NCERT Solutions for Class 11 Science Maths Chapter 1 - Sets
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NCERT Solutions for Class 11 Science Maths Chapter 1 Sets are provided here with simple step-by-step explanations. These solutions for Sets are extremely popular among class 11 Science students for Maths Sets Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 11 Science Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 12:

Question 1:

Write the following sets in the roaster from
(i) A = {x : x ∈ R, 2x + 11 = 15}
(ii) B = {x | x² = x, x ∈ R}
(iii) C = {x | x is a positive factor of a prime number p}

Answer:

(i) Given: A = {x : x ∈ R, 2x + 11 = 15}
$\Rightarrow 2 x + 11 = 15 \Rightarrow 2 x = 4 \Rightarrow x = 2$
Hence, A = {2}

(ii) Given, B = {x | x² = x, x ∈ R}
$x^{2} = x \Rightarrow x^{2} - x = 0$
$\Rightarrow x (x - 1) = 0 ∴ x = 0, 1$
Hence, B = {0, 1}

(iii) Given, C = {x | x is a positive factor of a prime number p}
Since p is a prime number so it must have only two positive factors i.e. 1 and p
Hence, C = {1, p}

Page No 13:

Question 2:

Write the following sets in the roaster form :
(i) D = {t | t³ = t, t ∈ R}
(ii) $E = \{w |\frac{w - 2}{w + 3} = 3, w \in R\}$
(iii) F = {x | x⁴ – 5x² + 6 = 0, x ∈ R}

Answer:

(i) Given: D = {t | t³ = t, t ∈ R}
$\Rightarrow t^{3} = t \Rightarrow t^{3} - t = 0 \Rightarrow t (t^{2} - 1) = 0$
$\Rightarrow t (t + 1) (t - 1) = 0 ∴ t = - 1, 0, 1$
Hence, D = {-1, 0, 1}

(ii) Given, $E = \{w |\frac{w - 2}{w + 3} = 3, w \in R\}$
$\frac{w - 2}{w + 3} = 3 \Rightarrow w - 2 = 3 w + 9$
$\Rightarrow - 2 - 9 = 3 w - w \Rightarrow 2 w = - 11 ∴ w = \frac{- 11}{2}$
Hence, E = $\{\frac{- 11}{2}\}$

(iii) Given, F = {x | x⁴ – 5x² + 6 = 0, x ∈ R}
$\Rightarrow$ x⁴ – 5x² + 6 = 0
$\Rightarrow (x^{2} - 3) (x^{2} - 2) = 0 ∴ x = \pm \sqrt{3}, \pm \sqrt{2}$
Hence, F = $\{- \sqrt{3}, - \sqrt{2}, \sqrt{2}, \sqrt{3}\}$

Page No 13:

Question 3:

If Y = {x | x is a positive factor of the number 2^{p – 1} (2^p– 1), where 2^p – 1 is a prime number}. Write Y in the roaster form.

Answer:

Given that $2^{p} - 1$ is a prime number
$\Rightarrow$ it has two factors i.e. 1 and $2^{p} - 1$ .
Factors of $2^{p - 1}$ are $1, 2, 2^{2}, 2^{3}, . . ., 2^{p - 1}$ .
Hence, $Y = \{1, 2, 2^{2}, 2^{3}, . . ., 2^{p - 1}, (2^{p} - 1)\}$

Page No 13:

Question 4:

State which of the following statements are true and which are false. Justify your answer.
(i) 35 ∈ {x | x has exactly four positive factors}.
(ii) 128 ∈ {y | the sum of all the positive factors of y is 2y}
(iii) 3 ∉ {x | x⁴ – 5x³ + 2x² – 112x + 6 = 0}
(iv) 496 ∉ {y | the sum of all the positive factors of y is 2y}.

Answer:

(i) Factors of 35 are 1, 5, 7 and 35 which are all positive.
Since 35 is an element of the set.
Hence, the statement is true.

(ii) Factors of 128 are 1, 2, 4, 8, 16, 32, 64 and 128 which are all positive.
Sum of all these factors is 255 which is not equal to $2 \times 128$ .
Hence, the statement is false.

(iii) Let f(x) = x⁴ – 5x³ + 2x² – 112x + 6 = 0
f(3) = $3^{4} - 5 \times 3^{3} + 2 \times 3^{2} - 112 \times 3 + 6 = - 366 \neq 0$
Hence, the statement is true.

(iv) Prime factorisation of 496 is $2^{4} \times 31$ .
So, factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496 which are all positive.
Sum of these factors is 992 which is equal to $2 \times 496$ so 496 belongs to the set.
Hence, the statement is false.

Page No 13:

Question 5:

Given L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}
Verify that L – (M â‹ƒ N) = (L – M) ∩ (L – N)

Answer:

Given: L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}
L – (M $\cup$ N) = {1, 2, 3, 4} – {1, 3, 4, 5, 6} = {2}
(L – M) $\cap$ (L – N) = [{1, 2, 3, 4} – {3, 4, 5, 6}] $\cap$ [{1, 2, 3, 4} – {1, 3, 5}]
= {1, 2} $\cap$ {2} = {2}
Hence, L – (M $\cup$ N) = (L – M) $\cap$ (L – N)

Page No 13:

Question 6:

If A and B are subsets of the universal set U, then show that
(i) A ⊂ A â‹ƒ B
(ii) A ⊂ B ⇔ A â‹ƒ B = B
(iii) (A â‹‚ B) ⊂ A

Answer:

(i) To prove: $A \subset (A \cup B)$
Let x $\in$ A
$\Rightarrow$ x $\in$ A or x $\in$ B
$\Rightarrow$ x $\in$ ( $A \cup B$ )
$\Rightarrow$ $A \subset (A \cup B)$

(ii) To prove: $A \subset B \Leftrightarrow A \cup B = B$
Let $x \in (A \cup B)$
$\Rightarrow x \in A$ or $x \in B$
$\Rightarrow x \in B$ (Since A ⊂ B)
$∴ (A \cup B) \subset B$
And it is obvious that $B \subset (A \cup B)$
Hence, $A \cup B = B$
Also, let $y \in A$
$\Rightarrow y \in (A \cup B)$
$\Rightarrow y \in B$ (Since A â‹ƒ B = B)
$∴ A \subset B$
Hence, $A \subset B \Leftrightarrow A \cup B = B$ .

(iii) To prove: $(A \cap B) \subset A$
Let $x \in (A \cap B)$
$\Rightarrow x \in A$ and $x \in B$
$\Rightarrow x \in A \Rightarrow (A \cap B) \subset A$
Hence, $(A \cap B) \subset A$

Page No 13:

Question 7:

Given that N = {1, 2, 3, ... , 100}. Then write
(i) the subset of N whose elements are even numbers.
(ii) the subset of N whose element are perfect square numbers.

Answer:

Given that N = {1, 2, 3,...,100}
(i) Let the required subset of N is A such that A = {x| x is an even number}
Therefore, A = {2, 4, 6, 8,...,100}

(ii) Let the required subset of N is B such that B = {y| y is a perfect square number}
Therefore, B = {1, 4, 9, 16,...,100}

Page No 13:

Question 8:

If X = {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by

(i) 4n

(ii) n + 6

(iii) $\frac{n}{2}$

(iv) n – 1

Answer:

X = {1, 2, 3}
Since n represents the members of X so n = 1, 2, 3

(i) Let A = {4n | n $\in$ X} = {4, 8, 12}
(ii) Let B = {n + 6 | n $\in$ X} = {7, 8, 9}
(iii) Let C = $\{\frac{n}{2} | n \in X\}$ = $\{\frac{1}{2}, 1, \frac{3}{2}\}$
(iv) Let D = {n − 1 | n $\in$ X} = {0, 1, 2}

Page No 13:

Question 9:

If Y = {1, 2, 3, ... 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.
(i) a ∈ Y but a² ∉ Y
(ii) a + 1 = 6, a ∈ Y
(iii) a is less than 6 and a ∈ Y

Answer:

Given: Y = {1, 2, 3, ... 10}

(i) Let A = $\{a | a \in Y but a^{2} \notin Y\} = \{4, 5, 6, 7, 8, 9, 10\}$

(ii) Let B = $\{a | a + 1 = 6, a \in Y\}$ = {5}

(iii) Let C = $\{a | a < 6, a \in Y\} = \{1, 2, 3, 4, 5\}$

Page No 13:

Question 10:

A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20}
B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Answer:

Here we see that
$A \cap B = \{6, 12\} B \cap C = \{15\} A \cap B \cap C = \emptyset$
Based on the above, the Venn diagram is as follows:

Page No 14:

Question 11:

Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationship among sets U, G, B and S.

Answer:

Given:
U = set of all boys and girls
G = set of all girls
B = set of all boys
S = set of all students who take swimming
Since some but not all the students take swimming so, the Venn diagram is as follows-

Page No 14:

Question 12:

For all sets A, B and C, show that (A – B) ∩ (C – B) = A – (B â‹ƒ C)
Determine whether each of the statement in Exercises 13 – 17 is true or false. Justify your answer.

Answer:

Let $x \in (A - B) \cap (A - C)$
$\Rightarrow x \in (A - B) and x \in (A - C) \Rightarrow (x \in A and x \notin B) and (x \in A and x \notin C)$
$\Rightarrow x \in A and (x \notin B and x \notin C) \Rightarrow x \in A and x \notin (B \cup C)$
$\Rightarrow x \in A - (B \cup C) \Rightarrow (A - B) \cap (A - C) \subset A - (B \cup C) . . . . . (i)$

Now, let $y \in A - (B \cup C)$
$\Rightarrow y \in A and y \notin (B \cup C) \Rightarrow y \in A and (y \notin B and y \notin C)$
$\Rightarrow (y \in A and y \notin B) and (y \in A and y \notin C) \Rightarrow y \in (A - B) and y \in (A - C)$
$\Rightarrow y \in (A - B) \cap (A - C) \Rightarrow A - (B \cup C) \subset (A - B) \cap (A - C) . . . (ii)$
From (i) and (ii),
$A - (B \cup C) = (A - B) \cap (A - C)$ .

Page No 14:

Question 13:

For all sets A and B, (A – B) â‹ƒ (A ∩ B) = A

Answer:

LHS = $(A - B) \cup (A \cap B) = \{(A - B) \cup A\} \cap \{(A - B) \cup B\} = A \cap (A \cup B) = A$ = RHS
Hence, the given statement is true.

Page No 14:

Question 14:

For all sets A, B and C, A – (B – C) = (A – B) – C

Answer:

Venn diagram of $(B - C)$ , $A - (B - C)$ , $(A - B)$ and $(A - B) - C$ are as follows:

We observe that $A - (B - C)$ and $(A - B) - C$ are not same so, the statement is false.

Page No 14:

Question 15:

For all sets A, B and C, if A ⊂ B, then A ∩ C ⊂ B ∩ C

Answer:

$Let x \in (A \cap C)$
$\Rightarrow x \in A and x \in C$
$\Rightarrow x \in B and x \in C$ ( $∵ A \subset B$ )
$\Rightarrow x \in (B \cap C) \Rightarrow (A \cap C) \subset (B \cap C)$
Hence, the given statement is true.

Page No 14:

Question 16:

For all sets A, B and C, if A ⊂ B, then A â‹ƒ C ⊂ B â‹ƒ C

Answer:

$Let x \in (A \cup C) \Rightarrow x \in A or x \in C$
$\Rightarrow x \in B or x \in C (∵ A \subset B)$
$\Rightarrow x \in B \cup C \Rightarrow (A \cup C) \subset (B \cup C)$
Hence, the given statement is true.

Page No 14:

Question 17:

For all sets A, B and C, if A ⊂ C and B ⊂ C, then A â‹ƒ B ⊂ C.

Answer:

$Let x \in (A \cup B) \Rightarrow x \in A or x \in B$
$\Rightarrow x \in C or x \in C (∵ A \subset C and B \subset C)$
$\Rightarrow x \in C \Rightarrow (A \cup B) \subset C$
Hence, the given statement is true.

Page No 14:

Question 18:

Using properties of sets prove the given statement.
For all sets A and B, A â‹ƒ (B – A) = A â‹ƒ B

Answer:

Here, we need to prove that $A \cup (B - A) = A \cup B$
$LHS = A \cup (B - A) = A \cup (B \cap A')$
$= (A \cup B) \cap (A \cup A') = (A \cup B) \cap U = (A \cup B) = RHS$
Hence proved.

Page No 14:

Question 19:

Using properties of sets prove the given statement.
For all sets A and B, A – (A – B) = A ∩ B

Answer:

$LHS = A - (A - B) = A - (A \cap B')$
$= A \cap (A \cap B')' = A \cap \{A' \cup (B')'\}$
$= A \cap (A' \cup B) = (A \cap A') \cup (A \cap B)$
$= \emptyset \cup (A \cap B) = (A \cap B) = RHS$
Hence proved.

Page No 14:

Question 20:

Using properties of sets prove the given statement.
For all sets A and B, A – (A ∩ B) = A – B

Answer:

$LHS = A - (A \cap B) = A \cap (A \cap B)' (∵ A - B = A \cap B')$
$= A \cap (A' \cup B') [∵ (A \cap B)' = A' \cup B'] = (A \cap A') \cup (A \cap B')$
$= \emptyset \cup (A \cap B') = (A \cap B') = A - B = RHS$
Hence proved.

Page No 14:

Question 21:

Using properties of sets prove the given statement.
For all sets A and B, (A â‹ƒ B) – B = A – B

Answer:

$\begin{array}{rcl} LHS & = & (A \cup B) - B \\ = & (A \cup B) \cap B' [∵ (P - Q) = P \cap Q'] \\ = & (A \cap B') \cup (B \cap B') \\ = & (A \cap B') \cup \emptyset [∵ (P \cap P') = \emptyset] \\ = & (A \cap B') \\ = & A - B \\ = & RHS \end{array}$
LHS = RHS
Hence proved.

Page No 14:

Question 22:

Using properties of sets prove the given statement.
Let $T = \{x |\frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x}\}$ . Is T an empty set? Justify your answer.

Answer:

Given: $T = \{x |\frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x}\}$
$∴ \frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x} \Rightarrow \frac{x + 5 - 5 x + 35}{x - 7} = \frac{4 x - 40}{13 - x}$
$\Rightarrow \frac{- 4 x + 40}{x - 7} = \frac{4 x - 40}{13 - x} \Rightarrow (4 x - 40) (x - 13) = (4 x - 40) (x - 7)$
$\Rightarrow (4 x - 40) (x - 13 - x + 7) = 0 \Rightarrow - 6 (4 x - 40) = 0 \Rightarrow (4 x - 40) = 0 \Rightarrow x = \frac{40}{4} \Rightarrow x = 10$
$\Rightarrow T = \{10\}$
Hence, T is not an empty set.

Page No 14:

Question 23:

Let A, B and C be sets. Then show that A ∩ (B â‹ƒ C) = (A ∩ B) â‹ƒ (A ∩ C)

Answer:

$Let x \in A \cap (B \cup C) \Rightarrow x \in A and x \in (B \cup C) \Rightarrow x \in A and (x \in B o r x \in C)$
$\Rightarrow (x \in A and x \in B) o r (x \in A and x \in C) \Rightarrow x \in (A \cap B) or x \in (A \cap C)$
$\Rightarrow x \in (A \cap B) \cup (A \cap C) \Rightarrow A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C) . . . . . (1)$

Now,
$Let y \in (A \cap B) \cup (A \cap C) \Rightarrow y \in (A \cap B) or y \in (A \cap C) \Rightarrow (y \in A and y \in B) or (y \in A and y \in C)$
$\Rightarrow y \in A and (y \in B or y \in C) \Rightarrow y \in A and y \in (B \cup C)$
$\Rightarrow y \in A \cap (B \cup C) \Rightarrow (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C) . . . . . (2)$

From (1) and (2), we get
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Hence proved.

Page No 14:

Question 24:

Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed
(i) in English and Mathematics but not in Science
(ii) in Mathematics and Science but not in English
(iii) in Mathematics only
(iv) in more than one subject only

Answer:

Given:
$n (U) = 100, n (E) = 15, n (M) = 12, n (S) = 8, n (E \cap M) = 6,$ $n (M \cap S) = 7, n (E \cap S) = 4 and n (E \cap M \cap S) = 4$
Here U is the set of all the students, M is the set of students who passed in Mathematics, E is the set of students who passed in English and S is the set of students who passed in Science.

(i) n(English and Mathematics but not in Science)
$\begin{array}{rcl} n (E \cap M \cap S') & = & n (E \cap M) - n (E \cap M \cap N) \\ = & 6 - 4 \\ = & 2 \end{array}$

(ii) n(Mathematics and Science but not in English)
$\begin{array}{rcl} n (M \cap S \cap E') & = & n (M \cap S) - n (M \cap S \cap E) \\ = & 7 - 4 \\ = & 3 \end{array}$

(iii) n(Mathematics only)
$\begin{array}{rcl} n (Only Math) & = & n (M) - n (M \cap S) - n (E \cap M) + n (E \cap M \cap S) \\ = & 12 - 7 - 6 + 4 \\ = & 3 \end{array}$

(iv) n(more than one subject only)
= n(at least two subjects)
= $n (E \cap M \cap S') + n (E \cap M' \cap S) + n (E' \cap M \cap S) + n (E \cap M \cap S)$
= 2 + 0 + 3 + 4
= 9

Page No 14:

Question 25:

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither?

Answer:

Given: $n (U) = 60, n (C) = 25, n (T) = 20 and n (C \cap T) = 10$ , where U is the set of all students, C is the set of students playing cricket and T is the set of students playing tennis.
$\begin{array}{rcl} ∴ n (C \cup T) & = & n (C) + n (T) - n (C \cap T) \\ = & 25 + 20 - 10 \\ = & 35 \end{array}$
Therefore, the number of students who play either cricket or tennis is 35
$\begin{array}{rcl} The number of students who play neither & = & n (U) - n (C \cup T) \\ = & 60 - 35 \\ = & 25 \end{array}$

Hence, the number of students who play neither are 25.

Page No 15:

Question 26:

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.

Answer:

Given: $n (U) = 200, n (M) = 120, n (P) = 90, n (C) = 70, n (M \cap P) = 40, n (P \cap C) = 30, n (C \cap M) = 50$ , where M is the set of students who study mathematics, P is the set of students who study physics and C is the set of students who study chemistry.
The number of students who study none of the given subjects is 20.
So, $n (U) - n (M \cup P \cup C) = 20$

$\begin{array}{rcl} \Rightarrow & n (M \cup P \cup C) = n (U) - 20 \\ = & 200 - 20 \\ = & 180 \end{array}$

$\Rightarrow n (M) + n (P) + n (C) - n (M \cap P) - n (P \cap C) - n (C \cap M) + n (M \cap P \cap C) = n (M \cup P \cup C) \Rightarrow 120 + 90 + 70 - 40 - 30 - 50 + n (M \cap P \cap C) = 180 \Rightarrow 160 + n (M \cap P \cap C) = 180$
$\begin{array}{rcl} \Rightarrow & (M \cap P \cap C) = 180 - 160 \\ = & 20 \end{array}$
Hence, the number of students who study all the three subjects is 20.

Page No 15:

Question 27:

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find
(a) The number of families which buy newspaper A only.
(b) The number of families which buy none of A, B and C.

Answer:

Given: $n (U) = 10000, n (A) = 40 %, n (B) = 20 %, n (C) = 10 %, n (A \cap B) = 5 %, n (B \cap C) = 3 %, n (C \cap A) = 4 %, n (A \cap B \cap C) = 2 %$ , where A is the set of families buying newspaper A, B is the set of families buying newspaper B, C is the set of people buying newspaper C and U is the set of total families in the town.

(a) Percentage of families which buy newspaper A only
$= n (A) - n (A \cap B) - n (A \cap C) + n (A \cap B \cap C) = 40 - 5 - 4 + 2 = 33$
Therefore, the number of families which buy newspaper A only = 33% of 10000 = 3300.

(b) Percentage of families which buy none of the given newspaper
$= n (U) - n (A \cup B \cup C)$
$= n (U) - [n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (C \cap A) + n (A \cap B \cap C)] = 100 - [40 + 20 + 10 - 5 - 3 - 4 + 2] = 40$
Therefore, the number of families which buy none of the newspaper = 40% of 10000 = 4000.

Page No 15:

Question 28:

In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows:
French = 17, English = 13, Sanskrit = 15
French and English = 09, English and Sanskrit = 4
French and Sanskrit = 5, English, French and Sanskrit = 3.

Find the number of students who study
(i) French only
(ii) English only
(iii) Sanskrit only
(iv) English and Sanskrit but not French
(v) French and Sanskrit but not English
(vi) French and English but not Sanskrit
(vii) at least one of the three languages
(viii) none of the three languages

Answer:

Given: $n (U) = 50, n (F) = 17, n (E) = 13, n (S) = 15, n (F \cap E) = 9, n (E \cap S) = 4, n (S \cap F) = 5, n (E \cap F \cap S) = 3$ , where U, E, F and S are the sets of all the students, students studying English, French and Sanskrit respectively.
(i) Number of students studying French only $= n (F) - n (F \cap E) - n (F \cap S) + n (E \cap F \cap S)$
$= 17 - 9 - 5 + 3 = 6$

(ii) Number of students studying English only = $n (E) - n (E \cap F) - n (E \cap S) + n (E \cap F \cap S)$
$= 13 - 9 - 4 + 3 = 3$

(iii) Number of students studying Sanskrit only = $n (S) - n (S \cap F) - n (S \cap E) + n (E \cap F \cap S)$
$= 15 - 5 - 4 + 3 = 9$

(iv) Number of students studying English and Sanskrit but not French = $n (E \cap S) - n (E \cap F \cap S)$
$= 4 - 3 = 1$

(v) Number of students studying French and Sanskrit but not English = $n (F \cap S) - n (E \cap F \cap S)$
$= 5 - 3 = 2$

(vi) Number of students studying French and English but not Sanskrit = $n (F \cap E) - n (E \cap F \cap S)$
$= 9 - 3 = 6$

(vii) Number of students studying at least one of the three languages
$= n (E \cap F \cap S)$
$= n (E) + n (F) + n (S) - n (E \cap F) - n (F \cap S) - n (E \cap S) + n (E \cap F \cap S)$
$= 17 + 13 + 15 - 9 - 4 - 5 + 3 = 30$

(viii) Number of students studying none of the three languages
$= n (U) - n (E \cup F \cup S) = 50 - 30 = 20$

Page No 15:

Question 29:

Choose the correct answers from the given four options.
Suppose A₁, A₂, ..., A₃₀ are thirty sets each having 5 elements and B₁, B₂, ... , B_nare n sets each with 3 elements, let $\cup_{i = 1}^{30} A_{i} = \cup_{j = 1}^{n} B_{j} = S$ and each element of S belongs to exactly 10 of the A_i’s and exactly 9 of the B_i’s. then n is equal to
(A) 15
(B) 3
(C) 45
(D) 35

Answer:

Given: A₁, A₂, ... , A₃₀ are thirty sets each having 5 elements and B₁, B₂, ..., B_nare n sets each with 3 elements.
$A_{1} \cup A_{2} \cup A_{3} \cup . . . \cup A_{30} = 30 \times 5 (Each set has 5 elements)$
Since each element of S is repeated in any 10 $A_{i}' s$ so $n (S) = \frac{30 \times 5}{10} = 15$
Similarly, $n (S) = \frac{3 n}{9} = \frac{n}{3}$

Since both represents the number of elements in S so $\frac{n}{3} = 15 \Rightarrow n = 45$ .
Hence, the correct answer is option C.

Page No 15:

Question 30:

Choose the correct answers from the given four options.
Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n are, respectively,
(A) 4, 7
(B) 7, 4
(C) 4, 4
(D) 7, 7

Answer:

Since two finite sets have m and n elements so the number of their subsets are $2^{m}$ and $2^{n}$ .
According to the question,
$2^{m} - 2^{n} = 112 \Rightarrow 2^{n} (2^{m - n} - 1) = 2^{4} \times 7$
Comparing the factors, $2^{n} = 2^{4} \Rightarrow n = 4$
And,
$2^{m - n} - 1 = 7 \Rightarrow 2^{m - 4} = 8 \Rightarrow 2^{m - 4} = 2^{3} \Rightarrow m - 4 = 3 \Rightarrow m = 7$
Hence, the correct answer is option B.

Page No 16:

Question 31:

Choose the correct answers from the given four options.
The set (A ∩ B')' â‹ƒ (B ∩ C) is equal to
(A) A' â‹ƒ B â‹ƒ C
(B) A' â‹ƒ B
(C) A' â‹ƒ C'
(D) A' ∩ B

Answer:

Given: $(A \cap B')' \cup (B \cap C)$
$\Rightarrow (A' \cup (B')') \cup (B \cap C) \Rightarrow (A' \cup B) \cup (B \cap C)$
$\Rightarrow A' \cup \{B \cup (B \cap C)\} \Rightarrow A' \cup B$
Hence, the correct answer is option B.

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Question 32:

Choose the correct answers from the given four options.
Let F₁ be the set of parallelograms, F₂ the set of rectangles, F₃ the set of rhombuses, F₄ the set of squares and F₅ the set of trapeziums in a plane. Then F₁may be equal to
(A) F₂ ∩ F₃
(B) F₃ ∩ F₄
(C) F₂ â‹ƒ F₅
(D) F₂ â‹ƒ F₃ â‹ƒ F₄ â‹ƒ F₁

Answer:

As a fact, rectangles, squares and rhombuses are specific types of parallelograms only
so, the union of all these make a set of different types of parallelograms only.
Therefore, $F_{1} = F_{2} \cup F_{3} \cup F_{4} \cup F_{1}$
Hence, the correct answer is option D.

Page No 16:

Question 33:

Choose the correct answers from the given four options.
Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then
(A) S ∩ T ∩ C = Ï•
(B) S â‹ƒ T â‹ƒ C = C
(C) S â‹ƒ T â‹ƒ C = S
(D) S â‹ƒ T = S ∩ C

Answer:

Given: The sets T and C are intersecting and are inside the set S so the Venn diagram is as follows-

By the above Venn diagram, it is clear that $S \cup T \cup C = S$
Hence, the correct answer is option C.

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Question 34:

Choose the correct answers from the given four options.
Let R be set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then
(A) R = {(x, y) : 0 ≤ x ≤ a, 0 ≤ y ≤ b}
(B) R = {(x, y) : 0 ≤ x < a, 0 ≤ y ≤ b}
(C) R = {(x, y) : 0 ≤ x ≤ a, 0 < y < b}
(D) R = {(x, y) : 0 < x < a, 0 < y < b}

Answer:

Since R is the set of points inside the rectangle so no point on the sides can be the element of R.

Therefore, $R = \{(x, y) : 0 < x < a, 0 < y < b\}$
Hence, the correct answer is option D.

Page No 16:

Question 35:

Choose the correct answers from the given four options.
In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is
(A) 0
(B) 25
(C) 35
(D) 45

Answer:

Given: $n (U) = 60, n (C) = 25, n (T) = 20 and n (C \cap T) = 10$
Where U, C and T are the universal set, the set of students who play cricket and the set of students who play tennis respectively.

$\begin{array}{rcl} Number of students who play at least one of the games & = & n (C \cup T) \\ = & n (C) + n (T) - n (C \cap T) \\ = & 25 + 20 - 10 \\ = & 35 \end{array}$

$\begin{array}{rcl} ∴ The number of students who play neither & = & n (U) - n (C \cup T) \\ = & 60 - 35 \\ = & 25 \end{array}$

Hence, the correct answer is option B.

Page No 16:

Question 36:

Choose the correct answers from the given four options.
In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
(A) 210
(B) 290
(C) 180
(D) 260

Answer:

Given: $n (U) = 840, n (H) = 450, n (E) = 300 and n (H \cap E) = 200$
Where U, H and E are the universal set, the set of persons who read Hindi and the set of persons who read English respectively.

$\begin{array}{rcl} Number of persons who read at least one of them & = & n (H \cup E) \\ = & n (H) + n (E) - n (H \cap E) \\ = & 450 + 300 - 200 \\ = & 550 \end{array}$

$\begin{array}{rcl} ∴ The number of persons who read neither of them & = & n (U) - n (H \cup E) \\ = & 840 - 550 \\ = & 290 \end{array}$
Hence, the correct answer is option B.

Page No 16:

Question 37:

Choose the correct answers from the given four options.
If X = {8ⁿ – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N}. Then
(A) X ⊂ Y
(B) Y ⊂ X
(C) X = Y
(D) X â‹‚ Y = Ï•

Answer:

Given: X = {8ⁿ – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N}
Since 49n – 49 = 49(n – 1), that means elements of the set Y are divisible by 49 and can produce consecutive multiples of 49.
By mathematical induction, if 8ⁿ – 7n – 1 is divisible by 49.
Let P(n) = 8ⁿ – 7n – 1
Since P(1) = 0.
So, P(1) is true.

Let P(k) is true i.e. 8^k – 7k – 1 = 49c, where c ∈ N .....(1)

Now we will prove that p(k+1) is true.
$8^{k + 1} - 7 (k + 1) - 1 = 8^{k} \times 8 - 7 (k + 1) - 1 = (49 c + 7 k + 1) \times 8 - 7 k - 7 - 1$
$= 49 \times 8 c + 56 k + 8 - 7 k - 8 = 49 \times 8 c + 49 k = 49 (8 c + k)$
Thus, it is proved that 8ⁿ – 7n – 1 is divisible by 49 so, the elements of X are divisible by 49 but cannot produce consecutive multiples of 49. Clearly, all the elements of X is in Y but not every element of Y is in X.
Therefore, X ⊂ Y.
Hence, the correct answer is option A.

Page No 16:

Question 38:

Choose the correct answers from the given four options.
A survey shows that 63% of the people watch a News Channel whereas 76% watch another channel. If x% of the people watch both channel, then
(A) x = 35
(B) x = 63
(C) 39 ≤ x ≤ 63
(D) x = 39

Answer:

Given: $n (C_{1}) = 63 %, n (C_{2}) = 76 %, n (U) = 100 %$ and $n (C_{1} \cap C_{2}) = x %$
Where C_{1 and}C₂ are the sets of people watching respectively 1st and 2nd new channels.
We know,
$n (C_{1} \cup C_{2}) = n (C_{1}) + n (C_{2}) - n (C_{1} \cap C_{2}) = 63 + 76 - x = 139 - x$
As a fact, we know that $n (C_{1} \cup C_{2}) \leq n (U)$ and also $n (C_{1} \cap C_{2}) \leq$ elements of the smaller set.
So, we have
$139 - x \leq 100 \Rightarrow 39 \leq x$ .....(1)
And $n (C_{1} \cap C_{2}) \leq 63 \Rightarrow x \leq 63$ .....(2)

From (1) and (2), we get $39 \leq x \leq 63$
Hence, the correct answer is option C.

Page No 16:

Question 39:

Choose the correct answers from the given four options.
If sets A and B are defined as $A = {(x, y) | y = \frac{1}{x}, 0 \neq x \in R}, B = \{(x, y) |y = - x, x \in R\}$ , then
(A) A ∩ B = A
(B) A ∩ B = B
(C) A ∩ B = Ï•
(D) A â‹ƒ B = A

Answer:

Given: $A = \{(x, y) | y = \frac{1}{x}, 0 \neq x \in R\}$ and $B = \{(x, y) | y = - x, x \in R\}$
By equating the outputs of the functions given, we have
$\frac{1}{x} = - x \Rightarrow x^{2} = - 1$
But there is not such real value of x. That means there is no element common in the two sets.
Hence, the correct answer is option C.

Page No 17:

Question 40:

Choose the correct answers from the given four options.
If A and B are two sets, then A ∩ (A â‹ƒ B) equals
(A) A
(B) B
(C) Ï•
(D) A ∩ B

Answer:

Given expression: $A \cap (A \cup B)$
$= (A \cap A) \cup (A \cap B) = A \cup (A \cap B) = A [(A \cap B) \subset A]$
Hence, the correct answer is option A.

Page No 17:

Question 41:

Choose the correct answers from the given four options.
If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ... , 18} and N the set of natural numbers is the universal set, then A' â‹ƒ {(A â‹ƒ B) ∩ B'} is
(A) Ï•
(B) N
(C) A
(D) B

Answer:

Given: A = {1, 3, 5, 7, 9, 11, 13, 15, 17} and B = {2, 4, ... , 18}
Clearly, $A \cap B = \emptyset$
Now,
$A' \cup \{(A \cup B) \cap B'\} = A' \cup \{(A \cap B') \cup (B \cap B')\} = A' \cup \{(A \cap B') \cup \emptyset\}$

$= A' \cup (A \cap B') = (A' \cup A) \cap (A' \cup B') = N \cap (A' \cup B') [∵ (A' \cup B') \subset N]$

$= (A' \cup B') = (A \cap B)' [∵ (P' \cup Q') = (P \cap Q)'] = \emptyset' = N$
Hence, the correct answer is option B.

Page No 17:

Question 42:

Choose the correct answers from the given four options.
Let S = {x | x is a positive multiple of 3 less than 100} P = {x | x is a prime number less than 20}. Then n(S) + n(P) is
(A) 34
(B) 41
(C) 33
(D) 30

Answer:

Given: S = {x | x is a positive multiple of 3 less than 100} and P = {x | x is a prime number less than 20}
Therefore, S = {3, 6, 9, 12, ...,99}
$\Rightarrow$ n(S) = 33
P = {2, 3, 5, 7, 11, 13, 17, 19}
$\Rightarrow$ n(P) = 8
So, n(S) + n(P) = 33 + 8 = 41.
Hence, the correct answer is option B.

Page No 17:

Question 43:

Choose the correct answers from the given four options.
If X and Y are two sets and X' denotes the complement of X, then X ∩ (X â‹ƒ Y)' is equal to
(A) X
(B) Y
(C) Ï•
(D) X ∩ Y

Answer:

Given:
$X \cap (X \cup Y)' = X \cap (X' \cap Y') = (X \cap X') \cap (X \cap Y')$
$= \emptyset \cap (X \cap Y') = \emptyset$
Hence, the correct answer is option C.

Page No 17:

Question 44:

Fill in the blanks
The set {x ∈ R : 1 ≤ x < 2} can be written as ______________.

Answer:

Given set = {x ∈ R: 1 ≤  x < 2}
{x ∈ R: 1 ≤  x < 2} = {x | $x \in [1, 2)$ and $x \in R$ }

Hence, the set {x ∈ R: 1 ≤  x < 2} can be written as {x | $x \in [1, 2)$ and $x \in R$ }.

Page No 17:

Question 45:

Fill in the blanks
When A = Ï•, then number of elements in P(A) is ______________.

Answer:

Given: A = Ï•
So, the set is an empty set which means it has no element.
Therefore, the number of elements in $P (A) = 2^{0} = 1$
Hence, when A = Ï•, then the number of elements in P(A) is 1 .

Page No 17:

Question 46:

Fill in the blanks
If A and B are finite sets such that A ⊂ B, then n (A â‹ƒ B) = ______________.

Answer:

Given: A ⊂ B that means every element of A is also an element of B.
Then, $n (A \cup B) = n (B)$
Hence, If A and B are finite sets such that A ⊂ B, then n (A â‹ƒ B) = $n (B)$ .

Page No 17:

Question 47:

Fill in the blanks
If A and B are any two sets, then A – B is equal to ______________.

Answer:

$A - B$ is a set of all the elements of A except $A \cap B$ .

Hence, $A - B = A - (A \cap B)$ or $(A \cap B')$ .

Page No 17:

Question 48:

Fill in the blanks
Power set of the set A = {1, 2} is ______________.

Answer:

By the definition of power set, the power set of set A contains all the subsets of set A.
Therefore, the power set of $A = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$ .

Page No 17:

Question 49:

Fill in the blanks
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then the universal set of all the three sets A, B and C can be ______________.

Answer:

Given: A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}
Therefore, the universal set of all the three sets is $A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$ .

Page No 17:

Question 50:

Fill in the blanks
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}. Then
(i) (B â‹ƒ C)' is ______________.
(ii) (C – A)' is ______________.

Answer:

Given: U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}
(i) $(B \cup C)' = U - (B \cup C)$
$= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - [\{2, 4, 6, 7\} \cup \{2, 3, 4, 8\}] = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - \{2, 3, 4, 6, 7, 8\} = \{1, 5, 9, 10\}$

(ii) $(C - A)' = U - (C - A)$
$= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - [\{2, 3, 4, 8\} - \{1, 2, 3, 5\}] = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - \{4, 8\} = \{1, 2, 3, 5, 6, 7, 9, 10\}$

Page No 17:

Question 51:

Fill in the blanks
For all sets A and B, A – (A ∩ B) is equal to ______________.

Answer:

A – (A ∩ B) means the set containing all the elements of A after removing the common elements of A and B.
The following Venn diagram illustrates it:

Clearly, $A - (A \cap B) = A - B$ or $(A \cap B')$ .
Hence, for all sets A and B, A – (A ∩ B) is equal to $(A \cap B')$ .

Page No 17:

Question 52:

Match the following sets for all sets A, B and C

(i) ((A' â‹ƒ B') – A)'	(a) A – B
(ii) [B' â‹ƒ (B' – A)]'	(b) A
(iii) (A – B) – (B – C)	(c) B
(iv) (A – B) ∩ (C – B)	(d) (A × B) ∩ (A × C)
(v) A × (B ∩ C)	(e) (A × B) â‹ƒ (A × C)
(vi) A × (B â‹ƒ C)	(f) (A ∩ C) – B

Answer:

(i) $\{(A' \cup B') - A\}'$
$= \{(A \cap B)' \cap A'\}' = \{(A \cap B) \cup A\} = A$
Hence, (i) mathes with (b).
(ii) $\{B' \cup (B' - A)\}'$
$= \{B' \cup (B' \cap A')\}' = \{B' \cup (B \cup A)'\}' = \{B \cap (B \cup A)\} = B$
Hence, (ii) mathes with (c).
(iii) $(A - B) - (B - C)$
$= (A \cap B') - (B \cap C') = (A \cap B') \cap (B \cap C')' = (A \cap B') \cap (B' \cup C)$
$= \{A \cap (B' \cup C)\} \cap \{B' \cap (B' \cup C)\} = A \cap B' = A - B$
Hence, (iii) matches with (a).
(iv) (A – B) ∩ (C – B)
$= (A \cap B') \cap (C \cap B') = (A \cap C) \cap B' = (A \cap C) - B$
Hence, (iv) matches with (f).
(v) $A \times (B \cap C)$
$= (A \times B) \cap (A \times C)$
Hence, (v) matches with (d).
(vi) $A \times (B \cup C)$
$= (A \times B) \cup (A \times C)$
Hence, (vi) matches with (e).

Page No 18:

Question 53:

State True or False for the given statements.
If A is any set, then A ⊂ A.

Answer:

Since every set is a subset of itself so, A ⊂ A.
Hence, the given statement is true.

Page No 18:

Question 54:

State True or False for the given statements.
Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then B ⊄ M.

Answer:

Given: M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Since every element of the set B is also an element of the set A so, B and M are equal sets and also subsets of each other.
Hence, the given statement is false.

Page No 18:

Question 55:

State True or False for the given statements.
The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.

Answer:

Given sets are {1, 2, 3, 4} and {3, 4, 5, 6}.
Since every element of the two sets are not same so, they are not equal.
Hence, the given statement is false.

Page No 18:

Question 56:

State True or False for the given statements.
Q â‹ƒ Z = Q, where Q is the set of rational numbers and Z is the set of integers.

Answer:

Since every integer is also a rational number so, Z ⊂ Q.
Therefore, Q â‹ƒ Z = Q
Hence, the given statement is true.

Page No 18:

Question 57:

State True or False for the given statements.
Let sets R and T be defined as
R = {x ∈ Z | x is divisible by 2}
T = {x ∈ Z | x is divisible by 6}. Then T ⊂ R

Answer:

Given: R = {x ∈ Z | x is divisible by 2} and T = {x ∈ Z | x is divisible by 6}
Therefore, R = {2, 4, 6, 8, 10, 12, ...} and T = {6, 12, 18, 24, ...}
Clearly, every integer divisible by 6 is also divisible by 2 so, T ⊂ R.
Hence, the given statement is true.

Page No 18:

Question 58:

State True or False for the given statements.
Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}. Then A = B.

Answer:

Given: A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}
Since there are infinite real numbers in the interval [0, 2].
n(A) = 3 and n(B) = ∞.
Hence, the given statement is false.

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