Matrices Class 12 Maths 1 Exercise 2.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.3 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.3 Questions And Answers Maharashtra Board

Question 1.
Solve the following equations by the inversion method.
(i) x + 2y = 2, 2x + 3y = 3
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
This is of the form AX = B, where
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 1
∴ A-1 = \(\left[\begin{array}{rr}
-3 & 2 \\
2 & -1
\end{array}\right]\)
Now, premultiply AX = B by A-1, we get,
A-1(AX) = A-1B
∴ (A-1A)X = A-1B
∴ IX = A-1B
∴ X = \(=\left[\begin{array}{rr}
-3 & 2 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
∴ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(=\left[\begin{array}{r}
-6+6 \\
4-3
\end{array}\right]\) = \(=\left[\begin{array}{l}
0 \\
1
\end{array}\right]\)
By equality of matrices,
x = 0, y = 1 is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) x + y = 4, 2x – y = 5
Solution:
x + y = 4, 2x – y = 5
The given equations can be written in the matrix form as:
\(\left[\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
This is of the form AX = B ⇒ X ⇒ A-1B
A = \(\left[\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right]\)
|A| = -1 – 2 = -3 ≠ 0
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 5
By equality of matrices.
x = 3, y = 1

(iii) 2x + 6y = 8, x + 3y = 5
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
This is of the form AX = B, where
A = \(\left[\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
Let us find A-1.
|A| = \(\left|\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right|\) = 6 – 6 = 0
∴ A-1 does not exist.
Hence, x and y do not exist.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Solve the following equations by reduction method.
(i) 2x + y = 5, 3x + 5y = -3
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 2
By equality of matrices,
2x + y = 5 …(1)
7y = -21 …(2)
From (2), y = -3
Substituting y = -3 in (1), we get,
2x – 3 = 5
∴ 2x = 8 ∴ x = 4
Hence, x = 4, y = -3 is the required solution.

(ii) x + 3y = 2, 3x + 5y = 4.
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
1 & 3 \\
3 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
4
\end{array}\right]\)
By R2 – 3R1, we get
\(\left[\begin{array}{rr}
1 & 3 \\
0 & -4
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left(\begin{array}{r}
2 \\
-2
\end{array}\right)\)
∴ \(\left[\begin{array}{l}
x+3 \\
0-4 y
\end{array}\right]\) = \(\left[\begin{array}{r}
2 \\
-2
\end{array}\right]\)
By equality of matrices,
x + 3y = 2 …(1)
-4y = -2
From (2), y = \(\frac{1}{2}\)
Substituting y = \(\frac{1}{2}\) in (1), we get,
x + \(\frac{3}{2}\) = 2
∴ x = 2 – \(\frac{3}{2}=\frac{1}{2}\)
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) 3x – y = 1, 4x + y = 6
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 3
By equality of matrices,
12x – 4y = 4 … (1)
7y = 14 … (2)
From (2), y = 2
Substituting y = 2 in (1), we get,
12x – 8 = 4
∴ 12x = 12 ∴ x = 1
Hence, x = 1, y = 2 is the required solution.

(iv) 5x + 2y = 4, 7x + 3y = 5
Solution:
5x + 2y = 4 ………..(1)
7x + 3y = 5 …………(2)
Multiplying Eq. (1) with 7 and Eq. (2) with 5
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 6
Put y = -3 into Eq. (1)
5x + 2y = 4
5x + 2(-3) = 4
5x – 6 = 4
5x = 4 + 6
5x = 10
x = \(\frac{10}{5}\)
x = 2
Hence, x = 2, y = -3 is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
The cost of 4 pencils, 3 pens and 2 erasers is ₹ 60. The cost of 2 pencils, 4 pens and 6 erasers is ₹ 90, whereas the cost of 6 pencils, 2 pens and 3 erasers is ₹ 70. Find the cost of each item by using matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 eraser be ₹ x, ₹ y and ₹ z respectively.
Then, from the given conditions,
4x + 3y + 2z = 60
2x + 4y + 6z = 90, i.e., x + 2y + 3z = 45
6x + 2y + 3z = 70
These equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 4
By equality of matrices,
x + 2y + 3z = 45 …….(1)
– 5y – 10z = – 120 …….(2)
5z = 40
From (3), z = 8
Substituting z = 8 in (2), we get,
– 5y – 80 = -120
∴ – 5y = -40 ∴ y = 8
Substituting y = 8, z = 8 in (1), we get,
x + 16 + 24 = 45
∴ x + 40 = 45 ∴ x = 5
∴ x = 5, y = 8, z = 8
Hence, the cost is ₹ 5 for a pencil, ₹ 8 for a pen and ₹ 8 for an eraser.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 4.
If three numbers are added, their sum is 2. If 2 times the second number is subtracted from the sum of first and third numbers, we get 8 and if three times the first number is added to the sum of second and third numbers, we get 4. Find the numbers using matrices.
Solution:
Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 2
x + z – 2y = 8, i.e., x – 2y + 2 = 8
and y + z + 3x = 4, i.e., 3x + y + z = 4
Hence, the system of linear equations is
x + y + z = 2
x – 2y + z = 8
3x + y + z = 4
These equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 7
By equality of matrices,
x + y + z = 2 ……(1)
-3y = 6 ……(2)
– 2y – 2z = -2 ……..(3)
From (2), y = -2
Substituting y = -2 in (3), we get,
-2(-2) – 2z = -2
∴ -2z = -6 ∴ z = 3
Substituting y = -2, z = 3 in (1), we get,
x – 2 + 3 = 2 ∴ x = 1
Hence, the required numbers are 1, -2 and 3.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
The total cost of 3 T.V. sets and 2 V.C.R.s is ₹ 35000. The shop-keeper wants profit of ₹ 1000 per television and ₹ 500 per V.C.R. He can sell 2 T. V. sets and 1 V.C.R. and get the total revenue as ₹ 21,500. Find the cost price and the selling price of a T.V. sets and a V.C.R.
Solution:
Let the cost of each T.V. set be ₹ x and each V.C.R. be ₹ y. Then the total cost of 3 T.V. sets and 2 V.C.R.’s is ₹ (3x + 2y) which is given to be ₹ 35,000.
∴ 3x + 2y = 35000
The shopkeeper wants profit of ₹ 1000 per T.V. set and of ₹ 500 per V.C.R.
∴ the selling price of each T.V. set is ₹ (x + 1000) and of each V.C.R. is ₹ (y + 500).
∴ selling price of 2 T.V. set and 1 V.C.R. is
₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21,500.
∴ 2(x + 1000) + (y + 500) = 21500
∴ 2x + 2000 + y + 500 = 21500
∴ 2x + y = 19000
Hence, the system of linear equations is
3x + 2y = 35000
2x + y = 19000
These equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3 8
By equality of matrices,
2x + y = 19000 ……….(1)
-x = -3000 ……….(2)
From (2), x = 3000
Substituting x = 3000 in (1), we get,
2(3000) + y = 19000
∴ y = 13000
∴ the cost price of one T.V. set is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. set is ₹ 4000 and of one V.C.R. is ₹ 13500.

Class 12 Maharashtra State Board Maths Solution 

Matrices Class 12 Maths 1 Exercise 2.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.2 Questions And Answers Maharashtra Board

Question 1.
Find the co-factors of the elements of the following matrices
(i) \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Here, a11 = -11, M11 = 4
∴ A11 = (-1)1+1(4) = 4
a12 = 2, M12 = -3
∴ A12 = (-1)1+2(- 3) = 3
a21 = – 3, M21 = -2
∴ A21 = (- 1)2+1(2) = -2
a22 = 4, M22 = -1
∴ A22 = (-1)2+2(-1) = -1.

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
The co-factor of aij is given by Aij = (-1)i+jMij
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 1
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 2

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Find the matrix of co-factors for the following matrices
(i) \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Here, a11 = 1, M11 = -1
∴ A11 = (-1)1+1(-1) = -1
a12 = 3, M12 = 4
∴ A12 = (-1)1+2(4) = -4
a21 = 4, M21 = 3
∴ A21 = (-1)2+1(3) = -3
a22 = -1, M22 = 1
∴ A22 = (-1)2+1(1) = 1
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left(\begin{array}{rr}
-1 & -4 \\
-3 & 1
\end{array}\right)\)

(ii) \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 21
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 22
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 23
A11 = -14, A12 = -10, A13 = -6,
A21 = 6, A22 = -5, A23 = -3,
A31 = -2, A32 = -7, A33 = 1.
∴ the co-factor matrix
= \(\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]\) = \(\left[\begin{array}{rrr}
-14 & -10 & -6 \\
6 & -5 & -3 \\
-2 & -7 & 1
\end{array}\right]\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Here, a11 = 2, M11= 5
∴ A11 = (-1)1+1(5) = 5
a12 = -3, M12 = 3
∴ A12 = (-1)1+2(3) = -3
a21 = 3, M21 = -3
∴ A A21 = (-1)2+1(-3) = 3
a22 = 5, M22 = 2
∴ A22 = (-1)2+1 = 2
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left[\begin{array}{rr}
5 & -3 \\
3 & 2
\end{array}\right]\)
∴ adj A = \(\left(\begin{array}{rr}
5 & 3 \\
-3 & 2
\end{array}\right)\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 1
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 2
A11 = -3, A12 = -12, A13 = 6,
A21 = -1, A22 = 3, A23 = 2,
A31 = -11, A32 = -9, A33 = 1
∴ the co-factor matrix = \(\left[\begin{array}{lll}
\mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{15} \\
\mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\
\mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-3 & -12 & 6 \\
-1 & 3 & 2 \\
-11 & -9 & 1
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{rrr}
-3 & -1 & -11 \\
-12 & 3 & -9 \\
6 & 2 & 1
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\), verify that A (adj A) = (adj A) A = | A | ∙ I
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 3
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 4
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 5
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 6
From (1), (2) and (3), we get,
A(adj A) = (adj A)A = |A|∙I.
Note: This relation is valid for any non-singular matrix A.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
Find the inverse of the following matrices by the adjoint method
(i) \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right|\) = -2 + 15 = 13 ≠ 0
∴ A-1 exists.
First we have to find the co-factor matrix
= [Aij]2×2, where Aij = (-1)i+jMij
Now, A11 = (-1)1+1M11 = 2
A12 = (-1)1+2M12 = -(-3) = 3
A21 = (-1)2+1M21 = -5
A22 = (-1)2+2M22 = -1
Hence, the co-factor matrix
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 7

(ii) \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
|A| = \(\) = 6 + 8 = 14 ≠ 0
∴ A-1 exist
First we have to find the co-factor matrix
= [Aij] 2×2 where Aij = (-1)i+jMij
Now, A11 = (-1)1+1M11 = 3
A12 = (-1)1+2M = -4
A21 = (-2)2+1M21 = (-2) = 2
A22 = (-1)2+2M22 = 2
Hence the co-factor matrix
= \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
∴ A-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A) = \(\frac{1}{14}\left(\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right)\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 8
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 9
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 10
∴ A-1 = \(\frac{1}{3}\left[\begin{array}{rrr}
3 & 0 & 0 \\
-3 & 1 & 0 \\
9 & 2 & -3
\end{array}\right]\)

(iv) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
= 1(10 – 0) – 0 + 0
= 1(10) – 0 + 0
= 10 ≠ 0
∴ A-1 exists.
First we have to find the co-factor matrix
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 24
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 25
∴ A-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A)
= \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)
∴ A-1 = \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 6.
Find the inverse of the following matrices
(i) \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 11
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 12
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 13

(ii) \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 14
∴ A-1 = \(\left(\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right)\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 15
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 16
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 17

(iv) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 18
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 19
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.2 20

Class 12 Maharashtra State Board Maths Solution 

Matrices Class 12 Maths 1 Exercise 2.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.1 Questions And Answers Maharashtra Board

Question 1.
Apply the given elementary transformation on each of the following matrices.
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\), R1 ↔ R2
Solution:
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\)
By R1 ↔ R2, we get,
A ~ \(\left[\begin{array}{rr}
-1 & 3 \\
1 & 0
\end{array}\right]\)

Question 2.
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\), R1 → R1 → R2
Solution:
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\),
R1 → R1 → R2 gives,
B ~ \(\left[\begin{array}{rrr}
-1 & -6 & -1 \\
2 & 5 & 4
\end{array}\right]\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\), C1 ↔ C2; B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\), R1 ↔ R2. What do you observe?
Solution:
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\)
By C1 ↔ C2, we get,
A ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(1)
B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\)
By R1 ↔ R2, we get,
B ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(2)
From (1) and (2), we observe that the new matrices are equal.

Question 4.
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\), 2C2
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\), -3R1
Find the addition of the two new matrices.
Solution:
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\)
By 2C2, we get,
A ~ \(\left[\begin{array}{rrr}
1 & 4 & -1 \\
0 & 2 & 3
\end{array}\right]\)
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\)
By -3R1, we get,
B ~ \(\left[\begin{array}{rrr}
-3 & 0 & -6 \\
2 & 4 & 5
\end{array}\right]\)
Now, addition of the two new matrices
Maharashtra Board 12th Maths Solutions Chapter 2 Matrics Ex 2.1 1

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\), 3R3 and then C3 + 2C2.
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By 3R3, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
9 & 9 & 3
\end{array}\right]\)
By C3 + 2C2, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
9 & 9 & 3+2(9)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)

Question 6.
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\), C3 + 2C2 and then 3R3. What do you conclude from Ex. 5 and Ex. 6 ?
Solution:
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\)
By C3 + 2C2, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
3 & 3 & 1+2(3)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
3 & 3 & 7
\end{array}\right)\)
By 3R3, we get
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)
We conclude from Ex. 5 and Ex. 6 that the matrix remains same by interchanging the order of the elementary transformations. Hence, the transformations are commutative.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 7.
Use suitable transformation on \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) into an upper triangular matrix.
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
By R2 – 3R1, we get,
A ~ \(\left[\begin{array}{rr}
1 & 2 \\
0 & -2
\end{array}\right]\)
This is an upper triangular matrix.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 8.
Convert \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\) into an identity matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
By R2 – 2R1, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 5
\end{array}\right]\)
By \(\left(\frac{1}{5}\right)\)R2, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]\)
By R1 + R2, we get,
A ~ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
This is an identity matrix.

Question 9.
Transform \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper triangular matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\)
By R2 – 2R1 and R3 – 3R1, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 5 & -2
\end{array}\right]\)
By R3 – \(\left(\frac{5}{3}\right)\)R2, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 0 & -\frac{1}{3}
\end{array}\right)\)
This is an upper triangular matrix.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Miscellaneous Exercise 1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.

12th Maths Part 1 Mathematical Logic Miscellaneous Exercise 1 Questions And Answers Maharashtra Board

Question 1.
Select and write the correct answer from the given alternatives in each of the following questions:
i) If p ∧ q is false and p ∨ q is true, the ________ is not true.
(A) p ∨ q
(B) p ↔ q
(C) ~p ∨ ~q
(D) q ∨ ~p
Solution:
(b) p ↔ q.

(ii) (p ∧ q) → r is logically equivalent to ________.
(A) p → (q → r)
(B) (p ∧ q) → ~r
(C) (~p ∨ ~q) → ~r
(D) (p ∨ q) → r
Solution:
(a) p → (q → r) [Hint: Use truth table.]

(iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~p ∧ ~q) → (~p ∨ ~q)
(D) (~p ∨ ~q) → (~p ∧ ~q)
Solution:
(c) (~p ∧ ~q) → (~p ∨ ~ q)

(iv) If p ∧ q is F, p → q is F then the truth values of p and q are ________.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
Solution:
(b) T, F

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) The negation of inverse of ~p → q is ________.
(A) q ∧ p
(B) ~p ∧ ~q
(C) p ∧ q
(D) ~q → ~p
Solution:
(a) q ∧ p

(vi) The negation of p ∧ (q → r) is ________.
(A) ~p ∧ (~q → ~r)
(B) p ∨ (~q ∨ r)
(C) ~p ∧ (~q → ~r)
(D) ~p ∨ (~q ∧ ~r)
Solution:
(d) ~p ∨ (q ∧ ~r)

(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ǝ x ∈ A such that x + 3 = 8
(B) Ǝ x ∈ A such that x + 2 < 9
(C) Ɐ x ∈ A, x + 6 ≥ 9
(D) Ǝ x ∈ A such that x + 6 < 10
Solution:
(c) Ǝ x ∈ A, x + 6 ≥ 9.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Which of the following sentences are statements in logic? Justify. Write down the truth
value of the statements :
(i) 4! = 24.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(ii) π is an irrational number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is ‘F’. ….[T ∧ F ≡ F]

(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.

(v) cos2θ – sin2θ = cos2θ for all θ ∈ R.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

Question 3.
Write the truth values of the following statements :
(i) \(\sqrt {5}\) is an irrational but \(3\sqrt {5}\) is a complex number.
Solution:
Let p : \(\sqrt {5}\) is an irrational.
q : \(3\sqrt {5}\) is a complex number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(ii) Ɐ n ∈ N, n2 + n is even number while n2 – n is an odd number.
Solution:
Let p : Ɐ n ∈ N, n2 + n is an even number.
q : Ɐ n ∈ N, n2 – n is an odd number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F].

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) Ǝ n ∈ N such that n + 5 > 10.
Solution:
Ǝ n ∈ N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n ≥ 6, where n ∈ N, satisfy n + 5 > 10).

(iv) The square of any even number is odd or the cube of any odd number is odd.
Solution:
Let p : The square of any even number is odd.
q : The cube of any odd number is odd.
Then the symbolic form of the given statement is p ∨ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ∨ q is T. … [F ∨ T ≡ T].

(v) In ∆ ABC if all sides are equal then its all angles are equal.
Solution:
Let p : ABC is a triangle and all its sides are equal.
q : Its all angles are equal.
Then the symbolic form of the given statement is p → q
If the truth value of p is T, then the truth value of q is T.
∴ the truth value of p → q is T. … [T → T ≡ T].

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) Ɐ n ∈ N, n + 6 > 8.
Solution:
Ɐ n ∈ N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 ∈ N, n = 2 ∈ N do not satisfy n + 6 > 8).

Question 4.
If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of each of the following statement :
(i) Ǝ x ∈ A such that x + 8 = 15.
Solution:
True

(ii) Ɐ x ∈ A, x + 5 < 12.
Solution:
False

(iii) Ǝ x ∈ A, such that x + 7 ≥ 11.
Solution:
True

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) Ɐ x ∈ A, 3x ≤ 25.
Solution:
False

Question 5.
Write the negations of the following :
(i) Ɐ n ∈ A, n + 7 > 6.
Solution:
The negation of the given statements are :
Ǝ n ∈ A, such that n + 7 ≤ 6.
OR Ǝ n ∈ A, such that n + 7 ≯ 6.

(ii) Ǝ x ∈ A, such that x + 9 ≤ 15.
Solution:
Ɐ x ∈ A, x + 9 > 15.

(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 6.
Construct the truth table for each of the following :
(i) p → (q → p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 1

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 2

(iii) ~(~p ∧ ~q) ∨ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 3

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 4

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 5

Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(p → q) ∧ ~q)] → ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 6
All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q)] → ~p is a tautology.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) [(p ∨ q) ∧ ~p] ∧ ~q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 7
All the entries in the last column of the above truth table are F.
∴ [(p ∨ q) ∧ ~p] ∧ ~q is a contradiction.

(iii) (p → q) ∧ (p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 8
All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

(iv) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 9
All the entries in the last column of the above truth table are T.
∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) [(p ∧ (p → q)] → q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 10
All the entries in the last column of the above truth table are T.
∴ [(p ∧ (p → q)] → q is a tautology.

(vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 11
All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) is a tautology.

(vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 12
The entries in the last column are neither T nor all F.
∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(viii) (p → q) ∨ (q → p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 13
All the entries in the last column of the above truth table are T.
∴ (p → q) ∨ (q → p) is a tautology.

Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p ∨ q) is T and (p ∧ q) is T
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 14
Since p ∨ q and p ∧ q both are T, from the table the truth values of both p and q are T.

(ii) (p ∨ q) is T and (p ∨ q) → q is F
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 15
Since the truth values of (p ∨ q) is T and (p ∨ q) → q is F, from the table, the truth values of p and q are T and F respectively.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 16
Since the truth values of (p ∧ q) is F and (p ∧ q) → q is T, from the table, the truth values of p and q are either T and F respectively or F and T respectively or both F.

Question 9.
Using truth tables prove the following logical equivalences :
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 17
The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 18
The entries in the columns 5 and 7 are identical.
∴ (p ∧ q) → r ≡ p → (q → r).

Question 10.
Using rules in logic, prove the following :
(i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)
Solution:
By the rules of negation of biconditional,
~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q
∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q … (Negation of disjunction)
≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
(p ∨ q) ∧ ~ p
≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)
≡ F ∨ (q ∧ ~p) … (Complement Law)
≡ q ∧ ~ p … (Identity Law)
≡ ~p ∧ q …(Commutative Law)
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
~ (p ∨ q) ∨ (~p ∧ q)
≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)
≡ ~p ∧ (~q ∨ q) … (Distributive Law)
≡ ~ p ∧ T … (Complement Law)
≡ ~ p … (Identity Law)
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 11.
Using the rules in logic, write the negations of the following :
(i) (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 disjunction)
≡ {~ p ∧ ~q) ∨ (~q ∧ r) … (Negation of negation)
≡ (~q ∧ ~p) ∨ (~q ∧ r) … (Commutative law)
≡ (~ q) ∧ (~ p ∨ r) … (Distributive Law)

(ii) 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 disjunction)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(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)

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

Question 12.
Express the following circuits in the symbolic form. Prepare the switching table :
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 19
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q: the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is :
(p ∧ q) ∨ (~p) ∨ (p ∧ ~q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 21

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 20
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed.
Then the symbolic form of the given statement is : (p ∨ q) ∧ (p ∨ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 22

Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 23
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p: the switch S1‘ is closed or the switch S1 is open
~ q: the switch S2‘ is closed or the switch S2 is open.
Then the given circuit in symbolic form is :
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~q)
Using the laws of logic, we have,
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~ q)
= (p ∧ ~q) ∨ [(~p ∧ q) ∨ (~p ∧ ~q) …(By Complement Law)
= (p ∧ ~q) ∨ [~p ∧ (q ∨ ~q)} (By Distributive Law)
= (p ∧ ~q) ∨ (~p ∧ T) …(By Complement Law)
= (p ∧ ~q) ∨ ~ p …(By Identity Law)
= (p ∨ ~p) ∧ (~q ∨ ~p) …(By Distributive Law)
= ~q ∨ ~p …(By Identity Law)
= ~p ∨ ~p …(By Commutative Law)
Hence, the simplified circuit for the given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 24

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 25
Solution:
(ii) Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
s : the switch S4 is closed
t : the switch S5 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open
~ s : the switch S4‘ is closed or the switch S4 is open
~ t : the switch S5‘ is closed or the switch S5 is open.
Then the given circuit in symbolic form is
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]
Using the laws of logic, we have,
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~ t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]
= [(p∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … (By De Morgan’s Law)
= (p ∧ q) ∨ [ ~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … (By Distributive Law)
= (p ∧ q) ∨ F … (By Complement Law)
= p ∧ q … (By Identity Law)
Hence, the alternative simplified circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 26

Question 14.
Check whether the following switching circuits are logically equivalent – Justify.
(A)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 27
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
(A) The symbolic form of the given switching circuits are
p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) respectively.
By Distributive Law, p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Hence, the given switching circuits are logically equivalent.

(B)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 28
Solution:
The symbolic form of the given switching circuits are
(p ∨ q) ∧ (p ∨ r) and p ∨ (q ∧ r)
By Distributive Law,
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Hence, the given switching circuits are logically equivalent.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 29
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed, or the switch S1 is open
~q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form Of the given circuit is :
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
≡ (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) y (p ∧ ~q ∧ r) …(By Commutative Law)
≡ (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Complement Law)
≡ F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [(~p ∨ p) ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) … (By Distributive Law)
≡ [T ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) = (q ∧ r) ∨ (p ∧ ~q ∧ r) …(By Complement Law)
≡ (q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [q ∨ (p ∧ ~q)] ∧ r … (By Distributive Law)
≡ [q ∨ p) ∧ ((q ∨ ~q)] ∧ r … (By Distributive Law)
≡ [(q ∨ p) ∧ T] ∧ r …(By Complement Law)
≡ (q ∨ p) ∧ r … (By Identity Law)
≡ (p ∨ q) ∧ r …(By Commutative Law)
∴ the alternative arrangement of the new circuit with minimum switches is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 30

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 16.
Simplify the following so that the new circuit circuit.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 31
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given switching circuit is :
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
Using the laws of logic, we have,
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
≡ (~p ∨ q ∨ p ∨ ~q) ∨ (p ∨ q)
≡ [(~p ∨ p) ∨ (q ∨ ~q)] ∨ (p ∨ q) … (By Commutative Law)
≡ (T ∨ T) ∨ (p ∨ q) … (By Complement Law)
≡ T ∨ (p ∨ q) … (By Identity Law)
≡ T … (By Identity Law)
∴ the current always flows whether the switches are open or closed. So, it is not necessary to use any switch in the circuit.
∴ the simplified form of given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 32

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 33
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open.
Then, the symbolic form of the given switching circuit is : [p ∨ (~ q) ∨ (~ r)] ∧ [p ∨ (q ∧ r)]
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 34
From the table, the’ final column’ and the column of p are identical. Hence, the given circuit is equivalent to the simple circuit with only one switch S1.
the simplified form of the given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 35

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.5 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.5 Questions And Answers Maharashtra Board

Question 1.
Express the following circuits in the symbolic form of logic and write the input-output table.
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 1
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed or the switch S1is open
~q : the switch S2‘ is closed or the switch S2 is open
~r : the switch S3‘ is closed or the switch S3 is open
l : the lamp L is on
(i) The symbolic form of the given circuit is : p ∨ (q ∧ r) = l
l is generally dropped and it can be expressed as : p ∨ (q ∧ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 7

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 2
Solution:
The symbolic form of the given circuit is : (~ p ∧ q) ∨ (p ∧ ~ q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 8

(iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 3
Solution:
The symbolic form of the given circuit is : [p ∧ (~q ∨ r)] ∨ (~q ∧ ~ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 9

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 4
Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ q ∧ (r ∨ ~p).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 10

(v)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 5
Solution:
The symbolic form of the given circuit is : [p ∨ (~p ∧ ~q)] ∨ (p ∧ q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 11

(vi)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 6
Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ (q ∨ r) ∧ (r ∨ p)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 12

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Construct the switching circuit of the following :
(i) (~p∧ q) ∨ (p∧ ~r)
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open.
Then the switching circuits corresponding to the given statement patterns are :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 13

(ii) (p∧ q) ∨ [~p ∧ (~q ∨ p ∨ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 14

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 15

(iv) (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 16

(v) p ∨ (~p ) ∨ (~q) ∨ (p ∧ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 17

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) (p ∧ q) ∨ (~p) ∨ (p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 18

Question 3.
Give an alternative equivalent simple circuits for the following circuits :
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 19
Solution:
(i) Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch Si is open Then the symbolic form of the given circuit is :
p ∧ (~p ∨ q).
Using the laws of logic, we have,
p ∧ (~p ∨ q)
= (p ∧ ~ p) ∨ (p ∧ q) …(By Distributive Law)
= F ∨ (p ∧ q) … (By Complement Law)
= p ∧ q… (By Identity Law)
Hence, the alternative equivalent simple circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 20

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 21
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~q : the switch S2‘ is closed or the switch S2 is open
~r : the switch S3‘ is closed or the switch S3 is open.
Then the symbolic form of the given circuit is :
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p).
Using the laws of logic, we have
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p)
≡ [p ∧ (q ∨ r)] ∨ [ ~(r ∨ q) ∧ p] …. (By De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] … (By Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)) … (By Distributive Law)
≡ p ∧ T … (By Complement Law)
≡ p … (By Identity Law)
Hence, the alternative equivalent simple circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 22

Question 4.
Write the symbolic form of the following switching circuits construct its switching table and interpret it.
i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 23
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is :
(p ∨ ~q) ∨ (~p ∧ q)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 24
Since the final column contains all’ 1′, the lamp will always glow irrespective of the status of switches.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 25
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~p : the switch S1 is closed or the switch S1 is open.
~q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is : p ∨ (~p ∧ ~q) ∨ (p ∧ q)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 26
Since the final column contains ‘0’ when p is 0 and q is ‘1’, otherwise it contains ‘1′.
Hence, the lamp will not glow when S1 is OFF and S2 is ON, otherwise the lamp will glow.

iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 27
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~q : the switch S2‘ is closed or the switch S2 is open
~r: the switch S3‘ is closed or the switch S3 is open.
Then the symbolic form of the given circuit is : [p ∨ (~q) ∨ r)] ∧ [p ∨ (q ∧ r)]
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 28
From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which S1 is ‘ON’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
Obtain the simple logical expression of the following. Draw the corresponding switching circuit.
(i) p ∨ (q ∧ ~ q)
Solution:
Using the laws of logic, we have, p ∨ (q ∧ ~q)
≡ p ∨ F … (By Complement Law)
≡ p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S1 is closed
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 29

(ii) (~p ∧ q) ∨ (~p ∧ ~q) ∨ (p ∧ ~q)]
Solution:
Using the laws of logic, we have,
(~p ∧ q) ∨ (~p ∨ ~q) ∨ (p ∧ ~q)
≡ [~p ∧ (q ∨ ~q)] ∨ (p ∧ ~ q)… (By Distributive Law)
≡ (~p ∧ T) ∨ (p ∧ ~q) … (By Complement Law)
≡ ~p ∨ (p ∧ ~q) … (By Identity Law)
≡ (~p ∨ p) ∧ (~p ∧~q) … (By Distributive Law)
≡ T ∧ (~p ∧ ~q) … (By Complement Law)
≡ ~p ∨ ~q … (By Identity Law)
Hence, the simple logical expression of the given expression is ~ p ∨ ~q.
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open,
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 30

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) [p (∨ (~q) ∨ ~r)] ∧ (p ∨ (q ∧ r)
Solution:
Using the laws of logic, we have,
[p ∨ (~ (q) ∨ (~r)] ∧ [p ∨ (q ∧ r)]
= [p ∨ { ~(q ∧ r)}] ∧ [p ∨ (q ∧ r)] … (By De Morgan’s Law)
= p ∨ [~(q ∧ r) ∧ (q ∧ r) ] … (By Distributive Law)
= p ∨ F … (By Complement Law)
= p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S1 is closed
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 31

(iv) (p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ∨ (p ∧ q ∧ r)
Question is Modified
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r)∨ (p ∧ q ∧ r)
Solution:
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r)
= (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Commutative Law)
= (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Complement Law)
= F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~ p ∨ p) ∧ (q ∧ r) … (By Distributive Law)
= T ∧ (q ∧ r) … (By Complement Law)
= q ∧ r … (By Identity Law)
Hence, the simple logical expression of the given expression is q ∧ r.
Let q : the switch S2 is closed
r : the switch S3 is closed.
Then the corresponding switching circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5 32

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.4 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.4 Questions And Answers Maharashtra Board

Question 1.
Using rules of negation write the negations of the following with justification.
(i) ~q → p
Solution:
The negation of ~q → p is
~(~q → p) ≡ ~ q ∧ ~p…. (Negation of implication)

(ii) p ∧ ~q
Solution:
The negation of p ∧ ~q is
~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)
≡ ~ p ∨ q … (Negation of negation)

(iii) p ∨ ~q
Solution:
The negation of p ∨ ~ p is
~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)
≡ ~ p ∧ q … (Negation of negation)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) (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 disjunction)
≡ (~ p ∧ q) ∧ ~ r … (Negation of negation)

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

(vi) ~(p ∧ q) ∨ (p ∨ ~q)
Solution:
The negation of ~(p ∧ q) ∨ (p ∨ ~q) is
~[~(p ∧ q) ∨ (p ∨ ~q)] ≡ ~[~(p ∧ q)] ∧ ~(p ∨ ~q) … (Negation of disjunction)
≡ ~[~(p ∧ q)] ∧ [ p ∧ ~(~q)] … (Negation of disjunction)
≡ (p ∧ q) ∧ (~ p ∧ q) … (Negation of negation)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

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

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

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Rewrite the following statements without using if .. then.
(i) If a man is a judge then he is honest.
Solution:
Since p → ≡ ~p ∨ q, the given statements can be written as :
A man is not a judge or he is honest.

(ii) It 2 is a rational number then \(\sqrt {2}\) is irrational number.
Solution:
2 is not a rational number or \(\sqrt {2}\) is irrational number.

(iii) It f(2) = 0 then f(x) is divisible by (x – 2).
Solution:
f(2) ≠ 0 or f(x) is divisible by (x – 2).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Without using truth table prove that :
(i) p ↔ q ≡ (p∧ q) ∨ (~ p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p ↔ q) ∧ (q ↔ p) … (Biconditional Law)
≡ (~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) ∨ F] ∨ [F ∨ (q ∧ p)] … (ComplementLaw)
≡ (~ p ∧ ~ q) ∨ (q ∧ p) … (Identity Law)
≡ (~ p ∧ ~ q) ∨ (p ∧ q) … (Commutative Law)
≡ (p ∧ q) ∨ (~p ∧ ~q) … (Commutative Law)
≡ RHS.

(ii) (p ∨ q) ∧ (p ∨ ~q) ≡ p
Solution:
LHS = (p ∨ q) ∧ (p ∨ ~q)
≡ p ∨ (q ∧ ~q) … (Distributive Law)
≡ p ∨ F … (Complement Law)
≡ p … (Identity Law)
≡ RHS.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

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

(iv) ~[(p ∨ ~q) → (p ∧ ~q)] ≡ (p ∨ ~q) ∧ (~p ∨ q)
Solution:
LHS = ~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~p ∨ q)… (Negation of negation)
≡ RHS.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.3 Questions And Answers Maharashtra Board

Question 1.
If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following.
(i) Ǝ x ∈ A such that x – 8 = 1
Solution:
Clearly x = 9 ∈ A satisfies x – 8 = 1. So the given statement is true, hence its truth value is T.

(ii) Ɐ x ∈ A, x2 + x is an even number
Solution:
For each x ∈ A, x2 + x is an even number. So the given statement is true, hence its truth value is T.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) Ǝ x ∈ A such that x2 < 0
Solution:
There is no x ∈ A which satisfies x2 < 0. So the given statement is false, hence its truth value is F.

(iv) Ɐ x ∈ A, x is an even number
Solution:
x = 3 ∈ A, x = 5 ∈ A, x = 7 ∈ A, x = 9 ∈ A, x = 11 ∈ A do not satisfy x is an even number. So the given statement is false, hence its truth value is F.

(v) Ǝ x ∈ A such that 3x + 8 > 40
Solution:
Clearly x = 11 ∈ A and x = 12 ∈ A satisfies 3x + 8 > 40. So the given statement is true, hence its truth value is T.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) Ɐ x ∈ A, 2x + 9 > 14
Solution:
For each x ∈ A, 2x + 9 > 14. So the given statement is true, hence its truth value is T.

Question 2.
Write the duals of each of the following.
(i) p ∨ (q ∧ r)
Solution:
The duals of the given statement patterns are :
p ∧ (q ∨ r)

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

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

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) p ∧ ~q
Solution:
p ∨ ~q

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

(vi) ~p ∧ (~q ∧ (p ∨ q) ∧ ~r)
Solution:
~p ∨ (~q ∨ (p ∧ q) ∨ ~r)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) [~(p ∨ q)] ∧ [p ∨ ~(q ∧ ~s)]
Solution:
[ ~(p ∧ q)] ∨ [p ∧ ~(q ∨ ~s)]

(viii) c ∨ {p ∧ (q ∨ r)}
Solution:
t ∧ {p ∧ (q Ar)}

(ix) ~p ∨ (q ∧ r) ∧ t
Solution:
~p ∧ (q ∨ r) ∨ c

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(x) (p ∨ q) ∨ c
Solution:
(p ∧ q) ∧ t

Question 3.
Write the negations of the following.
(i) x + 8 > 11 or y – 3 = 6
Solution:
Let p : x + 8 > 11, q : y — 3 = 6.
Then the symbolic form of the given statement is p ∨ q.
Since ~(p ∨ q) ≡ ~p ∧ ~q, the negation of given statement is :
‘x + 8 > 11 and y – 3 ≠ 6’ OR
‘x + 8 ≮ 11 and y – 3 ≠ 6’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) 11 < 15 and 25 > 20
Solution:
Let p: 11 < 15, q : 25 > 20.
Then the symbolic form of the given statement is p ∧ q.
Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of given statement is :
’11 ≮ 15 or 25 > 20.’ OR
’11 ≯ 15 or 25 ≮ 20.’

(iii) Qudrilateral is a square if and only if it is a rhombus.
Solution:
Let p : Quadrilateral is a square.
q : It is a rhombus.
Then the symbolic form of the given statement is p ↔ q.
Since ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of given statement is :
‘ Quadrilateral is a square but it is not a rhombus or quadrilateral is a rhombus but it is not a square.’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) It is cold and raining.
Solution:
Let p : It is cold.
q : It is raining.
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 not cold or not raining.’

(v) If it is raining then we will go and play football.
Solution:
Let p : It is raining.
q : We will go.
r : We play football.
Then the symbolic form of the given statement is p → (q ∧ r).
Since ~[p → (q ∧ r)] ≡ p ∧ ~(q ∧ r) ≡ p ∧ (q ∨ ~r), the negation of the given statement is :
‘It is raining and we will not go or not play football.’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) \(\sqrt {2}\) is a rational number.
Solution:
Let p : \(\sqrt {2}\) is a rational number.
The negation of the given statement is
‘ ~p : \(\sqrt {2}\) is not a rational number.’

(vii) All natural numbers are whole numers.
Solution:
The negation of the given statement is :
‘Some natural numbers are not whole numbers.’

(viii) Ɐ n ∈ N, n2 + n + 2 is divisible by 4.
Solution:
The negation of the given statement is :
‘Ǝ n ∈ N, such that n2 + n + 2 is not divisible by 4.’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ix) Ǝ x ∈ N such that x – 17 < 20
Solution:
The negation of the given statement is :
‘Ɐ x ∈ N, x – 17 ≯ 20.’

Question 4.
Write converse, inverse and contrapositive of the following statements.
(i) If x < y then x2 < y2 (x, y ∈ R)
Solution:
Let p : x < y, q : x2 < y2.
Then the symbolic form of the given statement is p → q.
Converse : q → p is the converse of p → q.
i.e. If x2 < y2, then x < y.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If x ≯ y, then x2 ≯ y2. OR
If x ≮ y, then x2 ≮ y2.
Contrapositive : ~q → p is the contrapositive of
p → q i.e. If x2 ≯ y2, then x ≯ y. OR
If x2 ≮ y2, then x ≮ y.

(ii) A family becomes literate if the woman in it is literate.
Solution:
Let p : The woman in the family is literate.
q : A family become literate.
Then the symbolic form of the given statement is p → q
Converse : q → p is the converse of p → q.
i.e. If a family become literate, then the woman in it is literate.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If the woman in the family is not literate, then the family does not become literate.
Contrapositive : ~q → ~p is the contrapositive of p → q. i e. If a family does not become literate, then the woman in it is not literate.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) If surface area decreases then pressure increases.
Solution:
Let p : The surface area decreases.
q : The pressure increases.
Then the symbolic form of the given statement is p → q.
Converse : q → p is the converse of p→ q.
i.e. If the pressure increases, then the surface area decreases.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If the surface area does not decrease, then the pressure does not increase.
Contrapositive : ~q → ~p is the contrapositive of p → q.
i.e. If the pressure does not increase, then the surface area does not decrease.

(iv) If voltage increases then current decreases.
Solution:
Let p : Voltage increases.
q : Current decreases.
Then the symbolic form of the given statement is p → q.
Converse : q →p is the converse of p → q.
i.e. If current decreases, then voltage increases.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If voltage does not increase, then current does not decrease.
Contrapositive : ~q → ~p, is the contrapositive of p → q.
i.e. If current does not decrease, then voltage doesnot increase.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.2 Questions And Answers Maharashtra Board

Question 1.
Construct the truth table for each of the following statement patterns:
(i) [(p → q) ∧ q] → p
Solution :
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 1

(ii) (p ∧ ~q) ↔ (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 2

(iii) (p ∧ q) ↔ (q ∨ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 3

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) p → [~(q ∧ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 4

(v) ~p ∧ [(p ∨ ~q ) ∧ q]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 5

(vi) (~p → ~q) ∧ (~q → ~p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 6

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) (q → p) ∨ (~p ↔ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 7

(viii) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 8

(ix) p → [~(q ∧ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 4

(x) (p ∨ ~q) → (r ∧ p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 9

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Using truth tables prove the following logical equivalences.
(i) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 10
The entries in the columns 4 and 6 are identical.
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(ii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 11
The entries in the columns 3 and 7 are identical.
∴ ~(p ∨ q) ∧ (~p ∧ q) = ~p.

(iii) p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 12
The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)].

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) p → (q → p) ≡ ~p → (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 13
The entries in the columns 4 and 7 are identical.
∴ p → (q → p) ≡ ~p → (p → q).

(v) (p ∨ q ) → r ≡ (p → r) ∧ (q → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 14
The entries in the columns 5 and 8 are identical.
∴ (p ∨ q ) → r ≡ (p → r) ∧ (q → r).

(vi) p → (q ∧ r) ≡ (p → q) ∧ (p → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 15
The entries in the columns 5 and 8 are identical.
∴ p → (q ∧ r) ≡ (p → q) ∧ (p → r).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 16
The entries in the columns 5 and 8 are identical.
∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r).

(viii) [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 17
The entries in the columns 3 and 7 are identical.
∴ [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r.

(ix) ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 18
The entries in the columns 6 and 9 are identical.
∴ ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency.
(i) (p ∧ q) → (q ∨ p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 19
All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → (q ∨ p) is a tautology.

(ii) (p → q) ↔ (~p ∨ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 20
All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~p ∨ q) p is a tautology.

(iii) [~(~p ∧ ~q)] ∨ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 21
The entries in the last column of the above truth table are neither all T nor all F.
∴ [~(~p ∧ ~q)] ∨ q is a contingency.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) [(p → q) ∧ q)] → p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 22
The entries in the last column of the above truth table are neither all T nor all F.
∴ [(p → q) ∧ q)] → p is a contingency

(v) [(p → q) ∧ ~q] → ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 23
All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q] → ~p is a tautology.

(vi) (p ↔ q) ∧ (p → ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 24
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.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) ~(~q ∧ p) ∧ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 25
The entries in the last column of the above truth table are neither all T nor all F.
∴ ~(~q ∧ p) ∧ q is a contingency.

(viii) (p ∧ ~q) ↔ (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 26
All the entries in the last column of the above truth table are F.
∴ (p ∧ ~q) ↔ (p → q) is a contradiction.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ix) (~p → q) ∧ (p ∧ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 27
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.

(x) [p → (~q ∨ r)] ↔ ~[p → (q → r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 28
All the entries in the last column of the above truth table are F.
∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.1 Questions And Answers Maharashtra Board

Question 1.
State which of the following sentences are statements. Justify your answer. In case of the statement, write down the truth value :
(i) 5 + 4 = 13.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(ii) x – 3 = 14.
Solution:
It is an open sentence, hence it is not a statement.

(iii) Close the door.
Solution:
It is an imperative sentence, hence it is not a statement.

(iv) Zero is a complex number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) Please get me breakfast.
Solution:
It is an imperative sentence, hence it is not a statement.

(vi) Congruent triangles are also similar.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vii) x2 = x.
Solution:
It is an open sentence, hence it is not a statement,

(viii) A quadratic equation cannot have more than two roots.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ix) Do you like Mathematics ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(x) The sun sets in the west.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xi) All real numbers are whole numbers.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(xii) Can you speak in Marathi ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(xiii) x2 – 6x – 7 = 0, when x = 7.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(xiv) The sum of cuberoots of unity is zero.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xv) It rains heavily.
Solution :
It is an open sentence, hence it is not a statement.

Question 2.
Write the following compound statements symbolically:
(i) Nagpur is in Maharashtra and Chennai is in Tamil Nadu.
Solution:
Let p : Nagpur is in Maharashtra.
q : Chennai is in Tamil Nadu.
Then the symbolic form of the given statement is P∧q.

(ii) Triangle is equilateral or isosceles,
Solution:
Let p : Triangle is equilateral.
q : Triangle is isosceles.
Then the symbolic form of the given statement is P∨q.

(iii) The angle is right angle if and only if it is of measure 90°.
Solution:
Let p : The angle is right angle.
q : It is of measure 90°.
Then the symbolic form of the given statement is p↔q

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) Angle is neither acute nor obtuse.
Solution:
Let p : Angle is acute.
q : Angle is obtuse.
Then the symbolic form of the given statement is
~p ∧ ~q.

(v) If ∆ ABC is right angled at B, then m∠A + m∠C = 90°.
Solution:
Let p : ∆ ABC is right angled at B.
q : m∠A + m∠C = 90°.
Then the symbolic form of the given statement is p → q

(vi) Hima Das wins gold medal if and only if she runs fast.
Solution:
Let p : Hima Das wins gold medal
q : She runs fast.
Then the symbolic form of the given statement is p ↔ q.

(vii) x is not irrational number but it is a square of an integer.
Solution:
Let p : x is not irrational number
q : It is a square of an integer
Then the symbolic form of the given statement is p ∧ q
Note : If p : x is irrational number, then the symbolic form of the given statement is ~p ∧ q.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Write the truth values of the following :
(i) 4 is odd or 1 is prime.
Solution:
Let p : 4 is odd.
q : 1 is prime.
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 v q is F. … [F ∨ F = F]

(ii) 64 is a perfect square and 46 is a prime number.
Solution:
Let p : 64 is a perfect square.
q : 46 is a prime number.
Then the symbolic form of the given statement is p∧q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(iii) 5 is a prime number and 7 divides 94.
Solution:
Let p : 5 is a prime number.
q : 7 divides 94.
Then the symbolic form of the given statement is p∧q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(iv) It is not true that 5 – 3i is a real number.
Solution:
Let p : 5 – 3i is a real number.
Then the symbolic form of the given statement is ~ p.
The truth values of p is F.
∴ the truth values of ~ p is T. … [~ F ≡ T]

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) If 3 × 5 = 8, then 3 + 5 = 15.
Solution:
Let p : 3 × 5 = 8.
q : 3 + 5 = 15.
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 ≡ T]

(vi) Milk is white if and only if sky is blue.
Solution:
Let p : Milk is white.
q : Sky is blue
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]

(vii) 24 is a composite number or 17 is a prime number.
Solution :
Let p : 24 is a composite number.
q : 17 is a prime number.
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]

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 4.
If the statements p, q are true statements and r, s are false statements, then determine the truth values of the following:
(i) p ∨ (q ∧ r)
Solution:
Truth values of p and q are T and truth values of r and s are F.
p ∨ (q ∧ r) ≡ T ∨ (T ∧ F)
≡ T ∧ F ≡ T
Hence the truth value of the given statement is true.

(ii) (p → q) ∨ (r → s)
Solution:
(p → q) ∨ (r → s) ≡ (T → T) ∨ (F → F)
≡ T ∨ T ≡ T
Hence the truth value of the given statement is true.

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

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

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) (~r ↔ p) → (~q)
Solution:
(~r ↔ p) → (~q) ≡ (~F ↔ T) → (~T)
≡ (T ↔ T) → F
≡ T → F ≡ F
Hence the truth value of the given statement is false.

(vi) [~p ∧ (~q ∧ r) ∨ (q ∧ r) ∨ (p ∧ r)]
Solution:
[~p ∧ (~q ∧ r)∨(q ∧ r)∨(p ∧ r)]
≡ [~T ∧ (~T ∧ F)] ∨ [(T ∧ F) V (T ∧ F)]
≡ [F ∧ (F ∧ F)] ∨ [F V F]
≡ (F ∧ F) ∨ F
≡ F ∨ F ≡ F
Hence the truth value of the given statement is false.

(vii) [(~ p ∧ q) ∧ (~ r)] ∨ [(q → p) → (~ s ∨ r)]
Solution:
[(~ p ∧ q) ∧ (~ r)] ∨ [(q → p) → (~ s ∨ r)]
≡ [(~T ∧ T) ∧ (~F)] ∨ [(T → T) → (~F ∨ F)]
≡ [(F ∧ T) ∧ T] ∨ [T → (T ∨ F)]
≡ (F ∧ T) ∨ (T → T)
≡ F ∨ T ≡ T
Hence the truth value of the given statement is true.

(viii) ~ [(~p ∧ r) ∨ (s → ~q)] ↔ (p ∧ r)
Solution :
~ [(~p ∧ r) ∨ (s → ~q)] ↔ (p ∧ r)
≡ ~ [(~T ∧ F) ∨ (F → ~T)] ↔ (T ∧ F)
≡ ~ [(F ∧ F) ∨ (F → F)] ↔ F
≡ ~ (F ∨ T) ↔ F
≡ ~T ↔ F
≡ F ↔ F ≡ T
Hence the truth value of the given statement is true.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
Write the negations of the following :
(i) Tirupati is in Andhra Pradesh.
Solution:
The negations of the given statements are :
Tirupati is not in Andhra Pradesh.

(ii) 3 is not a root of the equation x2 + 3x – 18 = 0.
Solution:
3 is a root of the equation x2 + 3x – 18 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) \(\sqrt {2}\) is a rational number.
Solution:
\(\sqrt {2}\) is not a rational number.

(iv) Polygon ABCDE is a pentagon.
Solution:
Polygon ABCDE is not a pentagon.

(v) 7 + 3 > 5.
Solution :
7 + 3 > 5.

Class 12 Maharashtra State Board Maths Solution 

Foreign Trade of India Question Answer Class 12 Economics Chapter 10 Maharashtra Board

Balbharti Maharashtra State Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India Textbook Exercise Questions and Answers.

Std 12 Economics Chapter 10 Question Answer Foreign Trade of India Maharashtra Board

Class 12 Economics Chapter 10 Foreign Trade of India Question Answer Maharashtra Board

Economics Class 12 Chapter 10 Question Answer Maharashtra Board

1. Choose the correct option:

Question 1.
Types of foreign trade
a) Import trade
b) Export trade
c) Entrepot trade
d) Internal trade
Options:
1) a and b
2) a, b and c
3) a, b, c and d
4) None of these
Answer:
2) a, b and c

Maharashtra Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India

Question 2.
Export trends of India’s foreign trade includes
a) Engineering goods
b) Gems and Jewellery
c) Textiles and ready-made garments
d) Gold
Options:
1) a and c
2) a, b and c
3) b, c and d
4) None of these
Answer:
2) a, b and c

Question 3.
Role of foreign trade is
a) To earn foreign exchange
b) To encourage investment
c) Lead to division of labour
d) Bring change in composition of exports
Options:
1) a and b
2) a, b and c
3) b and d
4) None of these
Answer:
2) a, b and c

2. Identify and explain the concepts from the given illustrations:

Question 1.
India purchased petroleum from Iran.
Answer:
Concept: Import trade
Explanation: Import trade means purchase of goods and services by one country from another country.

Question 2.
Maharashtra purchased wheat from Punjab.
Answer:
Concept: Internal/Home/Domestic trade Explanation : Internal trade is also known as home trade or domestic trade. This trade is within the country. It is between two or more states of the country.

Question 3.
England imported cotton from India, made readymade garments from it and sold them to Malaysia.
Answer:
Concept: Entrepot trade
Explanation : It means purchase of goods and services from one country and selling the same to another country.

Question 4.
Japan sells smart phones to Myanmar.
Answer:
Concept: Export trade
Explanation : It means sell of goods and services by one country to another country.

3. Distinguish between the following:

Question 1.
Internal trade and International trade.
Answer:

Internal / Domestic / Home trade External / Foreign / International trade
(a) It means exchange of goods and services within the country. (a) It means exchange of goods and services between two or more countries.
(b) The goods and services are produced and sold within the country. (b) The goods and services are produced in one country and sold in other country.
(c) E.g. Kashmir apples sold in Maharashtra. (c) E.g. Kashmir apples sold in Dubai.

Question 2.
Trends in imports and Trends in exports of foreign trade.
Answer:

Trends in imports Trends in exports
(a) It means year wise numerical changes in imports of a country. (a) It means year wise numerical changes in exports of a country.
(b) India’s major imported goods are – petroleum, gold, fertilizers, iron and steel, etc. (b) India’s major exported goods are engineering goods, petroleum and chemical products, gems and jewellery, etc.
(c) Petroleum has highest import percent of 22.6 in 2016-17. (c) Engineering goods has highest export percent 23.7 in 2016-17.

Question 3.
Balance of payments and Balance of trade.
Answer:

Balance in payment Balance in trade
(a) It means systematic recording of all international economic transactions of that country during a year. (a) It means the difference between the value of a country’s exports and imports in a year.
(b) It is a broad concept. (b) It is narrow concept.

4. Answer the following:

Question 1.
Explain the concept of foreign trade and its types.
Answer:

Foreign trade is the exchange of goods and c services between two or more countries, Foreign trade is the trade across the j boundaries of a country.

There are three important types of foreign trade.

  • Import trade : It is a buying of goods and services from other country by home country. Excessive import can have a negative impact on home country. E.g. India buying petroleum from Iraq, Kuwait, etc.
  • Export trade : It is selling of goods and services by home country to another country.
    Excessive export can have a positive impact on the home country. E.g. India exporting tea and spices to USA, China, etc.
  • Entrepot trade : It means buying of goods and services from one country and selling them to another country. E.g. England importing cotton from India, making readymade garments from it and selling them to Malaysia.

Maharashtra Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India

Question 2.
Explain any four features of composition of Indias foreign trade.
Answer:
There are many changes in India’s foreign trade from last seven decades (70 years)

  • Gross National Income : India’s foreign trade has great significance for its GNP. It increased upto 48.8% in the year 2016-17.
  • Change in composition of exports : After independence there was change in the composition of India’s export trade from primary products to manufactured goods.
  • Change in composition of imports :
    After independence there was change in the composition of India’s import trade from consumer goods to capital goods.
  • Development of new ports : India’s foreign trade is handled mainly by Mumbai, Calcutta and Chennai ports. India has developed more new ports at Kandla, Cochin, Vishakhapatnam.
  • Oceanic trade : Most of India’s foreign trade is by sea. About 68% of India’s trade is by sea.

Question 3.
Explain the trend in India’s imports.
Answer:
India is importing various goods from other countries. Following are the major imported goods of India :

  • Petroleum : It has largest share in India’s import. In the year 2016-17, it has 22.6% share in India’s total import.
  • Gold: After petroleum, the second most imported item is gold. In the year 2011, ) India’s import of gold was $53.9 billion and in the year 2018-19 it declined upto $32.8 billion.
  • Fertilizers: The share of fertilizers in import expenditure declined from 4.1% in 1990-91 to only 1.3% in 2016-17.
  • Iron and Steel: In the year 2016-17, the share of iron and steel in India’s total import was 2.1%.

5. State with reasons whether you agree or disagree with the following statements:

Question 1.
During British nile, indigenous handicrafts suffered a severe blow.
Answer:
Yes, I do agree with this statement.

  • During the British rule India was exporting raw materials to England and was importing final goods from England.
  • Indian handicraft was unable to face competition with imported goods from England.
  • An imported goods were cheaper as compared to handicraft goods.
  • The demand for machine made cheap commodity had raised in Indian market.
  • That’s why Indian handicraft industries suffered during the British rule.

Question 2.
Trade is an engine of growth for an economy.
Answer:
Yes, I agree with this statement.

  • Trade permits a more efficient allocation of national resources.
  • Foreign trade provide foreign exchange which can be used to import modern machinery and technology from advanced countries.
  • Foreign trade encourages producers to produce more goods for export.
  • It leads to an increase in total investment in an economy.
  • Thus, we can say, trade is an engine to growth for an economy.

Question 3.
Foreign trade leads to division of labour and specialization at world level.
Answer:
Yes, I agree with this statement.

  • Some countries have abundant natural resources.
  • These countries should export raw material and import finished goods from countries which are advanced in skilled man power
  • Under specialisation specific work is given to the workers within a production process.
  • Specialisation can increase the productivity of a firm or economy.
  • Eg. Incase of car manufacturing company, some workers will design the cars, some workers will work on different section of assembly line, some workers will work on j testing cars, some workers will work on marketing of cars.

Maharashtra Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India

6. Observe the following table and answer the questions geven below it.

Direction of Indias imports
Maharashtra Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India 1

Questions:

Question 1.
Which organisation has the least share in the direction of India’s imports in 2015-16?
Answer:
Eastern Europe has the least share in the direction of India’s import.

Question 2.
Which orgamsation has maximum share in India’s direction of imports in 1990-91?
Answer:
OECD [Organisation for Economic Co-operation and Development has maximum share in India’s direction of imports in 1990-91.

Question 3.
Expand the abbreviations of OECD and OPEC
Answer:
OECD : Organisation for Economic Co-operation and Development.
OPEC : Organisation of Petroleum Exporting Countries.

Question 4.
State your opinion regarding the direction of India’s imports from 1990-91 to 2015-16.
Answer:
In the year 1990-91, OECD (54.0%) and in the year 2015-16, Developing nations (43.2%) has the highest share in the direction of India’s imports. India should encourage industries those are producing import substitute goods,which will help to reduce import from developing nations and help to save foreign exchange.

Question 5.
How much is the percentage of increase in the imports of developing nations in 2015-16 as
compared to 1990-9 1?
Answer:
There is 24.6% increase in the imports of developing nations.

7. Answer in detail :

Question 1.
Explain the meaning and role of foreign trade.
Answer:
Trade means buying and selling of goods and services. Foreign trade means when goods and services are exchanged between two or more countries.
According to Wasserman and Hultman “International trade consists of transaction between residents of different countries”.
Role of foreign trade :

  • Brings reputation and helps earn goodwill : Exporting country can earn reputation and goodwill in the international market. Eg. Japan in electronic goods- Panasonic, Canon, Sony, Hitachi. Germany in Automobile – BMW, Audi, Mercedes- Benz, Volkswagen, Porsche. USA in food- McDonalds, KFC, USA in computers – Dell HP, IBM.
  • Division of labour and specialisation: It helps to increase the productivity of a firm or economy. Under specialisation specific work is given to the workers within a production process. Eg. Some workers will design cars, some workers will work on assembly lines, some workers will work on testing cars, some workers will work on marketing of cars.
  • To earn foreign exchange: Foreign trade is playing very important role in earning foreign exchange. This foreign exchange can be used to import advanced technology and machinery from developed countries.
  • Encourages investment : Foreign trade leads to an increase in total investment in an economy. The rise in investment help to produce more goods and services for export.
  • Availability of multiple choices : Due to availability of imported goods, it helps to improve standard of living of the people in the country.
  • Stability in price level : Foreign trade helps to control the changes in price level by keeping demand and supply position stable,
  • Helpful during natural calamities : Foreign trade enables a country to import food grains and medicines from other countries to help the affected people.
  • Optimum allocation and utlization of resources : Due to foreign trade those goods are produced which have demand in international market. There is maximum allocation and utlisation of resources to produce more goods and services for export.
  • Promotes world peace : Foreign trade is bringing countries closer which leads to better understanding, co-operation and integration.

Maharashtra Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India

Question 2.
Explain the recent trends in India’s exports.
Answer:
Export means selling of goods and services by home country to another country. Excessive export can have a positive impact on the home country.

(i) Engineering Goods : Engineering goods includes transport equipment, automobiles and auto components, machinery and instruments. India’s top export item is engineering goods accounting for 22.5% in India’s total export in 2014-15 and this share has increased upto 25% in the year 2017-18. India is exporting engineering goods to Sri Lanka, UAE and USA.

(ii) Petroleum Products : India’s refining capacity increased significantly since 2001-02 due to which India turned a net exporter of petroleum refinery products. In the year 2013-14 the share of petroleum products in total export was 20.1% and in the year 2016-17 it declined upto 11.7%.

(iii) Chemicals and chemical products:
It includes drugs (Medicines) and pharmaceuticals. This is one sector where India is highly competitive on both quality and pricing factor. India became global hub for pharma production. India is exporting its chemicals and chemical products to USA, China and Germany. The share of this item was 10.4% in 2014-15.

(iv) Gems and Jewellery: Gems and Jewellery plays an important role in earning the foreign exchange for India. In the year 2014¬15 the share of Gems and Jewellery was 13.3% in India’s total export and it declined upto 5.32% in the year 2018-19.

(v) Textiles and readymade garments :
India’s readymade garments have huge demand in the international market. India is exporting textiles to USA, China and Bangladesh. India is exporting readymade garments to USA, UAE and UK. In the year 2014-15 India’s export of textile and garments was 11.3% of total export of India and it has declined upto 6.3% in the year 2016-17.

Intext Questions

Try this : (Text Book Page No. 94)

Name the goods exported to and imported from India to China and Japan in recent years
Answer:

Goods exported by India Goods imported by India
To China : From China :
raw materials and industrial inputs like organic chemicals, mineral fuels, cotton, ores, plastic materials, etc. electronic items, machinery, and plastic items.
To Japan : From Japan :
fisheries products, wheat, tea, coffee, species and herbs. mineral   fuels, machinery and food items.

Find out: (Text Book Page No. 95)

Find the recent share of India’s foreign trade in Gross National Income.
Answer:
India’s foreign trade accounts for 48.8% of her Gross National Income.

Maharashtra Board Class 12 Economics Solutions Chapter 10 Foreign Trade of India

Find out: (Text Book Page No. 97)

List the countries coming under OPEC and OECD.
Answer:
The countries coming under OPEC (Organisation of Petroleum Exporting Countries) are :
(a) Algeria, (b) Angola, (c) Congo, (d) Equatorial Guinea, (e) Gabon, (f) Iran, (g) Iraq, (h) Kuwait, (i) Libya, (j) Nigeria, (k) Saudi Arabia (1) United Arab Emirates, 0 Venezuela.

12th Std Economics Questions And Answers: