FPM.

Answer 19

4200 FPM.
Correct
12,000 ft/ 20 miles= 600 ft per mile
600 ft per mile X 7miles a minute = 4200.

Question 20

You are flying at FL370 in cruise flight and the OAT increases 5 degrees C, and the headwind increases 5kts. What happens to your TAS and GS?
TAS increases by 4kts, therefore GS increases by 1kt.
TAS increases by 6kts, therefore GS increases by 1kt.
TAS increases by 10kts, therefore GS increases by 1kt.

Answer 20

TAS increases by 6kts, therefore GS increases by 1kt.
Correct
TAS increases approx. 1.2kts/1o C increase. So 5°C x 1.2kts = 6KTAS increase
6KTAS-5KTS headwind = 1kt GS increase.

Question 21

You’re at FL370 traveling at 420KTAS with a 60Kt tailwind. ATC clears you to descent to 10000ft at VOR passage which is 104nm away. What is your rate of descent?
  2100ft/min.
  2600ft/min.
  2400ft/min.
  2000 ft/min.

Answer 21

2100ft/min.
Correct
27000 feet, 480KGS, 104nm. 104NM / 8NM/min = 13min
27000 feet / 13min = 2100ft/min (round up).

Question 22

You are flying Mach 0.7 with a 50kt tailwind. What is your ground speed?
  450 kts GS.
  460 kts.
  455 kts.
  470 kts GS.

Answer 22

470 kts GS.
Correct
0.7IMN = 7NM/min x 60min/hr. = 420KTAS + 50kts wind = 470kts GS.

Question 23

You’re at 1500 AGL in a plane that slows at 10 kts per mile. Your approach speed is 130 and you’re going 230. How far from the field to you want to begin slowing down, assuming a 3 degree glideslope?
14 miles.
10 miles.
15 miles.

Answer 23

15 miles.

Question 24

Your aircraft is burning nine thousand pounds of fuel per hour. How long can you hold with 7500 pounds of fuel remaining.
  50 minutes.
  40 minutes.
  45 minutes.
  30 minutes.

Answer 24

50 minutes.
Correct
9000/60 = 150 LBS/Min, 7500LBS / 150 LBS-min = 50 minutes.

Question 25

On the 10DME arc you need a 2NM lead radial, how many radials is that?
  4 radials.
  3 radials.
  6 radials.
  12 radials.

Answer 25

12 radials.
Correct
60-to-1 Rule, at 60NM, 1 Radial = 1NM, so at 10NM from the VOR there are 6 radials/NM, multiplied by 2NM and you get 12 radials.

Question 26

If you are at FL370 and descending to FL200 at a VSI of 2500fpm. How long will it take to descend?

6. 4min.
7. 8min.
8. 8min.
9. 6min.

Answer 26

6.8min.
Correct
(37000-20000) ft / 2500 ft / min = 6.8min.

Question 27

Winds are 340/28. RWY 1 in use. What is your crosswind?
  9.
  14.
  16.
  18.

Answer 27

14.
Correct
Rule of six should be applied here 1/2 of wind=14.

Question 28

You have to descend 12,000’ in 20 miles, your speed is .7 Mach. What should your VSI be?
4100 FPM.
3800 FPM.
4200 FPM.

Answer 28

4200 FPM.
Correct
12,000 ft/ 20 miles= 600 ft per mile
600 ft per mile X 7miles a minute = 4200.

Question 29

You’re at 1500 AGL in a plane that slows at 10 kts per mile. Your approach speed is 130 and you’re going 230. How far from the field to you want to begin slowing down, assuming a 3 degree glideslope?
14 miles.
15 miles.
10 miles.

Answer 29

15 miles.

Question 30

One dot off on the localizer at 6nm equals what offset?
.08nm or approximately 800ft.
.04nm or approximately 600ft.
.05nm or approximately 300ft.

Answer 30

.05nm or approximately 300ft.
Correct
VOR 2 degrees per dot with full scale deflection equal to 10 degrees
ILS localizer 0.5 degrees per dot with full scale deflection 2.5 degrees
ILS glidepath 0.14 degrees per dot with full scale deflection 0.7 degrees
So the old 60-to-1 rule comes into play here again. At 6mn there are 10 degrees/NM. So 1 degree is = to 0.1NM, and 0.5 degrees is = to 0.05NM which is about 300 feet.
ROT: 50’/NM per dot deviation (so 6NM = 300 ft.)
ROT: 24’/NM per dot deviation off glidepath.

Question 31

You are on a visual approach at Denver and are at 7000’ MSL, the TDZE is 5200’. When would you descend for a 3 degree glidepath?
  10NM.
  3NM.
  6NM.
  5NM.

Answer 31

6NM.
Correct
1800ft / 300ft/nm = 6NM.

Question 32

You are flying at FL370 in cruise flight and the OAT increases 5 degrees C, and the headwind increases 5kts. What happens to your TAS and GS?
TAS increases by 6kts, therefore GS increases by 1kt.
TAS increases by 4kts, therefore GS increases by 1kt.
TAS increases by 10kts, therefore GS increases by 1kt.

Answer 32

TAS increases by 6kts, therefore GS increases by 1kt.
Correct
TAS increases approx. 1.2kts/1o C increase. So 5°C x 1.2kts = 6KTAS increase
6KTAS-5KTS headwind = 1kt GS increase.

Question 33

You are 6nm from the rwy on final and the localizer shows 1 degree of deflection. How far from the runway course are you?
.1nm (or 600 ft).
.4nm (or 300 ft).
.1nm (or 900 ft).

Answer 33

.1nm (or 600 ft).
Correct
VOR 2 degrees per dot with full scale deflection equal to 10 degrees
ILS localizer 0.5 degrees per dot with full scale deflection 2.5 degrees
ILS glidepath 0.14 degrees per dot with full scale deflection 0.7 degrees
So the old 60-to-1 rule comes into play here again. At 6mn there are 10 degrees/NM. So 1 degree is = to 0.1NM, and 0.5 degrees is = to 0.05NM which is about 300 feet.
ROT: 50’/NM per dot deviation (so 6NM = 300 ft.)
ROT: 24’/NM per dot deviation off glidepath.

Question 34

At mach 0.6, what is the VSI in a 1degree descent?
  420 FPM.
  350 FPM.
  600 FPM.
  400 FPM.

Answer 34

600 FPM.
Correct
0.6 IMN = 6NM/min
1 degree = 100ft/NM
VSI = 6NM/min x 100ft/NM = 600ft/min.

Question 35

If you are at FL370 and descending to FL200 at a VSI of 2500fpm. How long will it take to descend?

6. 8min.
7. 4min.
8. 8min.
9. 6min.

Answer 35

6.8min.
Correct
(37000-20000) ft / 2500 ft / min = 6.8min.

Question 36

What is your drift correction if you have a 35kt crosswind and are flying at 0.7IMN
  5 deg.
  9 deg.
  6 deg.
  4 deg.

Answer 36

5 deg.
Correct
Drift = Crosswind component / TAS in NM/min. 35kts/7nm/min = 5 deg.

Question 37

You’re at FL370 traveling at 420KTAS with a 60Kt tailwind. ATC clears you to descent to 10000ft at VOR passage which is 104nm away. What is your rate of descent?
  2400ft/min.
  2000 ft/min.
  2100ft/min.
  2600ft/min.

Answer 37

2100ft/min.
Correct
27000 feet, 480KGS, 104nm. 104NM / 8NM/min = 13min
27000 feet / 13min = 2100ft/min (round up).

Question 38

You are established on the 20DME arc to the north and you are going to intercept the 120R inbound to the VOR. Your flying 300KTAS, what is your lead point?
  129 radial.
  139 radial.
  125 radial.
  128 radial.

Answer 38

129 radial.
Correct
Turn radius = TAS/60min/hr – 2, so 300 / 60 – 2 = 3NM
Lead point = 60/DME x Turn radius (NM) = 60 / 20NM x 3NM = 9 degrees
Lead Radial = Radial +/- Lead radial. In this case we are approaching R120 from the south (north heading) so we would start a left hand turn at R129 to intercept the R120 inbound.

Question 39

Landing runway 09 and winds are 050/30kts. What is the crosswind component?
  18KTS.
  20KTS.
  22KTS.
  15KTS.

Answer 39

20KTS.
Correct
The rule of sixths… Relative to the runway
10deg=1/6,
20deg=1/3,
30deg=1/2,
40deg=4/6
50deg=5/6
60deg-90deg=6/6.
050deg is 40deg relative to 090. So 4/6 of the wind is crosswind. 4/6 x 30 = 20kts.

Question 40

You are flying 470KTAS and have a 100Kt tailwind. How long does it take to go 870NM?
  1+20.
  1+15.
  1+30.
  1+45.

Answer 40

1+30.
Correct
870NM / 570NM / hr. = 1.5 hours (rounding helps, 900/600=1.5)

Question 41

You are at FL270, flying at .75IMN, and are 25 DME from the VOR. What descent gradient and VSI is required to be at 12000 feet over the VOR?
  4500ft/min and 6 degrees.
  4800ft/min and 6 degrees.
  4100ft/min and 6 degrees.
  4600ft/min and 6 degrees.

Answer 41

4500ft/min and 6 degrees.
Correct
15000ft/25NM = 600ft/NM (remember 100ft/NM = 1 deg gradient), so 600ft/nm is 6 deg gradient.
.75IMN = 7.5NM/min x 600ft/NM = 4500fpm descent rate.

Question 42

If you are on an ILS approach at 180KTS, what is your rate of descent?
  600 FPM.
  350 FPM.
  900 FPM.
  400FPM.

Answer 42

900 FPM.
Correct
Take ½ your Ground Speed and add a zero. 90+0 = 900 FPM.

Question 43

You are at FL350 going Mach .84, 112nm from XYZ VOR. You are cleared to cross 12nm prior to XYZ VOR at 10000 feet and 250KIAS. Indicated airspeed in the descent is 300KIAS. When do you start your descent?
  92nm.
  88nm.
  95nm.
  85nm.

Answer 43

92nm.
Correct
25000ft of altitude to lose / 1000 x 3 =75nm. Reduce speed by 50kts at a rate of 10kts / 1 NM = 5nm. So far we have 80nm. We must be at 10000 feet by 12nm prior to the VOR, so we’ll add the 12 to the 80nm for a total of 92nm from the VOR.

Question 44

What is the higher TAS?
  260KCAS at FL300 and -30deg C.
  350KTAS at FL200 and -20deg C.
  250KCAS at FL370 and -40deg C.
  350KTAS at 10000ft.

Answer 44

250KCAS at FL370 and -40deg C.

Correct
A quick ROT to use is KTAS = KIAS + ALT/200. So for 250 + 37000/200 = 250 + 185 = 435KTAS. 260 + 30000/200 = 410KTAS.

Question 45

You are flying at 20000 feet with a thunderstorm 20nm in front of you. The radar look up angle is 5 degrees. How high is the thunderstorm?
  25,000 ft.
  30,000 ft.
  21,000 ft.
  28,000 ft.

Answer 45

30,000 ft.
Correct
1 degree of radar elevation is equal to 100ft at 1nm.
5deg x 100feet x 20nm = 10000 feet. (Add to 20000ft to get 30000ft)

Question 46

You are at 90DME at FL300 inbound to the VOR. You are requested to cross the station at 3000ft. If you are at .6IMN, when do you start down and what is descent gradient?
  3 degrees / 1800fpm / 80nm.
  3 degrees / 1800fpm / 90nm.
  4 degrees / 1600fpm/ 90nm.
  6 degrees / 1800fpm / 90nm.

Answer 46

3 degrees / 1800fpm / 90nm.
Correct
27000/90NM = 300ft/NM 0r 3 deg gradient
300ft/NM x 6NM/min = 1800fpm.
** One sheet had 81NM (27K x 3 = 81NM) and a descent gradient of 3.3 deg (27000ft/81NM = 330ft/NM = 3.3 deg gradient)

Question 47

Flying direct to the station on the 180 degree radial, at 37000 feet, and 70 DME from the VOR. Approach tells you to cross 15DME on the 360 Radial at 10000 feet. What is your descent rate?
317 FT/NM (No airspeed given to convert it to FPM) or approx. 3 degree gradient.
340 FT/NM (No airspeed given to convert it to FPM) or approx. 3 degree gradient.
320 FT/NM (No airspeed given to convert it to FPM) or approx. 6 degree gradient.
Correct
You need to lose 27000 feet and have 85NM to do it. 27000 ft/85NM = 317ft/NM
If you are given an airspeed in NM/min you can multiply it by 317ft/NM to get FPM.

Answer 47

317 FT/NM (No airspeed given to convert it to FPM) or approx. 3 degree gradient.

Question 48

What would have a faster Mach speed?
  300KIAS at 5K on a standard day.
  250KIAS at 15K on a standard day.
  250KIAS at FL400 OAT -49 degrees C.
  300KIAS at 10K OAT 6 degrees C.

Answer 48

250KIAS at FL400 OAT -49 degrees C.
Correct
The coldest temperature is going to produce the lowest local speed of sound and thus the highest IMN. 250 at FL400 = 450KTAS, 300 at 10K = 350KTAS, 250 at 15K = 325KTAS, 300 at 5K = 325KTAS

Question 49

On a 3 degree glide slope 140kts GS at 700 FPM rate of descent, the head winds increase, what adjustment must be made?
Increase descent rate.
Reduce rate of descent.
Continue at the current rate.

Answer 49

Reduce rate of descent.
Correct
140KGS = 700FPM, 20kt headwind reduces Kts GS to 120. 600FPM.

Question 50

Two airplanes are traveling in the same direction 50nm apart. The one in front is going .76 IMN, and the one in the rear is going .86IMN. When will the one in the rear overtake the one in the front?
  45 Minutes.
  50minutes.
  35 Minutes.
  40 Minutes.

Answer 50

50minutes.
Correct
7.6NM/min and 8.6NM/min = 1NM/min advantage x 50 NM = 50 minutes.

Question 51

The bearing pointer moves from 5 degrees in front of the wing to 5 degrees behind the wing in 8 minutes, you are doing 360kts. How far from the station are you?
  240NM.
  265NM.
  288NM.
  280NM.

Answer 51

288NM.
Correct
360 NM/hr / 60 min/hr = 6 NM/min. 6NM / min x 8 min = 48 NM. 48NM / 10 degrees = 4.8 degrees / NM x 60 = 288nm. At 60nm, 1 degree = 1nm. At 120nm 1 degree = 2nm. At 180nm 1 degree = 3nm, at 240nm 1 degree = 4nm. So the answer makes sense. At 288nm 1 degree = 4.8nm

Question 52

Your DME is inoperative. You turn 90 degrees to the station and time for 1min. The needle falls from 5 degrees above the wingline to 5 degrees below the wingline. You are doing 480KTAS. What is your distance from the station?
  52NM.
  40NM.
  48NM.
  50NM.

Answer 52

48NM.
Correct
480KTAS (assuming no wind) = 8NM/min in 1 minute, you have traveled 8NM.
8NM / 10 degrees = 0.8 NM per degree
60 x 0.8 NM per degree = 48NM.

Question 53

You are at FL370, 0.86IMN. You need to cross the outer marker at 1600 feet. When do you start your descent? Your descent profile assumes idle power and .86IMN transitioning to 300KIAS, then 250KIAS at 10000ft. Assume the aircraft slows down at 10KTS/NM.
105NM FROM THE OUTER MARKER (118 from the Airfield).
110NM FROM THE OUTER MARKER (115 from the Airfield).
120NM FROM THE OUTER MARKER (115 from the Airfield).

Answer 53

110NM FROM THE OUTER MARKER (115 from the Airfield).
Correct
Round 1600 to 2000 37000-2000/1000 x 3 = 105NM
300KIAS-250KIAS = 50KTS / 10 KTS/NM = 5NM
Descent + Speed reduction = 110NM
Outer marker approx. 5nm from runway. So add 5NM if req.

Question 54

Your aircraft is burning nine thousand pounds of fuel per hour. How long can you hold with 7500 pounds of fuel remaining.
  40 minutes.
  50 minutes.
  30 minutes.
  45 minutes.

Answer 54

50 minutes.
Correct
9000/60 = 150 LBS/Min, 7500LBS / 150 LBS-min = 50 minutes.

Question 55

There is a heavy aircraft at the outer marker at five miles flying 120KTS, you are seven miles behind the heavy aircraft flying thirty knots faster, where will you be when the heavy aircraft touches down?
  5.5nm from runway.
  5.75nm from the runway.
  4.75nm from runway.
  6nm from the runway.

Answer 55

5.75nm from the runway.
Correct
120KTS=2NM/min. 150KTS=2.5NM/min. It takes the heavy jet 2.5 mins. to land. So we are twelve miles from the runway (seven miles behind the heavy at five miles.) In 2.5 mins we have flown 6.25nm (2.5NM/min x 2.5min = 6.25NM). 12NM-6.25NM=5.75NM fr.

Question 56

MDA for the LOC 27 is 450’AGL, Time from FAF to MAP is 3:00. How long after the FAF will your VDP occur?

2: 30.
2: 15.
2: 45.
2: 10.

Answer 56

2:15.
Correct
Take 10% of the MDA (AGL) and subtract it from the total time. That is your time on the approach until reaching the VDP 450 x 10% = 45 sec. 3:00 – 0:45 = 2:15

Question 57

MDA for the LOC 27 is 450’AGL, How far from the runway is the VDP?

2. 0NM.
3. 8NM.
4. 4NM.
5. 5NM.

Answer 57

1.5NM.
Correct
450 ft/ 300ft/nm = 1.5nm (300 ft/nm is roughly a 3 degree glide path).
Visual Descent Point (VDP): A defined point on the final approach course of a non-precision straight-in approach from which a normal decent from the MDA to the runway touch-down zone may commence, provided the runway or approach lights are clearly visible to pilot.
To calculate a 3 degree decent angle from the VDP t the runway, divide the groundspeed by 2, then multiple the result by 10 (100kts (GS) 2 x 10 = 500 fpm.
EEPP p. 77.

Question 58

The LOC course inbound is 360 degrees. The FAF is defined by the XYZ VOR 090 radial, at 20 DME. Your bearing pointer shows that the VOR is on a 288 degree bearing from your present position on the LOC course. How far away is the FAF?
  4.5nm.
  6nm.
  3nm.
  4nm.

Answer 58

6nm.
Correct
You are on the 108 radial and have 18 radials to go until the 090 radial while flying a course of 360 deg. If 1 radial (or degree) is equal to 1nm at 60 miles from a VOR, then 2 radials (or degrees) equal a mile at 30nm from the VOR, and 3 radials equal a mile at 20nm from the VOR. So 18 degrees / (3 degrees / NM) = 6NM.

Question 59

You are flying at 0.86 IMN and you slow down. At what speed must you notify ATC?

0. 84 IMN (5% Change).
1. 82IMN (5% Change).
2. 83 IMN (5% Change).

Answer 59

0.82IMN (5% Change).
Correct
8.6NM/min x 60 = 516KTAS-10KTAS=506KTAS/60=8.4NM/min=0.84IMN
5% of 516KTAS is 26 so 516-26=490KTAS/60=8.2NM/min=0.82IMN
Any change in the average TRUE AIRSPEED (at cruising altitude) when it varies by 5% or 10kts (Whichever is GREATER) from that filed in the flight plan. AIM 5-3-3

Question 60

Fuel burn in holding is 3000lb/hr. per engine. You have 5000lbs of fuel, how long can you hold?
  45min.
  50min.
  44min.
  35min.

Answer 60

50min
Correct
5000lbs / 6000lbs/hr. = 5/6 hr. or 50min.

Question 61

Inbound to the VOR heading 180 degrees at 300KTAS. You want to intercept the 20 DME arc, what is your lead point?
  18nm.
  23nm.
  21nm.
  22nm.

Answer 61

23nm.
Correct
Turn radius = 300 / 60 – 2 = 3NM
Lead point = Arc DME + turn radius

Question 62

One aircraft departs ATL traveling at .7IMN and a second aircraft departs ATL 30 minutes later traveling .8IMN. How long before the second aircraft overtakes the first?
  210minutes.
  180 Minutes.
  160 Minutes.
  230 Minutes.

Answer 62

210minutes.
Correct
7NM/min x 30min = 210NM head start
Every min the second airplane is airborne it will gain 1NM on the first plane.

Question 63

You are on a 090 bearing to the station and cleared to hold on the 060 bearing. What heading would you turn to after crossing the fix?
  240 degrees (060 bearing is the 240 radial).
  245 degrees (065 bearing is the 240 radial).
  230 degrees (030 bearing is the 240 radial).
  250 degrees (060 bearing is the 240 radia.

Answer 63

240 degrees (060 bearing is the 240 radial).
Correct
Bearings are TO the station, Radials are from the station. Essentially, cleared to hold at the XYZ VOR 240 radial. You are on the 270 radial (090 bearing TO the station) so at the fix you would turn right for a direct entry into holding.

Question 64

You are flying Mach 0.7 with a 50kt tailwind. What is your ground speed?
  450 kts GS.
  460 kts.
  455 kts.
  470 kts GS.

Answer 64

470 kts GS.
Correct
0.7IMN = 7NM/min x 60min/hr. = 420KTAS + 50kts wind = 470kts GS.

Question 65

You’re at 1500’ AGL, and you are flying at 230KIAS. Your approach speed is 130KIAS, and your aircraft slows at a rate of 10kts/NM. How far out from the runway do you need to slow down in order to shoot the ILS?
  15nm.
  22nm.
  24nm.
  18nm.

Answer 65

15nm.
Correct
100kt speed reduction / 10kts/NM = 10nm. The outer marker of a typical ILS is at 5nm, where you would want to be at approach speed. So,15nm from the airport.

Question 66

In standard holding on the 020R 40DME fix at 6000 feet with 20nm legs, where will the tail of your bearing pointer be on the outbound turn?
  014 degrees.
  12 degrees.
  16 degrees.
  18 degrees.

Answer 66

014 degrees.
Correct
You are flying inbound on the 020R, so the head of your bearing pointer is pointed at the VOR (200 deg). You begin your standard rate turn at 200KIAS (holding speed at 6000ft). To find the radius (90 deg) of your turn use the 1% of your GS (no wind so it’s 200Kts) which is 2nm. The diameter (180 deg) of the turn is twice that…so 4NM. To find how many radials you have gone in your outbound turn use the 60-to-1 rule. At 40DME 1nm = 1.5radials. So in your outbound right hand turn, you ended up 6 radials to the north or on the 014R (4nm turn x 1.5 radials / NM @ 40NM) = 6 radials. So assuming you are now heading 020 deg on the outbound leg with no wind, the head is still pointing at the VOR, so the tail of your bearing pointer will be pointing to 014 degrees (the radial you are on at the completion of the outbound turn). Remember the tail will fall slightly in the outbound leg and the head will rise.

Question 67

You are at 25000 ft, there is a Level 5 thunderstorm 80nm in front of you. You tilt your radar up 1.5 degrees and the cell disappears. How high is the top of the thunderstorm?
37000 ft.
31000 ft.
35000 ft.
32000 ft.
Correct
1 degree of radar elevation is equal to 100ft at 1nm.
1.5deg x 100feet x 80nm = 150 x 80 = 12000 feet. (Add to 25000ft to get 37000)

Answer 67

37000 ft.

1 degree of radar elevation is equal to 100ft at 1nm.
1.5deg x 100feet x 80nm = 150 x 80 = 12000 feet. (Add to 25000ft to get 37000)

Question 68

On the 10DME arc you need a 2NM lead radial, how many radials is that?
  4 radials.
  6 radials.
  12 radials.
  3 radials.

Answer 68

12 radials.
Correct
60-to-1 Rule, at 60NM, 1 Radial = 1NM, so at 10NM from the VOR there are 6 radials/NM, multiplied by 2NM and you get 12 radials.

Question 69

You are on a northerly heading and the VOR bearing pointer is showing a bearing of 260. Eight minutes later the bearing is showing 280. Your traveling 420KTAS and the wind is 360 @ 60KTS. How far away is the VOR?
  140NM.
  144NM.
  130NM.
  120NM.

Answer 69

144NM.
Correct
420KTAS – 60KTS headwind = 360KTS GS. 360KTS/60min = 6NM/min
6NM/min x 8 min = 48NM / 20 degrees = 2.4 NM per degree
Apply the 60-to-1 rule so 60 x 2.4 = 144nm.