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? [1999] plzz explain

1) 10
2) 17
3) 12
4) 22

• Total=I+II+III-{favored exactly 2}- 2*{favored exactly 3}+{favored neither}
  
  As {those who favored neither of proposal}=100-78=22 then: 100=50+30+20-{favored exactly 2 proposals}-2*5+22 --> {favored exactly 2 proposals}=12.
  
  Now, those who favored more than one of the 3 proposals equals to {favored exactly 2 proposals}+{favored exactly 3 proposals}=12+5=17
• 10 years agoHelpfull: Yes(1) No(2)
• The below solution is correct.Use vane diagrams to solve.(Favored exactly 3) *2 is subtracted because it comes thrice during addition of the intersections where as it should be added only once within a 100%.
• 9 years agoHelpfull: Yes(0) No(0)
• Let A=favoured proposal III
  
  B=favoured proposal III
  
  C=favoured proposal III
  
  Here, n(A U B U C)=78
  
  We Knows,
  
  n(AUBUC)= n(A)+n(B)+n(C)-n(A ∩ B)-n(A ∩ C)-n(A ∩ B)+n(A ∩ B ∩ C)
  
  78 = 50 + 30 + 20 - n(A ∩ B) - n(A ∩ C) - n(A ∩ B) + 5.
  
  n(A ∩ B) + n(A ∩ C) + n(A ∩ B) = 27
  
  
  
  For those who favored exactly 2 proposals we have to subtract n(A ∩ B ∩ C) from n(A ∩ B) , n(A ∩ C) and n(A ∩ B)
  
  n(A ∩ B) + n(A ∩ C) + n(A ∩ B) - 3*n(A ∩ B ∩ C)
  
  =27-15 =12
  
  
  
  For those who favored exactly 3 proposals=5
  
  Now, those who favored more than one of the 3 proposals = 12 + 5 =17
  
  
  
  Hence,Option (C)17 is the correct choice.
• 7 years agoHelpfull: Yes(0) No(0)