ASE P2 Practice Exam: Questions & Answers
1
Determine the density altitude for these conditions:
Altimeter setting = 30.35
Runway temperature = +25°F
Airport elevation = 3,894 ft. MSL - ANSWER:2,000 ft. MSL
2
What is the effect of a temperature increase from 30 to 50 °F on the density altitude if the pressure
altitude remains at 3,000 feet MSL? - ANSWER:1,300-foot increase.
Increasing the temperature from 30°F to 50°F, given a constant pressure altitude of 3,000 ft., requires
you to find the 3,000-ft. line on the density altitude chart at the 30°F level. At this point, the density
altitude is approximately 1,650 feet. Then move up the 3,000-ft. line to 50°F, where the density altitude
is approximately 2,950 feet. There is an approximate 1,300-ft. increase (2,950 - 1,650 feet). Note that
50°F is just about standard and pressure altitude is very close to density altitude.
3
What is the effect of a temperature increase from 35 to 50°F on the density altitude if the pressure
altitude remains at 3,000 feet MSL? - ANSWER:1,000-foot increase.
4
What is the effect of a temperature decrease and a pressure altitude increase on the density altitude
from 90°F and 1,250 feet pressure altitude to 55°F and 1,750 feet pressure altitude? - ANSWER:1,700-
foot decrease.
The requirement is the effect of a temperature decrease and a pressure altitude increase on density
altitude. First, find the density altitude at 90°F and 1,250 ft. (approximately 3,600 feet). Then find the
density altitude at 55°F and 1,750 ft. pressure altitude (approximately 1,900 feet). Next, subtract the two
numbers. Subtracting 1,900 ft. from 3,600 ft. equals a 1,700-ft. decrease in density altitude.
5
Determine the pressure altitude at an airport that is 3,563 feet MSL with an altimeter setting of 29.96. -
ANSWER:3,527 feet MSL.
,Note that the question asks only for pressure altitude, not density altitude. Pressure altitude is
determined by adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for nonstandard pressure. This
is the true altitude plus or minus the pressure altitude conversion factor (based on current altimeter
setting). On the chart, an altimeter setting of 30.0 requires you to subtract 73 ft. to determine pressure
altitude (note that at 29.92, nothing is subtracted because that is pressure altitude). Since 29.96 is
halfway between 29.92 and 30.0, you need only subtract 36 (-73/2) from 3,563 ft. to obtain a pressure
altitude of 3,527 ft. (3,563 - 36). Note that a higher-than-standard barometric pressure means pressure
altitude is lower than true altitude.
6
Determine the pressure altitude at an airport that is 1,386 feet MSL with an altimeter setting of 29.97. -
ANSWER:1,341 feet MSL.
Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg. This is the true altitude
plus or minus the pressure altitude conversion factor (based on current altimeter setting). Since 29.97 is
not a number given on the conversion chart, you must interpolate. Compute 5/8 of -73 (since 29.97 is
5/8 of the way between 29.92 and 30.0), which is 45. Subtract 45 ft. from 1,386 ft. to obtain a pressure
altitude of 1,341 feet. Note that if the altimeter setting is greater than standard (e.g., 29.97), the
pressure altitude (i.e., altimeter set to 29.92) will be less than true altitude.
7
What is the effect of a temperature increase from 25 to 50° F on the density altitude if the pressure
altitude remains at 5,000 feet? - ANSWER:1,650-foot increase.
Increasing the temperature from 25°F to 50°F, given a pressure altitude of 5,000 ft., requires you to find
the 5,000-ft. line on the density altitude chart at the 25°F level. At this point, the density altitude is
approximately 3,800 ft. Then move up the 5,000-ft. line to 50°F, where the density altitude is
approximately 5,400 ft. There is about a 1,600-ft. increase (5,400 ft. - 3,800 ft.). As temperature
increases, so does density altitude; i.e., the atmosphere becomes thinner (less dense). Because a 1,600-
foot increase is not an answer choice, 1,650-foot increase would be the best answer.
8
Determine the pressure altitude with an indicated altitude of 1,380 feet MSL with an altimeter setting of
28.22 at standard temperature. - ANSWER:2,991 feet MSL.
Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for
nonstandard pressure. This is the indicated altitude of 1,380 ft. plus or minus the pressure altitude
conversion factor (based on the current altimeter setting).
On the right side of Fig. 8 is a pressure altitude conversion factor schedule. Add 1,533 ft. for an altimeter
setting of 28.30 and 1,630 ft. for an altimeter setting of 28.20. Using interpolation, you must subtract
,20% of the difference between 28.3 and 28.2 from 1,630 ft. (1,630 - 1,533 = 97 × .2 = 19). Since 1,630 -
19 = 1,611, add 1,611 ft. to 1,380 ft. to get the pressure altitude of 2,991 feet.
9
Determine the density altitude for these conditions:
Altimeter setting = 29.25
Runway temperature = +81°F
Airport elevation = 5,250 ft MSL - ANSWER:8,500 feet MSL.
With an altimeter setting of 29.25" Hg, about 626 ft. (579 plus 1/2 the 94-ft. pressure altitude conversion
factor difference between 29.2 and 29.3) must be added to the field elevation of 5,250 ft. to obtain the
pressure altitude, or 5,876 feet. Note that barometric pressure is less than standard and pressure
altitude is greater than true altitude. Next, convert pressure altitude to density altitude. On the chart,
find the point at which the pressure altitude line for 5,876 ft. crosses the 81°F line. The density altitude
at that spot shows somewhere in the mid-8,000s of feet. The closest answer choice is 8,500 feet. Note
that, when temperature is higher than standard, density altitude exceeds pressure altitude.
10
Determine the approximate ground roll distance required for takeoff.
OAT = 38°C
Pressure altitude = 2,000 ft
Takeoff weight = 2,750 lb
Headwind component = Calm - ANSWER:1,150 feet.
Begin on the left section of Fig. 40 at 38°C (see outside air temperature at the bottom). Move up
vertically to the pressure altitude of 2,000 feet. Then proceed horizontally to the first reference line.
Since takeoff weight is 2,750, move parallel to the closest guideline to 2,750 pounds. Then proceed
horizontally to the second reference line. Since the wind is calm, proceed again horizontally to the right-
hand margin of the diagram (ignore the third reference line because there is no obstacle, i.e., ground roll
is desired), which will be at 1,150 feet.
11
Determine the total distance required for takeoff to clear a 50-foot obstacle.
OAT = Std
Pressure altitude = Sea level
Takeoff weight = 2,700 lb
, Headwind component = Calm - ANSWER:1,400 feet.
Begin in the left section of Fig. 40 by finding the intersection of the sea level pressure altitude and
standard temperature (15°C) and proceed horizontally to the right to the first reference line. Then
proceed parallel to the closest guideline, to 2,700 pounds. From there, proceed horizontally to the right
to the third reference line. You skip the second reference line because the wind is calm. Then proceed
upward parallel to the closest guideline to the far right side. To clear the 50-ft. obstacle, you need a
takeoff distance of about 1,400 feet. NOTE: This question was previously released by the FAA and the
FAA's objective is for you to select the "most correct" answer from the choices given. The actual answer
is 1,250 feet, but since 1,400 feet is the closest answer, it should be chosen as correct.
12
Determine the total distance required for takeoff to clear a 50-foot obstacle.
OAT = Std
Pressure altitude = 4,000 ft
Takeoff weight = 2,800 lb
Headwind component = Calm - ANSWER:1,750 feet.
The takeoff distance to clear a 50-ft. obstacle is required. Begin on the left side of the graph at standard
temperature (as represented by the curved line labeled "ISA"). From the intersection of the standard
temperature line and the 4,000-ft. pressure altitude, proceed horizontally to the right to the first
reference line, and then move parallel to the closest guideline to 2,800 pounds. From there, proceed
horizontally to the right to the third reference line (skip the second reference line because there is no
wind), and move upward following equidistantly between the diagonal lines all the way to the far right.
You are at 1,750 ft., which is the takeoff distance to clear a 50-ft. obstacle.
13
Determine the approximate ground roll distance required for takeoff.
OAT = 32°C
Pressure altitude = 2,000 ft
Takeoff weight = 2,500 lb
Headwind component = 20 kts - ANSWER:650 feet.
Begin with the intersection of the 2,000-ft. pressure altitude curve and 32°C in the left section of Fig. 40.
Move horizontally to the right to the first reference line, and then parallel to the closest guideline to
2,500 lb. Then move horizontally to the right to the second reference line, and then parallel to the
1
Determine the density altitude for these conditions:
Altimeter setting = 30.35
Runway temperature = +25°F
Airport elevation = 3,894 ft. MSL - ANSWER:2,000 ft. MSL
2
What is the effect of a temperature increase from 30 to 50 °F on the density altitude if the pressure
altitude remains at 3,000 feet MSL? - ANSWER:1,300-foot increase.
Increasing the temperature from 30°F to 50°F, given a constant pressure altitude of 3,000 ft., requires
you to find the 3,000-ft. line on the density altitude chart at the 30°F level. At this point, the density
altitude is approximately 1,650 feet. Then move up the 3,000-ft. line to 50°F, where the density altitude
is approximately 2,950 feet. There is an approximate 1,300-ft. increase (2,950 - 1,650 feet). Note that
50°F is just about standard and pressure altitude is very close to density altitude.
3
What is the effect of a temperature increase from 35 to 50°F on the density altitude if the pressure
altitude remains at 3,000 feet MSL? - ANSWER:1,000-foot increase.
4
What is the effect of a temperature decrease and a pressure altitude increase on the density altitude
from 90°F and 1,250 feet pressure altitude to 55°F and 1,750 feet pressure altitude? - ANSWER:1,700-
foot decrease.
The requirement is the effect of a temperature decrease and a pressure altitude increase on density
altitude. First, find the density altitude at 90°F and 1,250 ft. (approximately 3,600 feet). Then find the
density altitude at 55°F and 1,750 ft. pressure altitude (approximately 1,900 feet). Next, subtract the two
numbers. Subtracting 1,900 ft. from 3,600 ft. equals a 1,700-ft. decrease in density altitude.
5
Determine the pressure altitude at an airport that is 3,563 feet MSL with an altimeter setting of 29.96. -
ANSWER:3,527 feet MSL.
,Note that the question asks only for pressure altitude, not density altitude. Pressure altitude is
determined by adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for nonstandard pressure. This
is the true altitude plus or minus the pressure altitude conversion factor (based on current altimeter
setting). On the chart, an altimeter setting of 30.0 requires you to subtract 73 ft. to determine pressure
altitude (note that at 29.92, nothing is subtracted because that is pressure altitude). Since 29.96 is
halfway between 29.92 and 30.0, you need only subtract 36 (-73/2) from 3,563 ft. to obtain a pressure
altitude of 3,527 ft. (3,563 - 36). Note that a higher-than-standard barometric pressure means pressure
altitude is lower than true altitude.
6
Determine the pressure altitude at an airport that is 1,386 feet MSL with an altimeter setting of 29.97. -
ANSWER:1,341 feet MSL.
Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg. This is the true altitude
plus or minus the pressure altitude conversion factor (based on current altimeter setting). Since 29.97 is
not a number given on the conversion chart, you must interpolate. Compute 5/8 of -73 (since 29.97 is
5/8 of the way between 29.92 and 30.0), which is 45. Subtract 45 ft. from 1,386 ft. to obtain a pressure
altitude of 1,341 feet. Note that if the altimeter setting is greater than standard (e.g., 29.97), the
pressure altitude (i.e., altimeter set to 29.92) will be less than true altitude.
7
What is the effect of a temperature increase from 25 to 50° F on the density altitude if the pressure
altitude remains at 5,000 feet? - ANSWER:1,650-foot increase.
Increasing the temperature from 25°F to 50°F, given a pressure altitude of 5,000 ft., requires you to find
the 5,000-ft. line on the density altitude chart at the 25°F level. At this point, the density altitude is
approximately 3,800 ft. Then move up the 5,000-ft. line to 50°F, where the density altitude is
approximately 5,400 ft. There is about a 1,600-ft. increase (5,400 ft. - 3,800 ft.). As temperature
increases, so does density altitude; i.e., the atmosphere becomes thinner (less dense). Because a 1,600-
foot increase is not an answer choice, 1,650-foot increase would be the best answer.
8
Determine the pressure altitude with an indicated altitude of 1,380 feet MSL with an altimeter setting of
28.22 at standard temperature. - ANSWER:2,991 feet MSL.
Pressure altitude is determined by adjusting the altimeter setting to 29.92" Hg, i.e., adjusting for
nonstandard pressure. This is the indicated altitude of 1,380 ft. plus or minus the pressure altitude
conversion factor (based on the current altimeter setting).
On the right side of Fig. 8 is a pressure altitude conversion factor schedule. Add 1,533 ft. for an altimeter
setting of 28.30 and 1,630 ft. for an altimeter setting of 28.20. Using interpolation, you must subtract
,20% of the difference between 28.3 and 28.2 from 1,630 ft. (1,630 - 1,533 = 97 × .2 = 19). Since 1,630 -
19 = 1,611, add 1,611 ft. to 1,380 ft. to get the pressure altitude of 2,991 feet.
9
Determine the density altitude for these conditions:
Altimeter setting = 29.25
Runway temperature = +81°F
Airport elevation = 5,250 ft MSL - ANSWER:8,500 feet MSL.
With an altimeter setting of 29.25" Hg, about 626 ft. (579 plus 1/2 the 94-ft. pressure altitude conversion
factor difference between 29.2 and 29.3) must be added to the field elevation of 5,250 ft. to obtain the
pressure altitude, or 5,876 feet. Note that barometric pressure is less than standard and pressure
altitude is greater than true altitude. Next, convert pressure altitude to density altitude. On the chart,
find the point at which the pressure altitude line for 5,876 ft. crosses the 81°F line. The density altitude
at that spot shows somewhere in the mid-8,000s of feet. The closest answer choice is 8,500 feet. Note
that, when temperature is higher than standard, density altitude exceeds pressure altitude.
10
Determine the approximate ground roll distance required for takeoff.
OAT = 38°C
Pressure altitude = 2,000 ft
Takeoff weight = 2,750 lb
Headwind component = Calm - ANSWER:1,150 feet.
Begin on the left section of Fig. 40 at 38°C (see outside air temperature at the bottom). Move up
vertically to the pressure altitude of 2,000 feet. Then proceed horizontally to the first reference line.
Since takeoff weight is 2,750, move parallel to the closest guideline to 2,750 pounds. Then proceed
horizontally to the second reference line. Since the wind is calm, proceed again horizontally to the right-
hand margin of the diagram (ignore the third reference line because there is no obstacle, i.e., ground roll
is desired), which will be at 1,150 feet.
11
Determine the total distance required for takeoff to clear a 50-foot obstacle.
OAT = Std
Pressure altitude = Sea level
Takeoff weight = 2,700 lb
, Headwind component = Calm - ANSWER:1,400 feet.
Begin in the left section of Fig. 40 by finding the intersection of the sea level pressure altitude and
standard temperature (15°C) and proceed horizontally to the right to the first reference line. Then
proceed parallel to the closest guideline, to 2,700 pounds. From there, proceed horizontally to the right
to the third reference line. You skip the second reference line because the wind is calm. Then proceed
upward parallel to the closest guideline to the far right side. To clear the 50-ft. obstacle, you need a
takeoff distance of about 1,400 feet. NOTE: This question was previously released by the FAA and the
FAA's objective is for you to select the "most correct" answer from the choices given. The actual answer
is 1,250 feet, but since 1,400 feet is the closest answer, it should be chosen as correct.
12
Determine the total distance required for takeoff to clear a 50-foot obstacle.
OAT = Std
Pressure altitude = 4,000 ft
Takeoff weight = 2,800 lb
Headwind component = Calm - ANSWER:1,750 feet.
The takeoff distance to clear a 50-ft. obstacle is required. Begin on the left side of the graph at standard
temperature (as represented by the curved line labeled "ISA"). From the intersection of the standard
temperature line and the 4,000-ft. pressure altitude, proceed horizontally to the right to the first
reference line, and then move parallel to the closest guideline to 2,800 pounds. From there, proceed
horizontally to the right to the third reference line (skip the second reference line because there is no
wind), and move upward following equidistantly between the diagonal lines all the way to the far right.
You are at 1,750 ft., which is the takeoff distance to clear a 50-ft. obstacle.
13
Determine the approximate ground roll distance required for takeoff.
OAT = 32°C
Pressure altitude = 2,000 ft
Takeoff weight = 2,500 lb
Headwind component = 20 kts - ANSWER:650 feet.
Begin with the intersection of the 2,000-ft. pressure altitude curve and 32°C in the left section of Fig. 40.
Move horizontally to the right to the first reference line, and then parallel to the closest guideline to
2,500 lb. Then move horizontally to the right to the second reference line, and then parallel to the
