Updated: Jul 19
Let's look at even more of the most challenging Private Pilot Test questions. You can also access the full FlightInsight Test Bank to get ready for the real thing!
For now, let's have an in-depth look at another five questions that give all of us the most trouble:
(Refer to the Figure) Determine the total distance required to land over a 50-foot obstacle. Pressure altitude = 5,000 ft Headwind = 8 kts Temperature = 41°F Runway = Hard surface
· 837 feet.
· 956 feet.
· 1,076 feet.
We'll have to use the chart to determine how much runway is needed to clear a 50-foot obstacle and land in the given conditions. The third section of the chart, with 5,000 feet and 41°F listed, is appropriate for the question. Since we need to clear a 50 foot obstacle and land, we'll use the 1,195 feet figure.
Next, notice in Note 1 on the chart that for every 4 knots of headwind, we'll decrease landing distance 10%. We have not 4 knots of headwind, but 8 knots, so instead of reducing by 10%, we'll reduce by 20%.
20% of 1195 is 239, and 1195 - 239 = 956 feet.
The two incorrect choices, 837 feet and 1,076 feet, involve the wrong calculations by reducing the 1,195 figure from the chart by either 30% or 10%, respectively, instead of 20%.
(Refer to the Figure) 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?
· 1,000 feet.
· 1,100 feet.
· 1,300 feet.
Unfortunately, questions involving this density altitude chart are an exercise in careful line drawing and sharp eyes. If we start from 35°F (make sure not to use Celsius), and draw a line up to the 3,000 foot pressure altitude line, we can move to the left and find a density altitude of about 1,850 feet. If we do the same for 50°F, we'll get a density altitude of 2,850 feet. A difference of 1,000 feet.
The bad news is, it's very easy to misread the lines you draw and be off by 100 or more feet, and to make matters worse, the answer choices are close enough together to make eliminating obvious wrong ones difficult. Also, the E6-B isn't much help for precise density altitude calculations.
However, all of these questions on the FAA test which reference the density altitude chart use Fahrenheit for temperature, so if you can just remember that a 15°F change in temperature results in a 1,000 foot change in density altitude, you can apply that rate to any problem you get in this topic and you might have a fighting chance!
(Refer to the Figure) Determine the approximate manifold pressure setting with 2,450 RPM to achieve 65 percent maximum continuous power at 6,500 feet with a temperature of 36°F higher than standard.
· 19.8 inches Hg.
· 20.8 inches Hg.
· 21.0 inches Hg.
A lot of us don't train in aircraft with constant speed propellers, so the concept of manifold pressure may be foreign. Just remember that there are two power indications in aircraft equipped with constant speed propellers: RPM (tachometer) and manifold pressure.
We're at 6,500 feet. The closest altitudes are 6,000 and 8,000 feet, so we'll look at both of those rows.
Since temperature is 36°F higher than standard, the right part of the chart is appropriate. The 6,000 and 8,000 rows in the right part of the chart read 21.0" and 20.8" respectively. At 6,500 feet, we're a lot closer to the 21.0 figure than to the 20.8 figure, so we'll go with the 21.0 inches Hg.
(Refer to the Figure) The NALF Fentress (NFE) Airport is in what type of airspace?
· Class C.
· Class E.
· Class G.
This one is surprisingly hard. We've learned that Class B is a solid blue circle, C is a solid magenta circle, D is a segmented blue circle, and E and G are determined by a combination of shaded areas and other markings, but now we have a segmented magenta circle around Fentress, which we don't see in any traditional airspace markings.
At first glance, it looks like a towered controlled airport, such as a Class D, even though those circles would be blue, and Class D isn't an answer choice. The only other choice for a tower controlled airport would be Class C, but this airspace is actually Class E all the way down to the surface.
Typically, Class E starts at either 700 AGL or 1200 AGL, with some exceptions. However, sometimes it's necessary to bring the protections Class E gives aircraft all the way down to the surface, and that's what's been done here at Fentress, as indicated by the segmented magenta circle.
How far will an aircraft travel in 7.5 minutes with a ground speed of 114 knots?
· 14.25 NM.
· 15 NM.
· 14.5 NM.
Your E6-B skills will be on display for this question. Use the slide rule side of the E6-B and align the 60 on the inner ring with 114 on the outer ring. This is how far the aircraft travels in one hour. To find how far it travels in 7.5 minutes, find the half way mark between 70 and 80 on the inner ring, and read 14.25 on the outer ring. See the E6-B below.
Alternatively, use math to find the answer. distance equals speed times time. So distance equals 114 knots times 7.5 minutes, or .125 hours. 114 * .125 = 14.25 miles.
Have a great holiday and don't study too hard! Check back here after the New Year when the new Private Pilot Knowledge Test Prep Course will be available online.
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