Latest Elitmus Aptitude Question SOLUTION: 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 o

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.

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 .

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

please explain it clearly..

try using the fo0rmula,,p(AUBUC)..thus you ll get the ans as 51%

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.

draw ven diagram..and construct 4 diff eqn..and slv it..