A = {-2,0,1,3,5}, B = {-1,0,2,5,6}
C = {-1,2,5,6,7}
BUC = {-1,0,2,5,6,7}
A-(BUC) = {-2,1,3}.............(1)
(A-B) = {-2,1,3}
(A-C) = {-2,0,1,3}
(A-B)∩(A-C) = {-2,1,3}............(2)
From (1) and (2), it is verified that
A-(BUC) = (A-B)⋂(A-C)

Let A, B and C represent sets of students who speak Tamil, English and French respectively.
Given, n(AUBUC) = 100, n(A) = 85, n(B) = 40, n(C) = 20
n(A⋂B) = 32, n(B⋂C) = 13, n(A⋂C) = 10
We know that,
n(A U B U C) = n(A) + n(B) + n(C) - n(A⋂B) - n(B⋂C) - n(A⋂C) + n(A⋂B⋂C)
100 = 85 + 40 +20+ -32 -13-10 + n(A⋂B⋂C)
Then, n(A⋂B⋂C) = 100 - 90 = 10
Therefore, 10 students speak all the three languages

Let C, F and H represent sets of students who play Cricket, Football and Hockey respectively.
Then , n(C) = 240, n(F) = 180, n(H) = 164, n(C ∩ F) = 42,
n(F ∩ H) = 38, n(C ∩ H) = 40, n(C ∩ F ∩ H) = 16.
Let us represent the given data in a Venn diagram.
(i) The number of students in the college = 174 + 26 + 116 + 22 + 102 + 24 + 16 = 480
(ii) The number of students who play only one game = 174 + 116 + 102 = 392

 

The roster form of sets P, Q and R are
P = {1,2, 3, 4,5,6,7,8,9}, Q = {1, 3,5,7,9} and R = {2, 3, 5,7,11,13}
First, we find (Q ⋂ R) = {3, 5, 7}
Then, P −(Q ⋂ R) = {1, 2, 4 ,6, 8, 9}........ (1)
Next, P–Q = {2, 4, 6, 8} and P–R = {1, 4, 6, 8, 9}
and so, (P −Q) U (P −R) = {1,2, 4, 6, 8, 9}............ (2)
Hence from (1) and (2), it is verified that P −(Q ∩ R) = (P −Q) U (P −R).
Finding the elements of set Q
Given, x = 2n + 1
n = 0 ⟶ x = 2(0) + 1 = 0 + 1 = 1
n = 1 ⟶ x = 2(1) + 1 = 2 + 1 = 3
n = 2 ⟶ x = 2(2) + 1 = 4 + 1 = 5
n = 3 ⟶ x = 2(3) + 1 = 6 + 1 = 7
n = 4 ⟶ x = 2(4) + 1 = 8 + 1 = 9
Therefore, x takes values such as 1, 3, 5, 7 and 9.

n(U) = 125
y = 2x
z = x + 10
ஃ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 - 45
4x = 80
x = 20
ஃ y = 2x = 2 \(\times\) 20 = 40
z = x +10 = 20 + 10 = 30
Hence x = 20; y = 40; z = 30

(2)=(5)
ஃ (A∩B)'=A'UB'

U = {4,7,8,10,11,12,15,16}
A = {7,8,11,12,}, B = {4,8,12,15}
De Morgan's Laws for complementation.
(AUB)' = A' ∩ B'
AUB = {4, 7, 8, 11, 12, 15}
(AUB)' = {4, 7, 8,10,11,12,15, 16} - {4, 7, 8,11,12, 15} = {10, 16}...(1)
A' = {4, 10, 15,16}; B' = {7, 10, 11,16}
A'∩B' = {10,16}...(2)
From (1) and (2) it is verified that
(AUB)' = A'∩B'

ஃ A-(B∩C)=(A-B)U(A-C)
Hence it is proved.

K = {a,b,d,e,f}, L = {b,c,d,g} and M = {a,b,c,d,h}
(i) K\(\cup\)(L\(\cap \)M)
L\(\cap \) M = {b,c,d,g} \(\cap \) {a,b,c,d,h} = {b,c,d}
K\(\cup\)(L\(\cap \)M) = {a,b,d,e,f} \(\cup\) {b,c,d} = {a,b,c,d,e,f}
(ii) K∩(L\(\cup\)M)
L\(\cup\)M = {a,b,c,d,g,h}
K⋂(L \(\cup\) M) = {a,b,d,e,f) \(\cap \) {a,b,c,d,g,h} = {a,b,d}
(iii) (K \(\cup\) L) \(\cap \) (K\(\cup\)M)
K \(\cup\) L = {a,b,c,d,e,f,g}
K \(\cup\) M = {a,b,c,d,e,f,h}
(K \(\cup\) L) \(\cap \) (K \(\cup\) M) = {a,b,c,d,e,f}
(iv) (K \(\cap \) L) \(\cup\) (K \(\cap \) M)
(K \(\cap \) L) = {b,d}
(K \(\cap \) M) = {a,b,d}
(K \(\cap \) L) \(\cup\) (K \(\cap \) M) = {b,d} \(\cup\) {a,b,d} = {a,b,d}

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.