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

From (1) and (2), it is verified that \(\left( A\cup B \right) '\)=\(A'\cap B'\)

From (1) and (2), we get A -(BUC) = (A-B)∩(A-C). Hence it is verified.

A = {x : x ∈ Z, -2 < x ≤ 4} = {-1,0,1,2,3,4}
B = {x : x ∈ W, x ≤ 5} = {0,1,2,3,4,5}
C = {-4,-1,0,2,3,4}
AU(B⋂C)
B⋂C = {0,1,2,3,4,5} ∩ {-4,-1,0,2,3, 4} = {0,2, 3, 4}
AU(B∩C)
B⋂C = {0,1,2,3,4,5} ∩ {-4,-1,0,2,3,4} = {0,2,3,4}
AU(B⋂C) = {-1,0,1,2,3,4} U (0,2,3,4}
= {-1,0,1,2,3,4}........(1)
(A∩B)U(A∩C)
A⋂B = {0,1,2,3,4}
A⋂C = {-1,0,2,3,4}
(A⋂B)⋃(A⋂C) = {0,1,2,3,4} U {-1,0,2,3,4}
= {-1,0,1,2,3,4}..........(2)
From (1) and (2), it is verified that
AU(B∩C) = (AUB)∩(A⋃C)

Let S, C and B represent sets of families who owned Scooter, Car and Bicycle respectively.
Given, n(U)=600 n(S) =\(\frac{3}{5}\) \(\times\) 600 = 360
n(C) = \(\frac{1}{3}\) \(\times\) 600 = 200 n(B)=\(\frac{1}{4}\) \(\times\) 600 = 150
n(S⋂C⋂B)=\(\frac{2}{15}\) \(\times\) 600 = 80
From Venn diagram,
(i) The number of families owned atleast two types of vehicles = 40 + 6 + 10 + 80 = 136
(ii) The number of families owned no vehicle
= 600 – (owned atleast one vehicle)
= 600- (230 + 40 + 74 + 6 + 54 + 10 + 80)
= 600 - 494 = 106

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

A = {b,c,e,g,h}
B = {a,c,d,g,i}
C = {a,d,e,g,h}
B ∩ C = {a,d,g}
A - (B ∩ C) = {b,c,e,g,h} - {a,d,g} = {b,c,e,h}...(1)
A-B = {b,c,e,g,h} - {a,c,d,g,i} = {b,e,h}
A-C = {b,c,e,g,h} - {a,d,e,g,h} = {b,c}
(A - B)⋃(A - C) = {b,c,e,h}...(2)
From (1) and (2) it is verified that
A-(B ⋂ C) = (A - B) U (A - C)

(i) 
(ii) 
(iii) 
(iv) 

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}