SSC higher math exercise 1.1 solution part 2
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21. Given that A = {x : 2 < x ≤ 5, x ∈ ∇},
B = {x : 1 ≤x < 3, x ∈ ∇} and C = {2, 4, 5}. Express the following sets in such set notation:
(a) A ∩ B (b) A′ ∩ B′ and (c) A′ ∩ B
Solution: Given, A = {x : 2 < x ≤ 5, x ∈ ∇},
B = {x : 1 ≤x < 3, x ∈ ∇} and C = {2, 4, 5}
(a) A ∩ B = {x : 2 < x ≤ 5, x ∈ ∇} ∩ {x : 1 ≤x < 3, x ∈ ∇}
= {x : 2 < x < 3, x ∈ ∇}
(b) Here, U = ∇
∴ A ∪ B = {x : 2 < x ≤ 5, x ∈ ∇} ∪ {x : 1 ≤x < 3, x ∈ ∇}
= {x : 1 ≤ x ≤ 5, x ∈ ∇}
According to De Morgan’s law,
A′ ∩ B′ = (A ∪ B)′
= U – (A ∪ B)
= ∇ – {x : 1 ≤ x ≤ 5, x ∈ ∇}
= {x : x < 1 or x > 5, x ∈ ∇}
(c) Here, U = ∇
A′ = U – A
∴ A′ = ∇ – {x : 1 ≤ x ≤ 5, x ∈ ∇}
= {x : x ≤ 2 or x > 5, x ∈ ∇}
B = {x : 1 ≤x < 3, x ∈ ∇}
∴ A′ ∩ B = {x : x ≤ 2 or x > 5, x ∈ ∇} ∩ {x : 1 ≤x < 3, x ∈ ∇}
= {x : 1≤ x ≤ 2, x ∈ ∇}
22. Given that U = {x : x < 10, x ∈ ∇}, A = {x : 1 < x ≤ 4} and B = {x : 3 ≤ x < 6}. Express the following sets in such set notation:
(a) A ∩ B (b) A′ ∩ B (c) A ∩ B′ and (d) A′ ∩ B′
Solution:
Given, U = {x : x < 10, x ∈ ∇}
A = {x : 1 < x ≤ 4}
and B = {x : 3 ≤ x < 6}
(a) A ∩ B = {x : 1 < x ≤ 4} ∩ {x : 3 ≤ x < 6}
= {x: 3 ≤ x ≤ 4}
[N.B.: Answer in the textbook is wrong.]
(b) A′ = U – A
= {x : x < 10, x ∈ ∇} – {x : 1 < x ≤ 4}
= {x : x ≤ 1 or 4 < x < 10}
∴ A′ ∩ B = {x : x ≤ 1 or 4 < x < 10} ∩ {x : 3 ≤ x < 6}
= {x : 4 < x < 6}
(c) B′ = U – B
= {x : x < 10, x ∈ ∇} – {x : 3 ≤ x < 6}
= {x : x < 3 or 6 ≤ x < 10}
∴ A – B′ = {x : 1 < x ≤ 4} – {x : x < 3 or 6 ≤ x < 10}
= {x : 1 < x < 3}
(d) A ∪ B = {x : 1 < x ≤ 4} ∪ {x : 3 ≤ x < 6}
= {x : 1 < x < 6}
According to the law of DeMorgans,
A′ ∩ B′ = (A ∪ B)′
= U – (A ∪ B)
= {x : x < 10, x ∈ ∇} – {x : 1 < x < 6}
= {x : x ≤ 1 or 6 ≤ x < 10}
[N.B.: Answer in the textbook is wrong.]
23. The sets A and B have been given below. Find in each case A∪B and verify that A ⊂ (A ∪ B) and B ⊂ (A ∪ B)
i. A = {–2, –1, 0, 1, 2} and B = {–3, 0, 3}
ii. A = {x : x ∈ N, x < 10 and x is a multiple of 2}
and B = {x : x ∈ N, x < 10 and x is a multiple of 3}.
Solution:
(i) Given, A = {–2, – 1, 0 1, 2}
and B = {–3, 0, 3}
∴ A ∪ B = {–2, – 1, 0, 1, 2} ∪ {–3, 0, 3}
= {–3, – 2, – 1, 0, 1, 2, 3} (Ans.)
∴ A ⊂ (A ∪ B) and B ⊂ (A ∪ B) (Verified)
(ii) A = {x : x ∈ N, x < 10 and x is a multiple of 2}
B = { x : x ∈ N, x < 10 and x is a multiple of 3}
∴ A = {2, 4, 6 8} and B = {3, 6, 9}
∴ A ∪ B = {2, 4, 6, 8} ∪ {3, 6, 9}
= {2, 3, 4, 6, 8, 9} (Ans.)
∴ A ⊂ (A ∪ B) and B ⊂ (A ∪ B) (Verified)
24. Find A∩B by using the sets below and verify that (A ∩ B) ⊂ A and (A ∩ B) ⊂ B
(i) A = {0, 1, 2, 3, 5}; B = {–1, 0, 2}
(ii) A = {a, b, c, d}, B = {b, x, c, y}
Solution:
(i) Given, A = {0, 1, 2, 3, 5}
B = {–1, 0, 2}
∴ A ∩ B = {0, 1, 2, 3, 5} ∩ {–1, 0, 2}
∴ A ∩ B = {0, 2} (Ans.)
∴ (A ∩ B) ⊂ A and (A ∩ B) ⊂ B (Verified)
(ii) Given, A = {a, b, c, d}
B = {b, x, c, y}
∴ A ∩ B = {a, b, c, d} ∩ {b, x, c, y}
= {b, c} (Ans.)
∴ (A ∩ B) ⊂ A and (A ∩ B) ⊂ B (Verified)
25. Among the girls of Anwara College, a survey was conducted about their reading habits of the magazines of the Bichitra, the Sandhani and the Purbani.
It was found that 60% of the girls read the Bichitra, 50% read the Sandhani, 50% read the Purbani, 30% read the Bichitra and the Sandhani, 30% read the Bichitra and the Purbani, 20% read the Sandhani and the Purbani, while 10% read all three magazines.
(i) What percentage of the girls does not read any of the three magazines?
(ii) What percentage of the girls read just two of the above magazines?
Solution: Let, the set of all girls be U, the set of girls who read Bichitra is B, the set of girls who read Sandhani is S, the set of girls who read Purbani is P.
∴ In percentage n(U) =100%, n(B) = 60%, n(S) = 50%, n(P) = 50%, n (B ∩ S) = 30%, n (B ∩ P) = 30%,
n (P ∩ S) = 20%,
n (P ∩ B ∩ S) = 10%
(i) The set of girls who read at least one of the magazines is (B ∪ P ∪ S) [Shown in the Venn diagram]
∴ The number of girls who do not read any of the three magazines is n(U) – n (B ∪ P ∪ S) [white part of the Venn diagram]
Now, n (B ∪ P ∪ S) = n (B) + n (P) + n (S) – n (B ∩ P)
– n (B ∩ S) – n (P ∩ S) + n (B ∩ P ∩ S)
= 60% + 50% + 50% – 30% – 30% – 20% + 10%
= 90%
∴ The percentage of the girls, who do not read any of the three magazines,
n (U) – n (B ∪ P ∪ S)
= 100% – 90%
= 10% Ans. 10%
(ii) The number of girls, who read only the Bichitra and the Purbani, = n [(B ∩ P) \ S]
= n (B ∩ P) – n (B ∩ P ∩ S) [Shown in the Venn diagram]
= 30% – 10%
= 20%
Set of girls who read only the Bichitra and the Purbani.
The number of girls who read only the Bichitra and the Sandhani = n [(B ∩ S)\P]
= n (B ∩ S) – n (B ∩ P ∩ S) [Shown in the Venn diagram]
= 30% – 10% = 20%
Set of girls who read only the Bichitra and the Sandhani.
The number of girls who read only the Sandhani and the Purbani = n {(P ∩ S)\B)}
= n (P ∩ S) – n (P ∩ B ∩ S) [Shown in the Venn diagram]
= 20% – 10% = 10%
Set of girls who read only the Purbani and the Sandhani.
∴ The percentage of the girls who read just two magazines = 20% + 20% + 10% = 50% [Shown in the Venn diagram ]
Set of girls read just two magazines
Ans. 50%
26. A = {x : x ∈ ∇ and x² – (a +b)x + ab = 0},
B = {1, 2} and C = {2, 4, 5}
a. Find the elements of the set A.
b. Show that, P(B ∩ C) = P(B) ∩ P(C)
c. Prove that, A × (B ∪ C) = (A × B) ∪ (A × C)
Solution to the question no. 26
a) A = {x : x ∈ ∇ and x² – (a + b) x + ab = 0}
= {x : x ∈ ∇and x² – ax – bx + ab = 0}
= {x : x ∈ ∇ and x (x – a) – b (x – a) = 0}
= {x : x ∈ ∇ and (x – a) (x – b) = 0}
= {x : x ∈ ∇ and x = a, b}
∴ The elements of the set A are a and b.
b) Given, B = {1, 2}, C = {2, 4, 5}
P(B) = {{1}, {2}, {1, 2}, Ø}
P(C) = {{2,}, {4}, {5}, {2, 4}, {2, 5}, {4, 5}, {2, 4, 5}, Ø}
∴ P(B) ∩ P(C) = {{1}, {2}, {1, 2}, Ø} ∩ {{2,}, {4}, {5}, {2, 4}, {2, 5}, {4, 5}, {2, 4, 5}, Ø} = {{2},Ø}
Again, B ∩ C = {1, 2} ∩ {2, 4, 5}
= {2}
∴ P(B ∩ C) = {{2}, Ø}
∴ P(B ∩ C) = P(B) ∩ P(C). (Shown)
c) Given, B = {1, 2}, C = {2, 4, 5}
and A = {a, b} [From ‘a’]
B ∪ C = {1, 2} ∪ {2, 4, 5} = {1, 2, 4, 5}
L.S. = A × (B ∪ C) = {a, b} × {1, 2, 4, 5}
= {(a, 1), (a, 2), (a, 4), (a, 5), (b, 1), (b, 2), (b, 4), (b, 5)}
Again, A × B = {a, b} × {1, 2}
= {(a, 1), (a, 2), (b, 1), (b, 2)}
and A × C = {a, b} × {2, 4, 5}
= {(a, 2), (a, 4), (a, 5), (b, 2), (b, 4), (b, 5)}
R.S. = (A × B) ∪ (A × C)
= {(a, 1), (a, 2), (b, 1), (b, 2)} ∪ {(a, 2), (a, 4), (a, 5),
(b, 2), (b, 4), (b, 5)}
= {(a, 1), (a, 2), (a, 4 ), (a, 5), (b, 1), (b, 2), (b, 4), (b, 5)}
∴ A × (B ∪ C) = (A × B) ∪ (A × C) (Proved)
27. Out of 100 students of a class, 42 students play football, 46 play cricket and 39 play hockey. Among them 13 play football and cricket, 14 play cricket and hockey and 12 play football and hockey. Besides 7 students play none of these games –
a. Show the set of students who play the above three games or none of the games in Venn diagram.
b. Find how many students play all three games.
c. How many students play only one game ? How many play just two of the games?
Solution to the question no. 27
a)
Let, the set of all students be U, the set of students who play football is F, the set of students who play cricket is C, the set of students who play hockey is H.
n (U) = 100
n (F) = 42, n (C) = 46, n (H) = 39
n (F ∩ C) = 13, n (C ∩ H) = 14, n (F ∩ H) = 12
n (F ∩ C ∩ H)′ = 7
b) We know,
n (F ∪ C ∪ H)′ = n (U) – n(F ∪ C ∪H)
or, 7 = 100 – n (F ∪ C ∪ H)
∴ n (F ∪ C ∪H) = 93
Now, n (F ∪ C ∪ H) = n (F) + n (C) + n (H) – n (F ∩ C)
– n (F ∩ H) – n (C ∩ H) + n (F ∩ C ∩ H)
or, 93 = 42 + 46 + 39 – 13 – 12 – 14 + n (F ∩ C ∩ H)
or, n (F ∩ C ∩ H) + 88 = 93
∴ n (F ∩ C ∩ H) = 5
∴ The number of students who play all three games is 5.
Ans. 5 students.
c) Only play football = n (F)– n (F ∩ C) – n (F ∩ H) + n (F ∩ C ∩ H)
= 42 – 13 – 12 + 5
= 22 students
Only play cricket = n (C) – n (F ∩ C) – n (C ∩ H)
+ n (F ∩ C ∩ H)
= 46 – 13 – 14 + 5
= 24 students
Only play hockey = n (H) – n(H ∩ C) – n (H ∩ F)
+ n (F ∩ C ∩ H)
= 39 – 14 – 12 + 5
= 18 students
∴ Only play one game = 22 + 24 + 18 = 64 students
Only play football and cricket = n (F ∩ C) – n (F ∩ C ∩ H)
= 13 – 5
= 8 students
Only play cricket and hockey = n (C ∩ H) – n (F ∩ C ∩ H)
= 14 – 5
= 9 students
