Help with a tricky brainteaser question

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III.
If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?
Choose one answer.

A. 5

B. 12

C. 17

D. 22
Show you working out and understanding

P( I ∪ II ∪ III) = P(I) + P(II) + P(III) - P(I ∩ II) - P(I ∩ III) - P(II ∩ III) + P(! ∩ II ∩ III) = 78
50 + 30 + 20 - P(I ∩ II) - P(I ∩ III) - P(II ∩ III) + 5 = 78
P(I ∩ II) + P(I ∩ III) + P(II ∩ III) = 27

The percent that vote for all three proposals are included 3 times in this total (once for each probability in the sum).

27 - 3*5 = 12. 12 is the percent of people that favoured exactly 2 proposals.

The percent of people that favoured more than one proposal is 12 + 5 = 17 (favoured 2 or 3 proposals)

The answer is C. 17

This question is actually not that hard. All you really needed was what percent favoured only one of the proposals. In this case it is 78%. Subtract that from 100% and you get 22%. That is your answer. The proposals I, II, III weren't needed. The question asks what percent favoured more than one, and that includes the given 5% because three is greater than one.
The answer is D.

P(I) + P(II) + P(III) - P(2 of the 3) + P(all3) = 78
50+30+20 - P(2 of the 3) + 5 = 78
P(2 of the 3) = 27
P(more than 1) = 27-5 = 22% [D] <---------

5% of those asked favoured all three of the proposals