in an opinion poll,78% of those asked were in favour of at least one of the three participants A,B,C.50% OF THOSE ASKED favoured A.30% favoured B.20%c.If 5% of those favoured all of the three,what percentage favoured exactly one?

• 61%
  
  5% were in favour of all
  Let x% be in favour of A and B
  y% be in favour of B and C
  z % be in favour of A and C
  
  => 78 - 5 - (x + y + z) = (50-5-x-z) + (30-5-x-y) + (20-5-y-z)
  => (x + y + z) = 12%
  
  => (78 - (x+y+z)-5) = 61% favoured exactly one.
• 12 years agoHelpfull: Yes(47) No(11)
• In the above question, it is mentioned that 78% of asked favored at least one of the three....so it implies
  (A U B U C) = 78%
  (A ∩ B) ----> This implies, in favor of both A and B.
  (B ∩ C) ----> This implies, in favor of both B and C.
  (A ∩ C) ----> This implies, in favor of both A and C.
  (A ∩ B ∩ C) ----> This implies, in favor of all three A, B and C.
  let us assume (A ∩ B) = x%
  (A ∩ C) = y%
  (B ∩ C) = z%
  it is given (A ∩ B ∩ C) = 5%
  Now,
  
  50% were in favor of A, it means
  50% = only A + (A ∩ B) + (A ∩ C) - (A ∩ B ∩ C) [ subtracting (A ∩ B ∩ C) because it has been counted in
  (A ∩ B) and (A ∩ C) twice so I subtracted it once from
  the result ]
  so, (only A) + x + y - 5 = 50
  (only A) = 55 - x - y
  Similarly,
  30% were in favor of B, it means
  30% = only B + (A ∩ B) + (B ∩ C) - (A ∩ B ∩ C)
  so, (only B) + x + z - 5 = 50
  (only B) = 35 - x - z
  Similarly,
  20% were in favor of C, it means
  20% = only C + (A ∩ C) + (B ∩ C) - (A ∩ B ∩ C)
  so, (only C) + y + z - 5 = 20
  (only C) = 25 - x - z
  
  now we know
  (A U B U C) = (A) + (B) + (C) - (A ∩ B) - (A ∩ C) - (B ∩ C) + (A ∩ B ∩ C)
  78 = 50 + 30 + 20 - x - y - z + 5
  x + y + z = 105 - 78
  x + y + z = 27
  
  now % favor exactly one
  (only A) + (only B) + (only C) = 55 - x - y + 35 - x - z + 25 - y - z
  = 115 - 2( x + y + z ) --------------------------- [ x + y + z = 27 ]
  = 115 - 2 * 27
  = 115 - 54
  = 61
  so, the answer is 61% of asked people favoured exactly one.
  I hope I tried my best to make you understand this .
• 7 years agoHelpfull: Yes(8) No(0)
• 61
  a=only A
  b=only B
  c=only C
  d=A&B
  e=only B&C
  f=only A&C
  g=only A&B&C
  h=not(opinion)
  a+b+c+2*(d+e+f)+3g+n=122
  a+d+f+g=50
  b+d+e+g=30
  c+f+g+e=20
  g=5
  n=22
  now,
  a+b+c+2(d+e+f)=85
  we know,
  a+b+c+d+e+f+g=78
  put g=5
  we get
  a+b+c+d+e+f=73
  from here,
  (a+b+c+2(d+e+f))-(a+b+c+d+e+f)=d+e+f=85-73=12
  2*(d+e+f)=24
  a+b+c=85-(2*(d+e+f))=85-24=61
  exactly one=a+b+c=61
• 11 years agoHelpfull: Yes(2) No(4)
• please explain it clearly..
• 10 years agoHelpfull: Yes(2) No(0)
• try using the fo0rmula,,p(AUBUC)..thus you ll get the ans as 51%
• 9 years agoHelpfull: Yes(1) No(6)
• 51%
  
  5% were in favour of all
  Let x% be in favour of A and B
  y% be in favour of B and C
  z % be in favour of A and C
  
  => 78 + 5 - (x + y + z) = (50+5-x-z) + (30+5-x-y) + (20+5-y-z)
  => (x + y + z) = 32%
  
  => (78 - (x+y+z)+5) = 51% favoured exactly one.
  
• 9 years agoHelpfull: Yes(1) No(2)
• draw ven diagram..and construct 4 diff eqn..and slv it..
• 9 years agoHelpfull: Yes(0) No(0)