The following temperatures have been observed over a station at 12:00 UTC. Assume the station is at MSL.
Height  Temperature  
20 000 ft  12  °C  
18 000 ft  11  °C  
16 000 ft  10  °C  
14 000 ft  10  °C  
12 000 ft  6  °C  
10 000 ft  2  °C  
8 000 ft  +2  °C  
6 000 ft  +6  °C  
4 000 ft  +12  °C  
2 000 ft  +15  °C  
surface  +15  °C 

A
The layer between 16 000 and 18 000 ft is absolutely unstable.

B
Assuming that the MSL pressure is 1 013.25 hPa the true altitude of an aircraft would actually be higher than the indicated altitude.

C
The temperature at 10 000 ft is in agreement with the temperature in the International Standard Atmosphere.

D
The height of the freezing level over the station is approximately 12 000 ft.
First step: Calculate the ISA temperature for each of the heights given.
20 000 ft  15 – 2 x 20 = 25ºC 
18 000 ft  15 – 2 x 18 = 21ºC 
16 000 ft  15 – 2 x 16 = 17ºC 
14 000 ft  15 – 2 x 14 = 13ºC 
12 000 ft  15 – 2 x 12 = 9ºC 
10 000 ft  15 – 2 x 10 = 5ºC 
8 000 ft  15 – 2 x 8 = 1ºC 
6 000 ft  15 – 2 x 6 = 3ºC 
4 000 ft  15 – 2 x 4 = 7º C 
2 000 ft  15 – 2 x 2 = 11ºC 
Surface  15ºC 
 Looking at the question text, we conclude that the actual temperature is warmer than ISA. And we know that, disregarding pressure, for temperatures WARMER THAN ISA; TRUE ALTITUDE > INDICATED ALTITUDE
“The temperature at 10 000 ft is in agreement with the temperature in the International Standard Atmosphere.” Incorrect – The actual temperature at 10 000 ft is 2ºC; ISA temperature at that same level is 5ºC. ISA +3ºC
“The height of the freezing level over the station is approximately 12 000ft.” incorrect – The freezing level is approximately at 9 000 ft. (8 000 ft = 2ºC, therefore 0ºC will be at 9 000 ft)
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