12
SETS AND VENN
DIAGRAMS
Section I : Sets
1. Describe the following sets in roster form :
(i) {x / x = n 2 , n ∈ N , 2 ≤ n ≤ 5}
(ii) {x / x is composite number and 11 < x < 25}
(iii) {x / x ∈ W, x is divisible by 4 and 6, x ≤ 100}
(iv) {x / x is two digit number whose sum of digits is 7}
n


, n ∈ N and n ≤ 5
(v)  x / x =
n+3


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2n + 1


(vi)  x / x =
, n ∈ W and n ≤ 10 
2n + 3


(vii) {x / x ∈ W , x 2 < 20}
(viii)
{x / x = 5 p, P ∈ I and x 2 < 400}
1


(ix)  x / x = , n ∈ N and n ≤ 5
n


(x) {x / 3x – 5 < 15, x ∈ W }
(xi) {First four planets of our solar system}.
Ans. (i) {4, 9, 16, 25}
(ii) {12, 14, 15, 16, 18, 20, 21, 22, 24}
(iii) {0, 12, 24, 36, 48, 60, 72, 84, 96}
(iv) {16, 25, 34, 43, 52, 61, 70}
1 2 3 4 5
(v)  , , , , 
4 5 6 7 8
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21 
1 3 5 7
(vi)  , , , , ⋅⋅⋅ 
23 
3 5 7 9
(vii) {– 4, – 3, – 2, – 1, 0, 1, 2, 3, 4}
(viii) {– 15, – 10, – 5, 0, 5, 10, 15}
 1 1 1 1
(ix) 1, , , , 
 2 3 4 5
(x) {0, 1, 2, 3, 4, 5, 6}
(xi) {Mercury, Venus, Earth, Mars}
2. Write the following sets in the builder form :
(i) {11, 13, 17, 19, 23, 29, 31, 37}
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1
 1 1
(ii) 1, , , ⋅⋅⋅, 
9
 2 3
(iii) {21, 23, 25, 27, 29, 31, 33, 35, 37}
1 3 5 7

(iv)  , , , , ⋅⋅⋅
3 5 7 9

(v) {– 10, – 5, 0, 5, 10, 15, …, 100}
(vi) {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.
Ans. (i) {x / x is a prime number, 10 < x < 40}.
1


(ii)  x / x = , n ∈ N and n < 10 
n


(iii) {x / x = 2n – 1, n ∈ N, 11 < n < 19}
n


, n is odd natural number 
(iv)  x / x =
n+2


2n + 1


, n ∈W 
or
x / x =
2n + 3


(v) {x / x = 5 p, p ∈ I and – 2 ≤ p ≤ 20}
(vi) {x / x ∈ N and x is a factor of 48}.
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3. Find, which of the following sets are singleton set :
(i) The set of points of intersection of two non-parallel straight
lines on the same plane.
(ii) A = {x : 7x – 3 = 11}
(iii) B = {y : 2y + 1 < 3 and y ∈ W}
Ans.(i) The set of points of intersection of two non-parallel straight
lines on the same plane is a singleton set.
(ii) A = {x : 7x – 3 = 11}
7x – 3 = 11
7x = 11 + 3
⇒
14
=2
7x
=
14
x
=
⇒
⇒
7
A = {2}
∴
Hence, the given set A has only one element, so it is a singleton set.
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(iii) B = {y : 2y + 1< 3 and y ∈ W }
2y + 1 < 3
2y + 1 – 1 < 3 – 1 (Subtracting 1 from both sides)
⇒
2y < 2
⇒
2
y<
(Dividing both sides by 2)
⇒
2
y<1
⇒
B = {0}
∴
Hence, it is a singleton set.
4. State whether the following pairs of sets are equivalent or not :
(i) A = {x : x ∈ N and 11 ≥ 2x –1} and
B = {y : y ∈ W and 3 ≤ y ≤ 9}
(ii) Set of whole numbers and set of multiples of 3.
(iii) P = {5, 6, 7, 8} and M = {x : x ∈ W and x ≤ 4}
Ans. (i) A = {x : x ∈ N and 11 ≥ 2x – 1}
11 ≥ 2x – 1
11 + 1 ≥ 2x – 1 + 1
( Adding 1 to both sides)
⇒
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12
≥x ⇒ 6 ≥x
2
A = {1, 2, 3, 4, 5, 6}
∴
n(A) = 6
∴
B = {y : y ∈ W and 3 ≤ y ≤ 9}
3 ≤ y≤9
∴
B = {3, 4, 5, 6, 7, 8, 9}
∴
n (B) = 7
∴
∴ Cardinal number of set A = 6 and cardinal number of
set B = 7
Hence, set A and set B are not equivalent.
(ii) Set of whole numbers and set of multiples of 3 are equivalent
because both these sets have infinite number of elements.
(iii)
P = {5, 6, 7, 8}
n (P) = 4
M = {x : x ∈ W and x ≤ 4}
M = {0, 1, 2, 3, 4}
n(M) = 5
Cardinal number of set P = 4
and Cardinal number of set M = 5. Hence, these sets are not
equivalent.
5. (A) Let A = {all quadrilaterals}, B = {all rectangles}, C = {all
squares} and D = {all rhombuses} in a plane. State, whether the
following statements are true or false.
(i) B ⊂ C ⊂ A
(ii) C ⊂ B ⊂ A
⇒ 12 ≥ 2x ⇒
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(iii) C ⊂ D ⊂ A
(v) A ⊇ B ⊇ C
(iv) D ⊂ C ⊂ A
(vi) A ⊆ B ⊆ C
Ans.(i) False (ii) True (iii) True (iv) False (v) True (vi) False
(B) Let A = {all triangles}, B = {all isosceles triangles} and C = {all
equilateral triangles}. State whether the following statements are
true or false.
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(i) B ⊂ C ⊂ A
(ii) C ⊂ B ⊂ A
Ans. (i) False
(ii) True
6. Let A be the set of letters in the word, ‘seed’. Find:
(i) A
(ii) n (A)
(iii) Number of subsets of A
(iv) Number of proper subsets
Ans. (i) A = {s, e, d}
(ii) n (A) = number of elements = 3
(iii) Number of subsets of A = 23 = 8
{ }, {s}, {e}, {d}, {s, e}, {s, d}, {e, d} {s, e, d}.
(iv) Number of proper subsets of A
= 2n – 1= 23 – 1 = 8 – 1= 7
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because { } or φ is not a proper subset.
7. (A) Find the power set of each of the following sets :
(i) A = {0, 5}
(ii) B = {7, 9}
(iii) C = {2, 4, 6}
Ans. (i) P (A) = { φ , {0}, {5}, {0, 5}}.
(ii) P (B) = { φ , {7}, {9}, {7, 9}}.
(iii) P (C) = { φ , {2}, {4}, {6}, {2, 4}, {2, 6}, {4, 6}, {2, 4, 6}}.
(B) Let A = {1, {2}}. Find the power set of A.
Ans. (i) P (A) = { φ , {1}, {{2}}, {1, {2}}.
8. Let
A = { x : x 2 ∈ ξ},
ξ = {x : x ∈ N , x < 50},
B = {x : x = n 2 ,
n ∈ N } and C = {x : x is a factor of 36}. List the elements of each
of the sets A, B and C. Also, state whether each of the following
statements is true or false :
(ii) A = B
(iii) A ↔ B
(i) A ⊆ B
(iv) B ↔ C
(v) n (A) < n (C)
Ans.
ξ = {x : x ∈ N , x < 50} = {1, 2, 3, 4, 5, …, 49, 50}
2
A = {x : x ∈ξ} = {1, 2, 3, 4, 5, 6, 7}
Math Class VIII
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B = {x : x = n2, n ∈ N} = {1, 4, 9, 16, 25, 36, 49}
C = {x : x is a factor of 36}
= {1, 2, 3, 4, 6, 9, 12, 18, 36}
(i) False
(ii) False
(iii) True
(iv) False
(v) True
9. Let A = {letters of BOMBAY} and B = {letters of CALCUTTA}
(i) Are these sets disjoint or overlapping ?
(ii) Are these sets equal ?
(iii) Are these sets equivalent ?
(iv) Is any of these sets, subset of the other ?
(v) Describe a universal set for this problem.
Ans.
A = {letters of BOMBAY} = {A, B, M, O,Y}
B = {letters of CALCUTTA} = {A, C, L, T, U}.
(i) These sets are overlapping.
(ii) These sets are not equal.
(iii) These sets are equivalent.
(iv) Neither A ⊆ B nor B ⊆ A .
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(v) {Letters of English alphabet}.
10. Let ξ = {x : x ∈W , x < 15}, A = {multiples of 2}, B = {multiples of 3},
C = {multiples of 5} and D = {multiples of 6}
State whether each of the following statements is true or false :
(iii) C = D
(i) C and D are disjoint
(ii) C ↔ D
(iv) A ↔ B
(v) A ⊇ B
(iv) D ⊆ A
(vii) D ⊆ B
(viii) C ⊆ B
Ans. The sets ξ, A, B, C and D in roster form are
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
A = {2, 4, 6, 8, 10, 12, 14}
B = {3, 6, 9, 12}
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(i)
(iii)
(iv)
(v)
(vi)
(viii)
C = {5, 10}
D = {6, 12}
True
(ii) True
False, because C and D are disjoint sets.
False, because A has more elements than B
False, because B is not contained in A.
True
(vii) True
False, because C is not contained in B.
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11. Given the universal set = {x : x ∈ N and x < 20}, find :
A = {x : x = 3p ; p ∈ N}
Ans. Universal set U = {x : x ∈ N and x < 20}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}
A = {x : x = 3p ; p ∈ N}
x = 3p
If p = 1, then x = 3 × 1 = 3
If p = 2, then x = 3 × 2 = 6
If p = 3, then x = 3 × 3 = 9
If p = 4, then x = 3 × 4 = 12
If p = 5, then x = 3 × 5 = 15
If p = 6, then x = 3 × 6 = 18
∴ A = {3, 6, 9, 12, 15, 18}
12. Find the proper subsets of {x : x2 – 9x – 10 = 0}.
Ans.
x2 – 9x – 10 = 0 ⇒ x2 – 10x + x – 10 = 0
=0
⇒ x(x – 10) + 1(x – 10) = 0 ⇒ (x – 10) (x + 1)
If, (x –10) = 0 ⇒ x = 10 and, if (x + 1) = 0 ⇒ x = – 1
∴ Given set = {–1, 10}
∴ Proper subsets of this set = {–1}, {10}, {–1, 10}
13. Let ξ = {1, 2, 3, …, 9}, A = {1, 2, 3, 4, 6, 7} and B = {4, 6, 8}. Find :
(i) A′
(ii) A ∩ B (iii) B – A (iv) ( A ∩ B)' (v) A′ ∪ B′
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Ans. ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4, 6, 7} and B = {4, 6, 8}
(i) A′ = ξ – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 7} = {5, 8, 9}
(ii) A ∩ B ={1, 2, 3, 4, 5, 6, 7} ∩ {4, 6, 8} = {4, 6}
(iii) B – A = {4, 6, 8} – {1, 2, 3, 4, 5, 6, 7} = { }
(iv) ( A ∩ B)' = ξ – ( A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 6}
= {1, 2, 3, 5, 7, 8, 9}
(v) A′ ∪ B′ = {5, 8, 9} ∪ {1, 2, 3, 5, 7, 9} = {1, 2, 3, 5, 7, 8, 9}
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14. Let ξ = {x : x ∈ W, x ≤ 10}, A = {x : x ≥ 5}and B = {x : 3 ≤ x < 8}.
Verify that :
(ii) ( A ∩ B)' = A '∪ B '
(i) ( A ∪ B)' = A '∩ B '
Ans.The given sets in the roster form are
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {5, 6, 7, 8, 9, 10}
B = {3, 4, 5, 6, 7}
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(i) LHS = ( A ∪ B)'
A ∪ B = {5, 6, 7, 8, 9, 10} ∪ {3, 4, 5, 6, 7}
= {3, 4, 5, 6, 7, 8, 9, 10}
∴ ( A ∪ B)′ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
– {3, 4, 5, 6, 7, 8, 9, 10} = {1, 2}
RHS = A '∩ B '
A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
= {1, 2, 3, 4}
B ' = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
= {0, 1, 2, 8, 9, 10}
A '∩ B ' = {1, 2, 3, 4} ∩ {0, 1, 2, 8, 9, 10} = {1, 2}
Hence,
( A ∪ B)' = A '∩ B '
(ii) LHS = ( A ∩ B)'
A ∩ B = {5, 6, 7, 8, 9, 10} ∩ {3, 4, 5, 6, 7} = {5, 6, 7}
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( A ∩ B)' = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} –{5, 6, 7} = {0, 1, 2,
3, 4, 8, 9, 10}
R.H.S. = A '∪ B '
A ' = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
= {0, 1, 2, 3, 4}
B ' = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
= {0, 1, 2, 8, 9, 10}
∴ A '∪ B ' = {0, 1, 2, 3, 4} ∪ {0, 1, 2, 8, 9, 10}
= {0, 1, 2, 3, 4, 8, 9, 10}
Hence, ( A ∩ B)' = A '∪ B '
15. Given A = {0, 1, 2, 4, 5}, B = {0, 2, 4, 6, 8} and C = {0, 3, 6, 9}.
Show that :
(i) A ∪ ( B ∪ C ) = ( A ∪ B) ∪ C
i. e., the union of sets is associative.
(ii) A ∩ ( B ∩ C ) = ( A ∩ B) ∩ C
i.e., the intersection of sets is associative.
Ans.A = {0, 1, 2, 4, 5} and B = {0, 2, 4, 6, 8} and C = {0, 3, 6, 9}
(i) B ∪ C = {0, 2, 4, 6, 8} ∪ {0, 3, 6, 9} = {0, 2, 3, 4, 6, 8, 9}
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A ∪ ( B ∪ C ) = {0, 1, 2, 4, 5} ∪ {0, 2, 3, 4, 6, 8, 9}
…(I)
A ∪ ( B ∪ C ) = {0, 1, 2, 3, 4, 5, 6, 8, 9}
⇒
A ∪ B = {0, 1, 2, 4, 5} ∪ {0, 2, 4, 6, 8} = {0, 1, 2, 4, 5, 6, 8}
∴ ( A ∪ B) ∪ C = {0, 1, 2, 4, 5, 6, 8} ∪ {0, 3, 6, 9}
⇒ ( A ∪ B) ∪ C = {0, 1, 2, 3, 4, 5, 6, 8, 9}
From (I) and (II), we have
…(II)
A ∪ ( B ∪ C ) = ( A ∪ B) ∪ C.
(ii) B ∩ C = {0, 2, 4, 6, 8} ∩ {0, 3, 6, 9} = {0, 6}
Now, A ∩ ( B ∩ C ) = {0, 1, 2, 4, 5} ∩ {0, 6}
⇒
Math Class VIII
A ∩ ( B ∩ C ) = {0}
(I)
9
Question Bank
A ∩ B = {0, 1, 2, 4, 5} ∩ {0, 2, 4, 6, 8} = {0, 2, 4}
∴
(II)
( A ∩ B) ∩ C = {0, 2, 4} ∩ {0, 3, 6, 9}
⇒ ( A ∩ B) ∩ C = {0}
From (I) and (II) we have
A ∩ ( B ∩ C ) = ( A ∩ B) ∩ C
16. If A = {x ∈ W : 5 < x < 10}, B = {3, 4, 5, 6, 7} and C = {x = 2n;
n ∈ N and n ≤ 4}. Find :
(i) A ∩ ( B ∪ C )
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(ii) ( B ∪ A) ∩ ( B ∪ C )
(iii) B ∪ ( A ∩ C )
(iv) ( A ∩ B) ∪ ( A ∩ C )
Name the sets which are equal.
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A = {x ∈ W : 5 < x < 10} = {6, 7, 8, 9}
B = {3, 4, 5, 6, 7}
C = {x = 2n; n ∈ N and n ≤ 4}
x = 2n
If n = 1, then x = 2 × 1 = 2
If n = 2, then x = 2 × 2 = 4
If n = 3, then x = 2 × 3 = 6
If n = 4, then x = 2 × 4 = 8
∴ C = {2, 4, 6, 8}
(i) B ∪ C = {3, 4, 5, 6, 7} ∪ {2, 4, 6, 8} = {2, 3, 4, 5, 6, 7, 8}
A ∩ ( B ∪ C ) = {6, 7, 8, 9} ∩ {2, 3, 4, 5, 6, 7, 8}
Ans.
⇒
A ∩ ( B ∪ C ) = {6,7,8}
(ii) B ∪ A = {3, 4, 5, 6, 7} ∪{6, 7, 8, 9} = {3, 4, 5, 6, 7, 8, 9}
( B ∪ A) ∩ ( B ∪ C ) = {3, 4, 5, 6, 7, 8, 9} ∩ {2, 3, 4, 5, 6, 7, 8}
= {3, 4, 5, 6, 7, 8}
(iii) ( A ∩ C ) = {6, 7, 8, 9} ∩ {2, 4, 6, 8} = {6, 8}
B ∪ ( A ∩ C ) = {3, 4, 5, 6, 7} ∪ {6, 8} = {3, 4, 5, 6, 7, 8}
Math Class VIII
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(iv) A ∩ B= {6, 7, 8, 9} ∩ {3, 4, 5, 6, 7} = {6, 7}
∴ ( A ∩ B) ∪ ( A ∩ C ) = {6, 7} ∪ {6, 8}
= {6, 7, 8}
17. Let ξ = {x : x ∈ N, x < 10}, A = {odd numbers}, B = {even numbers}
and C = {prime numbers}. List the elements of the following sets :
(ii) B '
(iii) C '
(i) A '
(iv) A ∪ B
(v) A ∩ B
(vi) A ∪ C
(vii) B ∪ C
(viii) A – B
(ix) A – C
(x) A ∩ ( B ∪ C )
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Ans. ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 3, 5, 7, 9}
B = {2, 4, 6, 8}
C = {2, 3, 5, 7}
(i) A ' = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5, 7, 9} = {2, 4, 6, 8}
(ii) B ' = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}
(iii) C ' = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7} = {1, 4, 6, 8, 9}
(iv) A ∪ B = {1, 2, 3, 5, 7, 9} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
(v) A ∩ B = {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8} = φ or { }
(vi) A ∪ C = {1, 3, 5, 7, 9} ∪ {2, 3, 5, 7} = {1, 2, 3, 5, 7, 9}
(vii) B ∪ C = {2, 4, 6, 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}
(viii) A – B= {1, 3, 5, 7, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}
(ix) A – C= {1, 3, 5, 7, 9} – {2, 3, 5, 7} = {1, 9}
(x) A ∩ ( B ∪ C ) = {1, 3, 5, 7, 9} ∩ {2, 3, 4, 5, 6, 7, 8} = {3, 5, 7}
18. Let ξ = {x : x ∈ W, x ≤ 15}, P = {multiples of 2}, Q = {multiples
of 3} and R = {multiples of 5}. Show that :
(i) ( P ∪ Q)' = P′ ∩ Q '
(ii) ( P ∩ R)' = P '∪ R '
(iii) P ∩ (Q ∪ R) = ( P ∩ Q) ∪ ( P ∩ R).
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Question Bank
Ans. ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
P = {2, 4, 6, 8, 10, 12, 14}
Q = {3, 6, 9, 12, 15}
R = {5, 10, 15}
(i) L H S = ( P ∪ Q)'
∴ P ∪ Q = {2, 4, 6, 8, 10, 12, 14} ∪ {3, 6, 9, 12, 15}
= {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}
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∴ ( P ∪ Q)′ = ξ – ( P ∪ Q) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15} – {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}
= {0, 1, 5, 7, 11, 13}
R H S = P ′ ∩ Q′
∴ P ′ = {0, 1, 2, 3, ....., 15} – {2, 4, 6, 8, 10, 12, 14}
= {0, 1, 3, 5, 7, 9, 11, 13, 15}
Q′ = {0, 1, 2, 3, ....., 15} – {3, 6, 9, 12, 15}
= {0, 1, 2, 4, 5, 7, 8, 10, 11, 13, 14}
∴ P′ ∩ Q′ = {0, 1, 3, 5, 7, 9, 11, 13, 15}
∩ {0, 1, 2, 4, 5, 7, 8, 10, 11, 13, 14}
= {0, 1, 7, 5, 11, 13}
Hence, ( P ∪ Q)′ = P′ ∩ Q′
(Proved)
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(ii) L H S = ( P ∩ R)′
∴ P ∩ R = {2, 4, 6, 8, 10, 12, 14} ∩ { 5, 10, 15} = {10}
∴ ( P ∩ R)′ = ξ – ( P ∩ R) = {0, 1, 2, 3, .....,15} – {10}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15}
R H S = P ′ ∪ R′
∴ P′ = {0, 1, 2, 3, .....,15} – {2, 4, 6, 8, 10, 12, 14}
= {0, 1, 3, 5, 7, 9, 11, 13, 15}
R′ = ξ – R
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
– {5, 10, 15}
Math Class VIII
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Question Bank
= {0, 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14}
P ′ ∪ R′ = {0, 1, 2, 3, 5, 7, 9, 11, 13, 15}
∪ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15}
Hence, ( P ∩ R)′ = P′ ∪ R′
(iii) L H S = P ∩ (Q ∪ R) = {2, 4, 6, 8, 10, 12, 14}
∩ {3, 5, 6, 9, 10, 12, 15}
= {6, 10, 12}
R H S = ( P ∩ Q) ∪ ( P ∩ R )
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∴ P ∩ Q = {2, 4, 6, 8, 10, 12, 14} ∩ {3, 6, 9, 12, 15} = {6, 12}
∴ P ∩ R = {2, 6, 8, 10, 12, 14} ∩ {5, 10, 15} = {10}
∴ ( P ∩ Q) ∪ ( P ∩ R) = {6, 12} ∪ {10} = {6, 10, 12}
Hence, P ∩ (Q ∪ R) = ( P ∩ Q) ∪ ( P ∩ R).
19. If n (A) = 20, n (B) = 16 and n( A ∪ B) = 30, find n( A ∩ B).
Ans.Given that,
n (A) = 20, n(B) = 16 and n( A ∪ B) = 30
Then n( A ∩ B) = ?
We know that n( A ∪ B) = n( A) + n( B) – n( A ∩ B)
⇒
⇒
⇒
Hence,
30 = 20 + 16 – n( A ∩ B)
30 = 36 – n( A ∩ B)
n( A ∩ B) = 36 – 30 ⇒ n( A ∩ B) = 6
n( A ∩ B ) = 6
20. If A = {x : x 2 ≤ 16}, B = { x :
x
– 2 < 3} and the universal set is W,
3
the set of whole numbers.
(i) Find sets A and B.
(ii) Verify : A′ ∩ B = B – ( A ∩ B)
Ans.
W = {0, 1, 2, 3, 4, 5, 6, 7, 8,…}
Math Class VIII
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Question Bank
A = {x : x 2 ≤ 16}
x 2 ≤ 16
⇒
x≤4
A = {0, 1, 2, 3, 4}
∴
 x

B =  x : – 2 < 3
 3

x
x
< 3+ 2
–2<3 ⇒
3
3
x
< 5 ⇒ x < 15
⇒
3
∴ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
(i) A = {0, 1, 2, 3, 4}
B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
(ii) A′ = {5, 6, 7, 8, …}
A′ ∩ B = {5, 6, 7, 8,.......}
∩ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
A′ ∩ B = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
…(I)
⇒ A ∩ B = {0, 1, 2, 3, 4}
∩ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
= {0, 1, 2, 3, 4}
B – (A ∩ B) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
– {0, 1, 2, 3, 4}
... (II)
⇒ B – ( A ∩ B) = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
From (I) and (II), we can say that
A′ ∩ B = B – ( A ∩ B)
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21. If ξ = {x : x ≤ 12, x ∈ N }; A = {x : x is an even number}; B = {m : m
divisible by 3} and C = {x : 3 < x ≤ 9}; then verify that :
(i) A – ( B ∪ C )′ = A – ( B′ ∩ C ′)
(ii) A – ( B′ ∪ C ′) = A – ( B ∩ C )′
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Question Bank
Ans. ξ = {x : x ≤ 12; x ∈ N } = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A = {x : x is an even number}= {2, 4, 6, 8, 10, 12}
B = {m : m is divisible by 3}= {3, 6, 9, 12}
C = {x : 3 < x ≤ 9} = {4, 5, 6, 7, 8, 9}
(i)
B ∪ C = {3, 6, 9, 12} ∪ {4, 5, 6, 7, 8, 9}
= {3, 4, 5, 6, 7, 8, 9, 12}
( B ∪ C )′ = {1, 2, 10, 11}
Z
B
A – ( B ∪ C )′ = {2, 4, 6, 8, 10, 12} – {1, 2, 10, 11}
⇒
A – ( B ∪ C )′ = {4, 6, 8, 12}
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…(I)
B′ = {1, 2, 4, 5, 7, 8, 10, 11}
C ′ = {1, 2, 3, 10, 11, 12}
B′ ∩ C ′ = {1, 2, 4, 5, 7, 8, 10, 11}
∩ {1, 2, 3, 10, 11, 12}
= {1, 2, 10, 11}
∴
A – ( B′ ∩ C ′) = {2, 4, 6, 8, 10, 12} – {1, 2, 10, 11}
⇒ A – ( B′ ∩ C ′) = {4, 6, 8, 12}
From (I) and (II), we have
...(II)
A – ( B ∩ C )′ = A – ( B′ ∩ C ′)
(ii)
∴
B′ ∪ C ′ = {1, 2, 4, 5, 7, 8, 10, 11}
∪ {1, 2, 3, 10, 11, 12}
= {1, 2, 3, 4, 5, 7, 8, 10, 11, 12}
A – ( B′ ∪ C ′) = {2, 4, 6, 8, 10, 12}
– {1, 2, 3, 4, 5, 7, 8, 10, 11, 12}
…(I)
⇒ A – ( B′ ∪ C ′) = {6}
Now,
B ∩ C = {3, 6, 9, 12} ∩ {4, 5, 6, 7, 8, 9} = {6, 9}
( B ∩ C )′ = {1, 2, 3, 4, 5, 7, 8, 10, 11, 12}
∴
∴ A – ( B ∩ C )′ = {2, 4, 6, 8, 10, 12} – {1, 2, 3, 4, 5, 7, 8, 10, 11, 12}
Math Class VIII
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Question Bank
A – ( B ∩ C )′ = {6}
⇒
From (I) and (II), we have
…(II)
A – ( B′ ∪ C ′) = A – ( B ∩ C )′
22. Given ξ = {x : x is a natural number between 25 and 45};
A = {x : x is an even number} and B = {x : x is a multiple of 3}.
Find :
(i) n (A) + n (B)
(ii) n( A ∪ B) + n( A ∩ B)
(iii) n(A – B)
(iv) n( A′ ∩ B)
Z
B
Is n(A) + n(B) = n( A ∪ B) + n( A ∩ B) ?
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Is n(A – B) = n( A′ ∩ B) ?
Ans. ξ = {26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
39, 40, 41, 42, 43, 44}
A = {26, 28, 30, 32, 34, 36, 38, 40, 42, 44}
B = {27, 30, 33, 36, 39, 42}
(i) n(A) = 10, n(B) = 6
n(A) + n(B) = 10 + 6 = 16
∴
(ii) We know that
n( A ∪ B) = n( A) + n( B) – n( A ∩ B)
⇒ n( A ∪ B) + n( A ∩ B) = n( A) + n( B)
⇒ n( A ∪ B) + n( A ∩ B) = 16
(iii) A – B = {26, 28, 30, 32, 34, 36, 38, 40, 42, 44}
– {27, 30, 33, 36, 39, 42}
= {26, 28, 32, 34, 38, 40, 44}
∴ n(A – B) = 7
(iv) A′ = {27, 29, 31, 33, 35, 37, 39, 41, 43}
A′ ∩ B = {27, 29, 31, 33, 35, 37, 39, 41, 43} ∩ {27, 30, 33, 36, 39, 42}
= {27, 33, 39}
Math Class VIII
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Question Bank
∴
Yes,
No,
n( A′ ∩ B) = 3
n(A) + n(B) = n(A ∪ B) + n(A ∩ B)
n(A – B) ≠ n( A′ ∩ B)
23. If n(ξ) = 30, n(A) = 22, n(B) =15 and n( A ∪ B) = 25; find :
(i) n( A ∩ B) (ii) n( A′)
(iii) n( B′)
Ans. n(ξ) = 30
n(A) = 22
(i) We know that
n(B) = 15
n( A ∪ B) + n( A ∩ B) = n( A) + n( B)
⇒
n( A ∪ B) = 25
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⇒ 25 + n( A ∩ B) = 22 + 15
⇒
(iv) n( A ∩ B)′
n( A ∩ B) = 22 + 15 – 25
n( A ∩ B) = 37 – 25 ⇒ n( A ∩ B) = 12
n( A′) = n(ξ) – n( A)
(ii)
⇒
n( A′) = 30 – 22 ⇒ n( A′) = 8
n( B′) = n(ξ) – n( B)
(iii)
⇒
n( B′) = 30 –15 ⇒ n( B′) = 15
n( A ∩ B) = 12
(iv)
n( A ∩ B)′ = n(ξ) – n( A ∩ B)
⇒
n( A ∩ B)′ = 30 – 12 ⇒ n( A ∩ B)′ = 18
24. If n( A ∪ B) = 40, n( A ∩ B) = 8 and n (A – B) = n (B – A), find:
(i) n (A)
(ii) n (B)
Ans. Given
n( A ∪ B) = 40 , n( A ∩ B) = 8
and n (A – B)= n (B – A)
We know that,
∴ n (A – B) + n (B – A) + n (A ∩ B) = n (A ∪ B)
⇒ n ( A – B ) + n ( A – B ) + n( A ∩ B ) = n ( A ∪ B )
Math Class VIII
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Question Bank
⇒
⇒
⇒
2 n (A – B) + 8 = 40
2 n (A – B)= 40 – 8 ⇒ 2 n (A – B) = 32
32
n( A – B ) =
⇒ n (A – B) = 16
2
n (A – B) = n (B – A) = 16
∴
(i) n( A) = n( A – B) + n( A ∩ B) = 16 + 8 = 24
(ii) n( B) = n( B – A) + n( A ∩ B) = 16 + 8 = 24
Z
B
25. If n(ξ) = 40, n( A′) = 15, n( B) = 12 and n (( A ∩ B)′) = 32, find :
(ii) n( B′)
(i) n(A)
(iii) n( A ∩ B)
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(iv) n( A ∪ B)
(v) n(A – B)
(vi) n(B – A)
Ans. n(ξ) = 40 , n( A′) = 15 , n (B) = 12 and
n(( A ∩ B)′) = 32
(i) n( A) = n(ξ) – n( A′) = 40 – 15 = 25
(ii) n( B′) = n(ξ) – n( B) = 40 – 12 = 28
(iii) n( A ∩ B) = n(ξ) – n(( A ∩ B)′) = 40 – 32 = 8
(iv) n( A ∪ B) = n( A) + n( B) – n( A ∩ B) = 25 + 12 – 8 = 25 + 4 = 29
(v) n( A – B) = n( A) – n( A ∩ B) = 25 – 8 = 17
(vi) n( B – A) = n( B) – n( A ∩ B) = 12 – 8 = 4
26. If n (A – B) = 12, n (B – A) = 16 and n( A ∩ B) = 5, find :
(i) n (A)
(ii) n (B)
(iii) n( A ∪ B)
n( A ∩ B ) = 5
Ans. n (A – B) = 12, n (B – A) = 16
and
(i) n( A) = n( A – B) + n( A ∩ B) [∵ n( A – B) = n( A) – n( A ∩ B)]
= 12 + 5 = 17
(ii) n( B) = n( B – A) + n( A ∩ B) [∵ n( B – A) = n( B) – ( A ∩ B)]
= 16 + 5 = 21
(iii) n( A ∪ B) = n( A) + n( B) – n( A ∩ B) = 17 + 21 – 5 = 38 – 5 = 33
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Question Bank
Section II : Venn diagrams
1. From the adjoining Venn diagram, find the following sets:
(i) ξ
(ii) A ∩ B
(iii) A ∩ B ∩ C
(iv) C ′
(v) A – C
(vi) B – C
(vii) C – B
(viii) ( A ∪ B)′
(ix) ( A ∪ B ∪ C )′
Ans. (i) ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(ii) A ∩ B = {8, 0, 5}
Z
B
ξ
(iii) A ∩ B ∩ C = {0, 5}
A
B
10
8
0
5
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(iv) C′ = {2, 7, 8, 9, 10, 11, 12}
(v) A – C = {8, 10}
(vi) B – C = {7, 8, 11}
(vii) C – B = {3, 4, 6}
(viii) ( A ∪ B)′ = {2, 4, 6, 9, 12}
2
3
4
11
7
1
6
C
(ix) ( A ∪ B ∪ C )′ = {2, 9, 12}
2. From the given diagram find:
A
c
b
a
(i) A ∪ B
(ii) A′ ∩ B
e
f
d
(iii) A – B
(iv) B – A
h
(v) ( A ∪ B)′
Ans. (i) A ∪ B = {a, c, d, e} ∪ {b, c, e, f}
⇒ A ∪ B = {a, b, c, d, e, f}
A′ = {b, f, g, h}
(ii)
A′ ∩ B = {b, f, g, h} ∩ {b, c, e, f}
A′ ∩ B = {b, f}
⇒
(iii)
A – B = {a, c, d, e} – {b, c, e, f} = {a, d}
(iv)
B – A = {b, c, e, f} – {a, c, d, e} = {b, f}
A ∪ B = {a, b, c, d, e, f}
(v)
( A ∪ B)′ = {h, g}
∴
Math Class VIII
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12
9
B
ξ
g
Question Bank
3. From the given diagram, find:
(i) A′
(ii) B′
(iii) A′ ∪ B′
(iv) ( A ∩ B)′
Is A′ ∪ B′ = ( A ∩ B)′ ?
Also, verify if A′ ∩ B′ = ( A ∪ B)′
9
ξ
8
A
3
6
1
4
5
2
B
7
10
Ans. (i) A = {1, 3, 4, 6} ∴ A′ = {2, 5, 7, 8, 9, 10}
(ii) B = {1, 2, 5} ∴ B ′ = {3, 4, 6, 7, 8, 9, 10}
(iii) A′ ∪ B′ = {2, 5, 7, 8, 9, 10} ∪ {3, 4, 6, 7, 8, 9, 10}
= {2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) A ∩ B = {1, 3, 4, 6} ∩ {1, 2, 5} = {1}
∴ ( A ∩ B)′ = {2, 3, 4, 5, 6, 7, 8, 9, 10}
From Part (iii) and Part (iv) we conclude
A′ ∪ B′ = ( A ∩ B)′
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A′ ∩ B′ = {2, 5, 7, 8, 9, 10} ∩ {3, 4, 6, 7, 8, 9, 10}
A′ ∩ B′ = {7, 8, 9, 10}
... (I)
A ∪ B = {1, 3, 4, 6} ∪ {1, 2, 5}
= {1, 2, 3, 4, 5, 6}
...(II)
( A ∪ B)′ = {7, 8, 9, 10}
∴
From (I) and (II), we have
ξ
A
B
A′ ∩ B′ = ( A ∪ B)′
a
e
d
g
4. Use the given diagram to find:
b c
f
h
j
p
i
(i) A ∪ ( B ∩ C )
k
(ii) B – (A – C)
l
n
C
m
(iii) A – B
(iv) A ∩ B′
Is A ∩ B′ = A – B ?
B ∩ C = {d, e, f, g, h, j} ∩ {h, i, j, k, l} = {h, j}
Ans. (i)
A ∪ ( B ∩ C ) = {a, b, c, d, g, h, i} ∪ {h, j}
∴
= {a, b, c, d, g, h, i, j}
(ii)
A – C = {a, b, c, d, g, h, i} – {h, i, j, k, l}
= {a, b, c, d, g}
Now
⇒
Now
Math Class VIII
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Question Bank
B – (A – C) = {d, e, f, g, h, j} – {a, b, c, d, g} = {e, f, h, j}
(iii)
A – B = {a, b, c, d, g, h, i} – {d, e, f, g, h, j}
A – B = {a, b, c, i}
...(I)
⇒
(iv)
B′ = {a, b, c, i, k, l, m, n, p}
A ∩ B′ = {a, b, c, d, g, h, i} ∩ {a, b, c, i, k, l, m, n, p}
A ∩ B′ = {a, b, c, i}
…(II)
⇒
From (I) and (II) we can conclude A ∩ B′ = A – B
5. Draw a Venn-diagram to show the relationship between two overlapping sets A and B. Now shade the region representing :
(ii) A ∪ B
(iii) B – A
(i) A ∩ B
A∩ B = ξ
Ans.(i)
B
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A
(ii)
(iii)
A∪ B =
ξ
B–A=
ξ
B
A
A
B
6. Draw a Venn-diagram to show the relationship between sets A and
B; such that A ⊆ B . Now shade the region representing :
(i) A ∪ B
Ans.(i)
(ii) B ′ ∩ A
A∪ B = ξ
(iii) A ∩ B
A
Math Class VIII
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(iv) ( A ∪ B)′
B
Question Bank
(ii)
B′ ∩ A =
ξ
B
A
(iii)
A∩ B =
B
ξ
A
(iv)
( A ∪ B)′ =
B
ξ
A
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B
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7. Two sets A and B are such that A ∩ B = φ . Draw a Venn-diagram to
show the relationship between A and B. Shade the region representing :
(ii) ( A ∪ B)′
(iii) B – A
(i) A ∪ B
Ans. (i)
A∪ B =
ξ
A
B
(ii)
( A ∪ B)′ =
ξ
A
B
(iii)
B–A =
A
ξ
B
8. (A) State the sets represented by the shaded portion of following
Venn-diagrams :
Math Class VIII
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Question Bank
(i)
ξ
(iii)
ξ
A
B
A
B
(ii)
ξ
A
B
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B
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Ans. (i) ( A ∪ B)′
(ii) B – A or A′ ∩ B
(iii) ( B – A)′
(B) In each of the given diagrams, shade the region which represents the
set given underneath the diagram :
(ii) ξ
(i) ξ
A
A
B
(A ∩ B)′
(B – A)′
(iii)
ξ
P
B
Q
(P ∪ Q)′
Ans. (i)
( B – A)′ =
ξ
(ii)
( A ∩ B)′ =
ξ
Math Class VIII
A
B
A
B
23
Question Bank
(iii)
( P ∪ Q)′ =
ξ
P
Q
9. In the adjoining figure, A and B are two subsets of the universal set ξ
such that B ⊂ A ⊂ ξ, n( A) = 41, n( B) = 25 and n(ξ) = 50. Find:
A
(i) n ( A)′ (ii) n( B′) (iii) n( A – B)
Ans. (i) n( A′) = n(ξ) – n( A) = 50 – 41 = 9
Z
B
ξ
B
(ii) n( B′) = n(ξ) – n( B) = 50 – 25 = 25
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(iii) n(A – B) = n( A) – n( A ∩ B) = n( A) – n( B)
[∵ n(A ∩ B = n(B)]
= 41 – 25 = 16
10. If ξ = {x : x ∈ N and x ≤ 20}, A = {x : x is a multiple of 4}, B = {x : x
is a multiple of 6} and C = {x : x is a factor of 36}. Draw a Venndiagram to show that the relationship between the given sets.
Hence, find : (i) A ∩ C
(ii) A – B
(iii) A ∩ B ∩ C
ξ
17
Ans. ξ = {x : x ∈ N and x ≤ 20}
A
B
8
7
16 20
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
12
6
5
13, 14, 15, 16, 17, 18, 19, 20}
4
18
10
3
9
1
A = {x : x is a multiple of 4}
13
11
2
14
15
C
19
= {4, 8, 12, 16, 20}
B = {x : x is a multiple of 6} = {6, 12, 18}
C = {x : x is a factor 36} = {1, 2, 3, 4, 6, 9, 12, 18}
(i) A ∩ C = {4, 12}
(ii) A – B = {4, 8, 16, 20}
(iii) A ∩ B ∩ C = {12}
11. In a class of 60 pupils, 28 play hockey, 33 play cricket and 14 play
none of these games. Draw Venn diagram to find :
(i) Pupils play both the games.
(ii) How many play hockey only?
(iii) How many play cricket only?
Math Class VIII
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Question Bank
Ans.Let H for hockey and C for cricket
n(H) = 25, n(C) = 33
and 14 play none of these games
i.e.,
n( H ∪ C ) = 60 – 14 = 46
(i)
n( H ∩ C ) = n( H ) + n(C ) – n( H ∪ C )
= 28 + 33 – 46 = 61 – 46 = 15
(ii) n( H ) – n( H ∩ C ) = 28 –15 = 13
(iii) n(C ) – n( H ∩ C ) = 33 –15 = 18
12. In a club, three-tenths of its members play cards only, four-tenths
play carrom only. If 15 members play none of these games and 90
play both, find using Venn diagram, the total number of members
in the club.
Ans. Let number of members in the club be x.
Z
B
L
A
A
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3
x
10
4
Number of members play carrom only = x
10
3x 4 x
ξ
+
+ 90 + 15 = x
∴
Y
X
10 10
3x
4x
3x 4 x
10 90
10
–
x–
= 105
⇒
10 10
15
10 x – 3 x – 4 x
= 105
⇒
10
105 × 10
3x
= 105 ⇒ x =
⇒
⇒ x = 350
3
10
Hence, total number of members in the club are 350.
13. In a colony, two-fifths of the families read the magazine ‘Femina’
and three-fourths of the families read the magazine ‘Filmfare’. If
40 families read none of these two magazines and 100 families
read both, use Venn-diagram to find the number of families in the
colony.
Number of members play card only =
Math Class VIII
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Question Bank
Ans. Let number of families in the colony be x.
ξ
2
X
x
Y
5
3x
2x
– 100 100
– 100
3
4
5
Families reading’Filmfare’ = x
4
The Venn diagram is as given alongside
40
 2x

 3x

–100
100
–
100
+
+



 + 40 = x
∴
5
4




2x
3x
– 100 + 100 +
– 100 + 40 = x
⇒
5
4
2 x 3x
2 x 3x
– 60 = x
+
+
– x = 60
⇒
⇒
5
4
5
4
23 x – 20 x
8 x + 15 x – 20 x
= 60
⇒
⇒
= 60
20
20
60 × 20
3x
= 400
= 60 x =
⇒
3
20
Hence, total number of families in the colony is 400.
14. In a class of 50 boys, 35 like horror movies, 30 like war movies
and 5 like neither. Find the number of those that like both.
Ans.Let H denote for horror movies and W denote for war movies
∴ n (H) = 35, n(W) = 30 and 5 like neither i.e.,
n( A ∪ B) = 50 – 5 = 45.
Hence, number of boys like both n( H ∩ W )
n( H ) + n(W ) – n( H ∪ W ) = 35 + 30 – 45 = 65 – 45 = 20.
15. In a certain locality of Delhi there are 1000 families. A survey
showed that 504 subscribe to ‘The Hindustan Times’ daily newspaper and 478 subscribe to ‘The Times of India’ and 106 subscribe to both. Find the number of families which do not subscribe
to any of these newspapers.
Ans.Total number of families are 1000
Let A be the set of families who subscribe ‘The Hindustan Time’,
B is the set of families who subscribe ‘The Times of India’.
Families reading ‘Femina’ =
Z
B
L
A
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Question Bank
Then, n(A) = 504, n(B) = 478 and n(A ∩ B) =106
∴ Those families who subscribe to either the ‘Hindustan Times’ or
to the ‘Times of India’ or both
= n( A ∪ B) = n( A) + n( B) – n( A ∩ B)
= 504 + 478 –106 = 982 – 106 = 876
Hence, those families who subscribe to none 1000 – 876 = 124.
16. In a class of 90 students, 50 students got distinction in Mathematics, 42 got distinction in Science and 24 students got distinction in
both the subjects. Represent this information by a Venn diagram.
Hence, find the number of students getting distinction in
(i) Mathematics only
(ii) Science only
(iii) any of the two subjects
(iv) neither Mathematics nor Science
Ans.Let U be all the students in class.
X be the students who gots distinction in Mathematics
Y be the students who gots distinction in Science
n (U) = 90
∴
U
Y
X
22
n(X) = 50
26
18
24
n (Y) = 42
Z
B
L
A
A
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D
N
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n( X ∩ Y ) = 24
Thus, number of students getting distinction in
(i) Maths only = 26
(ii) Science only = 18
(iii) any of the two subjects = 26 + 24 + 18 = 68
(iv) neither Mathematics nor Science = 22
17. In a group of 60 persons, 45 speak Bengali, 28 speak English and
all the persons speak at least one language. Find how many people
speak both Bengali and English. Draw a Venn diagram.
Ans. Let U = group of persons
Math Class VIII
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Question Bank
X = Those persons who speak Bengali
Y = Those persons who speak English
n (U) = 60
n (X) = 45
n (Y) = 28
and
n( X ∪ Y ) = 60
U
X
45
Y
13
28
n( X ∩ Y ) = n( X ) + n(Y ) – n( X ∪ Y )
= 45 + 28 – 60 = 13.
18. Forty-three persons went to a canteen, which sells soup and tea. If 18
persons took soup only; 8 took tea only and 5 took nothing. Use
Venn diagram to find :
X
U
Y
(i) how many took both ?
18
8
12
(ii) how many took ‘soup’ ?
5
(iii) how many took ‘tea’ ?
Ans. Total number of persons who visited canteen = 43
Let U represents the set of persons who visited canteen.
Number of persons who took soup only = 18
Number of persons who took tea only = 8
Number of persons who took nothing = 5
Number of persons who took either of two = 43 – 5 = 38
Number of persons who took both = 38 – (18 + 8) = 38 – 26 = 12
(i) Number of those who took both = 12
(ii) Number of those who took soup = 18 + 12 = 30
(iii) Number of those who took tea = 8 + 12 = 20
19. If ξ is the set of boys in your school and B is the set of boys who play
badminton. Draw Venn diagram showing that some of boys do not
ξ
play badminton.
B
If n(ξ) = 40 and n( B′) = 17; find :
(i) how many do not play badminton ?
(ii) how many play badminton?
Z
B
L
A
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Question Bank
Ans.
ξ = Set of boys in the school
B = Set of boys who play badminton
B′ = Set of boys who do not play badminton
= Shaded Portion
Now, if
n(ξ) = 40
n( B′) = 17
Then (i) number of Boys who do not play badminton = 17
(ii) Number of boys who play badminton = 40 – 17 = 23
20. Let ξ = {all triangles drawn in plane},
I = {isosceles triangles}
and R = {right angled triangles}
Draw a Venn diagram to show these sets in their correct relationship.
Shade the region representing I ∩ R and write the measures of the
angles of the triangles of this region.
ξ
I
R
Ans. The given sets are
ξ = {all triangles drawn in a plane}
I = {isosceles triangles}
I∩R
R = {right angled triangles}
Measures of angles of the triangles in the region I ∩ R = 45°, 45°, 90°.
21. In a city, 50% people read newspaper A, 45% read newspaper B, and
25% read neither A nor B. What percentage of people read both the
newspapers A as well as B ?
Ans. Let A = Those people who read newspaper A
B = Those people who read newspaper B
Here, n(ξ) = 100% , n( A) = 50% , n( B) = 45% ,
and
n( A ∩ B)′ = 25%
n( A ∪ B) = n(ξ) – n(( A ∪ B)′)
∴
=100% – 25% = 75%
We know that n( A ∪ B) = n( A) + n( B) – n( A ∩ B)
75% = 50% + 45% – n( A ∩ B)
⇒
Z
B
L
A
A
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U AT
D
N
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Question Bank
n( A ∩ B) = 95% – 75%
⇒
n( A ∩ B) = 20%
⇒
22. Preeti and Rashmi contested the selection for the post of head girl of
the school, for which the students of classes 10th, 11th and 12th voted.
If three-seventh of the students voted for preeti only; three-seventh
for Rashmi; fifty for both and 50 students did not use their votes,
find :
(i) the total number of students in classes 10th, 11th and 12th,
(ii) the number of students, who voted for Preeti and
(iii) the number of students, who voted for Rashmi only.
Ans. Let total number of students in classes 10th, 11th, and 12th be x
Let ξ represents the set of students in classes 10th, 11th and 12th
Z
B
L
A
A
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D
N
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3x
7
Number of students who voted for both (Preeti and Rashmi) = 50
3x
Number of students who voted for Rashmi =
7
 3x

=
Number of students who voted for Rashmi only  – 50 
 7

Number of those who did not use their voted = 50
From Venn diagram,
ξ
X
Y
Number of students who voted for Preeti only =
3x
3x
+ 50 +
– 50 + 50
7
7
3x 3 x
–
= 50 – 50 + 50
x–
7
7
7 x – 3x – 3x
= 50
7
7x – 6x
x
= 50 ⇒ = 50
7
7
x = 50 × 7 = 350
x=
⇒
⇒
⇒
⇒
Math Class VIII
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3x
7
50
3x
–50
7
50
Question Bank
(i) Total number of students in classes 10th, 11th and 12th = 350
(ii) Number of students who voted for Preeti
3x
3 × 350
=
+ 50 =
+ 50
7
7
= 3 × 50 + 50 = 150 + 50 = 200
(iii) Number of students who voted for Rashmi only
3x
3 × 350
– 50 =
– 50
=
7
7
= 150 – 50 = 100
23. The students of a certain school have a choice of three games : Tennis, Badminton and Cricket. The following table gives the percentage of students who play some or all the games :
Games
% of
students
Z
B
L
A
A
L ION
U AT
D
N
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T
IN
Tennis Badminton
35
30
Tennis
and
Badminton
10
Badminton Cricket Cricket
and
and
only
Cricket
Tennis
10
8
30
All
Games
3
Draw a Venn diagram and use it to determine the percentage of students who
(i) play Tennis only
(ii) play Badminton only
(iii) play Cricket
(iv) do not play any of the games.
Ans. Let X = Those students who play Tennis
Y = Those students who play Badminton
Z = Those students who play Cricket. u
Y
Then from given data
15
20
7
13
X
3
n (U) = 100%
7
5
n (X) = 35%
30
Z
n (Y) = 30%
n ( X ∩ Y ) = 10%
n (Y ∩ Z ) = 10%
n ( X ∩ Z ) = 8%
Math Class VIII
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Question Bank
n (only Z) = 30%
n ( X ∩ Y ∩ Z ) = 3%
Using Venn diagram
(i) Play Tennis only = 20% (ii) Play Badminton only = 13%
(iii) Play Cricket = 45%
(iv) Do not play any of the games = 15%
24. A and B are two overlapping sets such that n ( A ∩ B) = x + 4,
n (A – B) = 4x – 8 and n (B) = 3x + 8. Find x, if n ( A ∪ B) = 70.
Also find n ( A′ ∩ B).
ξ
A
L
A
A
L ION
U AT
D
N
© E ER
T
IN
n ( A ∪ B) = 70
Ans.
Z
B
4x – 8
n ( A ∩ B) = x + 4
x+4
B
2x + 4
n ( A – B) = 4x – 8
n (B) = 3x + 8
n (B – A) = n (B) – n ( A ∩ B)
= 3x + 8 – (x + 4) = 3x + 8 – x – 4 = 2x + 4
Using Venn diagram, we have
4x – 8 + x + 4 + 2x + 4 = 70
4x + 2x + x = 70 + 8 – 4 – 4
⇒
∴
⇒
7x = 70 ⇒ x =
70
= 10
7
n( A′ ∩ B) = n (only B)
= n (B – A) = 2x + 4
= 2 × 10 + 4 = 20 + 4 = 24
Math Class VIII
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Question Bank