n(A - B) = 32 + x, n(B A) = 5x
n(A \(\cap\) B) = x
From the Venn diagram:
Given n(A) = n(B)
32 + x + x = x + 5x
32 + 2x = 6x
32 = 6x - 2x
32 = 4x
x = \({32\over 4}=8\)
The value of x = 8.

Let M and S represent the student failed in Mathematics and Science.
Given: Number of students passed in Mathematics is 50%
∴ Number of students failed in Mathematics = 100 - 50% = 50%
n(M) = 50%

Number of students passed in Science is 70%
∴ Number of students failed in Science = 100 - 70% - 30%
n(S) = 30%
Number of students failed in both the subjects is 10%
n(M\(\cap\)S) = 10%
n(M\(\cup\)S) = n(M) + n(S) - n(M\(\cap\)S)
= 50 + 30 - 10
= 80 - 10 = 70
Given: 70% of the students failed in atleast anyone of the subject
∴ 30% of the students passed in atleast anyone of the subjects.
30 students passed mean, the total number of students is 100.
∴ 300 students passed means, the total number of students = \(\frac{100 \times 30}{30}\)= 1000
Total number of students appeared in the examination is = 1000

Let the people who like tea be T. 
Let the people who like coffee be C 
By using formula 
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10 
The number of people who like both coffee and tea = 10.
(ii) The number of people who do not like tea 
n(T) = n(U) – n(T) = 45 – 35 = 10
(iii) The number of people who do not like coffee 
n(C) = n(U) – n(C) = 45 – 20 = 25.

Let T be the set of families buy Tamil news Papers
Let E be the set of families buy English news Papers
n(TUE) = 100
n(T) = 65 and n(E) = 55
Let "X" be the number families buy both newspapers.
By using venn-diagram
From the venn-diagrarn we get
65 - x + x + 55 - x = 100
120 - x = 100
120 -100 = x
20 = x
(i) Families buy both newspapers = 20
(ii) Families buy Tamil newspapers only = 65 - x
= 65 - 20
= 45
(iii) Families buy English newspapers only = 55 - x
= 55 - 20
= 35.

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.