The first point to note is that your Annexes are not up to the current amendment state as used by the UK CAA. But they are the ones that were used for these questions when first used.
The second point tio note is that in many graphical questions you will not get exactly the correct answer. But you should be close enough to select the correct option.
Q247) (For this question use annex 032-6569A or Performance
Manual SEP 1 Figure 2.4)
With regard to the landing chart for the single engine
aeroplane determine the landing distance from a height
of 50 ft .
Given :
O.A.T : 27 °C
Pressure Altitude: 3000 ft
Aeroplane Mass: 2900 lbs
Tailwind component: 5 kt
Flaps: Landing position (down)
Runway: Tarred and Dry
A) approximately : 1120 feet
B) approximately : 1700 feet
C) approximately : 1370 feet
D) approximately : 1850 feet (mark scheme answer)
Using the current version of this graph I get 1790 ft. this is closest to option D 1859 ft.
Q218) (For this question use annex 032-6581A or Performance
Manual SEP 1 Figure 2.3)
Using the climb performance chart, for the single engine
aeroplane, determine the ground distance to reach a
height of 2000 ft above the reference zero inthe
following conditions:
Given :
O.A.T. at take-off: 25°C
Airport pressure altitude: 1000 ft
Aeroplane mass: 3600 lbs
Speed: 100 KIAS
Wind component: 15 kts Headwind
A) 24 637 ft
B) 18 832 ft
C) 18 347 ft (mark scheme answer)
D) 21 505 ft
Putting the data into the graph gives ROC = 1130 fpm
Height gain to get to 2000ft above RZ = 2000 - 50(screen height) = 1950 ft
Dividing height gain by ROC gives a climb time of 1.725663717 minutes
Putting the mean altitude (2000 ft and mean OAT 23 deg C) into the CRP 5 gives TAS = 105 kts
Ground distance = time x ground speed
If we assume nil wind then ground speed is approximately equal to TAS = 105 knots giving distance = 105 kts 101.3 ft/min/knot x 1.725663717 minutes = 18355 ft.
Using the 15 kt headwind unfactored (ground speed = 105 - 15 = 90 knots)
This gives distance = 90 x 101.3 x 1.725663717 = 15732 ft
Note, This is how the CAP 698 state you should do it for the SEP1.
Uisng 50% of the 15 knot headwind (ground speed = 105 - 7.5 = 97.5)
This gives 97.5 x 101.3x 1.725663717 = 17044 ft
All of these answers are lower than the lowest option (which is the correct option). So you can identify 18347 as the correct option.
Q232) (For this question use annex 032-4733A or Performance
Manual MRJT 1 Figure 4.28)
What is the minimum field length required for the worst
wind situation, landing a twin jet aeroplane with the antiskid
inoperative?
Elevation: 2000 ft
QNH: 1013 hPa
Landing mass: 50 000 kg
Flaps: as required for minimum landing distance
Runway condition: dry
Wind:
Maximum allowable tailwind: 15 kt
Maximum allowable headwind: 50 kt
A) 2600 m.
B) 2700 m.
C) 2900 m.
D) 3100 m. (mark scheme answer)
The worst case wind is 15 knot tailwind.
The least distance will be with flap 40
Go into the graph at 50000 Kg mass and move horizontally to the 1000 ft pressure altitude in the top right set of lines (Anti-skid inop)
Then go vertially down through the flpa sacles (bacuse you are using flap 40) and straight through surface condition (because runway is dry) down to 15 kt tailwind.
Floow the sloping lines up to the right to the reference line then go vertically down to come out at 3100 m.