X\(\cup \)Y = {5, 6, 7} \(\cup \) {5, 7, 9, 10}
= { 5, 6, 7, 9, 10}
X\(\cap\)Y = {5, 6, 7} \(\cap\) {5, 7, 9, 10}
= {5, 7}
X\(\Delta\)Y = (X\(\cup \)Y) \(\cup \) (X\(\cap\)Y)
= {5, 6, 7, 9, 10} - {5, 7}
= {6, 9, 10}
X-Y = {5, 6, 7} - {5, 7, 9, 10}
= {6}
Y-X = {5, 7, 9, 10} - {5, 6, 7}
= {9, 10}
X\(\Delta\)Y = (X-Y)\(\cup \) (Y-X)
= {6} \(\cup \) {9, 10}
= { 6, 9, 10}

Let M be the set of all students take part in music.
Let D be the set of all students take part in drama.
n( M) = 25, n(D) = 30 and n(M\(\cap \)D) = 8
By using venn diagram
From the venn diagram we get.
(i) Number of students take part in only music = 17
(ii) Number of students take part in only drama = 22
(iii) Total number of students in the class = 17 + 8 + 22 = 47

Alter :
(i) The number of students who take part in only music
= n(M) - n(M\(\cap \)D)
= 25 - 8 = 17
(ii) The number of students who take part in only drama
= n(D) - n(M\(\cap \)D)
= 30 - 8 = 22
(iii) The total number of students
n(M∪D) = n(M) + n(D) - n(M\(\cap \)D)
= 25 + 30 - 8 = 47

A′

Given A = {1, 2, 6}, B = {2, 3, 4}
A∪B = {1,2,3, 4,6}.

PRINCIPAL
B = {P, R, I, N, C, A, L}

Let R be the set of children forgotten their rulers.
Let P be the set of children forgotten their pencil.
n(R) = 20, n(P) = 17, n(R\(\cap \)P) = 5
By using venn-diagram

From the venn -diagram we get
(i) Number of children forgotten only pencil = 12
(ii) Number of children forgotten only ruler 15
(iii) Total number of children in the class 15 + 5 + 12 = 32

Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M U S) = 40, n (M) = 26, n (S) = 24
n (M U S) = n (M) + n (S) - n (M n S)
40 = 26 + 24 - n (M n S)
n (M n S) = 26 + 24 - 40
= 50 - 40 = 10
∴ Number of students opted for Mathematics and Science = 10

(i) A = {1,2,5,6,7}
(ii) B = {3,4,5,6}
(iii) A ∩ B = {5,6}
(iv) A U B = {1,2,3,4,5,6,7}
(v) A-B = {1,2,7}
(vi) B - A = {3,4}
(vii) (A-B) ∩ (B -A) = { }.

Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A \(\cap\) B) = 50
n(A \(\cup\) B) = 500 .........(1)
n(A) + n(B) - n(A \(\cap\) B) = 400 + 200 - 50
= 600 - 50
= 550  ...........(2)
From (1) and (2) we get
\(n(A\cup B)\ne n(A) +n(B)-n(A\cap B)\)
\(\therefore\) The given data is incorrect.

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.