Students appreciate clear and concise ISC Maths Class 11 Solutions Chapter 1 Sets Chapter Test that guide them through exercises.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Chapter Test

Question 1.
Decide, among the following sets, which are subsets of which.
A = {x : x is a solution of x2 – 8x + 12 = 0},
B = {2 ,4, 6},
C = {x : x is an even natural number),
D = {6).
Solution:
Given A = {x : x is a solution of x2 – 8x + 12 = 0}
Now, x2 – 8x + 12 = 0
⇒ (x – 2) (x – 6) = 0
⇒ (x – 2) (x – 6) = 0
⇒ x = 2, 6
and B = {2, 4, 6} and A = {2, 6}
Hence every member of set A be a member of set B
∴ A be a subset of B.
Thus A ⊂ B.
Since C = {x : x is an even natural number}
= {2, 4, 6, 8, ………………….}
and D = {6}

Question 2.
Let ξ = the set of all letters in the word ‘TAMILNADU’ and X = {x : x is a vowel and x ∈ ξ}.
(i) Write ξ and X in the roster form.
(ii) Tell n (ξ) and n(X).
(Ill) List all the proper subsets of X.
(iv) What is the cardinal number of the power set of X?
Solution:
(i) In roster form,
ξ = {T, A, M, I, L, N, D, U}
and X = {x : x is a vowel and x ∈ ξ}
= {A, I, U}

(ii) n (ξ) = no. of distinct elements in ξ = 8
and n (X) = Number of distinct elements in
X = 3

(iii) The proper subsets of X are given below:
Φ, {A), {I}, {U}, {A, I}, {A, U}, {I, U)

(iv) Power set of X = P(X)
= {Φ, {A}, {I}, {U}, {A, I}, {A, U}, {I, U}, {(A. I, U}}
∴ n (P (X)) = 23 = 8.

Question 3.
Let A be the set of letters in the word “POOR”. Write the power set of A.
Solution:
A = {P, O, R}
P(A) = {Φ, {P}, {O}, {R, {P, O}, {O, R}, {P, R}, {P, O, R}}
∴ Cardinal number of power set of A = n (P (A)) = 8

Question 4.
Find the power set of the following sets:
(i) {- 1, 0, 1}
(ii) {0, 1, {0, 1}}
Solution:
(i) Let A = {- 1, 0, 1}
∴ P (A) = {Φ, (- 1), {0}, {1}, {- 1, 0}, {0, 1}, {- 1, 1}, {- 1,0, 1}}

(ii) {0, 1, {0, 1}
Let A = {0, 1, {0, 1}}
∴ P(A) = {Φ, {0}, {1}, {{0, 1}}, {0, 1}, {0, {0, 1}}, {1, {0, 1}}, {0, 1, {0, 1}}}

Question 5.
If A = {2, 3, 5, 7, 8}, B = {1, 5, 9} and A’ = {1, 4, 6, 9}, verify that
(i) (A ∪ B)’ = A’ ∩ B’
(ii) B – A = A’ ∩ B
Solution:
Given A = {2, 3, 5, 7, 8) ;
B = {1, 5, 9} and
A’ = {1, 4, 6, 9}
Since A ∪ A’ = ξ
⇒ ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(i) A ∪ B = {1, 2, 3, 5, 7, 8, 9}
∴ (A ∩ B)’ = ξ – A ∪ B
= {4, 6} ………………(1)
B’ = ξ – B = {2, 3, 4, 6, 7, 8}
∴ A’ ∩ B’ = {1, 4, 6, 9} ∩ {2, 3, 4, 6, 7, 8}
= {4, 6} …………………(2)
∴ from (1) and (2) ; we have
(A ∪ B)’ = A’ ∩ B’

(ii) L.H.S. = B – A = {1, 9}
R.H.S. = A’ ∩B
= {1, 4, 6, 9} ∩ {1, 5, 9}
= {1, 9}
∴ B – A = A’ ∩ B

Question 6.
For all sels A, B and C, is A – (B – C) = (A – B) – C true? Justify your answer.
Solution:
(i) Let A = {1, 2},
B = {2, 3}
and C = {1, 3}
∴ B – C = {2}
Thus A – (B – C) = {1, 2} – {2} = {1} …………………(1)
Now, A – B = {1}
∴ (A – B) – C = {1} – {1, 3} = Φ ………………………(2)
∴ from (1) and (2) ; we have
∴ given statement is false.

Question 7.
If n (ξ) = 30, n(A’) = 15, n (B) = 5 and n (A ∩ B) = 3, find
(i) n (A)
(ii) n (A ∪ B)
(iii) n(A – B).
Solution:
Given n (ξ) = 30,
n (A’) = 15,
n (B) = 5
and n (A ∩ B) = 3

(i) n (A) = n(ξ) – n(A’)
= 30 – 15 = 15

(ii) n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
= 15 + 5 – 3 = 17

(iii) n (A – B) = n (A) – n (A ∩ B)
= 15 – 3 = 12

Question 8.
If n (ξ) = 40, n ((A ∪ B)’)= 12, n (A – B) = 10 and n (B – A) = 14, find
(i) n (A)
(ii) n (B)
(iii) n (A ∩ B).
Solution:
Given n (A) = 40,
n ((A ∪ B)’) = 12,
n (A – B) = 10
and n (B – A) = 14
Now n ((A ∪ B)’) = n (ξ) – n (A ∪ B)
⇒ 12 = 40 – n (A ∪ B)
⇒ n (A ∪ B) = 40 – 12 = 28
we know that,

(i) Since n (A – B) = 10
⇒ n (A) – n (A ∩ B) = 10
⇒ n (A) – 4 = 10
⇒ n(A) = 14

(ii) Since n (B – A) = 14
⇒ n (B) – n (A ∩ B) = 14
⇒ n(B) = 14 + 4 = 18

(iii) n (A ∪ B) = n (A – B) + n (A ∩ B) + n (B – A)
⇒ 28 = 10 + n (A ∩ B) + 14
⇒ n (A ∩ B) = 28 – 24 = 4

Question 9.
Two sets A and B are such that n (A ∪ B) = 18, n (A’ ∩ B) = 3 and n (A ∩ B’) = 5, find n (A ∩ B).
Solution:
Given n (A ∪ B) = 18,
n (A’ ∩ B) = 3
and n (A ∩ B’) = 5

we know that,
n (A’ ∩ B) = n (B – A)
and n (A ∩ B’) = n (A – B)
we know that,
n (A ∪ B) = n (A – B) + n (A ∩ B) + n (B – A)
⇒ 18 = 5 + n (A ∩ B) + 3
⇒ n (A ∩ B) = 18 – 8 = 10

Question 10.
Two sets A and B are such that n (A ∪ B) = 21, n (A’ ∩ B’) = 9 and n (A ∩ B) = 7, find n ((A ∩ B)’).
Solution:
Given n (A ∪ B) = 21 ;
n (A’ ∩ B’) = 9
and n (A ∩ B) = 7
Now n (A’ ∩ B’) = n ((A ∪ B) = 9
[using Demorgan’s law]
Here, ξ = (A ∪ B) ∪ (A ∪ B)’
⇒ n (ξ) = n (A ∪ B) + n ((A ∪ B)’)
⇒ n (ξ) = 21 + 9 = 30
∴ n (A ∩ B)’ = n (ξ) – n (A ∩ B)
= 30 – 7 = 23

Question 11.
If n (ξ) = 50, n (A) = 3x, n (B) = 2x and n (A ∩ B) = x = n ((A ∪ B)’), find
(i) the value of x
(ii) n (A – B).
Solution:
Given n (ξ) = 50,
n (A) = 3x,
n (B) = 2x,
n (A ∩ B) = x = n ((A ∪ B)’)

(i) Since x = n ((A ∪ B)’)
= n (ξ) – n (A ∪ B)
⇒ x = 50 – [n (A) + n (B) – n (A ∩ B)]
⇒ x = 50 – [3x + 2x – x]
⇒ x = 50 – 4x
⇒ 5x = 50
⇒ x = 10

(ii) n (A – B) = n(A) – n (A ∩ B)
= 3x – x
= 2x
= 2 × 10 = 20

Question 12.
If n (ξ) = 15, A and B are two sets such that A ⊂ B, n (A) = 8 and n (B) = 12, use Venn diagram to find the following:
(i) n (A’)
(ii) n (B’)
(iii) n (A ∩ B’)
(iv) n (A’ ∩ B).
Solution:
(i) Since A ⊂ B,
n (A) = 8 ;
n (B) = 12
and n (ξ) = 15

(i) n (A’) = 3 + 47
[or n (A’) = n (ξ) – n (A)
= 15 – 8 = 7]

(ii) n (B’) = 15 – 12 = 3
[or n (B’) = n (ξ) – n(B)
= 15 – 12 = 3]

(iii) n (A ∩ B)’ = 0
[or n (A ∩ B’) = n (A) – n(A ∩ B)
= 8 – 8 =0]

(iv) n (A’ ∩ B) = 4
[or n (A’ ∩ B) = n (B) – n (A ∩B)
= 12 – 8 = 4]

Question 13.
In an examination, 56 per cent of the candidates failed in English and 48 percent failed in Science. If 18 per cent failed in both English and Science, flnd the percentage of those who passed in both the subjects.
Solution:
Let ξ be the set of all candidates.
Let E and S be the set of candidates who passed in English and Science respectively.
Then n (ξ) = 100 % ;
n (E) = (100 – 56) % = 44 % ;
n (S) = (100 – 48) % = 52 %
Also n (E’ ∩ S’) = 18 %
⇒ n ((E ∪ S))’ = 18 %
⇒ n (ξ) – n (E ∪ S) = 18 %
⇒ 100 % – {n (E) + n (S) – n(E ∩ S)} = 18 %
⇒ 100% – {44 % + 52 % – n(E ∩ S)} = 18 %
⇒ 100 – 96 % + n (E ∩ S) = 18%
⇒ n (E ∩ S) = 14%
Hence, required % of students who passed in both the subjects = 14%

Question 14.
From amongst the 6000 literate individuals of a city, 50 % read newspaper A, 45 % read newspaper B and 25 % read neither A nor B. How many individuals read both the newspapers A as well as B?
Solution:
Let be the total no. of individuals.
LetA and B be the set of aIL individuals who read newspaper A and newspaper B.
∴ n (A) = 50 % ;
n (B) = 45 %
n (A’ ∩ B’) = 25 %
n ((A ∪ B)’) = 25 %
⇒ 100 % – n (A ∩ B) = 25 %
⇒ 100 % – [n (A) + n (B) – n (A ∩ B)] = 25 %
⇒ 100 % – [50 % + 45 % – n (A ∩ B)] = 25 %
⇒ 5 % + n (A ∩ B) = 25 %
⇒ n (A ∩ B) = 20 %
Hence total number of individuals who read newspaper A as well as B = 20% of 6000
= ($$\frac{20}{100}$$ × 6000) = 1200

Question 15.
In a beauty contest, half the number of judges voted for Miss A, $$\frac{2}{3}$$ of them voted for Miss B, 10 voted for both and 6 did not vote for either Miss A or Miss B. Find how many judges, In all, were present there.
Solution:
Let the total number of judges in all were present in a beauty contest = x
The no. ofjudges who voted for miss A = n (A) = $$\frac{x}{2}$$
and no. ofjudges who noted for miss B = n (B) = $$\frac{2}{3}$$ x
n (A ∩ B) = 10
and 6 = n ((A ∪ B))’
⇒ 6 = x – n (A ∪ B)
⇒ 6 = x – [n (A) + n (B) – n (A ∩ B)]
⇒ 6 = x – $$\left[\frac{x}{2}+\frac{2}{3} x-10\right]$$
⇒ – 4 = $$\frac{6 x-3 x-4 x}{6}$$
⇒ – 4 = – $$\frac{x}{6}$$
⇒ x = 4
Hence the required number of judges were present in beauty contest be 24.

Question 16.
In a group of 50 students, the number of students studying French, English and Sanskrit were found to be as follows :
French = 17, English = 13, Sanskrit = 15 ; French and English = 9, English and Sanskrit = 4, French and Sanskrit = 5 ; English, French aiid Sanskrit 3.
Find the number of students who study :
(i) French only
(ii) French and Sanskrit but not English
(iii) English only
(iv) French and English but not Sanskrit
(v) Sanskrit only
(vi) English and Sanskrit but not French
(vii) atleast one of the three languages
(viii) none of the three languages.
Solution:
Let ξ be the no. of students in a school.
Let F, E and S be the set of students studying French, English and Sanskrit.

Then the no. of students according to given information in the question are shown in different regions in adjoining venn diagram.

(i) Reqd. no. of students who study French only = 6

(ii) Reqd. no. of students who study french and Sanskrit but not English = 2

(iii) Reqd. no. of students who study English only = 3

(iv) Reqd. no. of students who study French and English but not Sanskrit = 6

(v) Reqd. no. of students who study Sanskrit only = 9

(vi) Reqd. no. of students who study English and Sanskrit but not french = 1

(vii) Reqd. no. of students who study atleast one of the three languages = 6 + 6 + 3 + 2 + 3 + 1 + 9 = 30

(viii) Reqd. number of students who study none of the three language = n (ξ) – no. of students who study atleast one of the three languages
= 50 – 30 = 20.

Question 17.
If A and B are two sets such that n (A) = 10 and n (B) = 7, then find:
(i) the least value of n (A ∩ B)
(ii) the greatest value of n (A ∩ B)
(ii,) the greatest value of n (A ∪ B)
(iv) the least value of n (A ∪ B).
Solution:
Given n (A) = 10 ;
n(B) = 7
we know that,
n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
⇒ n (A ∪ B) = 10 + 7 – n (A ∩ B)
⇒ n (A ∪ B) ≤ 17 ………………..(1)
[∵ n(A ∩ B) ≥ 0
∴ – n(A ∩ B) ≤ 0]
Now, equality sign helds in eqn. (1) only when (A ∩ B) = 0
i.e. A ∩ B = Φ
Now, A ⊂ A ∪ B and B ⊂ A ∪B
⇒ n (A) ≤ n (A ∪ B) and n(B) ≤ n (A∪B)
⇒ n (A ∪ B) ≥ max {n(A), n(B)}
⇒ n (A ∪ B) ≥ max {10, 7} = 10
⇒ 10 ≤ n (A ∪ B) …………………..(2)
From (1) and (2) ; we have
10 ≤ n (A ∪ B) ≤ 17
∴ least value of n (A ∪ B) = 10
greatest value of n (A ∪ B) = 17
Also A ∩ B ⊂ A and A ∩ B ⊂ B
⇒ n (A ∩ B) ≤ n(A) and n (A ∩ B) ≤ n(B)
⇒ n (A ∩ B) ≤ min. (n (A), n(B)}
⇒ n (A ∩ B) ≤ min {10, 7} = 7
It may be noted that equality sign hold when A ⊂ B or B ⊂ A.
Thus, greatest value of n (A ∩ B) = 7
also n (A ∩ B) ≥ 0
∴ least value of n (A ∩ B) = 0.
[when A ∩ B = Φ].