Posted: Wed Jun 27, 2012 2:07 pm Post subject: Badly Phrased, Contradictory, Doubtful or Wrong Questions
At 60o N the scale of a direct Mercator chart is 1: 3000 000. What is the scale at the equator?
a) 1 : 3 000 000
b) 1 : 3 500 000
c) 1 : 1 500 000
d) 1 : 6 000 000 <-- Correct
The underlined part is missing from the question
The shortest distance between 2 point of the surface of the earth is:
a) a great circle
b) the arc of a great circle
c) half the rhumb line distance
d) Rhumb line
(a) marked correct, whereas b) seems to be correct.
Great Circle is a circle on the surface of the earth whose centre and radius are those of the earth itself is called a Great Circle. It is called 'great' because a disc cut through the earth in the plane of the Great Circle would have the largest area that can be achieved. The shortest distance between two points on the Earth's surface is the shorter arc of the Great Circle joining the two points. Given two points on the Earth's surface, there will be only one Great Circle joining them (unless the points are diametrically opposed).
The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010oE and 030oW is 30o, is:
b) marked correct. However the wording for this question is not proper. Instead of the word "convergence angle" they should have used "convergency" because conversion angle is half of convergency. If conversion angle is given as 30 then convergency should be 60.
Convergency = Change of Longitude x Convergence Factor
Constant (Convergence Factor) = 60/40 is not 0.75
(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)SHA VOR (5243N 00853W) DME 41nm CRK VOR (5150N 00829W) DME 30nm. What is the position of the aircraft?
a) 5215N 00805W <-- Correct
b) 5205N 00915W
c) 5215N 00915W
d) 5225N 00810W
Incomplete question. Following statement is missing:
"Aircraft heading 270o(M). Both DME distances decreasing"
(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from CRK VOR (5151N 00830W) to position 5220N 00910W?
a) 322 M 39 nm
b) 330 M 41 nm
c) 330 M 39 nm <-- Marked Correct
d) 322 M 41 nm
Very close answers
(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)What is the radial and DME distance from SHA VOR (5243N 00853W) to BIRR airport (5311N 00754W)?
a) 068 M 42 nm
b) 060 M 40 nm
c) 068 M 40 nm
d) 060 M 42 nm
a) marked correct whereas the correct answer comes to be 068M 41nm if BIRR lat is 53°04' (like in other question banks) instead of 53°11'
On a Mercator chart, at latitude 60oN, the distance measured between W002o and E008ois 20 cm. The scale of this chart at latitude 60oN is approximately
a) 1 : 5 560 000
b) 1 : 278 000
c) 1 : 780 000
d) 1 : 556 000
c) marked correct however 2 is missing from the answer.
dep = dlong x cos lat = 600 x cos60 = 300
20 cm = 300 nm so 1cm = 15nm
15nm = 27.78 km or 2,778,000 cm so scale = 1:2,780,000
Chart scale is 1: 850 000
The chart distance between two points is 4 centimetres
Earth distance is approximately:
a) 4 NM
b) 74 NM
c) 100 NM
d) 40 NM
d) marked correct. the scale should be 1: 1,850,000 and not 1: 850,000 (1 is missing).
Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1: 200 000?
a) marked correct. A "0" is missing from the Scale. It should be 1:2,000,000
On a chart, the distance along a meridian between latitudes 45oN and 46oN is 6 cm. The scale of the chart is approximately:
a) 1 : 1 000 000
b) 1 : 850 000
c) 1 : 185 000
d) 1 : 18 500 000
b) marked correct. The corect answer is 1:1,850,000 i.e. 1 is missing from option "b".
If the true track from A to B is 090o, TAS is 460 knots, wind velocity is 360o/100 kts, variation is 10oE and deviation is -20; calculate the compass heading and ground speed.
a) 069o and 448 kts
b) 068o and 460 kts
c) 078o and 450 kts
d) 070o and 453 kts
a) marked correct. Option (a) correct with deviation of -2 and not -20. Clue: "There is probably no other question that mentions deviation of -20".
The rhumb line distance between points C (N6000.0 E00213.0) and D (N60000.0 W 00713.0) is:
a) 300 nm
b) 520 nm
c) 150 nm
d) 600 nm
a) marked correct. Whereas (c) is correct. departure = dlong x cos lat
Clue to identify this question: "rhumb line distance between points C and D"
If the headwid component is 50 kt, the FL is 330, temperature JSA -7oC and the ground speed is 496 kt, the Mach No. is:
a) marked correct. It will be correct if temperature is -70 not -7. Mach Number = TAS/LSS = 546/39 x square root of (273-70)
TAS = 470 kt
True HDG = 317o
W/V = 045o(T)/45 kt
Calculate the drift angle and GS
a) 3oR - 470 kt
b) 5oL - 270 kt
c) 5oL - 475 kt
d) 5oR - 475 kt
b) marked correct whereas (c) seems to be correct. Its a cross wind and GS cannot drop that low.
What is the Chlong (in degrees and minutes) from A (45N 1630E) to B (45N 15540W)?
d) marked correct. Typo in the question. Long for position A is 165°30'E and not 1630E.
An aircraft has a TAS of 300 knots and a safe endurance of 0 hours. If the wind component on the outbound leg is 50 knots head, what is the distance to the point of safe endurance?
a) 1500 nm
b) 1458 nm
c) 1544 nm
d) 1622 nm
b) marked correct. That means endurance is 10 hours. In the question there is a typo showing endurance 0 instead of 10.
By what amount must you change your rate of descent given a 10 knot increase in headwind on a 3o glideslope:
a) 40 feet per minute increase
b) 30 feet per minute increase
c) 50 feet per minute increase
d) 50 feet per minute decrease
c) marked correct whereas (d) is correct. Increase in headwind means that rate of descent has to be decreased in order to maintain the original glide path.
There is only one option that says "decrease" so thats the answer.
On a 3 degree glideslope ROD = Ground Speed x 5. e.g. at 100kts it will be 500fpm. If headwind increases by 10 kts then GS will be 90 and ROD will be 90x5 = 450 fpm. i.e. rate of descent must decrease by 50ft/min to maintain the required 3 degree glide slope.
(Refer to Jeppesen Student Manual - chart E(LO)1 or figure 061-11)You are at position 5340N 00840W. What is the QDR from the SHA VOR (5243N 00853W)?
b) marked correct, however the longitude for our position in the question is not given correct. According to other question banks it is 00800W and not 00840W.
Which of the following lists all the methods that can be used to enter Created Waypoints into the CDU of a B737-400 Electronic Flight Instrument System?
a) Identifier bearing/distance; place bearing/place bearing; latitude and longitude; waypoint name
b) Identifier bearing/distance; place bearing/place distance; along/across track displacement; latitude and longitude
c) Identifier bearing/distance; place distance/place distance; along track displacement; latitude and longitude
d) Identifier bearing/distance; place distance/place distance; along-track displacement; latitude and longitude
c) marked correct whereas (c) and (d) are identical. There is a typo here. (c) is actually:
"Identifier bearing/distance; place bearing/place bearing; along track displacement; latitude and longitude"
An aircraft travels from point A to point B, using the autopilot connected to the aircraft's inertial system. The co-ordinates of A (45oS 010oW) and B (45oS 030oW) have been entered. The true course of the aircraft on its arrival at B, to the nearest degree, is:
c) marked correct whereas (a) is correct.
Unique wording to identify this question: "An aircraft travels from point A to point B, using the autopilot connected to the aircraft's inertial system"
(Refer to figure 061-10) Which of the following beacons is 185 NM from AKRABERG (N6124 W00640)?
a) KIRKWALL (N5858 W00254)
b) STORNOWAY (N5815 W00617)
c) SUMBURGH (N5955 W00115)
d) SAXAVORD (N6050 W00050)
c) marked correct whereas (a) seems to be correct.
Doing it the plotting way by making a scale, Sumburgh comes out to be at 183.2nm from Akraberg (a diff of 1.8nm from 185) and Kirkwall 186.5nm (a diff of 1.5nm from 185) from Akraberg. More or less its the same but having one choice Kirkwall is close to 185.
Doing it the trigonometry way:
Akraberg to Kirkwall is 142 deg (ref true north). Reciprocal will be 322 and angle of the triangle becomes 360-322 = 38 deg
Change of latitude (adjacent of the triangle) betwen Akraberg and Kirkwall is 2.43 deg which is 146 nm.
Distance (Hyp) = 146 / Cos 38 = 185 nm.
Doing it with proper formulas, Sumburgh comes out to be at 182.7nm and Kirkwall 184.4nm from Akraberg which indicates that option (a) is more close to 185.
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