Do common sense has anything to do with this question? Because
corporation might forget the arbitrary 12 airplane rule
and have 11 times 20 = 220.
But perhaps they get a discount if all 12 hangars are used!
Playing by the rules of the question:-
Let a, b, c be the aeroplanes that carry
10, 15, and 20 passengers, respectively,
a + b + c = 12
10a +15b +20c = 220
We may rearrange these as
4a + 4b + 4c = 48
2a + 3b + 4c = 44
Now subtract
2a + b = 4
There appear to be 3 options
a = 0, b = 4 which means 220 = 0*10 + 4*15 + 8*20
a = 1, b = 2 which means 220 = 1*10 + 2*15 + 9*20
a = 2, b = 0 which means 220 = 2*10 + 0*15 + 7*20
If you dismiss the ones with zeros it may be that
the expected answer to "How many of each plane" is
One that carries 10, 2 that carries 15, and 9 that carries 20
