Total number of subscribers = 200

From the Venn diagram,
Number of members who subscribe only one magazine = a + b + c
Number of members who subscribe exactly two magazines = x + y + z
and 125 members subscribe atleast two magazines.
That is, x + y + z + r = 125         ..... (1)
Now, n(A U B U C) = 200, n(A) = 125, n(B) = 100, n(C) = 150, n(A∩B) = x + r
n(B∩C) = y + r, n(A∩C) = z + r, n(A∩B∩C) = r
We know that,
n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
200 = 125 + 100 + 150 – x – r – y – r – z – r + r
= 375 - (x + y + z + r) - r
= 375 -125    [∵ x + y + z + r = 125]
200 = 250 - r    ⇒ r = 50
From (1) x + y + z + 50 = 125
We get, x + y + z = 75
Therefore, number of members who subscribe exactly two magazines = 75.
From Venn diagram, (a + b + c) + (x +y + z + r) = 200 ........ (2)
substitute (1) in (2)
a + b + c + 125 = 200
a + b + c = 75
Therefore, number of members who subscribe only one magazine = 75.

A - Chess
B - Carrom
C - Table Tennis

n(A) = 22
n(B) = 21
n(C) = 15
n(A⋂C) = 10
n(B∩C) = 8
n(A∩B∩C) = 6
y = 22 - (x + 6 + 4) = 22 - (x + 10)
= 22 - x - 10 = 12 - x
z = 21 - (x + 6 + 2) = 21 - (8 + x)
= 21 - 8 - x = 13 - x
y + z + 3 + x + 2 + 4 + 6 = 35

x = 40 - 35 = 5
(i) Number of students who pay only chess and Carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 - x = 12 - 5 = 7
(iii) Number of students who play only carrom = 13 - x = 13 - 5 = 8

a = 600 - (120 -x + x + 80 -x)
= 600 - (200 -x)
= 600 - 200 + x = 400 + x
b = 350 - (120 -x + x + 100 -x)
= 350 - (220 - x)
= 350 - 230 + x = 130 + x
c = 280 - (80 - x + x + 100 - x)
2800 - (180 -x)
= 280 - 180 + x = 100 + x
Each farmer grew atleast one of the above three, the number of farmers who grew all the three is x.
= a + b + c + 120 - x + 100 - x + 80 - x + x = 1000
400 + x + 130 + x + 100 + x + 120 - x + 100 - x + 80 - x + x = 1000
ஃ 930 + x = 1000
x = 1000 - 930 = 70
ஃ 70 farmers grew all the three crops.

n(C) = \(800\times\frac{1}{2}=400\)

n(A∩B) = 70
n(B∩C) = 55
n(C∩A) = 60

(i) Number of people who use exactly two branded soaps
= 50 + 35 + 40 = 125
(ii) Number of people who use atleast one branded soaps
= 190 + 55 + 305 + 50 + 35 + 40 + 20 = 695
(iii) Number of people who do not use any of these brands
= 800 - [695] = 105

Tamil Newspaper buyers n(A) = 275
English Newspaper buyers n(B) = 150
Hindi Newspaper buyers n(C) = 45
n (A∪ B ∪ C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C)
Tamil and English Newspaper buyers n(A∩B) = 125
English and Hindi Newspaper buyers n(B∩C) = 17
Hindi and Tamil Newspaper buyers n(C∩A) = 5
All the three Newspaper buyers n(A∩B∩C) = 3

(i) Number of families buy only one newspaper = 148 + 11 + 26 = 185
(ii) Number of families buy atleast two news papers = 122 + 14 + 2 + 3 = 141
(iii) Total number of families in the colony = 148 + 11 + 26 + 122 + 14 + 2 + 3 = 326

\(n\left( A\cup B\cup C \right) =n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\)
(i) A = {a,c,e,f,h}, B = {c,d,e,f}, C = {a,b,c,f}
n(A) = 5, n(B) = 4, n(C) = 4
n(A⋂B) = 3
n(B⋂C) = 2
n(A∩C) = 3
n(A∩B∩C) = 2
A∩B = {c,e,f}
B∩C = {c,f}
A∩C = {a,c,f}
A∩B∩C = {c,f}
AUBUC = {a,c,d,e,f,b,h}
ஃ n(AUBUC) = 7............(1)
n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A⋂C)+n(A∩B∩C).........(2)
= 5 + 4 + 4 - 3 - 2 - 3 + 2 = 15 - 8 = 7
(1) = (2)
⇒ n(AUBUC) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A⋂C) + n(A∩B∩C)
Hence it is verified.
(ii) A = {1,3,5}, B = {2,3,5,6}, C = {1,5,6,7}
n(A) = 3, n(B) = 4, n(C) = 4
n(A∩B) = 2
n(B∩C) = 2
n(C⋂A) = 2
n(A⋂B⋂C) = 1
n(AUBUC) = 6
n(AUBUC) = n(A) + n(B) + n(C) - n(A⋂B) - n(B∩C) - n(A⋂C) + n(A∩B∩C)
6 = 3 + 4 + 4 - 2 - 2 - 2 + 1 = 12 - 6 = 6
Hence it is verified.

A = { y : y = \(\frac { a+1 }{ 2 } \), a\(\in \) W and a ≤ 5}
a = {0,1,2,3,4,5}
y = \(\frac{a+1}{2}=\frac{a+1}{2}=\frac{1}{2}\)
y = \(\frac{1+1}{2}=\frac{2}{2}=1\)
y = \(\frac{2+1}{2}=\frac{3}{2};y=\frac{3+1}{2}=\frac{4}{2}=2\)
y = \(\frac{4+1}{2}=\frac{5}{2};y=\frac{5+1}{2}=\frac{6}{2}=3\)
ஃ A = \(\{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3\}\)
B = { y : y = \(\frac{2n-1}{2}\), n∈W and n < 5}
n = {0,1,2,3,4}
y = \(\frac{2\times0-1}{2}=\frac{-1}{2}\)
y = \(\frac{2\times1-1}{2}=\frac{1}{2}; y=\frac{2\times2-1}{2}=\frac{3}{2}\)
y = \(\frac{2\times3-1}{2}=\frac{5}{2}; y=\frac{2\times4-1}{2}=\frac{7}{2}\)
ஃ B = \(\{-\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}\}\)
C = \(\{-1,-\frac{1}{2},1,\frac{3}{2},2\}\)
BUC = \(\{-1,-\frac{1}{2},\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},\frac{7}{2}\}\)
A-(BUC) = {3}...............(1)
A-B = {1,2,3}
A-C = \(\{\frac{1}{2},\frac{5}{2},3\}\)
(A-B)∩(A-C) = {3}...................(2)
From (1) and (2), it is verified that A-(BUC) = (A-B)∩(A-C)

A = {x : x = 6n, n ∈ W, n < 6}
x = 6n
n = {0,1,2,3,4,5}
x = 6 \(\times\) 0 = 0
x = 6 \(\times\) 1 = 6
x = 6 \(\times\) 2 =12
x = 6 \(\times\) 3 =18
x = 6 \(\times\) 4 = 24
x = 6 \(\times\) 5 = 30
ஃ A = {0,6,12,18,24,30}
B = {x : x = 2n, n∈N, 2n = {3,4,5,6,7,8,9}
x = 2n
x = 2 \(\times \) 3 = 6;
2\(\times\)4 = 8;
2 \(\times\)​​​​5 = 10 
2 \(\times\)6 = 12 
2\(\times\) 7 = 14 
2 \(\times\)8 = 16 
2 \(\times\)9 = 18
ஃ B = {6,8,10,12,14,16,18}
C = {x : x = 3n, n∈N, 4 ≤ n < 10
N = {4,5,6,7,8,9}
x = 3\(\times\)4 = 12 
x = 3 \(\times\) 5 = 15 
x = 3 \(\times\) 6 = 18
x = 3 \(\times\) 7 = 21 
x = 3 \(\times\) 8 = 24 
x = 3 \(\times\) 9 = 27
x = 2 \(\times\) 9 = 18
ஃ C = {12,15,18,21,24,27}
A-(B⋂C) = (A-B)U(A-C)
L.H.S = R.H.S
B⋂C = {12,18}
A-(B⋂C) = {0,6,12,18,24,30} - {12,18}
= {0,6,24,30} .........(1)
(A-B) = {0,24,30}
(A-C) = {0,6,30}
(A-B)U(A-C) = {0,6,24,30} .........(2)
From (1) and (2), it is verified that
A-(B⋂C) = (A-B)U(A-C).

(i) A-B = {3,4,6}
(ii) B-C = {-1,5,7}
(iii) A'UB'
A'UB'
A' = {1,2,0,-3,5,7,8}
B' = {-3,0,1,2,3,4,5,6}
A'UB' = {-3,0,1,2,3,4,5,6,7,8}
(iv) A'⋂B'
A'⋂B' = {-3,0,1,2}
(v) (BUC)'
BUC = {-3,-2,-1,0,3,5,7,8}
(BUC)' = U-(BUC)
= {-3,-2,-1,0,1,2,3,4,5,6,7,8} - {-3,-2,-1,0,3,5,7,8}
(BUC)' = {1,2,4,6}
(vi) A-(BUC) = {-2,-1,3,4,6} - {-3,-2,-1,0,3,5,7,8} = {4,6}
(vii) A-(B⋂C)
B⋂C = {-2,8}
A-(B⋂C) = {-2,-1,3,4,6} - {-2,8} = {-1,3,4,6}