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Matrices Class 12 Commerce Maths 1 Chapter 2 Exercise 2.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

Std 12 Maths 1 Exercise 2.2 Solutions Commerce Maths

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\) show that
(i) A + B = B + A
(ii) (A + B) + C = A + (B + C)
Solution:


From (1) and (2), we get
(A + B) + C = A + (B + C).

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\), then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
-3 & 7 & -8 \\
0 & -6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
9 & -1 & 2 \\
-4 & 2 & 5 \\
4 & 0 & -3
\end{array}\right]\), then find the matrix C such that A + B + C is a zero matrix.
Solution:
A + B + C = 0
∴ C = -A – B

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\), find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A – 4B + 5X = C
∴ 5X = C – 3A + 4B

Question 5.
If A = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\), find (A^T)^T.
Solution:
A = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
5 & 3 \\
1 & 2 \\
-4 & 0
\end{array}\right]\)
∴ (A^T)^T = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\) = A

Question 6.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\), find (A^T)^T.
Solution:
A = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rrr}
7 & -2 & 5 \\
3 & -4 & 9 \\
1 & 1 & 1
\end{array}\right]\)
∴ (A^T)^T = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\) = A

Question 7.
Find a, b, c if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symetric matrix.
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\)
Since, A is a symmetric matrix, a_ij = a_ji for all i and j

Question 8.
Find x, y, z if \(\left[\begin{array}{ccc}
0 & -5 i & x \\
y & 0 & z \\
\frac{3}{2} & -\sqrt{2} & 0
\end{array}\right]\) is a skew symmetric matrix.
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & -5 i & x \\
y & 0 & z \\
\frac{3}{2} & -\sqrt{2} & 0
\end{array}\right]\)
Since, A is skew-symmetric matrix,

Question 9.
For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
(i) \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Then A^T = \(\left[\begin{array}{rrr}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Since, A = A^T, A is a symmetric matrix.

(ii) \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
Solution:
Let B = \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
Then B^T = \(\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)\)
∴ B ≠ B^T
Also,
-B^T = \(\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)=\left(\begin{array}{rrr}
-2 & 5 & 1 \\
-5 & -4 & 6 \\
-1 & -6 & -3
\end{array}\right)\)
∴ B ≠ -B^T
Hence, B is neither symmetric nor skew-symmetric matrix.

(iii) \(\left[\begin{array}{ccc}
0 & 1+2 i & i-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\)
Solution:

Hence, C is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [a_ij]_3×3, where a_ij = i – j. State whether A is symmetric or skew-symmetric.
Solution:

Question 11.
Solve the following equations for X and Y, if 3X – Y = \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find matrices A and B, if 2A – B = \(\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:

Question 13.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:

By equality of matrices, we get
2x + y – 1 = 3 ……..(1)
and 4y = 18 ……….(2)
From (2), y = \(\frac{9}{2}\)
Substituting y = \(\frac{9}{2}\) in (1), we get
2x + \(\frac{9}{2}\) – 1 = 3
∴ 2x = 3 – \(\frac{7}{2}\) = \(\frac{-1}{2}\)
∴ x = \(\frac{-1}{4}\)
Hence, x = \(\frac{-1}{4}\) and y = \(\frac{9}{2}\).

Question 14.
If \(\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)
By equality of matrices,
2a + b = 2 ….. (1)
3a – b = 3 …… (2)
c + 2d = 4 …… (3)
2c – d = -1 …… (4)
Adding (1) and (2), we get
5a = 5
∴ a = 1
Substituting a = 1 in (1), we get
2(1) + b = 2
∴ b = 0
Multiplying equation (4) by 2, we get
4c – 2d = -2 …… (5)
Adding (3) and (5), we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (4), we get
2(\(\frac{2}{5}\)) – d = -1
∴ d = \(\frac{4}{5}\) + 1 = \(\frac{9}{5}\)
Hence, a = 1, b = 0, c = \(\frac{2}{5}\) and d = \(\frac{9}{5}\).

Question 15.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B:
July sales (in Rupees), Physics, Chemistry, Mathematics
A = \(\left[\begin{array}{lll}
5600 & 6750 & 8500 \\
6650 & 7055 & 8905
\end{array}\right]\) First Row Suresh / Second Row Ganesh
August Sales (in Rupees), Physics, Chemistry, Mathematics
B = \(\left[\begin{array}{ccc}
6650 & 7055 & 8905 \\
7000 & 7500 & 10200
\end{array}\right]\) First Row Suresh / Second Row Ganesh
(i) Find the increase in sales in Z from July to August 2017.
(ii) If both book shops get 10% profit in the month of August 2017,
find the profit for each bookseller in each subject in that month.
Solution:
The sales for July and August 2017 for Suresh and Ganesh are given by the matrices A and B as:

(i) The increase in sales (in ₹) from July to August 2017 is obtained by subtracting the matrix A from B.


Hence, the increase in sales (in ₹) from July to August 2017 for:
Suresh book shop: ₹ 1050 in Physics, ₹ 305 in Chemistry, and ₹ 405 in Mathematics.
Ganesh book shop: ₹ 350 in Physics, ₹ 445 in Chemistry, and ₹ 1295 in Mathematics.
(ii) Both the book shops get 10% profit in August 2017,
the profit for each bookseller in each subject in August 2017 is obtained by the scalar multiplication of matrix B by 10%,
i.e. \(\frac{10}{100}=\frac{1}{10}\)

Hence, the profit for Suresh book shop are ₹ 665 in Physics, ₹ 705.50 in Chemistry and ₹ 890.50 in Mathematics and for Ganesh book shop are ₹ 700 in Physics, ₹ 750 in Chemistry and ₹ 1020 in Mathematics.

12th Commerce Maths Book Pdf 

• Matrices Ex 2.1 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.2 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.3 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.4 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.5 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.6 Class 12 Commerce Maths Textbook Solutions
• Matrices Miscellaneous Exercise 2 Class 12 Commerce Maths Textbook Solutions

Matrices Class 12 Commerce Maths 1 Chapter 2 Exercise 2.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

Std 12 Maths 1 Exercise 2.1 Solutions Commerce Maths

Question 1.
Construct a matrix A = [a_ij]_3×2 whose elements aij isgiven by
(i) a_ij = \(\frac{(i-j)^{2}}{5-i}\)
Solution:

(ii) a_ij = i – 3j
Solution:
a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = -2
a₁₂ = 1 – 3(2) = 1 – 6 = -5
a₂₁ = 2 – 3(1) = 2 – 3 = -1
a₂₂ = 2 – 3(2) = 2 – 6 = -4
a₃₁ = 3 – 3(1) = 3 – 3 = 0
a₃₂ = 3 – 3(2) = 3 – 6 = -3
∴ A = \(\left[\begin{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
\end{array}\right]\)

(iii) a_ij = \(\frac{(i+j)^{3}}{5}\)
Solution:

Question 2.
Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular matrix:

Solution:
(i) Since, all the elements below the diagonal are zero, it is an upper triangular matrix.
(ii) This matrix has only one column, it is a column matrix.
(iii) This matrix has only one row, it is a row matrix.
(iv) Since, diagonal elements are equal and non-diagonal elements are zero, it is a scalar matrix.
(v) Since, all the elements above the diagonal are zero, it is a lower triangular matrix.
(vi) Since, all the non-diagonal elements are zero, it is a diagonal matrix.
(vii) Since, diagonal elements are 1 and non-diagonal elements are 0, it is an identity (or unit) matrix.

Question 3.
Which of the following matrices are singular or non-singular:
(i) \(\left[\begin{array}{ccc}
a & b & c \\
p & q & r \\
2 a-p & 2 b-q & 2 c-r
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
Solution:
Let C = \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
∴ |C| = \(\left|\begin{array}{rrr}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right|\)
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49
= 52 ≠ 0
∴ C is a non-singular matrix.

(iv) \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
Solution:
Let D = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
∴ |D| = \(\left|\begin{array}{rr}
7 & 5 \\
-4 & 7
\end{array}\right|\)
= 49 – (-20)
= 69 ≠ 0
∴ D is a non-singular matrix.

Question 4.
Find k, if the following matrices are singular:
(i) \(\left[\begin{array}{cc}
7 & 3 \\
-2 & K
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
7 & 3 \\
-2 & K
\end{array}\right]\)
Since, A is a singular matrix, |A| = 0
∴ \(\left|\begin{array}{rr}
7 & 3 \\
-2 & k
\end{array}\right|\) = 0
∴ 7k – (-6) = 0
∴ 7k = -6
∴ k = \(-\frac{6}{7}\)

(ii) \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{~K} & 1 \\
10 & 9 & 1
\end{array}\right]\)
Solution:
Let B = \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{~K} & 1 \\
10 & 9 & 1
\end{array}\right]\)
Since, B is a singular matrix, |B| = 0
∴ \(\left|\begin{array}{rrr}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right|\) = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6.

(iii) \(\left[\begin{array}{ccc}
K-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Solution:
Let C = \(\left[\begin{array}{ccc}
K-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Since, C is a singular matrix, |C| = 0
∴ \(\left|\begin{array}{crr}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right|\) = 0
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3(-6 – 1) = 0
∴ 8k – 8 – 20 – 21 = 0
∴ 8k = 49
∴ k = \(\frac{49}{8}\)

12th Commerce Maths Book Pdf 

• Matrices Ex 2.1 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.2 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.3 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.4 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.5 Class 12 Commerce Maths Textbook Solutions
• Matrices Ex 2.6 Class 12 Commerce Maths Textbook Solutions
• Matrices Miscellaneous Exercise 2 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Miscellaneous Exercise 1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.

Std 12 Maths 1 Miscellaneous Exercise 1 Solutions Commerce Maths

(I) Choose the correct alternative:

Question 1.
Which of the following is not a statement?
(a) Smoking is injurious to health
(b) 2 + 2 = 4
(c) 2 is only even prime number
(d) Come here
Answer:
(d) Come here

Question 2.
Which of the following is an open statement?
(a) x is a natural number
(b) Give me a glass of water
(c) Wish you best of luck
(d) Good morning to all
Answer:
(a) x is a natural number

Question 3.
Let p ∧ (q ∨ r) = (p ∧ q) ∨ (p ∧ r). Then this law is known as
(a) Commutative law
(b) Associative law.
(c) De Morgan’s law
(d) Distributive law
Answer:
(d) Distributive law

Question 4.
The false statement in the following is:
(a) p ∧ (~p) is a contradiction
(b) (p → q) ↔ (~q → ~p) is a contradiction
(c) ~(~p) ↔ p is a tautology
(d) p ∨ (~p) ↔ p is a tautology.
Answer:
(b) (p → q) ↔ (~q → ~p) is a contradiction

Question 5.
Consider the following three statements
p : 2 is an even number.
q : 2 is a prime number.
r : Sum of two prime numbers is always even.
Then, the symbolic statement (p ∧ q) → ~r means:
(a) 2 is an even and prime number and the sum of two prime numbers is always even.
(b) 2 is an even and prime number and the sum of two prime numbers is not always even.
(c) If 2 is an even and prime number, then the sum of two prime numbers is not always even.
(d) If 2 is an even and prime number, then the sum of two prime numbers is also even.
Answer:
(c) If 2 is an even and prime number, then the sum of two prime numbers is not always even.

Question 6.
If p : He is intelligent.
q : He is strong.
Then, symbolic form of statement: ‘It is wrong that, he is intelligent or strong’ is
(a) ~p ∨ ~p
(b) ~(p ∧ q)
(c) ~(p ∨ q)
(d) p ∨ ~q
Answer:
(c) ~(p ∨ q)

Question 7.
The negation of the proposition ‘If 2 is prime, then 3 is odd’, is
(a) If 2 is not prime, then 3 is not odd
(b) 2 is prime and 3 is not odd
(c) 2 is not prime and 3 is odd
(d) If 2 is not prime, then 3 is odd
Answer:
(b) 2 is prime and 3 is not odd

Question 8.
The statement (~p ∧ q) ∨ ~q is
(a) p ∨ q
(b) p ∧ q
(c) ~(p ∨ q)
(d) ~(p ∧ q)
Answer:
(d) ~(p ∧ q)
Hint:
(~p ∧ q) ∨ ~q = (~p ∨ ~q) ∧ (q ∨ ~q)
= (~p ∨ ~q) ∧ t
= ~p ∨ ~q
= ~(p ∧ q)

Question 9.
Which of the following is always true?
(a) ~(p → q) ≡ ~q → ~p
(b) ~(p ∨ q) ≡ ~p ∨ ~q
(c) ~(p → q) ≡ p ∧ ~q
(d) ~(p ∧ q) ≡ ~p ∧ ~q
Answer:
(c) ~(p → q) ≡ p ∧ ~q

Question 10.
~(p ∨ q) ∨ (~p ∧ q) is logically equivalent to
(a) ~p
(b) p
(c) q
(d) ~q
Answer:
(a) ~p
Hint:
~(p ∨ q) ∨ (~p ∧ q) ≡ (~p ∧ ~q) ∨ (~p ∧ q)
≡ ~p ∧ (~q ∨ q)
≡ ~p ∧ t
≡ ~p

Question 11.
If p and q are two statements, then (p → q) ↔ (~q → ~p) is
(a) contradiction
(b) tautology
(c) neither (a) nor (b)
(d) none of these
Answer:
(b) tautology

Question 12.
If p is the sentence ‘This statement is false’, then
(a) truth value of p is T
(b) truth value of p is F
(c) p is both true and false
(d) p is neither true nor false
Answer:
(d) p is neither true nor false

Question 13.
Conditional p → q is equivalent to
(a) p → ~q
(b) ~p ∨ q
(c) ~p → ~q
(d) p ∨ ~q
Answer:
(b) ~p ∨ q

Question 14.
Negation of the statement ‘This is false or That is true’ is
(a) That is true or This is false
(b) That is true and This is false
(c) This is true and That is false
(d) That is false and That is true
Answer:
(c) This is true and That is false

Question 15.
If p is any statement, then (p ∨ ~p) is a
(a) contingency
(b) contradiction
(c) tautology
(d) none of them
Answer:
(c) tautology

(II) Fill in the blanks:

Question 1.
The statement q → p is called as the ___________ of the statement p → q.
Answer:
Converse

Question 2.
Conjunction of two statements p and q is symbolically written as
Answer:
p ∧ q

Question 3.
If p ∨ q is true, then truth value of ~p ∨ ~q is ___________
Answer:
False

Question 4.
Negation of ‘some men are animal’ is ___________
Answer:
All men are not animal.
OR
No men are animals.

Question 5.
Truth value of if x = 2, then x² = -4 is ___________
Answer:
False

Question 6.
Inverse of statement pattern p → q is given by ___________
Answer:
~p → ~q

Question 7.
p ↔ q is false when p and q have ___________ truth values.
Answer:
Different

Question 8.
Let p : The problem is easy. r : It is not challenging. Then verbal form of ~p → r is ___________
Answer:
If the problem is not easy, then it is not challenging.

Question 9.
Truth value of 2 + 3 = 5 if and only if -3 > -9 is ___________
Answer:
T [Hint: T ↔ T = T]

(III) State whether each of the following is True or False:

Question 1.
Truth value of 2 + 3 < 6 is F.
Answer:
False

Question 2.
There are 24 months in a year is a statement.
Answer:
True

Question 3.
p ∧ q has truth value F if both p and q have truth value F.
Answer:
False

Question 4.
The negation of 10 + 20 = 30 is, it is false that 10 + 20 ≠ 30.
Answer:
False

Question 5.
Dual of (p ∧ ~q) ∨ t is (p ∨ ~q) ∨ c.
Answer:
False

Question 6.
Dual of ‘John and Ayub went to the forest’ is ‘John or Ayub went to the forest.’
Answer:
True

Question 7.
‘His birthday is on 29th February’ is not a statement.
Answer:
True

Question 8.
x² = 25 is true statement.
Answer:
False

Question 9.
The truth value of ‘√5 is not an irrational number’ is T.
Answer:
False

Question 10.
p ∧ t = p.
Answer:
True

(IV) Solve the following:

Question 1.
State which of the following sentences are statements in logic:
(i) Ice cream Sundaes are my favourite.
Solution:
It is a statement.

(ii) x + 3 = 8, x is variable.
Solution:
It is a statement.

(iii) Read a lot to improve your writing skill.
Solution:
It is an imperative sentence, hence it is not a statement.

(iv) z is a positive number.
Solution:
It is an open sentence, hence it is not a statement.

(v) (a + b)² = a² + 2ab + b² for all a, b ∈ R.
Solution:
It is a statement.

(vi) (2 + 1)² = 9.
Solution:
It is a statement.

(vii) Why are you sad?
Solution:
It is an interrogative sentence, hence it is not a statement.

(viii) How beautiful the flower is!
Solution:
It is an exclamatory sentence, hence it is not a statement.

(ix) The square of any odd number is even.
Solution:
It is a statement.

(x) All integers are natural numbers.
Solution:
It is a statement.

(xi) If x is a real number, then x² ≥ 0.
Solution:
It is a statement.

(xii) Do not come inside the room.
Solution:
It is an imperative sentence, hence it is not a statement.

(xiii) What a horrible sight it was!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question 2.
Which of the following sentences are statements? In case of a statement, write down the truth value:
(i) What is a happy ending?
Solution:
It is an interrogative sentence, hence it is not a statement.

(ii) The square of every real number is positive.
Solution:
It is a statement that is false, hence its truth value is F.

(iii) Every parallelogram is a rhombus.
Solution:
It is a statement that is true, hence its truth value is T.

(iv) a² – b² = (a + b)(a – b) for all a, b ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is T.

(v) Please carry out my instruction.
Solution:
It is an imperative sentence, hence it is not a statement.

(vi) The Himalayas is the highest mountain range.
Solution:
It is a statement that is true, hence its truth value is T.

(vii) (x – 2)(x – 3) = x² – 5x + 6 for all x ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is T.

(viii) What are the causes of rural unemployment?
Solution:
It is an interrogative sentence, hence it is not a statement.

(ix) 0! = 1.
Solution:
It is a statement that is true, hence its truth value is T.

(x) The quadratic equation ax² + bx + c = 0 (a ≠ 0) always has two real roots.
Solution:
It is a statement that is false, hence its truth value is F.

Question 3.
Assuming the first statement as p and second as q, write the following statements in symbolic form:
(i) The Sun has set and Moon has risen.
Solution:
Let p : The Sun has set.
q : Moon has risen.
Then the symbolic form of the given statement is p ∧ q.

(ii) Mona likes Mathematics and Physics.
Solution:
Let p : Mona likes Mathematics.
q : Mona likes Physics.
Then the symbolic form of the given statement is p ∧ q.

(iii) 3 is a prime number if 3 is a perfect square number.
Solution:
Let p : 3 be a prime number.
q : 3 is a perfect square number.
Then the symbolic form of the given statement is p ↔ q.

(iv) Kavita is brilliant and brave.
Solution:
Let p : Kavita is brilliant.
q : Kavita is brave.
Then the symbolic form of the given statement is p ∧ q.

(v) If Kiran drives a car, then Sameer will walk.
Solution:
Let p : Kiran drives a car.
q : Sameet will walk.
Then the symbolic form of the given statement is p → q.

(vi) The necessary condition for the existence of a tangent to the curve of the function is continuity.
Solution:
The given statement can be written as:
‘If the function is continuous, then the tangent to the curve exists.’
Let p : The function is continuous.
q : Tangent to the curve exists.
Then the symbolic form of the given statement is p → q.

(vii) To be brave is necessary and sufficient condition to climb Mount Everest.
Solution:
Let p : To be brave.
q : Climb Mount Everest.
Then the symbolic form of the given statement is p ↔ q.

(viii) x³ + y³ = (x + y)³, iff xy = 0.
Solution:
Let p : x³ + y³ = (x + y)³.
q : xy = 0.
Then the symbolic form of the given statement is p ↔ q.

(ix) The drug is effective though it has side effects.
Solution:
Let p : The drug is effective.
q : It has side effects.
Then the symbolic form of the given statement is p ∧ q.

(x) If a real number is not rational, then it must be irrational.
Solution:
Let p : A real number is not rational.
q : It must be irrational.
Then the symbolic form of the given statement is p → q.

(xi) It is not true that Ram is tall and handsome.
Solution:
Let p : Ram is tall.
q : Ram is handsome.
Then the symbolic form of the given statement is ~(p ∧ q).

(xii) Even though it is not cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is not cloudy and it is still raining,
Let p : It is not cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

(xiii) It is not true that intelligent persons are neither polite nor helpful.
Solution:
Let p : Intelligent persons are neither polite nor helpful.
Then the symbolic form of the given statement is ~p.

(xiv) If the question paper is not easy, then we shall not pass.
Solution:
Let p : The question paper is not easy.
q : We shall not pass.
Then the symbolic form of the given statement is p → q.

Question 4.
If p : Proof is lengthy.
q : It is interesting.
Express the following statements in symbolic form:
(i) Proof is lengthy and it is not interesting.
(ii) If the proof is lengthy, then it is interesting.
(iii) It is not true that the proof is lengthy but it is interesting.
(iv) It is interesting iff the proof is lengthy.
Solution:
The symbolic form of the given statements are:
(i) p ∧ ~q
(ii) p → q
(iii) ~(p ∧ q)
(iv) q ↔ p

Question 5.
Let p : Sachin win the match.
q : Sachin is a member of the Rajya Sabha.
r : Sachin is happy.
Write the verbal statement for each of the following:
(i) (p ∧ q) ∨ r
Solution:
Sachin wins the match and he is a member of the Rajya Sabha or Sachin is happy.

(ii) p → r
Solution:
If Sachin wins the match, then he is happy.

(iii) ~p ∨ q
Solution:
Sachin does not win the match or he is a member of the Rajya Sabha.

(iv) p → (q ∨ r)
Solution:
If Sachin wins the match, then he is a member of the Rajya Sabha or he is happy.

(v) p → q
Solution:
If Sachin wins the match, then he is a member of the Rajya Sabha.

(vi) (p ∧ q) ∧ ~r
Solution:
Sachin wins the match and he is a member of the Rajya Sabha but he is not happy.

(vii) ~(p ∨ q) ∧ r
Solution:
It is false that Sachin wins the match or he is a member of the Rajya Sabha but he is happy.

Question 6.
Determine the truth values of the following statements:
(i) 4 + 5 = 7 or 9 – 2 = 5.
Solution:
Let p : 4 + 5 = 7.
q : 9 – 2 = 5.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are F.
∴ the truth value of p ∨ q is F. …….[F ∨ F ≡ F]

(ii) If 9 > 1, then x² – 2x + 1 = 0 for x = 1.
Solution:
Let p : 9 > 1.
q : x² – 2x + 1 = 0 for x = 1.
Then the symbolic form of the given statement is p → q.
The truth values of both p and q are T.
∴ the truth value of p → q is T. …..[T → T ≡ T]

(iii) x + y = 0 is the equation of a straight line if and only if y² = 4x is the equation of the parabola.
Solution:
Let p : x + y = 0 is the equation of a straight line.
q : y² = 4x is the equation of the parabola.
Then the symbolic form of the given statement is p ↔ q.
The truth values of both p and q are T.
∴ the truth value of p ↔ q is T. …..[T ↔ T ≡ T]

(iv) It is not true that 2 + 3 = 6 or 12 + 3 = 5.
Solution:
Let p : 2 + 3 = 6.
q : 12 + 3 = 5.
Then the symbolic form of the given statement is ~(p ∨ q).
The truth values of both p and q are F.
∴ the truth value of ~(p ∨ q) is T. …..[~(F ∨ F) ≡ ~F ≡ T]

Question 7.
Assuming the following statements
p : Stock prices are high.
q : Stocks are rising.
to be true, find the truth values of the following:

(i) Stock prices are not high or stocks are rising.
Solution:
p and q are true, i.e. T.
∴ ~p and ~q are false, i.e. F.
The given statement in symbolic form is ~p ∨ q.
Since, ~T ∨ T ≡ F ∨ T ≡ T, the given statement is true.
Hence, its truth value is ‘T’.

(ii) Stock prices are high and stocks are rising if and only if stock prices are high.
Solution:
The given statement in symbolic form is (p ∧ q) ↔ p.
Since (T ∧ T) ↔ T ≡ T ↔ T ≡ T, the given statement is true.
Hence, its truth value is ‘T’.

(iii) If stock prices are high, then stocks are not rising.
Solution:
The given statement in symbolic form is p → ~q.
Since, T → ~T ≡ T → F ≡ F, the given statement is false.
Hence, its truth value is ‘F’.

(iv) It is false that stocks are rising and stock prices are high.
Solution:
The given statement in symbolic form is ~(q ∧ p).
Since, ~(T ∧ T) ≡ ~T ≡ F, the given statement is false.
Hence, its truth value is ‘F’.

(v) Stock prices are high or stocks are not rising iff stocks are rising.
Solution:
The given statement in symbolic form is (p ∨ ~q) ↔ q.
Since (T ∨ ~T) ↔ T ≡ (T ∨ F) ↔ T
≡ T ↔ T
≡ T, the given statement is true.
Hence, its truth value is ‘T’.

Question 8.
Rewrite the following statements without using conditional:
[Hint: P → q ≡ ~p ∨ q]
(i) If price increases, then demand falls.
(ii) If demand falls, then the price does not increase.
Solution:
Since, p → q ≡ ~p ∨ q, the given statements can be written as:
(i) Price does not increase or demand falls.
(ii) Demand does not fall or price does not increase.

Question 9.
If p, q, r are statements with truth values T, T, F respectively, determine the truth values of the following:
(i) (p ∧ q) → ~p
Solution:
Truth values of p, q, r are T, T, F respectively.
(p ∧ q) → ~p ≡ (T ∧ T) → ~T
≡ T → F
≡ F
Hence, the truth value of the given statement is false, i.e. F.

(ii) p ↔ (q → ~p)
Solution:
p ↔ (q → ~p) ≡ T ↔ (T → ~T)
≡ T ↔ (T → F)
≡ T ↔ F
≡ F
Hence, the truth value of the given statement is false, i.e. F.

(iii) (p ∧ ~q) ∨ (~p ∧ q)
Solution:
(p ∧ ~q) ∨ (~p ∧ q) ≡ (T ∧ ~T) ∨ (~T ∧ T)
≡ (T ∧ F) ∨ (F ∧ T)
≡ F ∨ F
≡ F
Hence, the truth value of the given statement is false, i.e. F.

(iv) ~(p ∧ q) → ~(q ∧ p)
Solution:
~(p ∧ q) → ~(q ∧ p) ≡ ~(T ∧ T) → ~(T ∧ T)
≡ ~T → ~T
≡ F → F
≡ T
Hence, the truth value of the given statement is true, i.e. T.

(v) ~[(p → q) ↔ (p ∧ ~q)]
Solution:
~[(p → q) ↔ (p ∧ ~q)]
≡ ~[(T → T) ↔ (T ∧ ~T)]
≡ ~[T ↔ (T ∧ F)]
≡ ~[T ↔ F]
≡ ~F
≡ T.
Hence, the truth value of the given statement is true, i.e. T.

Question 10.
Write the negations of the following:
(i) If ΔABC is not equilateral, then it is not equiangular.
Solution:
Let p : ΔABC is not equilateral.
q : It is not equiangular.
Then the symbolic form of the given statement is p → q.
Since, ~(p → q) ≡ p ∧ ~q, the negation of the given statement is:
‘ΔABC is not equilateral and it is equiangular.’

(ii) Ramesh is intelligent and he is hard working.
Solution:
Let p : Ramesh is intelligent.
q : He is hard working.
Then the symbolic form of the given statement is p ∧ q.
Since, ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is:
‘Ramesh is not intelligent or he is not hard-working.’

(iii) A angle is a right angle if and only if it is of measure 90°.
Solution:
Let p : An angle is a right angle.
q : It is of measure 90°.
Then the symbolic form of the given statement is p ↔ q.
Since, ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of the given statement is:
‘An angle is a right angle and it is not of measure 90° or an angle is of measure 90° and it is not a right angle.’

(iv) Kanchanjunga is in India and Everest is in Nepal.
Solution:
Let p : Kanchenjunga is in India.
q : Everest is in Nepal.
Then the symbolic form of the given statement is p ∧ q.
Since, ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is:
‘Kanchenjunga is not in India or Everest is not in Nepal.’

(v) If x ∈ A ∩ B, then x ∈ A and x ∈ B.
Solution:
Let p : x ∈ A ∩ B, q : x ∈ A, r : x ∈ B.
Then the symbolic form of the given statement is P → (q ∧ r)
Since, ~(p → q) ≡ p ∧ ~q and ~(p ∧ q)= ~p ∨ ~q,
the negation of the given statement is:
‘x ∈ A ∩ B and x ∉ A or x ∉ B.

Question 11.
Construct the truth table for each of the following statement patterns:
(i) (p ∧ ~q) ↔ (q → p)
Solution:
(p ∧ ~q) ↔ (q → p)

(ii) (~p ∨ q) ∧ (~p ∧ ~q)
Solution:
(~p ∨ q) ∧ (~p ∧ ~q)

(iii) (p ∧ r) → (p ∨ ~q)
Solution:
(p ∧ r) → (p ∨ ~q)

(iv) (p ∨ r) → ~(q ∧ r)
Solution:
(p ∨ r) → ~(q ∧ r)

(v) (p ∨ ~q) → (r ∧ p)
Solution:
(p ∨ ~q) → (r ∧ p)

Question 12.
What is a tautology? What is a contradiction? Show that the negation of a tautology is a contradiction and the negation of a contradiction is a tautology.
Solution:
Tautology: A statement pattern that has all the entries in the last column of its truth table as T is called a tautology.
For example:

In the above truth table for the statement p ∨ ~p,
we observe that all the entries in the last column are T.
Hence, the statement p ∨ ~p is a tautology.

Contradiction: A statement pattern that has all the entries in the last column of its truth table as F is called a contradiction.
For example:

In the above truth table for the statement p ∧ ~p,
we observe that all the entries in the last column are F.
Hence, the statement p ∧ ~p is a contradiction.

To show that the negation of a tautology is a contradiction and vice versa:
A tautology is true on every row of its truth table.
Since, ~T = F and ~F = T, when we negate a tautology, the resulting statement is false on every row of its table.
i.e. the negation of tautology is a contradiction.
Similarly, the negation of a contradiction is a tautology.

Question 13.
Determine whether the following statement patterns is a tautology or a contradiction or a contingency:
(i) [(p ∧ q) ∨ (~p)] ∨ [p ∧ (~q)]
Solution:
[(p ∧ q) ∨ (~p)] ∨ [p ∧ (~q)]

All the entries in the last column of the above truth table are T.
∴ [(p ∧ q) ∨ (~p)] ∨ [p ∧ (~q)] is a tautology.

(ii) [(~p ∧ q) ∧ (q ∧ r)] ∨ (~q)
Solution:
[(~p ∧ q) ∧ (q ∧ r)] ∨ (~q)

The entries in the last column of the above truth table are neither all T nor all F.
∴ [(~p ∧ q) ∧ (q ∧ r)] ∨ (~q) is a contingency.

(iii) [~(p ∨ q) → p] ↔ [(~p) ∧ (~q)]
Solution:
[~(p ∨ q) → p] ↔ [(~p) ∧ (~q)]

All the entries in the last column of the above truth table are F.
∴ [~(p ∨ q) → p] ↔ [(~p) ∧ (~q)] is a contradiction.

(iv) [~(p ∧ q) → p] ↔ [(~p) ∧ (~q)]
Solution:
[~(p ∧ q) → p] ↔ [(~p) ∧ (~q)]

The entries in the last column of the above truth table are neither all T nor all F.
∴ [~(p ∧ q) → p] ↔ [(~p) ∧ (~q)] is a contingency.

(v) [p → (~q ∨ r)] ↔ ~[p → (q → r)]
Solution:
[p → (~q ∨ r)] ↔ ~[p → (q → r)]

All the entries in the last column of the above truth table are F.
∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction.

Question 14.
Using the truth table, prove the following logical equivalences:
(i) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Solution:
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

The entries in columns 5 and 8 are identical.
∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

(ii) [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r
Solution:
[~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r

The entries in columns 3 and 7 are identical.
∴ [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r

(iii) p ∧ (~p ∨ q) ≡ p ∧ q
Solution:
p ∧ (~p ∨ q) ≡ p ∧ q

The entries in columns 5 and 6 are identical.
∴ p ∧ (~p ∨ q) ≡ p ∧ q

(iv) p ↔ q ≡ ~(p ∧ ~q) ∧ ~(q ∧ ~p)
Solution:
p ↔ q ≡ ~(p ∧ ~q) ∧ ~(q ∧ ~p)

The entries in columns 3 and 10 are identical.
∴ p ↔ q ≡ ~(p ∧ ~q) ∧ ~(q ∧ ~p)

(v) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
~p ∧ q ≡ (p ∨ q) ∧ ~p

The entries in columns 4 and 6 are identical.
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p

Question 15.
Write the converse, inverse, contrapositive of the following statements:
(i) If 2 + 5 = 10, then 4 + 10 = 20.
Solution:
Let p : 2 + 5 = 10.
q : 4 + 10 = 20.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q
i.e. If 4 + 10 = 20, then 2 + 5 = 10.
Inverse: ~p → ~q is the inverse of p → q
i.e. If 2 + 5 ≠ 10, then 4 + 10 ≠ 20.
Cotrapositive: ~q → ~p is the contrapositive of p → q,
i.e. If 4 +10 ≠ 20, then 2 + 5 ≠ 10.

(ii) If a man is a bachelor, then he is happy.
Solution:
Let p : A man is a bachelor.
q : He is happy.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q
i.e. If a man is happy, then he is a bachelor.
Inverse: ~p → ~q is the inverse of p → q
i.e. If a man is not a bachelor, then he is not happy.
Contrapositive: ~q → ~p is the contrapositive of p → q
i.e., If a man is not happy, then he is not a bachelor.

(iii) If I do not work hard, then I do not prosper.
Solution:
Let p : I do not work hard.
q : I do not prosper.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q
i.e. If I do not prosper, then I do not work hard.
Inverse: ~p → ~q is the inverse of p → q
i.e. If I work hard, then I prosper.
Contrapositive: ~q → ~p is the contrapositive of p → q
i.e. If I prosper, then I work hard.

Question 16.
State the dual of each of the following statements by applying the principle of duality:
(i) (p ∧ ~q) ∨ (~p ∧ q) ≡ (p ∨ q) ∧ ~(p ∧ q)
(ii) p ∨ (q ∨ r) ≡ ~[(p ∧ q) ∨ (r ∨ s)]
(iii) 2 is an even number or 9 is a perfect square.
Solution:
The duals are given by:
(i) (p ∨ ~q) ∧ (~p ∨ q) ≡ (p ∧ q) ∨ ~(p ∨ q)
(ii) p ∧ (q ∧ r) ≡ ~[(p ∨ q) ∧ (r ∧ s)]
(iii) 2 is an even number and 9 is a perfect square.

Question 17.
Rewrite the following statements without using the connective ‘If … then’:
(i) If a quadrilateral is a rhombus, then it is not a square.
(ii) If 10 – 3 = 7, then 10 × 3 ≠ 30.
(iii) If it rains, then the principal declares a holiday.
Solution:
Since, p → q ≡ ~p ∨ q the given statements can be written as:
(i) A quadrilateral is not a rhombus or it is not a square.
(ii) 10 – 3 ≠ 7 or 10 × 3 ≠ 30.
(iii) It does not rain or the principal declares a holiday.

Question 18.
Write the dual of each of the following:
(i) (~p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q)
(ii) (p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
(iii) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
(iv) ~(p ∨ q) ≡ ~p ∧ ~q.
Solution:
The duals are given by:
(i) (~p ∨ q) ∧ (p ∨ ~q) ∧ (~p ∨ ~q)
(ii) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
(iii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
(iv) ~(p ∧ q) ≡ ~p ∧ ~q

Question 19.
Consider the following statements:
(i) If D is a dog, then D is very good.
(ii) If D is very good, then D is a dog.
(iii) If D is not very good, then D is not a dog.
(iv) If D is not a dog, then D is not very good.
Identify the pairs of statements having the same meaning. Justify.
Solution:
Let p : D is a dog. and q : D is very good.
Then the given statements in the symbolic form are:
(i) p → q
(ii) q → p
(iii) ~q → ~p
(iv) ~p → ~q

The entries in columns (i) and (iii) are identical. Hence, these statements are equivalent.
∴ the statements (i) and (iii) have the same meaning.
Similarly, the entries in columns (ii) and (iv) are identical. Hence, these statements are equivalent.
∴ the statements (ii) and (iv) have the same meaning.

Question 20.
Express the truth of each of the following statements by Venn diagrams:
(i) All men are mortal.
Solution:
Let U : a set of all human being
A : set of all men
B : set of all mortals.
Then the Venn diagram represents the truth of the given statement is as below:

(ii) Some persons are not politicians.
Solution:
Let U : set of all human being
A : set of all persons
B : set of all politicians.
Then the Venn diagram represents the truth of the given statement is as follows:

(iii) Some members of the present Indian cricket are not committed.
Solution:
Let U : set of all human being
X : set of all members of present Indian cricket
Y : set of all committed members of the present Indian cricket.
Then the Venn diagram represents the truth of the given statement is as below:

(iv) No child is an adult.
Solution:
Let U : set of all human beings
C : set of all children
A : set of all adults.
Then the Venn diagram represents the truth of the given statement is as below:

Question 21.
If A = {2, 3, 4, 5, 6, 7, 8}, determine the truth value of each of the following statements:
(i) ∃ x ∈ A, such that 3x + 2 > 9.
Solution:
Clearly x = 3, 4, 5, 6, 7, 8 ∈ A satisfy 3x + 2 > 9.
So, the given statement is true, hence its truth value is T.

(ii) ∀x ∈ A, x² < 18.
Solution:
x = 5, 6, 7, 8 ∈ A do not satisfy x² < 18.
So the given statement is false, hence its truth value is F.

(iii) ∃x ∈ A, such that x + 3 < 11.
Solution:
Clearly x = 2, 3, 4, 5, 6, 7 ∈ A which satisfy x + 3 < 11.
So, the given statement is True, hence its truth value is T.

(iv) ∀x ∈ A, x² + 2 ≥ 5.
Solution:
x² + 2 ≥ 5 for all x ∈ A.
So, the given statement is true, hence its truth value is T.

Question 22.
Write the negations of the following statements:
(i) 7 is a prime number and the Taj Mahal is in Agra.
Solution:
Let p : 7 be a prime number.
q : Taj Mahal is in Agra.
Then the symbolic form of the given statement is p ∧ q.
Since, (p ∧ q) ≡ ~p ∨ ~q,
the negation of the given statement is:
‘7 is not a prime number or Taj Mahal is not in Agra.’

(ii) 10 > 5 and 3 < 8.
Solution:
Let p : 10 > 5.
q : 3 < 8.
Then the symbolic form of the given statement is P ∧ q.
Since, ~(p ∧ q) = ~p ∨ ~q, the negation of the given statement is:
’10 ≤ 5 or 3 ≥ 8′
OR
’10 ≯ 5 or 3 ≮ 8′

(iii) I will have tea or coffee.
Solution:
The negation of the given statement is:
‘I will not have tea and coffee.’

(iv) ∀n ∈ N, n + 3 > 9.
Solution:
The negation of the given statement is:
‘∃n ∈ N, such that n + 3 ≯ 9.’
OR
‘∃n ∈ N, such that n + 3 ≤ 9.’

(v) ∃x ∈ A, such that x + 5 < 11.
Solution:
The negation of the given statement is:
‘∀x ∈ A, x + 5 ≮ 1.’
OR
‘∀x ∈ A, x + 5 ≥ 11.’

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.6 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.7 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.8 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.9 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.10 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.10 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.10 Questions and Answers.

Std 12 Maths 1 Exercise 1.10 Solutions Commerce Maths

Question 1.
Express the truth of each of the following statements by Venn diagrams:
(i) Some hardworking students are obedient.
Solution:
Let U : set of all students
S : set of all hardworking students
O : set of all obedient students.
Then the Venn diagram represents the truth of the given statement is as below:

S ∩ O ≠ φ

(ii) No circles are polygons.
Solution:
Let U : set of closed geometrical figures in the plane
P : set of all polygons
C : set of all circles.
Then the Venn diagram represents the truth of the given statement is as follows:

P ∩ C = φ

(iii) All teachers are scholars and scholars are teachers.
Solution:
Let U : set of all human beings
T : set of all teachers
S : set of all scholars.
Then the Venn diagram represents the truth of the given statement is as below:

(iv) If a quadrilateral is a rhombus, then it is a parallelogram.
Solution:
Let U : set of all quadrilaterals
R : set of all rhombus
P : set of all parallelograms.
Then the Venn diagram represents the truth of the given statement is as below:

R ⊂ P

Question 2.
Draw the Venn diagrams for the truth of the following statements:
(i) Some share brokers are chartered accountants.
Solution:
Let U : set of all human beings
S : set of all share brokers
C : set of all chartered accountants.
Then the Venn diagram represents the truth of the given statement is as below:

S ∩ C ≠ φ

(ii) No wicket-keeper is a bowler in a cricket team.
Solution:
Let U : set of all human beings
W : set of all wicket keepers
B : set of all bowlers.
Then the Venn diagram represents the truth of the given statement is as follows:

W ∩ B = φ

Question 3.
Represent the following statements by Venn diagrams:
(i) Some non-resident Indians are not rich.
Solution:
Let U : set of all human beings
N : set of all non-resident Indians
R : set of all rich people.
Then the Venn diagram represents the truth of the given statement is as below:

N – R ≠ φ

(ii) No circle is a rectangle.
Solution:
Let U : set of all geometrical figures
C : set of all circles
R : set of all rectangles
Then the Venn diagram represents the truth of the given statement is as below:

C ∩ R = φ

(iii) If n is a prime number and n ≠ 2, then it is odd.
Solution:
Let U : set of all real numbers
P : set of all prime numbers n, where n ≠ 2
O : set of all odd numbers.
Then the Venn diagram represents the truth of the given statement is as below:

P ⊂ O

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.6 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.7 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.8 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.9 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.10 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.9 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.9 Questions and Answers.

Std 12 Maths 1 Exercise 1.9 Solutions Commerce Maths

Question 1.
Without using truth table, show that
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~(q ∨ p) …..(Conditional Law)
≡ [~p ∧ (~(q ∨ p)] ∨ [q ∧ (~q ∨ p] …..(Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] ……(Distributive Law)
≡ [(~p ∧ ~q) ∨ c] ∨ [c ∨ (q ∧ p)] …..(Complement Law)
≡ (~p ∧ ~q) ∨ (q ∧ p) ……(Identity Law)
≡ (~p ∧ ~q) ∨ (p ∧ q) ……(Commutative Law)
≡ (p ∧ q) ∨ (~p∧ q) ……(Commutative Law)
≡ RHS.

(ii) p ∧ [~p ∨ q) ∨ (~q)] ≡ p
Solution:
LHS = p ∧ [(~p ∨ q) ∨ (~q)]
≡ p ∧ [~p ∨ (q ∨ ~q)] ……(Associative Law)
≡ p ∧ [~p ∨ t] …….(Complement Law)
≡ p ∧ t ……(Identity Law)
≡ p ……(Identity Law)
= RHS.

(iii) ~[(p ∧ q) → ~(q)] ≡ p ∧ q
Solution:
LHS = ~[(p ∧ q) → ~(~q)]
≡ (p ∧ q) ∧ ~(~q) ……(Negation of implication)
≡ (p ∧ q) ∧ q …..(Negation of negation)
≡ p ∧ (q ∧ q) …..(Associative Law)
≡ P ∧ q ……(Idempotent Law)
= RHS

(iv) ~r → ~(p ∧ q) ≡ [~(q → r)] → (~p)
Solution:
LHS = ~r → ~(p ∧ q)
≡ ~q → (~p ∨ ~q) ……(De Morgan’s Law)
≡ ~(~r) ∨ (~p ∨~q) …..(Conditional Law)
≡ r ∨ (~p ∨ ~q) …..(Involution Law)
≡ r ∨ ~q ∨ ~p …..(Commutative Law)
≡ (~q ∨ r) ∨ (~p) ……(Commutative Law)
≡ ~(q → r) ∨ (~p) ……(Conditional Law)
≡ ~(q → r) → (~p) ……(Conditional Law)
= RHS.

(v) (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Solution:
LHS = (p ∨ q) → r
≡ ~(p → q) ∨ r ……..(Conditional Law)
≡ (~p ∧ ~q) ∨ r ……….(De Morgan’s Law)
≡ (~p ∨ r) ∧ (~q ∨ r) ………..(Distributive Law)
≡ (p → r) ∧ (q → r) …….(Conditional Law)
= RHS.

Question 2.
Using the algebra of statement, prove that:
(i) [p ∧ (q ∨ r)] ∨ [~r ∧ ~q ∧ p] ≡ p
Solution:
LHS = [p ∧ (q ∨ r)] ∨ [ ~r ∧ ~q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~r ∧ ~q) ∧ p] ……(Associative Law)
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ……(Commutative Law)
≡ [p ∧ (q ∨ r)] ∨ [ ~ (q ∨ r) ∧ p] ……(De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] ……(Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~ (q ∨ r) ] …..(Distributive Law)
≡ p ∧ t …….(Complement Law)
≡ p ……(Identity Law)
= RHS.

(ii) (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q) ≡ p ∨ ~q
Solution:
LHS = (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~q) ∨ (~p ∧ ~q)] ……(Associative Law)
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~q ∧ ~p)] …..(CommutativeLaw)
≡ (p ∧ q) ∨ [ ~q ∧ (p ∨ ~ p)] …..(Distributive Law)
≡ (p ∧ q) ∨ (~q ∧ t) ……(Complement Law)
≡ (p ∧ q) ∨ (~q) …….(Identity Law)
≡ (p ∨ ~q) ∧ (q ∨ ~q) ……(Distributive Law)
≡ (p ∨ ~q) ∧ t …….(Complement Law)
≡ p ∨ ~q …..(Identity Law)
= RHS.

(iii) (p ∨ q) ∧ (~p ∨ ~q) ≡ (p ∧ ~q) ∨ (~p ∧ q)
Solution:
LHS = (p ∨ q) ∧ (~p ∨ ~q)
≡ [p ∧ (~p ∨ ~q)] ∨ [q ∧ (~p ∨ ~q)] ……(Distributive Law)
≡ [(p ∧ ~p) ∨ (p ∧ ~q)] ∨ [q ∧ ~p) ∨ (q ∧ ~q)] ……(Distributive Law)
≡ [c ∨ (p ∧ ~q)] ∨ [(q ∧ ~p) ∨ c] ……(Complement Law)
≡ (p ∧ ~q) ∨ (q ∧ ~p) ……..(Identity Law)
≡ (p ∧ ~q) ∨ (~p ∧ q) ………(Commutative Law)
= RHS.

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.6 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.7 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.8 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.9 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.10 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.8 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.8 Questions and Answers.

Std 12 Maths 1 Exercise 1.8 Solutions Commerce Maths

Question 1.
Write the negation of each of the following statements:
(i) All the stars are shining if it is night.
Solution:
The given statement can be written as:
If it is night, then all the stars are shining.
Let p : It is night.
q : All the stars are shining.
Then the symbolic form of the given statement is p → q
Since, ~(p → q) ≡ p ∧ ~q,
the negation of the given statement is ‘It is night and all the stars are not shining.’

(ii) ∀ n ∈ N, n + 1 > 0.
Solution:
The negation of the given statement is
‘∃ n ∈ N, such that n + 1 ≤ 0.’

(iii) ∃ n ∈ N, such that (n² + 2) is odd number.
Solution:
The negation of the given statement is
‘∀ n ∈ N, n² + 2 is not an odd number.’

(iv) Some continuous functions are differentiable.
Solution:
The negation of a given statement is ‘All continuous functions are not differentiable.’

Question 2.
Using the rules of negation, write the negations of the following:
(i) (p → r) ∧ q
Solution:
The negation of (p → r) ∧ q is
~[(p → r) ∧ q] ≡ ~(p → r) ∨ (~q) …..[Negation of conjunction]
≡ (p ∧ ~r) ∨ (~q) ……[Negation of implication]

(ii) ~(p ∨ q) → r
Solution:
The negation of ~(p ∨ q) → r is
~[~(p ∨ q) → r] ≡ ~(p ∨ q) ∧ (~r) …..[Negation of implication]
≡ (~p ∧ ~q) ∧ (~r) ……[Negation of disjunction]

(iii) (~p ∧ q) ∧ (~q ∨ ~r)
Solution:
The negation of (~p ∧ q) ∧ (~q ∨ ~r) is
~[(~p ∧ q) ∧ (~q ∨ ~ r)] ≡ ~(~p ∧ q) ∨ ~(~q ∨ ~r) ……[Negation of conjunction]
≡ [~(~p) ∨ ~q] ∨ [~(~q) ∧ ~(~r)] … [Negation of conjunction and disjunction]
≡ (p ∨ ~q) ∨ (q ∧ r) …..[Negation of negation]

Question 3.
Write the converse, inverse, and contrapositive of the following statements:
(i) If it snows, then they do not drive the car.
Solution:
Let p : It snows.
q : They do not drive the car.
Then the symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If they do not drive the car, then it snows.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If it does not snow, then they drive the car.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If they drive the car, then it does not snow.

(ii) If he studies, then he will go to college.
Solution:
Let p : He studies.
q : He will go to college.
Then two symbolic form of the given statement is p → q.
Converse: q → p is the converse of p → q.
i.e. If he will go to college, then he studies.
Inverse: ~p → ~q is the inverse of p → q.
i.e. If he does not study, then he will not go to college.
Contrapositive: ~q → ~p is the contrapositive of p → q.
i.e. If he will not go to college, then he does not study.

Question 4.
With proper justification, state the negation of each of the following:
(i) (p → q) ∨ (p → r)
Solution:
The negation of (p → q) ∨ (p → r) is
~[(p → q) ∨ (p → r)] ≡ ~(p → q) ∧ ~(p → r) …..[Negation of disjunction]
≡ (p ∧ ~q) ∧ (p ∧ ~r) …[Negation of implication]

(ii) (p ↔ q) ∨ (~q → ~r)
Solution:
The negation of (p ↔ q) ∨ (~q → ~r) is
~[(p ↔ q) ∨ (~q → ~r)] ≡ ~(p ↔ q) ∧ ~(~q → ~r) …..[Negation of disjunction]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ [~q ∧ ~(~r)] ……[Negation of biconditional and implication]
≡ [(p ∧ ~q) ∨ (q ∧ ~p)] ∧ (~q ∧ r) ……[Negation of negation]

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~[(p → q) ∧ r] ≡ ~(p → q) ∨ (~r) …..[Negation of conjunction]
≡ (p ∧ ~q) ∨ (~r) …..[Negation of implication]

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.6 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.7 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.8 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.9 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.10 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.7 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.7 Questions and Answers.

Std 12 Maths 1 Exercise 1.7 Solutions Commerce Maths

Question 1.
Write the dual of each of the following:
(i) (p ∨ q) ∨ r
Solution:
(p ∧ q) ∧ r

(ii) ~(p ∨ q) ∧ [p ∨ ~(q ∧ ~r)]
Solution:
~(p ∧ q) ∨ [p ∧ ~(q ∨ ~r)]

(iii) p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r
Solution:
p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

(iv) ~(p ∧ q) ≡ ~p ∨ ~q
Solution:
~(p ∨ q) ≡ ~p ∧ ~q

Question 2.
Write the dual statement of each of the following compound statements:
(i) 13 is prime number and India is a democratic country.
Solution:
13 is prime number or India is a democratic country.

(ii) Karina is very good or everybody likes her.
Solution:
Karina is very good and everybody likes her.

(iii) Radha and Sushmita can not read Urdu.
Solution:
Radha or Sushmita can not read Urdu.

(iv) A number is real number and the square of the number is non-negative.
Solution:
A number is real number or the square of the number is non-negative.

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.6 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.7 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.8 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.9 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.10 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.6 Questions and Answers.

Std 12 Maths 1 Exercise 1.6 Solutions Commerce Maths

Question 1.
Prepare the truth tables for the following statement patterns:
(i) p → (~p ∨ q)
Solution:
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.

(ii) (~p ∨ q) ∧ (~p ∨ ~q)
Solution:

(iii) (p ∧ r) → (p ∨ ~q)
Solution:
Here are three statements and 4 connectives.
∴ there are 2 × 2 × 2 = 8 rows and 3 + 4 = 7 columns in the truth table.

(iv) (p ∧ q) ∨ ~r
Solution:

Question 2.
Examine, whether each of the following statement patterns is a tautology or a contradiction or a contingency:
(i) q ∨ [~(p ∧ q)]
Solution:

All the entries in the last column of the above truth table are T.
∴ q ∨ [~(p ∧ q)] is a tautology.

(ii) (~q ∧ p) ∧ (p ∧ ~p)
Solution:

All the entries in the last column of the above truth table are F.
∴ (~q ∧ p) ∧ (p ∧ ~p) is a contradiction.

(iii) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(iv) ~p → (p → ~q)
Solution:

All the entries in the last column of the truth table are T.
∴ p → (p → ~q) is a tautology.

Question 3.
Prove that each of the following statement pattern is a tautology:
(i) (p ∧ q) → q
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → q is a tautology.

(ii) (p → q) ↔ (~q → ~p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~q → ~p) is a tautology.

(iii) (~p ∧ ~q) → (p → q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∧ ~q) → (p → q) is a tautology.

(iv) (~p ∨ ~q) ↔ ~(p ∧ q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∨ ~q) ↔ ~(p ∧ q) is a tautology.

Question 4.
Prove that each of the following statement pattern is a contradiction:
(i) (p ∨ q) ∧ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.

(ii) (p ∧ q) ∧ ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∧ ~p is a contradiction.

(iii) (p ∧ q) ∧ (~p ∨ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∧ q) ∧ (~p ∨ ~q) is a contradiction.

(iv) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

Question 5.
Show that each of the following statement pattern is a contingency:
(i) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(ii) (p → q) ↔ (~p ∧ q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ↔ (~p ∧ q) is a contingency.

(iii) p ∧ [(p → ~q) → q]
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ p ∧ [(p → ~q) → q] is a contingency.

(iv) (p → q) ∧ (p → r)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ∧ (p → r) is a contingency.

Question 6.
Using the truth table, verify:
(i) p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).

(ii) p → (p → q) ≡ ~q → (p → q)
Solution:

The entries in columns 5 and 6 are identical.
∴ p → (p → q) ≡ ~q → (p → q)

(iii) ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q
Solution:

The entries in columns 5, 7 and 8 are identical.
∴ ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q.

(iv) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:

The entries in columns 3 and 7 are identical.
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 7.
Prove that the following pairs of statement patterns are equivalent:
(i) p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

(ii) p ↔ q and (p → q) ∧ (q → p)
Solution:

The entries in columns 3 and 6 are identical.
∴ p ↔ q ≡ (p → q) ∧ (q → p)

(iii) p → q and ~q → ~p and ~p ∨ q
Solution:

The entries in columns 5, 6 and 7 are identical.
∴ p → q ≡ ~q → ~p ≡ ~p ∨ q.

(iv) ~(p ∧ q) and ~p ∨ ~q
Solution:

The entries in columns 6 and 7 are identical.
∴ ~(p ∧ q) ≡ ~p ∨ ~q.

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.6 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.7 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.8 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.9 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.10 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Commerce Maths Textbook Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

Std 12 Maths 1 Exercise 1.5 Solutions Commerce Maths

Question 1.
Use qualifiers to convert each of the following open sentences defined on N, into a true statement:
(i) x² + 3x – 10 = 0
Solution:
∃ x ∈ N, such that x² + 3x – 10 = 0 is a true statement
(x = 2 ∈ N satisfy x² + 3x – 10 = 0)

(ii) 3x – 4 < 9
Solution:
∃ x ∈ N, such that 3x – 4 < 9 is a true statement.
(x = 1, 2, 3, 4 ∈ N satisfy 3x – 4 < 9)

(iii) n² ≥ 1
Solution:
∀ n ∈ N, n² ≥ 1 is a true statement.
(All n ∈ N satisfy n² ≥ 1)

(iv) 2n – 1 = 5
Solution:
∃ x ∈ N, such that 2n – 1 = 5 is a true statement.
(n = 3 ∈ N satisfy 2n – 1 = 5)

(v) y + 4 > 6
Solution:
∃ y ∈ N, such that y + 4 > 6 is a true statement.
(y = 3, 4, 5, … ∈ N satisfy y + 4 > 6

(vi) 3y – 2 ≤ 9
Solution:
∃ y ∈ N, such that 2y ≤ 9 is a true statement.
(y = 1, 2, 3 ∈ N satisfy 3y – 2 ≤ 9).

Question 2.
If B = {2, 3, 5, 6, 7}, determine the truth value of each of the following:
(i) ∀ x ∈ B, x is a prime number.
Solution:
(i) x = 6 ∈ B does not satisfy x is a prime number.
So, the given statement is false, hence its truth value is F.

(ii) ∃ n ∈ B, such that n + 6 > 12.
Solution:
Clearly n = 7 ∈ B satisfies n + 6 > 12.
So, the given statement is true, hence its truth value is T.

(iii) ∃ n ∈ B, such that 2n + 2 < 4.
Solution:
No element n ∈ B satisfy 2n + 2 < 4.
So, the given statement is false, hence its truth value is F.

(iv) ∀ y ∈ B, y² is negative.
Solution:
No element y ∈ B satisfy y² is negative.
So, the given statement is false, hence its truth value is F.

(v) ∀ y ∈ B, (y – 5) ∈ N.
Solution:
y = 2 ∈ B, y = 3 ∈ B and y = 5 ∈ B do not satisfy (y – 5) ∈ N.
So, the given statement is false, hence its truth value is F.

Maharashtra State Board Std 12th Commerce Maths Textbook Solutions 

• Mathematical Logic Ex 1.1 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.2 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.3 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.4 Class 12 Commerce Maths Textbook Solutions
• Mathematical Logic Ex 1.5 Class 12 Commerce Maths Textbook Solutions
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Std 12 History Chapter 1 Question Answer Renaissance in Europe and Development of Science Maharashtra Board

Balbharti Maharashtra State Board Class 12 History Solutions Chapter 12 India Transformed Part 2 Textbook Exercise Questions and Answers.

Class 12 History Chapter 1 Renaissance in Europe and Development of Science Question Answer Maharashtra Board

History Class 12 Chapter 1 Question Answer Maharashtra Board

1A. Choose the correct alternative and rewrite the statement.

Question 1.
In 1995, the health department of the Indian government launched the campaign, dubbed as __________
(a) Measles-Rubella
(b) Pulse Polio
(c) B.C.G.
(d) Triple vaccine
Answer:
(b) Pulse Polio

Question 2.
The first district in __________ to become completely literate was Ernakulum.
(a) Gujrat
(b) Kerala
(c) Karnataka
(d) Tamil Nadu
Answer:
(b) Kerala

1B. Find the incorrect pair from group ‘B’ and write the corrected one.

Question 1.

Answer:
SEESCAP – Institute for conservation of long-billed vultures

2. Write the names of historical places/persons/events.

Question 1.
Commission established vide Human Right Protection Act –
Answer:
National Human Rights Commission

Question 2.
The player who was awarded the highest title, ‘Bharat Ratna’ –
Answer:
Mr. Sachin Tendulkar

3. Complete the following concept map.

Question 1.

Answer:

4. Write short notes.

Question 1.
Speed Post
Answer:

• The Indian postal department started the service of speed post in 1986 which changed the nature of post service.
• This service was used by the majority of Indians, more than three crore letters and parcels were delivered from this service.
• The customer will get the message of successful delivery of their items.
• The postal department now offers courier services like passport delivery, business parcels, cash-on-delivery, logistics posts, and air freights.
• The post office even offers a packaging service. Over one lakh and fifty thousand post offices are offering the services like paying bills, sending festive cards and other objects.
• Since 2016, the postal department has started service of delivering Ganges water from Rishikesh and Gangotri at personal addresses.
• Buying personalised postage stamps with personal photographs and special schemes for philatelists are also available at ‘Post Shops’ opened at 80 post offices.

Question 2.
Air Pollution
Answer:

• The entire world, nowadays, is facing the problem of air pollution which is mainly done by vehicles.
• All most every state in India is facing the problem of air pollution.
• The ‘Centre for Science and Environment’, studied and proved that the increasing number of automobiles and vehicles of poor maintenance are the causes of air pollution in Delhi.
• The institution recommended the following suggestions to curb air pollution:
  • Use of CNG (Compressed Natural Gas) as fuel.
  • It was decided that vehicles without a ‘PUC’ certificate will not qualify for insurance and renewal of insurance.
• In this way, we can curb air pollution and make our environment clean and pure.

Question 3.
Eradication of Polio.
Answer:

• The government of India launched a campaign for the eradication of polio from India.
• The health department of the Indian government launched the campaign known as ‘Pulse Polio’ which was started with the joint sponsorship of‘World Health Organisation – WHO, Rotary International, UNICEF, and Indian Government.’
• The objective of the campaign was not to leave a single child under the age of five years without administering the polio vaccine.
• Awareness camps, home visits, and extensive advertising made this campaign successful.

5. Answer the following questions in detail.

Question 1.
Explain the sports policy of the Government of India.
Answer:
India is home to a diverse population playing and showing their talent in a variety of different sports. Every nation needs a well-knit sports policy.

• In 2001, the Government of India announced its sports policy. The main objective of this policy is:
  • To take sports to all parts of India.
  • To help the players to develop special skills.
  • To build supportive and fundamental sports facilities.
  • To help the National Sports Federation of India and associated institutions to search for sports talent.
  • To initiate co-operation from industries, corporate and private institutions for the cause of sports.
  • To create awareness of the importance of sports and interest in sports among people.
• In 2011, the Indian government announced a novel scheme for sports named ‘Come and Play’.
• Sports Authority of India gave permission to use five sports complexes in Delhi to local youths.
• They were also provided an opportunity to train under Sports Authority of India (SAI) coaches.
• The National Sports University was founded in Manipur in 2018.
• This university offers different courses from Bachelor and Masters to M.Phil. and Ph.D. Apart from this, sports universities also offer courses in sports, education, sports management, sports psychology, coaching, etc. Research in sports is also encouraged in the sports university.
• Khelo India.

Question 2.
Which programmes and facilities are created by the Ministry of Tourism of the Indian government to attract tourists?
Answer:
India has a rich cultural heritage. Several tourists visit India each year to see its cultural heritage, monuments, etc. The tourism industry is a continuously growing industry which also gives foreign exchange to the country. The Indian Government has adopted different policies to attract tourists. They are as follows:

• The government provides different facilities to the tourists who visit India. The three ministries of home affairs, tourism, and foreign affairs together created the facility of e-visa which included e-business visas, e-medical visas, and e-visa.
• A facility of giving information to tourists on mobile in Hindi and 10 other foreign languages, for 24 hours throughout the week was started.
• To avail of this facility, tourists have to dial 1363. The information regarding cruise tours, health, and sports tourism, eco-friendly tourism, adventurous sports tourism, film festivals are given through this service.
• Hotels with amenities and premium quality accommodation including luxurious services are available to the tourists.
• The Institutes which offer training courses in ‘Hospitality and Hotel Management’ have been established in major cities of India.
• An advertisement campaign called ‘Atulya Bharat’ was designed to attract tourists.
• A travel show entitled ‘GONORTHEAST’ was released on the Discovery channel to boost the tourism of beautiful places in the northeastern states of India.
• The government took the help of electronic and digital media channels like Discovery, BBC, History is showcasing various programs introducing India’s historical and cultural heritage.
• Swadesh’ and ‘Prasad’ schemes were launched by the Ministry of Tourism of Government of India to encourage visits to 95 pilgrimage and spiritual centers in India.
• The Ministry of Tourism, Government of India, and Federation of Associations in Indian Tourism and Hospitality (FAITH) organized Indian Tourism Mart – 2018. This was the first event based on the model of International Tourism Marts in other countries.

6. State your opinion.

Question 1.
Joint military practice sessions are beneficial for both participant countries.
Answer:

• Such sessions are extremely helpful for both countries because of the technological exchange that happens on these occasions.
• It also helps the armies of both the countries to know, learn and practice new methods of resolving problems.
• The process of modernization of arms gives impetus to further research.
• Due to the development of science and technology, there is increasing scope for the exchange of the latest technology to fight terrorism, to augment our own competencies, and optimum use of modern technology for the end of terrorism.
• The Indian army carried out exercises with different countries e.g., exercise with Oman army at Bakloh, there were combined exercises of Mangolian army and Jammu and Kashmir Rifles.

Question 2.
All of us have the responsibility of taking care of our heritage places.
Answer:

• India has an extraordinary and vast cultural heritage. It is in the form of ancient monuments, buildings, and other archaeological sites and remains.
• These monuments are the living witnesses of our golden historic era.
• It is our duty to preserve the monuments for the next generations. A little initiate from our side can save our heritage.
• The tourism industry generated foreign exchange on a large scale. Tourists come to India every year to see its cultural heritage. Therefore, it is our duty to preserve and protect our heritage.
• As a citizen of India, we should spread awareness among the people about the importance of the preservation of monuments. A little effort on our side can create desirable changes which will make past, present, and future generations of the country and the entire world proud of us.

Class 12 History Chapter 12 India Transformed Part 2 Intext Questions and Answers

Try to do this: (Textbook Page No. 100)

NRHM – Make a list of the benefits of the National Rural Health Mission to people.
Answer:

• The Indian government launched National Rural Health Mission (NRHM) in April 2005 with an aim to strengthen the health systems in rural and urban areas. The list of the benefits of the NRHM are as follows:
• It aims to provide equitable, affordable, and quality healthcare services.
• It has strengthened the healthcare infrastructure.
• It has brought down the maternal mortality rate among poor pregnant women.
• The prevalence of tobacco use and the number of tobacco users have been reduced.
• The Janani Shishu Suraksha Karyakram entitles pregnant women to give birth in public health institutions at no expense.
• The government launched different schemes for community participation under NRHM. Rogi Kalyan Samiti is responsible for maintaining the facilities and ensuring the provision of better facilities for the patients in the hospital.
• Established the Global Knowledge Hub for smokeless tobacco. It also issued an advisory to ban Electronic Nicotine Delivery Systems.
• After the implementation of various initiates under NHRM many states have shown improved progress in healthcare facilities.

Try to do this: (Textbook Page No. 100)

Make a list of solutions to reduce the levels of air pollution.
Answer:
Air pollution is the biggest threat to the environment and to all living species. Following are some of the solutions to reduce the level of air pollution:

• Public Transport: Encourage greater use of public transport i.e. the use of railways, bus services or metros, etc. Instead of using private vehicles, people should encourage to use public transport.
• Use of CNG Vehicles: Citizens should also be encouraged to use CNG vehicles as it is a much cleaner fuel than petrol or diesel. New registration should be discouraged by increasing registration charges of vehicles.
• Use bicycles: Using bicycles is the best way to reduce air pollution. The government should mark out bicycle lanes in residential colonies as well as on roads.
• Plant and care for trees: Trees filter pollutants and absorb carbon dioxide. Trees release oxygen into the atmosphere. The practice of planting trees provides more benefits to the environment.
• Use less energy: Use energy-efficient appliances. Turn off electrical appliances when not in use. Get an energy audit done and follow the advice.
• Avoid the use of crackers: Avoid the use of crackers during festivals and weddings as it creates a lot of air pollution and is harmful to birds and animals including human beings.
• Reuse, Reduce and Recycle: The three ‘Rs’ are the best way to reduce air pollution.
• Avoid using chemical products: Avoid using chemical products like paint, perfumes, sprays, etc as they contain harmful products. Try to use products with less chemical content.
• Prevention of forest fires and burning of garbage: Don’t burn garbage or leaves because it releases harmful smoke in the air which decreases the quality of air.

Try to do this: (Textbook Page No. 102)

1990 was the ‘International Year of Literacy’. Make a list of similarly declared international years for special causes and campaigns launched on the occasion.
Answer:
The following are the international years currently observed by the United Nations.
2024:

• International Year of Camelids

2022:

• International Year of Artisanal Fisheries and Aquaculture

2021:

• International Year of Peace and Trust
• International Year of Creative Economy for Sustainable Development
• International Year of Fruits and Vegetables
• International Year of Eliminations of Child Labour

2020:

• International Year of Planet Health
• International Year of the Nurse and Midwife

2019:

• International Year of Indigenous Languages
• International Year of Moderation
• International Year of Periodic Table of Chemical Elements

2017:

• International Year of Sustainable Tourism for Development.

2016:

• International Year of Pulses

2015:

• International Year of Light and Light-based Technologies
• International Year of Soils

2014:

• International Year of Solidarity with the Palestinian People
• International Year of Small Island Developing States
• International Year of Crystallography
• International Year of Family Farming

2013:

• International Year of Water cooperation
• International Year of Quinoa

2012:

• International Year of Cooperatives
• International Year of Sustainable Energy for All

2011:

• International Year of Forest
• International Year of Chemistry
• The International Year of African Descent

2010:

• The International Year of Biodiversity
• The International Year for the Rapprochement of cultures
• The International Year of Youth

2009:

• The International Year of Astronomy
• The International Year of Human Rights Learning
• The International Year of Natural Fibres
• The International Year of Reconciliation

2008:

• The International Year of Languages
• The International Year of Planet Earth
• The International Year of the Potato
• The International Year of Sanitation

2007-08:

• International Polar Year

2006:

• International Year of Deserts and Desertification

2005:

• International Year of Physics
• International Year of Sport and Physical Education
• International Year of Microcredit

2004:

• International Year of Rice
• International Year to Commemorate the Struggle against Slavery and its Abolition

2003:

• International Year of Freshwater
• Year of Kyrgyz Statehood

2002:

• International Year of Mountains
• International Year of Eco-tourism
• United Nations Year for Cultural Heritage

2001:

• International Year of Volunteers
• International Year of Mobilisation against Racism, Racial Discrimination, Xenophobia, and Related Intolerance
• United Nations Year of Dialogue among Civilisation

2000:

• International Year of Thanksgiving
• International Year for the Culture of Peace

Find out and tell us (Textbook Page No. 106)

Make a list of employment opportunities generated by the tourism industry.
Answer:
The travel and tourism industry in India is growing rapidly so many opportunities are available in the coming years in this field. Some of the opportunities that are available in the tourism industry are as follows:
(i) Hotels: Many job perspectives are available in the hotel industry. Some fields of hotel industries are

• Manager
• Operations
• Housekeeping
• Food and Beverage
• Front office
• Gardener
• Security officer/personnel etc.

(ii) Airlines: One can take up the following job in airlines

• Pilot
• Ground staff (Traffic Assistant, Counter staff, Booking, and Reservation)
• Flight Attendant

(iii) Tourism Department:

• Tour guides
• Tour planner
• Information assistants
• Reservation and counter staffs
• Sales and Marketing
• Interpreters
• Translators

(iv) Transportation Industry:
This is an ever-growing industry where one can have many job opportunities. Job opportunities are available in all types of transportation i.e.

• Railway service
• Bus service
• Cruise service or ferry service
• Private transportation – Cars, Rickshaws, Horse riding, etc.

(v) The employment opportunities are also available in the management of adventure sports, theme parks, amusement parks, water sports, mountaineering, children’s fantasy land, etc.

(vi) The other job opportunities in this field include Destination Manager, Itinerary Planner, Travel Agent, Foreign Exchange, Counselor, etc.

Find out and tell us (Textbook Page No. 106)

Suggest ways and means to enhance the heritage and historical tourism in India.
Answer:
Heritage and historical tourism tend to attract many tourists. Different ways and means to enhance, this type of tourism are as follows:

• Build better roads and access points to the heritage sites, packaged train travel, proper bus connections should be provided to the tourists.
• Aggressive tourism and marketing strategies are also necessary whether it is broadcasting the ‘Incredible India’ campaign abroad, holding different seminars, or offering Indian locations with facilities to promote foreign film productions in the country are some of the strategies.
• The Government of India has already started making a lot of improvements in this area. We need to capitalize on India as a destination. The government should make a documentary based on the heritage of India. One can even design a promotional campaign to enhance historical tourism.
• The historical places always have a story to tell. They offer amazing aesthetics and attract people from all over the world. If tourism is accompanied by mind-blowing hotels, resorts, good public transportation facilities, and delectable cuisine then it becomes an added advantage to the tourist.

Project (Textbook Page No. 108)

Make a list of various business opportunities available at tourist destinations.
Answer:
Introduction: Tourism is a continuously growing industry. By the early 21st century, international tourism had become one of the world’s most important activities. Various business opportunities are available at the tourist destinations and they are as follows:
(i) Travel Agency: Everyone needs a platform where customers can come to and take advice about tour packages of different places which travel agency provides. The travel agency helps tourists in planning their travel for which they charge their commission.

(ii) Hotel: Hotel business is the best opportunity from which you can gain a high rate of return as well as expand your chain of hotels in the future.

(iii) Online Travel Business: In the modern era, the way of doing business is changing. Now everything is available online. With the change in technology, the way of doing business has also changed. One can start an online business portal where customers have easy access to everything online.

(iv) Photography: Tourists always want to save the memories of the places they visit and a photographer helps them to restore their memories through their art of photography. There is a huge demand for photographers.

(v) Vehicle Renting: Some tourists like to plan their tours according to their own wishes. They don’t want any unknown person to be a part of their travel for safety reasons. Such tourists search for a vehicle that they can get on rent and go in the direction that they want. To invest in this business is profitable.

(vi) Tour Guide: A Tourist guide is a person who guides visitors in the language of their choice and interprets the cultural and natural history of the particular place. This is a challenging field with an increase in tours and travels.

(vii) Executive Chef: The executive chef is in charge of a restaurant’s kitchen and is responsible for managing the kitchen staff, planning the menu, and making sure that food hygiene is maintained. This is a well suitable job for people who are passionate about cooking.

Maharashtra State Board 12th Std History Textbook Solutions

• Renaissance in Europe and Development of Science Class 12 History Textbook Solutions
• European Colonialism Class 12 History Textbook Solutions
• India and European Colonialism Class 12 History Textbook Solutions
• Colonialism and the Marathas Class 12 History Textbook Solutions
• India: Social and Religious Reforms Class 12 History Textbook Solutions
• Indian Struggle against Colonialism Class 12 History Textbook Solutions
• Decolonisation to Political Integration of India Class 12 History Textbook Solutions
• World Wars and India Class 12 History Textbook Solutions
• World: Decolonisation Class 12 History Textbook Solutions
• Cold War Class 12 History Textbook Solutions
• India Transformed Part 1 Class 12 History Textbook Solutions
• India Transformed Part 2 Class 12 History Textbook Solutions