Category: Class 11

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.6 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.6

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.6 Text Book Back Questions and Answers

Question 1.
Find the equation of the parabola whose focus is the point F(-1, -2) and the directrix is the line 4x – 3y + 2 = 0.
Solution:
F(-1, -2)
l : 4x – 3y + 2 = 0
Let P(x, y) be any point on the parabola.
FP = PM
⇒ FP² = PM²
⇒ (x + 1)² + (y + 2)² = \(\left[\frac{4 x-3 y+2}{\sqrt{4^{2}+(-3)^{2}}}\right]^{2}\)
⇒ x² + 2x + 1 + y² + 4y + 4 = \(\frac{16 x^{2}+9 y^{2}+4-24 x y+16 x-12 y}{(16+9)}\)
⇒ 25(x² + y² + 2x + 4y + 5) = 16x² + 9y² – 24xy + 16x – 12y + 4
⇒ (25 – 16)x² + (25 – 9)y² + 24xy + (50 – 16)x + (100 + 12)y + 125 – 4 = 0
⇒ 9x² + 16y² + 24xy + 34x + 112y + 121 = 0

Question 2.
The parabola y² = kx passes through the point (4, -2). Find its latus rectum and focus.
Solution:
y² = kx passes through (4, -2)
(-2)² = k(4)
⇒ 4 = 4k
⇒ k = 1
y² = x = 4(\(\frac{1}{4}\))x
a = \(\frac{1}{4}\)
Equation of LR is x = a or x – a = 0
i.e., x = \(\frac{1}{4}\)
⇒ 4x = 1
⇒ 4x – 1 = 0
Focus (a, 0) = (\(\frac{1}{4}\), 0)

Question 3.
Find the vertex, focus, axis, directrix, and the length of the latus rectum of the parabola y² – 8y – 8x + 24 = 0.
Solution:
y² – 8y – 8x + 24 = 0
⇒ y² – 8y – 4² = 8x – 24 + 4²
⇒ (y – 4)² = 8x – 8
⇒ (y – 4)² = 8(x – 1)
⇒ (y – 4)² = 4(2) (x – 1)
∴ a = 2
Y² = 4(2)X where X = x – 1 and Y = y – 4

Question 4.
Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola (a) y² = 20x, (b) x² = 8y, (c) x² = -16y
Solution:
(a) y² = 20x
y² = 4(5)x
∴ a = 5

(b) x² = 8y = 4(2)y
∴ a = 2

(c) x² = -16y = -4(4)y
∴ a = 4

Question 5.
The average variable cost of the monthly output of x tonnes of a firm producing a valuable metal is ₹ \(\frac{1}{5}\) x² – 6x + 100. Show that the average variable cost curve is a parabola. Also, find the output and the average cost at the vertex of the parabola.
Solution:
Let output be x and average variable cost = y
y = \(\frac{1}{5}\) x² – 6x + 100
⇒ 5y = x² – 30x + 500
⇒ x² – 30x + 225 = 5y – 500 + 225
⇒ (x – 15)² = 5y – 275
⇒ (x – 15)² = 5(y – 55) which is of the form X² = 4(\(\frac{5}{4}\))Y
∴ Y average variable cost curve is a parabola
Vertex (0, 0)
x – 15 = 0; y – 55 = 0
x = 15; y = 55
At the vertex, output is 15 tonnes and average cost is ₹ 55.

Question 6.
The profit ₹ y accumulated in thousand in x months is given by y = -x² + 10x – 15. Find the best time to end the project.
Solution:
y = -x² + 10x – 15
⇒ y = -[x² – 10x + 5² – 5² + 15]
⇒ y = -[(x – 5)² – 10]
⇒ y = 10 – (x – 5)²
⇒ (x – 5)² = -(y – 10)
This is a parabola which is open downwards.
Vertex is the maximum point.
∴ Profit is maximum when x – 5 = 0 (or) x = 5 months.
After that profit gradually reduces.
∴ The best time to end the project is after 5 months.

Students can download 11th Business Maths Chapter 1 Matrices and Determinants Ex 1.1 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 1 Matrices and Determinants Ex 1.1

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.1 Text Book Back Questions and Answers

Question 1.
Find the minors and cofactors of all the elements of the following determinants.
(i) \(\left|\begin{array}{cc}
5 & 20 \\
0 & -1
\end{array}\right|\)
(ii) \(\left|\begin{array}{rrr}
1 & -3 & 2 \\
4 & -1 & 2 \\
3 & 5 & 2
\end{array}\right|\)
Solution:
(i) \(\left|\begin{array}{cc}
5 & 20 \\
0 & -1
\end{array}\right|\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Cofactor of 5 = A₁₁ = (-1)¹⁺¹ M₁₁ = 1 × -1 = -1
Cofactor of 20 = A₁₂ = (-1)¹⁺² M₁₂ = -1 × 0 = 0
Cofactor of 0 = A₂₁ = (-1)²⁺¹ M₂₁ = -1 × 20 = -20
Cofactor of -1 = A₂₂ = (-1)²⁺² M₂₂ = 1 × 5 = 5

(ii) \(\left|\begin{array}{rrr}
1 & -3 & 2 \\
4 & -1 & 2 \\
3 & 5 & 2
\end{array}\right|\)
Minor of 1 is M₁₁ = \(\left|\begin{array}{rr}
-1 & 2 \\
5 & 2
\end{array}\right|\) = -2 – 10 = -12
Minor of -3 is M₁₂ = \(\left|\begin{array}{ll}
4 & 2 \\
3 & 2
\end{array}\right|\) = 8 – 6 = 2
Minor of 2 is M₁₃ = \(\left|\begin{array}{rr}
4 & -1 \\
3 & 5
\end{array}\right|\) = 20 + 3 = 23
Minor of 4 is M₂₁ = \(\left|\begin{array}{rr}
-3 & 2 \\
5 & 2
\end{array}\right|\) = -6 – 10 = -16
Minor of -1 is M₂₂ = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 2
\end{array}\right|\) = 2 – 6 = -4
Minor of 2 is M₂₃ = \(\left|\begin{array}{rr}
1 & -3 \\
3 & 5
\end{array}\right|\) = 5 + 9 = 14
Minor of 3 is M₃₁ = \(\left|\begin{array}{cc}
-3 & 2 \\
-1 & 2
\end{array}\right|\) = -6 + 2 = -4
Minor of 5 is M₃₂ = \(\left|\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right|\) = 2 – 8 = -6
Minor of 2 is M₃₃ = \(\left|\begin{array}{cc}
1 & -3 \\
4 & -1
\end{array}\right|\) = -1 + 12 = 11
Cofactor of 1 is A₁₁ = (-1)¹⁺¹ M₁₁ = -12
Cofactor of -3 is A₁₂ = (-1)¹⁺² M₁₂ = -2
Cofactor of 2 is A₁₃ = (-1)¹⁺³ M₁₃ = 23
Cofactor of 4 is A₂₁ = (-1)²⁺¹ M₂₁ = -1 × -16 = 16
Cofactor of -1 is A₂₂ = (-1)²⁺² M₂₂ = -4
Cofactor of 2 is A₂₃ = (-1)²⁺³ M₂₃ = -14
Cofactor of 3 is A₃₁ = (-1)³⁺¹ M₃₁ = -4
Cofactor of 5 is A₃₂ = (-1)³⁺² M₃₂ = -1 × -6 = 6
Cofactor of 2 is A₃₃ = (-1)³⁺³ M₃₃ = 11

Question 2.
Evaluate \(\left|\begin{array}{rrr}
3 & -2 & 4 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
Solution:

= 3(0 – 2) + 2(6 – 1) + 4(4 – 0)
= -6 + 10 + 16
= 20

Question 3.
Solve: \(\left|\begin{array}{lll}
2 & x & 3 \\
4 & 1 & 6 \\
1 & 2 & 7
\end{array}\right|=0\)
Solution:

2(7 – 12) – x(28 – 6) + 3(8 – 1) = 0
2(-5) – x(22) + 3(7) = 0
-10 – 22x + 21 = 0
-22x + 11 = 0
-22x = -11
x = \(\frac{-11}{-22}=\frac{1}{2}\)

Question 4.
Find |AB| if A = \(\left[\begin{array}{rr}
3 & -1 \\
2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & 0 \\
1 & -2
\end{array}\right]\)
Solution:

∴ |AB| = \(\left|\begin{array}{rr}
8 & 2 \\
7 & -2
\end{array}\right|\) = -16 – 14 = -30

Question 5.
Solve: \(\left|\begin{array}{rrr}
7 & 4 & 11 \\
-3 & 5 & x \\
-x & 3 & 1
\end{array}\right|=0\)
Solution:

7(5 – 3x) – 4(-3 + x²) + 11(-9 + 5x) = 0
35 – 21x + 12 – 4x² – 99 + 55x = 0
-4x² – 21x + 55x + 35 + 12 – 99 = 0
-4x² + 34x – 52 = 0

Divide throughout by -2 we get
2x² – 17x + 26 = 0
(2x – 13) (x – 2) = 0
2x – 13 = 0 (or) x – 2 = 0
x = \(\frac{13}{2}\) (or) x = 2
∴ x = \(\frac{13}{2}\), x = 2

Question 6.
Evaluate: \(\left|\begin{array}{lll}
1 & a & a^{2}-b c \\
1 & b & b^{2}-c a \\
1 & c & c^{2}-a b
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & a & a^{2}-b c \\
1 & b & b^{2}-c a \\
1 & c & c^{2}-a b
\end{array}\right|\)

Question 7.
Prove that \(\left|\begin{array}{lll}
\frac{1}{a} & b c & b+c \\
\frac{1}{b} & c a & c+a \\
\frac{1}{c} & a b & a+b
\end{array}\right|=0\)
Solution:
\(\left|\begin{array}{lll}
\frac{1}{a} & b c & b+c \\
\frac{1}{b} & c a & c+a \\
\frac{1}{c} & a b & a+b
\end{array}\right|\)

Hence proved.

Question 8.
Prove that \(\left|\begin{array}{ccc}
-a^{2} & a b & a c \\
a b & -b^{2} & b c \\
a c & b c & -c^{2}
\end{array}\right|=4 a^{2} b^{2} c^{2}\)
Solution:
\(\left|\begin{array}{ccc}
-a^{2} & a b & a c \\
a b & -b^{2} & b c \\
a c & b c & -c^{2}
\end{array}\right|\)

= a²b²c² [-(0 – 4) + 0 + 0]
= 4a²b²c²

Students can download 11th Business Maths Chapter 1 Matrices and Determinants Ex 1.5 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 1 Matrices and Determinants Ex 1.5

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.5 Text Book Back Questions and Answers

Choose the Correct Answer.

Question 1.
The value of x if \(\left|\begin{array}{lll}
0 & 1 & 0 \\
x & 2 & x \\
1 & 3 & x
\end{array}\right|=0\) is
(a) 0, -1
(b) 0, 1
(c) -1, 1
(d) -1, -1
Answer:
(b) 0, 1
Hint:
0 – 1[x² – x] + 0 = 0
⇒ x² – x = 0
⇒ x(x – 1) = 0
⇒ x = 0 (or) x = 1

Question 2.
The value of \(\left|\begin{array}{lll}
2 x+y & x & y \\
2 y+z & y & z \\
2 z+x & z & x
\end{array}\right|\) is
(a) xyz
(b) x + y + z
(c) 2x + 2y + 2z
(d) 0
Answer:
(d) 0
Hint:
= \(\left|\begin{array}{lll}
2 x & x & y \\
2 y & y & z \\
2 z & z & x
\end{array}\right|\) C₁ → C₁ – C₃
= 0 (C₁ and C₂ are proportional)

Question 3.
The cofactor of -7 in the determinant \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\) is
(a) -18
(b) 18
(c) -7
(d) 7
Answer:
(b) 18
Hint:
A cofactor of -7 = \(\left|\begin{array}{rr}
2 & -3 \\
6 & 0
\end{array}\right|\)
= 0 + 18
= 18

Question 4.
If Δ = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2 \\
2 & 3 & 1
\end{array}\right|\) then \(\left|\begin{array}{lll}
3 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right|\) is
(a) Δ
(b) -Δ
(c) 3Δ
(d) -3Δ
Answer:
(b) -Δ
Hint:
\(\left|\begin{array}{lll}
3 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right|=-\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2 \\
2 & 3 & 1
\end{array}\right|\) R₁ ↔ R₂
= -Δ

Question 5.
The value of the determinant \(\left|\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right|^{2}\) is
(a) abc
(b) 0
(c) a²b²c²
(d) -abc
Answer:
(c) a²b²c²
Hint:
\(a^{2} b^{2} c^{2}\left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
= a²b²c² × 1²
= a²b²c²

Question 6.
If A is square matrix of order 3 then |kA| is:
(a) k|A|
(b) -k|A|
(c) k³|A|
(d) -k³|A|
Answer:
(c) k³|A|

Question 7.
adj (AB) is equal to:
(a) adj A adj B
(b) adj A^T adj B^T
(c) adj B adj A
(d) adj B^T adj A^T
Answer:
(c) adj B adj A

Question 8.
The inverse matrix of \(\left(\begin{array}{cc}
\frac{4}{5} & \frac{5}{12} \\
\frac{2}{5} & \frac{1}{2}
\end{array}\right)\) is
(a) \(\frac{7}{30}\left(\begin{array}{cc}
\frac{1}{2} & \frac{5}{12} \\
\frac{2}{5} & \frac{4}{5}
\end{array}\right)\)
(b) \(\frac{7}{30}\left(\begin{array}{cc}
\frac{1}{2} & \frac{-5}{12} \\
\frac{-2}{5} & \frac{1}{5}
\end{array}\right)\)
(c) \(\frac{30}{7}\left(\begin{array}{rr}
\frac{1}{2} & \frac{5}{12} \\
\frac{2}{5} & \frac{4}{5}
\end{array}\right)\)
(d) \(\frac{30}{7}\left(\begin{array}{rr}
\frac{1}{2} & \frac{-5}{12} \\
\frac{-2}{5} & \frac{4}{5}
\end{array}\right)\)
Answer:
(c) \(\frac{30}{7}\left(\begin{array}{rr}
\frac{1}{2} & \frac{5}{12} \\
\frac{2}{5} & \frac{4}{5}
\end{array}\right)\)

Question 9.
If A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) such that ad – bc ≠ 0 then A^-1 is:
(a) \(\frac{1}{a d-b c}\left[\begin{array}{cc}
d & b \\
-c & a
\end{array}\right]\)
(b) \(\frac{1}{a d-b c}\left[\begin{array}{ll}
d & b \\
c & a
\end{array}\right]\)
(c) \(\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
(d) \(\frac{1}{a d-b c}\left[\begin{array}{ll}
d & -b \\
c & a
\end{array}\right]\)
Answer:
(c) \(\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
Hint:

Question 10.
The number of Hawkins-Simon conditions for the viability of input-output analysis is:
(a) 1
(b) 3
(c) 4
(d) 2
Answer:
(d) 2

Question 11.
The inventor of input-output analysis is:
(a) Sir Francis Galton
(b) Fisher
(c) Prof. Wassily W. Leontief
(d) Arthur Cayley
Answer:
(c) Prof. Wassily W. Leontief

Question 12.
Which of the following matrix has no inverse?
(a) \(\left(\begin{array}{rr}
-1 & 1 \\
1 & -4
\end{array}\right)\)
(b) \(\left(\begin{array}{rr}
2 & -1 \\
-4 & 2
\end{array}\right)\)
(c) \(\left(\begin{array}{cc}
\cos a & \sin a \\
-\sin a & \cos a
\end{array}\right)\)
(d) \(\left(\begin{array}{rr}
\sin a & \sin a \\
-\cos a & \cos a
\end{array}\right)\)
Answer:
(b) \(\left(\begin{array}{rr}
2 & -1 \\
-4 & 2
\end{array}\right)\)
Hint:
So \(\left(\begin{array}{rr}
2 & -1 \\
-4 & 2
\end{array}\right)\) has no inverse.

Question 13.
Inverse of \(\left(\begin{array}{ll}
3 & 1 \\
5 & 2
\end{array}\right)\) is:
(a) \(\left(\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right)\)
(b) \(\left(\begin{array}{rr}
-2 & 5 \\
1 & -3
\end{array}\right)\)
(c) \(\left(\begin{array}{rr}
3 & -1 \\
-5 & -3
\end{array}\right)\)
(d) \(\left(\begin{array}{rr}
-3 & 5 \\
1 & -2
\end{array}\right)\)
Answer:
(a) \(\left(\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right)\)
Hint:
Let A = \(\left(\begin{array}{ll}
3 & 1 \\
5 & 2
\end{array}\right)\)
|A| = [6 – 5] = 1
adj A = \(\left[\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right]\)
∴ A^-1 = \(\left[\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right]\)

Question 14.
If A = \(\left(\begin{array}{rr}
-1 & 2 \\
1 & -4
\end{array}\right)\) then A (adj A) is:
(a) \(\left(\begin{array}{ll}
-4 & -2 \\
-1 & -1
\end{array}\right)\)
(b) \(\left(\begin{array}{rr}
4 & -2 \\
-1 & 1
\end{array}\right)\)
(c) \(\left(\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right)\)
(d) \(\left(\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right)\)
Answer:
(c) \(\left(\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right)\)
Hint:
A = \(\left(\begin{array}{rr}
-1 & 2 \\
1 & -4
\end{array}\right)\)
|A| = 4 – 2 = 2
We know that A (adj A) = |A| I
⇒ 2 \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right)\)

Question 15.
If A and B non-singular matrix then, which of the following is incorrect?
(a) A² = I implies A^-1 = A
(b) I^-1 = I
(c) If AX = B then X = B^-1A
(d) If A is square matrix of order 3 then |adj A| = |A|²
Answer:
(c) If AX = B then X = B^-1A
Hint:
If AX = B then X = A^-1B so, X = B^-1A is incorrect.

Question 16.
The value of \(\left|\begin{array}{rrr}
5 & 5 & 5 \\
4 x & 4 y & 4 z \\
-3 x & -3 y & -3 z
\end{array}\right|\) is:
(a) 5
(b) 4
(c) 0
(d) -3
Answer:
(c) 0
Hint:
= 4 × (-3) \(\left|\begin{array}{lll}
5 & 5 & 5 \\
x & y & z \\
x & y & z
\end{array}\right|\)
[Take out 4 from R₂ and -3 from R₃]
= 0 (∵ R₂ ≡ R₃)

Question 17.
If A is an invertible matrix of order 2 then det (A^-1) be equal
(a) det (A)
(b) \(\frac{1}{{det}(A)}\)
(c) 1
(d) 0
Answer:
(b) \(\frac{1}{{det}(A)}\)
Hint:
AA^-1 = I
|AA^-1| = |I|
|A| |A^-1| = 1
|A^-1| = \(\frac{1}{|\mathrm{A}|}\)
det A^-1 = \(\frac{1}{\det (A)}\)

Question 18.
If A is 3 × 3 matrix and |A| = 4 then |A^-1| is equal to:
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{16}\)
(c) 2
(d) 4
Answer:
(a) \(\frac{1}{4}\)
Hint:
|A^-1| = \(\frac{1}{|A|}=\frac{1}{4}\)

Question 19.
If A is a square matrix of order 3 and |A| = 3 then |adj A| is equal to:
(a) 81
(b) 27
(c) 3
(d) 9
Answer:
(d) 9
Hint:
|adj A| = |A|² = 3² = 9

Question 20.
The value of \(\left|\begin{array}{ccc}
x & x^{2}-y z & 1 \\
y & y^{2}-z x & 1 \\
z & z^{2}-x y & 1
\end{array}\right|\) is:
(a) 1
(b) 0
(c) -1
(d) -xyz
Answer:
(b) 0
Hint:

Question 21.
If A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), then |2A| is equal to:
(a) 4 cos 2θ
(b) 4
(c) 2
(d) 1
Answer:
(b) 4
Hint:
|2A| = 2² |A|
= 4 \(\left|\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right|\)
= 4 [cos²θ + sin²θ]
= 4 × 1
= 4

Question 22.
If Δ = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and A_ij is cofactor of a_ij, then value of Δ is given by:
(a) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃
(b) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
(c) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
(d) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Answer:
(d) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Question 23.
If \(\left|\begin{array}{ll}
x & 2 \\
8 & 5
\end{array}\right|=0\) then the value of x is:
(a) \(\frac{-5}{6}\)
(b) \(\frac{5}{6}\)
(c) \(\frac{-16}{5}\)
(d) \(\frac{16}{5}\)
Answer:
(d) \(\frac{16}{5}\)
Hint:
\(\left|\begin{array}{ll}
x & 2 \\
8 & 5
\end{array}\right|=0\)
5x – 16 = 0
⇒ x = \(\frac{16}{5}\)

Question 24.
If \(\left|\begin{array}{ll}
4 & 3 \\
3 & 1
\end{array}\right|\) = -5 then the value of \(\left|\begin{array}{rr}
20 & 15 \\
15 & 5
\end{array}\right|\) is:
(a) -5
(b) -125
(c) -25
(4) 0
Answer:
(b) -125
Hint:
\(\left|\begin{array}{rr}
20 & 15 \\
15 & 5
\end{array}\right|\)
= 5 × 5 \(\left|\begin{array}{ll}
4 & 3 \\
3 & 1
\end{array}\right|\)
= 5 × 5 × (-5)
= -125

Question 25.
If any three rows or columns of a determinant are identical then the value of the determinant is:
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(a) 0

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 1 Matrices and Determinants Ex 1.4

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.4 Text Book Back Questions and Answers

Question 1.
The technology matrix of an economic system of two industries is \(\left[\begin{array}{cc}
0.50 & 0.30 \\
0.41 & 0.33
\end{array}\right]\). Test whether the system is viable as per Hawkins Simon conditions.
Solution:
Technology matrix B = \(\left[\begin{array}{cc}
0.50 & 0.30 \\
0.41 & 0.33
\end{array}\right]\)
I – B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
0.50 & 0.30 \\
0.41 & 0.33
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0.50 & -0.30 \\
-0.41 & 0.67
\end{array}\right]\), the main diagonal elements are positive.
|I – B| = \(\left[\begin{array}{rr}
0.50 & -0.30 \\
-0.41 & 0.67
\end{array}\right]\)
= 0.335 – 0.123
= 0.212, positive
Since the main diagonal elements of I – B are positive and |I – B| is positive, Hawkins-Simon conditions are satisfied. Therefore, the given system is viable.

Question 2.
The technology matrix of ah economic system of two industries is \(\left[\begin{array}{rr}
0.6 & 0.9 \\
0.20 & 0.80
\end{array}\right]\). Test whether the system is viable as per Hawkins-Simon conditions.
Solution:
Technology matrix B = \(\left[\begin{array}{cc}
0.60 & 0.9 \\
0.20 & 0.80
\end{array}\right]\)
I – B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
0.60 & 0.9 \\
0.20 & 0.80
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0.4 & -0.9 \\
-0.20 & 0.20
\end{array}\right]\), the main diagonal elements are positive.
|I – B| = \(\left|\begin{array}{rr}
0.4 & -0.9 \\
-0.20 & 0.20
\end{array}\right|\)
= 0.4 × 0.20 – (-0.20) × (-0.9)
= 0.08 – 0.18
= -0.1, negative
Since |I – B| is negative one of the Hawkins-Simon condition is not satisfied. Therefore, the given system is not viable.

Question 3.
The technology matrix of an economic system of two industries is \(\left[\begin{array}{ll}
0.50 & 0.25 \\
0.40 & 0.67
\end{array}\right]\). Test whether the system is viable as per Hawkins-Simon conditions.
Solution:
Technology matrix B = \(\left[\begin{array}{ll}
0.50 & 0.25 \\
0.40 & 0.67
\end{array}\right]\)
I – B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
0.50 & 0.25 \\
0.40 & 0.67
\end{array}\right]\)
= \(\left[\begin{array}{rr}
0.50 & -0.25 \\
-0.40 & 0.33
\end{array}\right]\), the main diagonal elements are positive.
|I – B| = \(\left|\begin{array}{rr}
0.50 & -0.25 \\
-0.40 & 0.33
\end{array}\right|\)
= (0.50) (0.33) – (-0.40) (-0.25)
= 0.165 – 0.1
= 0.065 (positive)
Since the main diagonal elements of I – B are positive and |I – B| is positive, Hawkins-Simon conditions are satisfied. Therefore, the given system is viable.

Question 4.
Two commodities A and B are produced such that 0.4 tonne of A and 0.7 tonnes of B are required to produce a tonnes of A. Similarly 0.1 tonne of A and 0.7 tonne of B are needed to produce a tonnes of B. Write down the technology matrix. If 6.8 tonnes of A and 10.2 tones of B are required, find the gross production of both of them.
Solution:
Here the technology matrix is given under

The technology matrix is B = \(\left[\begin{array}{cc}
0.4 & 0.1 \\
0.7 & 0.7
\end{array}\right]\)
I – B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
0.4 & 0.1 \\
0.7 & 0.7
\end{array}\right]\) = \(\left[\begin{array}{rr}
0.6 & -0.1 \\
-0.7 & 0.3
\end{array}\right]\)
|I – B| = \(\left|\begin{array}{rr}
0.6 & -0.1 \\
-0.7 & 0.3
\end{array}\right|\)
= (0.6) (0.3) – (-0.1) (-0.7)
= 0.18 – 0.07
= 0.11
Since the main diagonal elements of I – B are positive and the value of |I – B| is positive, the system is viable.

Production of A is 27.82 tonnes and the production of B is 98.91 tonnes.

Question 5.
Suppose the inter-industry flow of the product of two industries is given as under.

Determine the technology matrix and test Hawkin’s-Simon conditions for the viability of the system. If the domestic changes to 80 and 40 units respectively, what should be the gross output of each sector in order to meet the new demands.
Solution:


The technology matrix B = \(\left[\begin{array}{ll}
\frac{1}{4} & \frac{2}{3} \\
\frac{1}{6} & \frac{1}{6}
\end{array}\right]\)
I – B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
\frac{1}{4} & \frac{2}{3} \\
\frac{1}{6} & \frac{1}{6}
\end{array}\right]\)
= \(\left[\begin{array}{rr}
\frac{3}{4} & -\frac{2}{3} \\
-\frac{1}{6} & \frac{5}{6}
\end{array}\right]\), elements of main diagonal are positive.

The main diagonal elements of I – B are positive and |I – B| is positive. Therefore the system is viable.

The output of industry X should be 181.62 and Y should be 84.32.

Question 6.
You are given the following transaction matrix for a two-sector economy.

(i) Write the technology matrix?
(ii) Determine the output when the final demand for the output sector 1 alone increases to 23 units.
Solution:


The main diagonal elements are positive and |I – B| is positive. Therefore the system is viable.

X = (I – B)^-1D, where

The output of sector 1 should be 34.16 and sector 2 should be 17.31.

Question 7.
Suppose the inter-industry flow of the product of two Sectors X and Y are given as under.

Find the gross output when the domestic demand changes to 12 for X and 18 for Y.
Solution:


Since the main diagonal elements of I – B are positive and |I – B| is positive the problem has a solution.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 1 Matrices and Determinants Ex 1.3

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.3 Text Book Back Questions and Answers

Question 1.
Solve by matrix inversion method: 2x + 3y – 5 = 0; x – 2y + 1 = 0.
Solution:
2x + 3y = 5
x – 2y = -1
The given system can be written as
\(\left[\begin{array}{rr}
2 & 3 \\
1 & -2
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
5 \\
-1
\end{array}\right]\)
AX = B
where A = \(\left[\begin{array}{rr}
2 & 3 \\
1 & -2
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{r}
5 \\
-1
\end{array}\right]\)
|A| = \(\left|\begin{array}{rr}
2 & 3 \\
1 & -2
\end{array}\right|\) = -4 – 3 = -7 ≠ 0
∴ A^-1 Exists.

∴ x = 1, y = 1

Question 2.
Solve by matrix inversion method:
(i) 3x – y + 2z = 13; 2x + y – z = 3; x + 3y – 5z = -8
(ii) x – y + 2z = 3; 2x + z = 1; 3x + 2y + z = 4
(iii) 2x – z = 0; 5x + y = 4; y + 3z = 5
Solution:
(i) The given system can be written as
\(\left[\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & -1 \\
1 & 3 & -5
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
13 \\
3 \\
-8
\end{array}\right]\)
AX = B
Where A = \(\left[\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & -1 \\
1 & 3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{r}
13 \\
3 \\
-8
\end{array}\right]\)
|A| = \(\left|\begin{array}{rrr}
3 & -1 & 2 \\
2 & 1 & -1 \\
1 & 3 & -5
\end{array}\right|\)
= 3(-5 + 3) – (-1) (-10 + 1) + 2 (6 – 1)
= 3(-2) + 1(-9) + 2(5)
= -6 – 9 + 10
= -5




\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
3 \\
-2 \\
1
\end{array}\right]\)
∴ x = 3, y = -2, z = 1.

(ii) The given system can be written as
\(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 0 & 1 \\
3 & 2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
3 \\
1 \\
4
\end{array}\right]\)
AX = B
where A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 0 & 1 \\
3 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
3 \\
1 \\
4
\end{array}\right]\)
|A| = \(\left|\begin{array}{rrr}
1 & -1 & 2 \\
2 & 0 & 1 \\
3 & 2 & 1
\end{array}\right|\)
= 1(0 – 2) – (-1)(2 – 3) + 2(4 – 0)
= -2 – (-1)(-1) + 2(4)
= -2 – 1 + 8
= 5



\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
-1 \\
2 \\
3
\end{array}\right]\)
x = -1, y = 2, z = 3.

(iii) The given system can be written as
\(\left[\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
4 \\
5
\end{array}\right]\)
AX = B
Where A = \(\left[\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
0 \\
4 \\
5
\end{array}\right]\)
|A| = \(\left|\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right|\)
= 2(3 – 0) – 0(15 – 0) – 1(5 – 0)
= 2(3) – 0(15) – 1(5)
= 6 – 0 – 5
= 1


∴ x = 1, y = -1, z = 2.

Question 3.
A salesperson Ravi has the following record of sales for the month of January, February, and March 2009 for three products A, B, and C. He has been paid a commission at a fixed rate per unit but at varying rates for products A, B and C.

Find the rate of commission payable on A, B and C per unit sold using matrix inversion method.
Solution:
Let x, y and z be the rate of commission for the three products A, B and C respectively.
9x + 10y + 2z = 800
15x + 5y + 4z = 900
6x + 10y + 3z = 850
The given system can be written as
\(\left[\begin{array}{rrr}
9 & 10 & 2 \\
15 & 5 & 4 \\
6 & 10 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
800 \\
900 \\
850
\end{array}\right]\)
AX = B
Where A = \(\left[\begin{array}{rrr}
9 & 10 & 2 \\
15 & 5 & 4 \\
6 & 10 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
800 \\
900 \\
850
\end{array}\right]\)
Now, |A| = \(\left|\begin{array}{rrr}
9 & 10 & 2 \\
15 & 5 & 4 \\
6 & 10 & 3
\end{array}\right|\)
= \(9\left|\begin{array}{rr}
5 & 4 \\
10 & 3
\end{array}\right|-10\left|\begin{array}{rr}
15 & 4 \\
6 & 3
\end{array}\right|+2\left|\begin{array}{rr}
15 & 5 \\
6 & 10
\end{array}\right|\)
= 9[15 – 40] – 10(45 – 24) + 2(150 – 30)
= 9[-25] – 10[21] + 2[120]
= -225 – 210 + 240
= -195


\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
17.948 \\
43.0769 \\
103.846
\end{array}\right]\)
\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
17.95 \\
43.08 \\
103.85
\end{array}\right]\)
∴ x = 17.95, y = 43.08, z = 103.85
The rate of commission of A, B and C are 17.95, 43.08 and 103.85 respectively.

Question 4.
The prices of three commodities A, B, and C are ₹ x, y, and z per unit respectively. P purchases 4 units of C and sells 3 units of A and 5 units of B. Q purchases 3 units of B and sells 2 units of A and 1 unit of C. R purchases 1 unit of A and sells 4 units of B and 6 units of C. In the process P, Q and R earn ₹ 6,000, ₹ 5,000 and ₹ 13,000 respectively. By using the matrix inversion method, find the prices per unit of A, B, and C.
Solution:
Take selling the units js positive earning and buying the units is negative earning.
Given that
3x + 5y – 4z = 6000
2x – 3y + z = 5000
-1x + 4y + 6z = 13000

The given statement can be written as
\(\left(\begin{array}{rrr}
3 & 5 & -4 \\
2 & -3 & 1 \\
-1 & 4 & 6
\end{array}\right)\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)=\left(\begin{array}{r}
6000 \\
5000 \\
13000
\end{array}\right)\)
AX = B
Where A = \(\left(\begin{array}{rrr}
3 & 5 & -4 \\
2 & -3 & 1 \\
-1 & 4 & 6
\end{array}\right)\), X = \(\left(\begin{array}{l}
x \\
y \\
z
\end{array}\right)\) and B = \(\left(\begin{array}{r}
6000 \\
5000 \\
13000
\end{array}\right)\)
X = A^-1B
|A| = \(\left|\begin{array}{rrr}
3 & 5 & -4 \\
2 & -3 & 1 \\
-1 & 4 & 6
\end{array}\right|\)
= 3(-18 – 4) – 5(12 + 1) – 4(8 – 3)
= 3(-22) – 5(13) – 4(5)
= -66 – 65 – 20
= -151


\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
3000 \\
1000 \\
2000
\end{array}\right]\)
The prices per unit of A, B and C are ₹ 3000, ₹ 1000 and ₹ 2000.

Question 5.
The sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add second and third to it, we get 46. By using the matrix inversion method find the numbers.
Solution:
Let the three numbers be x, y, and z.
x + y + z = 20
2x + y – z = 23
3x + y + z = 46
The given system can be written as




The numbers are 13, 2, and 5.

Question 6.
Weekly expenditure in an office for three weeks is given as follows. Assuming that the salary in all three weeks of different categories of staff did not vary, calculate the salary for each type of staff, using the matrix inversion method.

Solution:
Let ₹ x, ₹ y, ₹ z be the salary for each type of staff A, B and C.
4x + 2y + 3z = 4900
3x + 3y + 2z = 4500
4x + 3y + 4z = 5800
The given system can be written as
\(\left[\begin{array}{lll}
4 & 2 & 3 \\
3 & 3 & 2 \\
4 & 3 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
4900 \\
4500 \\
5800
\end{array}\right]\)
AX = B
where A = \(\left[\begin{array}{lll}
4 & 2 & 3 \\
3 & 3 & 2 \\
4 & 3 & 4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{c}
4900 \\
4500 \\
5800
\end{array}\right]\)
|A| = \(\left|\begin{array}{lll}
4 & 2 & 3 \\
3 & 3 & 2 \\
4 & 3 & 4
\end{array}\right|\)
= 4(12 – 6) – 2(12 – 8) + 3(9 – 12)
= 4(6) – 2(4) + 3(-3)
= 24 – 8 – 9
= 7




\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
700 \\
600 \\
300
\end{array}\right]\)
∴ Salary for each type of staff A, B and C are ₹ 700, ₹ 600 and ₹ 300.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.7

Samacheer Kalvi 11th Business Maths Algebra Ex 2.7 Text Book Back Questions and Answers

Choose the correct answer.

Question 1.
If nC₃ = nC₂ then the value of nC₄ is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5
Hint:
Given that nC₃ = nC₂
We know that if nC_x = nC_y then x + y = n or x = y
Here 3 + 2 = n
∴ n = 5

Question 2.
The value of n, when np₂ = 20 is:
(a) 3
(b) 6
(c) 5
(d) 4
Answer:
(c) 5
Hint:
nP₂ = 20
n(n – 1) = 20
n(n – 1) = 5 × 4
∴ n = 5

Question 3.
The number of ways selecting 4 players out of 5 is:
(a) 4!
(b) 20
(c) 25
(d) 5
Answer:
(d) 5
Hint:
5C₄ = 5C₁ = 5

Question 4.
If nP_r = 720(nC_r), then r is equal to:
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(c) 6
Hint:
Given nP_r = 720(nC_r)
\(\frac{n !}{(n-r) !}=720 \frac{n !}{r !(n-r) !}\)
1 = \(\frac{720}{r !}\)
r! = 720
r! = 6 × 5 × 4 × 3 × 2 × 1
r! = 6!
r = 6

Question 5.
The possible outcomes when a coin is tossed five times:
(a) 25
(b) 52
(c) 10
(d) \(\frac{5}{2}\)
Answer:
(a) 25
Hint:
Number of possible outcomes When a coin is tossed is 2
∴ When five coins are tossed (same as a coin is tossed five times)
Possible outcomes = 2 × 2 × 2 × 2 × 2 = 2⁵

Question 6.
The number of diagonals in a polygon of n sides is equal to:
(a) nC₂
(b) nC₂ – 2
(c) nC₂ – n
(d) nC₂ – 1
Answer:
(c) nC₂ – n

Question 7.
The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for all n ∈ N is:
(a) 2
(b) 6
(c) 20
(d) 24
Answer:
(d) 24
Hint:
Put n = 1 in n(n + 1) (n + 2) (n + 3)
= 1 × 2 × 3 × 4
= 24

Question 8.
If n is a positive integer, then the number of terms in the expansion of (x + a)ⁿ is:
(a) n
(b) n + 1
(c) n – 1
(d) 2n
Answer:
(b) n + 1

Question 9.
For all n > 0, nC₁ + nC₂ + nC₃ + …… + nC_n is equal to:
(a) 2n
(b) 2ⁿ – 1
(c) n²
(d) n² – 1
Answer:
(b) 2ⁿ – 1
Hint:
Sum of binomial coefficients 2n
i.e., nC₀ + nC₁ + nC₂ + nC₃ + ……. + nC_n = 2ⁿ
nC₁ + nC₂ + nC₃ + ……. + nC_n = 2ⁿ – nC₀ = 2ⁿ – 1

Question 10.
The term containing x³ in the expansion of (x – 2y)⁷ is:
(a) 3rd
(b) 4th
(c) 5th
(d) 6th
Answer: (c) 5th
Hint:
First-term contains x⁷.
The second term contains x⁶.
The fifth term contains x³.

Question 11.
The middle term in the expansion of \(\left(x+\frac{1}{x}\right)^{10}\) is:
(a) 10C₄ \(\left(\frac{1}{x}\right)\)
(b) 10C₅
(c) 10C₆
(d) 10C₇ x²
Answer:
(b) 10C₅
Hint:
x is x, a = \(\frac{1}{x}\), n = 10 which is even.
So the middle term is

Question 12.
The constant term in the expansion of \(\left(x+\frac{2}{x}\right)^{6}\) is:
(a) 156
(b) 165
(c) 162
(d) 160
Answer:
(d) 160
Hint:
Here x is x, a is \(\frac{2}{x}\) (Note that each term x will vanish)
∴ Constant term occurs only in middle term
n = 6
∴ middle term = \(t_{\frac{6}{2}+1}\) = t₃₊₁

Question 13.
The last term in the expansion of (3 + √2 )⁸ is:
(a) 81
(b) 16
(c) 8
(d) 2
Answer:
(b) 16
Hint:
(√2)⁸ = \(\left(2^{\frac{1}{2}}\right)^{8}\) = 2⁴ = 16

Question 14.
If \(\frac{k x}{(x+4)(2 x-1)}=\frac{4}{x+4}+\frac{1}{2 x-1}\) then k is equal to:
(a) 9
(b) 11
(c) 5
(d) 7
Answer:
(a) 9
Hint:
\(\frac{k x}{(x+4)(x-1)}=\frac{4}{x+4}+\frac{1}{2 x-1}\)
kx = 8x – 4 + x + 4
kx = 9x
k = 9

Question 15.
The number of 3 letter words that can be formed from the letters of the word ‘NUMBER’ when the repetition is allowed are:
(a) 206
(b) 133
(c) 216
(d) 300
Answer:
(c) 216
Hint:
Number of letters in NUMBER is 5
From 5 letters we can form 3 letter ways = 6 × 6 × 6 = 216.

Question 16.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:
(a) 18
(b) 12
(c) 9
(d) 6
Answer:
(a) 18
Hint:

To form a parallelogram we need 2 parallel lines from 4 and 2 intersecting lines from 3.
Number of parallelograms = 4C₂ × 3C₂
= \(\frac{4 \times 3}{2 \times 1} \times 3\)
= 18

Question 17.
There are 10 true or false questions in an examination. Then these questions can be answered in
(a) 240 ways
(b) 120 ways
(c) 1024 ways
(d) 100 ways
Answer:
(c) 1024 ways
Hint:
For each question, there are two ways of answering it.
for 10 questions the numbers of ways to answer = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2¹⁰
= 1024 ways

Question 18.
The value of (5C₀ + 5C₁) + (5C₁ + 5C₂) + (5C₂ + 5C₃) + (5C₃ + 5C₄) + (5C₄ + 5C₅) is:
(a) 2⁶ – 2
(b) 2⁵ – 1
(c) 2⁸
(d) 2⁷
Answer:
(a) 2⁶ – 2
Hint:
(5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅) + (5C₁ + 5C₂ + 5C₃ + 5C₄)
= 2⁵ + (5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅) – (5C₀ + 5C₅)
= 2⁵ + 2⁵ – (1 + 1) (∵ Adding and subtracting of 5C₀ and 5C₅)
= 2(2⁵) – 2 (∵ 5C₀ = 5C₅ = 1)
= 2⁶ – 2

Question 19.
The total number of 9 digit number which has all different digit is:
(a) 10!
(b) 9!
(c) 9 × 9!
(d) 10 × 10!
Answer:
(c) 9 × 9!
Hint:
Here we can use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
They are in 10 in total. We have to form a nine-digit number.

The first place from the left can be filled up by anyone of the digits other than zero in 9 ways. The second place can be filled up by anyone of the remaining (10 – 1) digits (including zero) in 9 ways, the third place in 8 ways, fourth place in 7 ways, fifth place in 6 ways, sixth place in 5 ways, seventh place in 4 ways, eighth place in 3 ways and ninth place in 2 ways.
∴ The number of ways of making 9 digit numbers = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 × 9!

Question 20.
The number of ways to arrange the letters of the word “CHEESE”:
(a) 120
(b) 240
(c) 720
(d) 6
Answer:
(a) 120
Hint: Here there are 6 letters.
The letter C occurs one time
The letter H occurs one time
The letter E occurs three times
The letter S occurs one time
Number of arrangements = \(\frac{6 !}{1 ! 1 ! 3 ! 1 !}=\frac{6 !}{3 !}=\frac{6 \times 5 \times 4 \times 3 !}{3 !}\) = 120

Question 21.
Thirteen guests have participated in a dinner. The number of handshakes that happened in the dinner is:
(a) 715
(b) 78
(c) 286
(d) 13
(b) 78
Hint:
To handshakes, we need two guests.
Number of selecting 2 guests from 13 is 13C₂ = \(\frac{13 \times 12}{2 \times 1}\) = 78

Question 22.
The number of words with or without meaning that can be formed using letters of the word “EQUATION”, with no repetition of letters is:
(a) 7!
(b) 3!
(c) 8!
(d) 5!
Answer:
(c) 8!
Hint:
There are 8 letters.
From 8 letters number of words is formed = 8P₈ = 8!

Question 23.
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
(a) 8
(b) 7
(c) 6
(d) 9
Answer:
(a) 8
Hint:
Sum of binomial coefficient = 256
i.e., 2ⁿ = 256
2ⁿ = 2⁸
n = 8

Question 24..
The number of permutation of n different things taken r at a time, when the repetition is allowed is:
(a) rⁿ
(b) n^r
(c) \(\frac{n !}{(n-r) !}\)
(d) \(\frac{n !}{(n+r) !}\)
Answer:
(b) n^r

Question 25.
The sum of the binomial coefficients is:
(a) 2ⁿ
(b) n²
(c) 2n
(d) n + 17
Answer:
(a) 2ⁿ

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.2

Samacheer Kalvi 11th Business Maths Algebra Ex 2.2 Text Book Back Questions and Answers

Question 1.
Find x if \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\)
Solution:
Given that \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\)
\(\frac{1}{6 !}+\frac{1}{7 \cdot 6 !}=\frac{x}{8 \cdot 7 \cdot 6 !}\)
Cancelling all 6! we get
\(\frac{1}{1}+\frac{1}{7}=\frac{x}{8 \times 7}\)
\(\frac{7+1}{7}=\frac{x}{8 \times 7}\)
\(\frac{8}{7}=\frac{x}{8 \times 7}\)
x = \(\frac{8}{7}\) × 7 × 8 = 64

Question 2.
Evaluate \(\frac{n !}{r !(n-r) !}\) when n = 5 and r = 2.
Solution:
\(\frac{n !}{r !(n-r) !}=\frac{5 !}{2 !(5-2) !}=\frac{5 !}{2 ! \times 3 !}=\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1}=10\)

Question 3.
If (n + 2)! = 60[(n – 1)!], find n.
Solution:
Given that (n + 2)! = 60(n – 1)!
(n + 2) (n + 1) n (n – 1)! = 60(n – 1)!
Cancelling (n – 1)! we get,
(n + 2)(n + 1)n = 60
(n + 2)(n + 1)n = 5 × 4 × 3
Both sides we consecutive product of integers
∴ n = 3

Question 4.
How many five digits telephone numbers can be constructed using the digits 0 to 9 If each number starts with 67 with no digit appears more than once?
Solution:
Given that each number starts at 67, we need a five-digit number. So we have to fill only one’s place, 10’s place, and 100th place. From 0 to 9 there are 10 digits. In these digits, 6 and 7 should not be used as a repetition of digits is not allowed. Except for these two digits, we have 8 digits. Therefore one’s place can be filled by any of the 8 digits in 8 different ways. Now there are 7 digits are left.

Therefore 10’s place can be filled by any of the 7 digits in 7 different ways. Similarly, 100th place can be filled in 6 different ways. By multiplication principle, the number of telephone numbers constructed is 8 × 7 × 6 = 336.

Question 5.
How many numbers lesser than 1000 can be formed using the digits 5, 6, 7, 8, and 9 if no digit is repeated?
Solution:
The required numbers are lesser than 1000.
They are one digit, two-digit or three-digit numbers.
There are five numbers to be used without repetition.
One digit number: One-digit numbers are 5.
Two-digit number: 10th place can be filled by anyone of the digits by 5 ways and 1’s place can be 4 filled by any of the remaining four digits in 4 ways.
∴ Two-digit number are 5 × 4 = 20.
Three-digit number: 100th place can be filled by any of the 5 digits, 10th place can be filled by 4 digits and one’s place can be filled by 3 digits.
∴ Three digit numbers are = 5 × 4 × 3 = 60
∴ Total numbers = 5 + 20 + 60 = 85.

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.1 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.1

Samacheer Kalvi 11th Business Maths Algebra Ex 2.1 Text Book Back Questions and Answers

Resolve into partial fractions for the following:

Question 1.
\(\frac{3 x+7}{x^{2}-3 x+2}\)
Solution:
Here the denominator x² – 3x + 2 is not a linear factor.
So if possible we have to factorise it then only we can split up into partial fraction.
x² – 3x + 2 = (x – 1) (x – 2)
\(\frac{3 x+7}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\) …….. (1)
Multiply both side by (x – 1) (x – 2)
3x + 7 = A(x – 2) + B(x – 1) …….. (2)
Put x = 2 in (2) we get
3(2) + 7 = A(2 – 2) + B(2 – 1)
6 + 7 = 0 + B(1)
∴ B = 13
Put x = 1 in (2) we get
3(1) + 7 = A(1 – 2) + B(1 – 1)
3 + 7 = A (-1) + 0
10 = A(-1)
∴ A = -10
Using A = -10 and B = 13 in (1) we get

Note: When the denominator is only two linear factors we can adopt the following method.

Question 2.
\(\frac{4 x+1}{(x-2)(x+1)}\)
Solution:
Let \(\frac{4 x+1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\) ……… (1)
Multiply both sides by (x – 2) (x + 1) we get
4x + 1 = A(x + 1) + B(x – 2) ……. (2)
Put x = -1 in (2) we get
4(-1) + 1 = A(-1 + 1) + B(-1 – 2)
-4 + 1 = A(0) + B(-3)
-3 = B(-3)
B = \(\frac{-3}{-3}\) = 1
Put x = 2 in (2) we get
4(2) + 1 = A(2 + 1) + B(2 – 2)
8 + 1 = A(3) + B(0)
9 = 3A
A = 3
Using A = 3, B = 1 in (1) we get
\(\frac{4 x+1}{(x-2)(x+1)}=\frac{3}{x-2}+\frac{1}{x+1}\)

Question 3.
\(\frac{1}{(x-1)(x+2)^{2}}\)
Solution:
Here the denominator has repeated factors. So we write
\(\frac{1}{(x-1)(x+2)^{2}}=\frac{A}{x-1}+\frac{B}{(x+2)}+\frac{C}{(x+2)^{2}}\) …… (1)
Multiply both sides by (x – 1) (x + 2)² we get
1 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1) …… (2)
Put x = 1 in (2) we get
1 = A(1 + 2)² + B(1 – 1) (1 + 2) + C(1 – 1)
1 = A(3²) + 0 + 0
1 = 9A
A = \(\frac{1}{9}\)
Put x = -2 in (2) we get
1 = A(-2 + 2)² + B(-2 – 1) (-2 + 2) + C(-2 – 1)
1 = 0 + 0 + C(-3)
C = \(\frac{-1}{3}\)
From (2) we have
1 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1)
0x² + 1 = A(x² + 4x + 4) + B(x² + x – 2) + C(x – 1)
Equating coefficient of x² on both sides we get
0 = A + B
0 = \(\frac{1}{9}\) (∴ A = \(\frac{1}{9}\))
B = \(-\frac{1}{9}\)
Using A = \(\frac{1}{9}\), B = \(-\frac{1}{9}\), C = \(-\frac{1}{3}\) in (1) we get,

Question 4.
\(\frac{1}{x^{2}-1}\)
Solution:

1 = A(x – 1) + B(x + 1) ……. (2)
Put x = 1 in (2) we get
1 = 0 + B(1 + 1)
1 = B(2)
B = \(\frac{1}{2}\)
Put x = -1 in (2) we get
1 = A(-1 – 1) + B(-1 + 1)
1 = -2A + 0
A = \(\frac{-1}{2}\)
Using A = \(\frac{-1}{2}\), B = \(\frac{1}{2}\) in (1) we get

Question 5.
\(\frac{x-2}{(x+2)(x-1)^{2}}\)
Solution:

x – 2 = A(x – 1)² + B(x + 2) (x – 1) + C(x + 2) ……(2)
Put x = 1 in (2) we get
1 – 2 = A(1 – 1)² + B(1 + 2) (1 – 1) + C(1 + 2)
-1 = 0 + 0 + 3C
C = \(-\frac{1}{3}\)
Put x = -2 in (2) we get
-2 – 2 = A(-2 – 1)² + B(-2 + 2) (-2 – 1) + C(-2 + 2)
-4 = A(-3)² + 0 + 0
-4 = 9A
A = \(\frac{-4}{9}\)
From (2) we have,
0x² + x – 2 = A(x – 1)² + B(x + 2) (x – 1) + C(x + 2)
Equating coefficients of x² on both sides we get
0 = A + B
0 = \(\frac{-4}{9}\) + B (∵ A = \(\frac{-4}{9}\))
B = \(\frac{4}{9}\)
Using A = \(\frac{-4}{9}\), B = \(\frac{4}{9}\), C = \(-\frac{1}{3}\) in (1) we get

Question 6.
\(\frac{2 x^{2}-5 x-7}{(x-2)^{3}}\)
Solution:

2x² – 5x – 7 = A(x – 2)² + B(x – 2) + C
2x² – 5x – 7 = A(x² – 4x + 4) + B(x – 2) + C …….. (2)
Put x = 2 in (2) we get
2(2²) – 5(2) – 7 = A(0) + B(0) + C
8 – 10 – 7 = 0 + 0 + C
-9 = C
C = -9
Equating coefficient of x2 on both sides of (2) we get
2 = A
A = 2
Equating coefficient of x on both sides of (2) we get
-5 = A(-4) + B(1)
-5 = 2(-4) + B(∵ A = 2)
-5 = -8 + B
B = 8 – 5 = 3
Using A = 2, B = 3, C = -9 in (1) we get

Question 7.
\(\frac{x^{2}-6 x+2}{x^{2}(x+2)}\)
Solution:
Here the denominator has three factors. So given fraction can be expressed as a sum of three simple fractions.
Let \(\frac{x^{2}-6 x+2}{x^{2}(x+2)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+2}\) …… (1)
Multiply both sides by x² (x + 2) we get

x² – 6x + 2 = Ax(x + 2) + B(x + 2) + C(x²) ……… (2)
Put x = 0 in (2) we get
0 – 0 + 2 = 0 + B(0 + 2) + 0
2 = B(2)
B = 1
Put x = -2 in (2) we get
(-2)² – 6(-2) + 2 = 0 + 0 + C(-2)²
4 + 12 + 2 = C(4)
18 = 4C
C = \(\frac{9}{2}\)
Comparing coefficient of x² on both sides of (2) we get,
1 = A + C
1 = A + \(\frac{9}{2}\)
A = 1 – \(\frac{9}{2}\) = \(\frac{2-9}{2}=\frac{-7}{2}\)
Using A = \(\frac{-7}{2}\), B = 1, C = \(\frac{9}{2}\) in (1) we get,
\(\frac{\left(x^{2}-6 x+2\right)}{x^{2}(x+2)}=\frac{-7}{2 x}+\frac{1}{x^{2}}+\frac{9}{2(x+2)}\)

Question 8.
\(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\)
Solution:
Here the quadratic factor x² + 1 is not factorisable.
Let \(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{(B x+C)}{x^{2}+1}\) ….. (1)
Multiply both sides by (x + 2) (x² + 1) we get,
x² – 3 = A(x² + 1) + (Bx + C) (x + 2)
Put x = -2 we get
(-2)² – 3 = [A(-2)² + 1] + 0
4 – 3 = A(4 + 1)
1 = 5A
A = \(\frac{1}{5}\)
Equating coefficient of x² on both sides of (2) we get
1 = A + B
1 = \(\frac{1}{5}\) + B
B = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Equating coefficients of x on both sides of (2) we get
0 = 2B + C
0 = 2(\(\frac{4}{5}\)) + C
C = \(\frac{-8}{5}\)
Using A, B, C’s values in (1) we get

Question 9.
\(\frac{x+2}{(x-1)(x+3)^{2}}\)
Solution:
Here the denominator has three factors. So given fraction can be expressed as a sum of three simple fractions.
Let \(\frac{x+2}{(x-1)(x+3)^{2}}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C}{(x+3)^{2}}\) ……. (1)
Multiply both sides by (x – 1) (x + 3)² we get
\(\frac{x+2}{(x-1)(x+3)^{2}}\) (x – 1) (x + 3)² = \(\frac{A}{x-1}\) (x – 1) (x + 3)² +
\(\frac{B}{x+3}\) (x – 1) (x + 3)² + \(\frac{C}{(x+3)^{2}}\) (x – 1) (x + 3)²
x + 2 = A(x + 3)² + B(x – 1) (x + 3) + C(x – 1) ……. (2)
Put x = 1 in (2) we get
1 + 2 = A(1 + 3)² + 0 + 0
3 = A(4)²
A = \(\frac{3}{16}\)
Put x = -3 in (2) we get
-3 + 2 = 0 + 0 + C(-3 – 1)
-1 = C(-4)
C = \(\frac{1}{4}\)
Comparing coefficient of x² on both sides of (2) we get,
0 = A + B
0 = \(\frac{3}{16}\) + B
B = \(-\frac{3}{16}\)
Using A = \(\frac{3}{16}\), B = \(-\frac{3}{16}\), C = \(\frac{1}{4}\) in (1) we get,
\(\frac{x+2}{(x-1)(x+3)^{2}}=\frac{3}{16(x-1)}-\frac{3}{16(x+3)}+\frac{1}{4(x+3)^{2}}\)

Question 10.
\(\frac{1}{\left(x^{2}+4\right)(x+1)}\)
Solution:
Here the quadratic factor x² + 4 is not factorisable.
Let \(\frac{1}{(x+1)\left(x^{2}+4\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+4}\) ……. (1)
Multiply both sides by (x + 1) (x² + 4) we get
1 = A(x² + 4) + (Bx + C) (x + 1) ……. (2)
Put x = -1 in (2) we get
1 = A((-1)² + 4) + 0
1 = A(1 + 4)
A = \(\frac{1}{5}\)
Equating coefficient of x² on both sides of (2) we get,
0 = A + B
0 = \(\frac{1}{5}\) + B
B = \(\frac{-1}{5}\)
Equating coefficient of x on both sides of (2) we get,
{∵ (Bx + C) (x + 1) = Bx² + Cx = Bx + C}
0 = B + C
0 = \(\frac{-1}{5}\) + C
C = \(\frac{1}{5}\)
Using A = \(\frac{1}{5}\), B = \(\frac{-1}{5}\), C = \(\frac{1}{5}\) we get,

Students can download 11th Business Maths Chapter 1 Matrices and Determinants Ex 1.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 1 Matrices and Determinants Ex 1.2

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.2 Text Book Back Questions and Answers

Question 1.
Find the adjoint of the matrix A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 4
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 4
\end{array}\right]\)
Adj A = \(\left[\begin{array}{rr}
4 & -3 \\
-1 & 2
\end{array}\right]\)

Question 2.
If A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\) then verify that A(adj A) = |A| I and also find A^-1.
Solution:
Given A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\)
|A| = \(\left|\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right|\)
= \(1\left|\begin{array}{ll}
4 & 3 \\
3 & 4
\end{array}\right|-3\left|\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right|+3\left|\begin{array}{ll}
1 & 4 \\
1 & 3
\end{array}\right|\)
= 1[16 – 9] – 3[4 – 3] + 3[3 – 4]
= 1(7) – 3(1) + 3(-1)
= 7 – 3 – 3
= 1



From (1) and (2), A(Adj A) = |A| I

Question 3.
Find the inverse of each of the following matrices:
(i) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 3
\end{array}\right]\)
(iii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
(iv) \(\left[\begin{array}{rrr}
-3 & -5 & 4 \\
-2 & 3 & -1 \\
1 & -4 & -6
\end{array}\right]\)
Solution:
(i) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)

(ii) \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 3
\end{array}\right]\)

(iii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)

(iv) \(\left[\begin{array}{rrr}
-3 & -5 & 4 \\
-2 & 3 & -1 \\
1 & -4 & -6
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{rr}
2 & 3 \\
1 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-1 & 4 \\
1 & -2
\end{array}\right]\), then verify adj(AB) = (adj B) (adj A).
Solution:


From (1) and (2), adj (AB) = (adj B) (adj A)

Question 5.
If A = \(\left[\begin{array}{rrr}
2 & -2 & 2 \\
2 & 3 & 0 \\
9 & 1 & 5
\end{array}\right]\) then, show that (adj A) A = O.
Solution:

Question 6.
If A = \(\left[\begin{array}{rrr}
-1 & 2 & -2 \\
4 & -3 & 4 \\
4 & -4 & 5
\end{array}\right]\) then, show that the inverse of A is A itself.
Solution:


∴ A^-1 = A
Hence proved.

Question 7.
If A^-1 = \(\left[\begin{array}{rrr}
1 & 0 & 3 \\
2 & 1 & -1 \\
1 & -1 & 1
\end{array}\right]\) then, find A.
Solution:
Given A^-1 = \(\left[\begin{array}{rrr}
1 & 0 & 3 \\
2 & 1 & -1 \\
1 & -1 & 1
\end{array}\right]\)
We know that (A^-1)^-1 = A
So we have to find inverse of A^-1

Question 8.
Show that the matrices A = \(\left[\begin{array}{lll}
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
\frac{4}{5} & \frac{-2}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{3}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-2}{5} & \frac{4}{5}
\end{array}\right]\) are inverses of each other.
Solution:
To prove that A and B are inverses of each other.
We have to prove that AB = BA = I.


Thus AB = BA = I
Hence A and B are inverses of each other.

Question 9.
If A = \(\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right]\), then verify that (AB)^-1 = B^-1A^-1
Solution:

Now we will find B^-1A^-1


From (1) and (2), (AB)^-1 = B^-1A^-1

Question 10.
Find λ if the matrix \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & \lambda & 4 \\
9 & 7 & 11
\end{array}\right]\) has no inverse.
Solution:

1[11λ – 28] – 1[22 – 36] + 3[14 – 9λ] = 0
11λ – 28 + 14 + 42 – 27λ = 0
-16λ + 28 = 0
-16λ = -28
λ = \(\frac{-28}{-16}=\frac{7}{4}\)

Question 11.
If X = \(\left[\begin{array}{rrr}
8 & -1 & -3 \\
-5 & 1 & 2 \\
10 & -1 & -4
\end{array}\right]\) and Y = \(\left[\begin{array}{rrr}
2 & 1 & -1 \\
0 & 2 & 1 \\
5 & p & q
\end{array}\right]\) then, find p, q if Y = X^-1
Solution:
Given that Y is the inverse of X.
∴ XY = I

6 – 3p = 0 and -9 – 3q = 0
6 = 3p and -9 = 3q
∴ p = 2; q = -3

Students can Download Tamil Nadu 11th Commerce Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Commerce Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – I

Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
……………………….. is a special type of bond issued in the currency other than the home currency.
(a) Government bond
(b) Foreign currency convertible bond
(c) Corporate bond
(d) Investment bond
Answer:
(b) Foreign currency convertible bond

Question 2.
……………………. acts as connective link between the producer and the consumer.
(a) Trade
(b) Industry
(c) Commerce
(d) Business
Answer:
(a) Trade

Question 3.
From the following which is not a recent trend in transportation?
(a) Metro rail
(b) Pipeline transport
(c) Bullock carts
(d) Ropeway transport
Answer:
(c) Bullock carts

Question 4.
Minimum how much amount can be transferred through RTGS?
(a) Any amount
(b) Rs 50,000
(c) Rs 2 lakhs
(d) Rs 5 Lakhs
Answer:
(c) Rs 2 lakhs

Question 5.
Goods are imported for the purpose of exporting to another country is termed as ……………………….
(a) Import trade
(b) Export trade
(c) Entrepot trade
(d) Internal trade
Answer:
(c) Entrepot trade

Question 6.
Agriculture income earned in india is ………………………..
(a) Fully taxable
(b) Fully exempted
(c) Not considered for income
(d) None of the above
Answer:
(b) Fully exempted

Question 7.
An …………………………. is a document prepared by the importer and sent to the exporter to buy the goods.
(a) Invoice
(b) Indent
(c) Enquiry
(d) Charter party
Answer:
(b) Indent

Question 8.
Which of the parties cannot demand performance of promise?
(a) Promisee
(b) Any of the joint promisees
(c) Death of a promisee, his legal representative
(d) Stranger to the contract
Answer:
(d) Stranger to the contract

Question 9.
The income paid to the (mercantile agent), broker is ………………………..
(a) Profit
(b) Brokerage
(c) Commission
(d) Salary
Answer:
(b) Brokerage

Question 10.
Ethics is important for ……………………….
(a) Top management
(b) Middle – level management
(c) Non – managerial employees
(d) All of them
Answer:
(d) All of them

Question 11.
……………………. is the oldest form of transport found in hilly areas, forest areas in remote places.
(a) Bullock carts
(b) Road transport
(c) Motor lorries
(d) Pathways transport
Answer:
(d) Pathways transport

Question 12.
The document which invites the public to buy the shares and debentures is known as …………………………
(a) Memorandum of Association
(b) Articles of Association
(c) Prospectus
(d) Partnership articles
Answer:
(c) Prospectus

Question 13.
Which one of the following is not correctly matched?
(a) Minimum two members – Partnership firm
(b) Minimum paid up capital Rs 100000 – Private company
(c) Voluntary Membership – Co – operatives
(d) 51% of paid up capital – Foreign company
Answer:
(d) 51% of paid up capital – Foreign company

Question 14.
ABC jointly promised to pay Rs. 50000 to D. Before performance of the contract, ‘C’ dies. Here the contract ………………………
(a) Becomes void on C’s death.
(b) Should be performed by A and B along with C’s legal representative.
(c) Should be performed by A and B alone.
(d) Should be renewed between A, B and D.
Answer:
(b) Should be performed by A and B along with C’s legal representative.

Question 15.
Airport Authority of India, is a public enterprise. Identify the form of organisation.
(a) Statutory corporations
(b) Departmental undertakings
(c) Multi – national corporations
(d) State – owned company
Answer:
(b) Departmental undertakings

Question 16.
Debenture holders are entitled to a fixed rate of ………………………
(a) Dividend
(b) Profits
(c) Interest
(d) Ratio
Answer:
(c) Interest

Question 17.
The State Bank of India came into being in ………………………
(a) 1995
(b) 1945
(c) 1955
(d) 1965
Answer:
(c) 1955

Question 18.
Match List – I with List – II and select the correct answer using the codes given below:

List – I	List – II
(i) Assembling industry	1. Sugar
(ii) Processing industry	2. Computer
(iii) Analytical industry	3. Fishery
(iv) Genetic industry	4. Oil industry

Answer:
Codes:

Question 19.
An ……………………… is a document prepared by the importer and sent to the exporter to buy the goods.
(a) Invoice
(b) Indent
(c) Enquiry
(d) Charter party
Answer:
(b) Indent

Question 20.
Point out the wrong statement from the following.
(a) Extractive industries extract products from natural resources.
(b) Genetic industries are engaged in breeding plants and animals for their use in further reproduction
(c) Manufacturing industries are engaged in producing goods through processing of raw material.
(d) Construction industries are involved in the successive stages for manufacturing sugar and paper, etc.
Answer:
(d) Construction industries are involved in the successive stages for manufacturing sugar and paper, etc.

PART – II

Answer any seven questions in which question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
It is the oldest form of organisation in public enterprises. It is run by the particular department. What is the name of the organisation? How is it controlled?
Answer:
Departmental form of organisation of managing state enterprises is the oldest form of organisation. Under departmental form of organisation, a public enterprise is run as a separate full – fledged ministry or as a major sub-division of a department of the Government.

Question 22.
Banks which do not provide regular service, and are controlled by stock exchange, What is the name of the bank? Write a short note about the bank?
Answer:
The name of the bank is Merchant bank. A commercial bank or its subsidiary may offer services like project counselling, underwriting etc for starting a company. It is called merchant banking.

Question 23.
What is meant by chartered company?
Answer:
Chartered companies are established by the King or Queen of a country. Powers and privilege of chartered company are specified in the charter. Power to cancel the charter is vested with King/Queen.

Question 24.
Some co – operative societies are formed for providing short term finance to the members. It is useful to the farmers especially. What is the name of the society?
Answer:
Cooperative credit societies are societies formed for providing short-term financial help to their members. Agriculturists, artisans, industrial workers, salaried employees, etc., form these credit societies.

Question 25.
Give any two functions of warehouses?
Answer:

1. Storage
2. Price stabilization
3. Equalization of demand and supply

Question 26.
Define the term‘factoring’?
Answer:
The term ‘factoring’ is derived from a Latin term ‘facere’ which means ‘to make or do’. Factoring is an arrangement wherein the trade debts of a company are sold to a financial institution at a discount.

Question 27.
State the meaning of Mail order business?
Answer:
Mail order business means that the buying and selling is through mail. There is no personal contact between the buyers and sellers in this type of trading.

Question 28.
Explain the term – Credit card?
Answer:
Banks issue credit cards to customers and other eligible persons. With this card, the holder can purchase goods and services on credit at any shop in India.

Question 29.
What is meant by consumer co-operatives?
Answer:
Consumer co – operatives are organised by consumers. The aim of the consumers co-operative is to supply the quality goods at fair prices. They also supply essential commodities through Public Distribution System (PDS).

Question 30.
What is mutual funds?
Answer:
An individual investor who wants to invest in equities and bond with a balance of risk and return generally can invest in mutual funds. Nowadays people invest in stock markets through a mutual fund.

PART – III

Answer any seven questions in which question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
What are the advantages of company? Explain any three?
Answer:
1. Large Capital:
A company can secure large capital compared to a sole trader or partnership. Large amount of capital is necessary for conducting business on a large scale.

2. Limited Liability:
The liability of a shareholder is limited. The risk of loss is limited to the unpaid amount on the face value of shares held. In the case of a company limited by shares, the liability of a shareholder is restricted to the unpaid amount on the shares held by him.

3. Transferability of Shares:
Transaction of Shares between two individuals is easy. So there is liquidity of investment. Any shareholder can easily convert his shares into money by selling his shares.

Question 32.
What is the necessity for Entrepot trade?
Answer:
Entrepot is necessary because of the following reasons:

1. The country may not have any accessible trade routes connecting the importing country.
2. The goods imported may require further processing or finishing before exporting, and these facilities may be lacking in the exporting or importing country.
3. There may not have any bilateral trade agreement between both the countries.

Question 33.
This is a loan taken by depositing document of title to the property with the banker. What is the name of the loan? Also how the banker grants the loan to the customers?
Answer:
The name of the loan is ‘hypothecation’. In this type of loan, the customer has to deposit the document of title to the property with the banker. But the physical possession of the property is with the borrower. If the loan is not paid, the lender will sell the property and can receive the loan amount.

Question 34.
What are the objectives of GST?
Answer:

1. To create a common market with uniform tax rate in India. (One Nation, One Tax, One Market)
2. To eliminate the cascading effect of taxes, GST allows set – off of prior taxes for the same transactions as input tax credit.
3. To boost Indian exports, the GST already collected on the inputs will be refunded and thus there will be no tax on all exports.
4. To increase the tax base by bringing more number of tax payers and increase tax revenue.
5. To simplify tax return procedures through common forms and avoidance of visiting tax departments.
6. To provide online facilities for payment of taxes and submission of forms.

Question 35.
Cargo vessels which do not follow set routes and timetable?
What is the kind and name of the transport?
Answer:
The name of the transport is Tramps. They are a kind of ocean going ships. They are essentially cargo vessels. They have no set routes. They do not follow any timetable. They sail only when they get sufficient load.

Question 36.
What is Gross Total Income?
Answer:
Income from the five heads, namely – Salaries, House Property, Profits and Gains of Business or Profession, Capital Gains, and Other Sources – is computed separately according to the provisions given in the Act. Income computed under these heads shall be aggregated after adjusting past and present losses and the total so arrived at is known as ‘Gross Total Income’.

Question 37.
The retailers having permanent establishment and deal in large scale are called fixed shop large scale retailers. There are various types of large scale retailers. What is telemarketing? Explain the types?
Answer:
Purchasing and selling of goods through telephones are called Telemarketing. It is divided into two types.

1. Telephonic marketing:
Potential customers are contacted through mobile to provide information about the products. Willing customers visit the office and place orders.

2. Television marketing:
In this method, customers are attracted by providing all information of product or service through TV demonstrations. Customers can contact through phone or website to place order.

Question 38.
Distinguish between hire purchase system and installment system of selling?
Answer:

Question 39.
Explain any three features of Self Help Groups?
Answer:
Self Help Group is a small informal voluntary association created for the purpose of enabling members to reap economic benefit out of mutual help, solidarity, and joint responsibility. Features of Self Help Groups:

1. The motto of every group members should be “saving first – credit latter”.
2. Self Help Group is homogeneous in terms of economic status.
3. The groups need not be registered.

Question 40.
What are the objectives of GST?
Answer:

1. To create a common market with uniform tax rate in India. (One Nation, One Tax, One Market)
2. To eliminate the cascading effect of taxes, GST allows set-off of prior taxes for the same transactions as input tax credit.
3. To boost Indian exports, the GST already collected on the inputs will be refunded and thus there will be no tax on all exports.
4. To increase the tax base by bringing more number of tax payers and increase tax revenue.
5. To simplify tax return procedures through common forms and avoidance of visiting tax departments.
6. To provide online facilities for payment of taxes and submission of forms.

PART – IV

Answer all the questions. [7 × 5 = 35]

Question 41 (a).
Write short notes on:

1. Owner’s funds
2. Borrowed funds

Answer:
1. Owner’s Funds:
Owner’s funds mean funds which are provided by the owner of the enterprises who may be an individual, or partners or shareholders of a company.

The profits reinvested in the business (ploughing back of profit or retained earnings) come under owner’s funds. These funds are not required to be refunded during the life time of business enterprise. It provides the owner the right to control the management of the enterprise.

2. Borrowed Funds:
The term ‘borrowed funds’ denotes the funds raised through loans or borrowings. For example debentures, loans from banks and financial institutions, public deposits, trade credit, lease financing, commercial papers, factoring, etc., represent borrowed funds.

1. These borrowed sources of funds provide specific period before which the fund is to be returned.
2. Borrower is under legal obligation to pay interest at given rate at regular intervals to the lender.
3. Generally borrowed funds are obtained on the security of certain assets like bonds, land, building, stock, vehicles, machinery, documents of title to the goods, and the like.

[OR]

(b) What are the advantages of sole trading?
Answer:

1. Easy Formation: No legal formalities are required to initiate a sole trading concern. Any person capable of entering into a contract can start it, provided he has the necessary resources for it.
2. Incentive to Work hard: There is a direct relationship between effort and reward. The fact that the entire profit can be taken by himself without sharing with anybody else induces him to work ceaselessly.
3. Small Capital: Small capital is an important as well as specific advantage of sole proprietorship. Sole proprietor can start business with small capital.
4. Credit Standing: Since his private properties are held liable for satisfying business debts, he can get more financial assistance from others.
5. Personal Contact with the Customers: Since sole proprietor knows each and every customer individually he can supply goods according to their taste and preferences. Thus he can cultivate personal relationship with the customers.
6. Flexibility: The sole trader can easily adjust himself to the changing requirements of his business.

Question 42 (a).
Explain the contents of Articles of Association?
Answer:
Contents of Articles of Association (AOA):

1. Amount of shares, capital, value and type of shares.
2. Rights of each class of shareholders regarding voting, dividend, return of capital
3. Rules regarding issue of shares and debentures
4. Procedures as well as regulations in respect of making calls on shares
5. Manner of transfer of shares
6. Declaration of dividends
7. Borrowing powers of the company
8. Rules regarding the appointment, remuneration, removal of directors
9. Procedure for conducting proxy, quorum, meetings, etc.
10. Procedures concerning keeping of books and audits
11. Seal of the company
12. Procedures regarding the winding up of the company.

[OR]

(b) What are the primary functions of commercial banks?
Answer:
The primary functions of a commercial bank are of three types. They are:

1. Accepting Deposits:
The basic deposit accounts offered by commercial banks are listed below:

A. Demand Deposits:
These deposits are repayable on demand on any day. These consist of –

• Savings Deposits: General public deposit their savings into this account. This account can be opened in one individual’s name or more than one name.
• Current Deposits: This account is suitable for business institutions. Individuals too can open this account. A higher minimum balance should be kept in this account.

B. Time Deposits:
These are repayable after a period. These include –

• Fixed Deposits (FD): Certain amount is deposited for a fixed period for a fixed rate of interest.
• Recurring Deposits (RD): Certain sum is deposited into the account every month for one year or five years or the agreed period. Interest rate is more than savings deposits and almost equal to fixed deposits.

2. Granting Loans and Advances:
Commercial banks lend money in order to earn interest.

A. Advances:

• Overdraft: It is a credit facility extended mostly to current account holding business community customers.
• Cash Credit: It is a secured credit facility given mostly to business institutions. Stock in hand, raw materials, other tangible assets, etc. are provided as collateral.
• Discounting of Bills: Business customers approach banks to discount the commercial bills of exchanges and provide money

B. Loans- Short term and medium term loans are provided by commercial banks against eligible collaterals to business concerns. Examples- housing loan, consumer loan, vehicle loan, educational Loan, jewel loan, etc.

3. Creation of Credit- Apart from the currency money issued by the RBI, the credit money in circulation created by commercial banks influence economic activities of a country to a large extent. Credit money of commercial banks is far greater in volume than the currency money.

Question 43 (a).
What are the disadvantages of MNC?
Answer:

1. Danger for Domestic Industries: MNCs, because of their vast economic power, pose a danger to domestic industries; which are still in the process of development. Domestic industries cannot face challenges posed by MNCs.
2. Transfer of Outdated Technology: Where MNCs transfer outdated technology to host nation, it serves no purpose.
3. No Benefit to Poor People: MNCs produce only those things, which are used by the rich. Therefore, poor people of host countries do not get, generally, any benefit, out of MNCs.
4. Danger to Independence: Initially MNCs help the Government of the host country, in a number of ways; and then gradually start interfering in the political affairs of the host country.
5. Deprivation of Job Opportunity of Local People: MNCs may not generate job opportunities to the people of home country.

[OR]

(b) Explain the organisational structure of RBI?
Answer:
The head office of the RBI is situated in Mumbai. This central office has 33 departments in 2017. It has four zonal offices in Mumbai, Delhi, Calcutta and Chennai functioning under local boards with deputy governors as their heads. It also has 19 regional offices and 11 sub-offices (2017).

The RBI is governed by a central board of directors. The 21 member board is appointed by the Government of India. It consists of:

1. One Governor and four deputy governors appointed for a period of four years,
2. Ten Directors from various fields
3. Two Government officials
4. Four Directors – one each from local boards.

Question 44 (a).
Distinguish between internal and international trade?
Answer:

[OR]

(b) State the features of departmental stores?
Answer:

1. Large Size: A department is a large scale retail showroom requiring a large capital investment by forming a joint stock company managed by a board of directors.
2. Wide Choice: It acts as a universal provider of a wide range of products from low priced to very expensive goods (Pin to Car) to satisfy all the expected human needs under one roof.
3. Departmentally organized: Goods offered for sale are classified into various departments.
4. Facilities provided: It provides a number of facilities and services to the customers such as restaurant, rest rooms, recreation, packing, free home delivery, parking, etc.
5. Centralised purchasing: All the purchases are made centrally and directly from the manufacturers and operate separate warehouses whereas sales are decentralised in different departments.

Question 45 (a).
What are the different types of transportation? Explain?
Answer:
Types of transportation:

(A) Land Transport:
Transport of people and goods by land vehicles is known as Land transport. It is also called as‘Surface Transport’.

• Pack Animals: Animals like horse, mule, donkey, camel, elephant etc., are used for carrying small loads in backward areas, hilly tracks, forest regions and deserts.
• Bullock Carts: It constitutes the predominant form of rural road transport in India for goods traffic and to some extent for passengers’ traffic.
• Road Transport: It is one of the most promising and potent means suitable for short and medium distances.
• Motor lorries and Buses: From the dawn of civilization, people have been endeavoring to form roads and use wheeled vehicles to facilitate transport of men and materials.
• Tramways: Tramways were initially horse drawn, later stearmpowered and now they are electrically operated.
• Railway Transport: The invention of steam engine by James Watt, revolutionized the mode of transport all over the world.
• Recent Trends in Transport: Metro Rail, Monorail, Bullet train, Pipeline Transport, Conveyor Transport, Ropeway transport and Hyper loop transport.

(B) Water Transport:

1. Inland Waterways: Inland Waterways comprise of rivers, canals and lakes. It is also known as internal water transport.
2. Ocean or Sea Transport: Ocean transport has been playing a significant role in development of economic, social and cultural relations among countries of the world.

• Coastal shipping
• Overseas shipping:
• Liner
• Tramps

(C) Air Transport:
Air transport is the fastest and the costliest mode of transport. Commercial air transport is now one of the most prominent modes of overseas transport. Domestic and International flights are the air travels.

[OR]

(b) Explain the principles of insurance?
Answer:
1. Utmost Good Faith:
According to this principle, both insurer and insured should enter into contract in good faith. Insured should provide all the information that impacts the subject matter. Insurer should provide all the details regarding insurance contract.

2. Insurable Interest:
The insured must have an insurable interest in the subject matter of insurance. Insurable interest means some pecuniary interest in the subject matter of the insurance contract.

3. Indemnity:
Indemnity means security or compensation against loss or damages. In insurance, the insured would be compensated with the amount equivalent to the actual loss and not the amount exceeding the loss.

This principle ensures that the insured does not make any profit out of the insurance. This principle of indemnity is applicable to property insurance alone.

4. Causa Proxima:
The word ‘Causa proxima’ means ‘nearest cause’. According to this principle, when the loss is the result of two or more cause, the proximate cause, i.e. the direct. The direct, the most dominant and most effective cause of loss should be taken into consideration. The insurance company is not liable for the remote cause.

5. Contribution:
The same subject matter may be insured with more than one insurer then it is known as ‘Double Insurance’. In such a case, the insurance claim to be paid to the insured must be shared on contributed by all insurers in proportion to the sum assured by each one of them.

6. Subrogation:
Subrogation means ‘stepping the shoes on others’. According to this principle, once the claim of the insured has been settled, the ownership right of the subject matter of insurance passes on to the insurer.

7. Mitigation: In case of a mishap, the insured must take off all possible steps to reduce or mitigate the loss or damage to the subject matter of insurance.

Question 46 (a).
Elucidate any five features of Income Tax?
Answer:
Features of Income Tax in India:
1. Levied as Per the Constitution Income tax is levied in India by virtue of entry No. 82 of list I (Union List) of Seventh Schedule to the Article 246 of the Constitution of India.

2. Levied by Central Government Income tax is charged by the Central Government on all incomes other than agricultural income. However, the power to charge income tax on agricultural income has been vested with the State Government as per entry 46 of list II, i.e., State List.

3. Direct Tax Income tax is direct tax. It is because the liability to deposit and ultimate burden are on same person. The person earning income is liable to pay income tax out of his own pocket and cannot pass on the burden of tax to another person.

4. Annual Tax Income tax is an annual tax because it is the income of a particular year which is chargeable to tax.

5. Tax on Person It is a tax on income earned by a person. The term ‘person’ has been defined under the Income tax Act. It includes individual, Hindu Undivided Family, Firm, Company, local authority, Association of person or body of Individual or any other artificial juridical persons. The persons who are covered under Income tax Act are called ‘assessees’.

[OR]

(b) Write down the functions of IMF?
Answer:

1. It acts as short term credit institution at the international level.
2. It provides machinery for ordinary adjustments of exchange rates.
3. It has a reservoir of currencies of the member countries from which a borrower can borrow currencies of other nations.
4. It promotes economic stability and global growth by encouraging countries adopt sound economic and financial policies.
5. It offers technical assistance and training to help member countries strengthen and implement effective policies. Technical assistance is offered in formulating banking, fiscal, monetary and exchange policies.
6. It helps member countries correct their imbalance in balance of payment.

Question 47 (a).
Describe the importance of international finance?
Answer:

1. International finance helps in calculating exchange rates of various currencies of nations and the relative worth of each and every nation in terms thereof.
2. It helps in comparing the inflation rates and getting an idea about investing in international debt securities.
3. It helps in ascertaining the economic status of the various countries and in judging the foreign market.
4. International Financial Reporting System (IFRS) facilitates comparison of financial statements made by various countries.
5. It helps in understanding the basics of international organisations and maintaining the balance among them.

[OR]

(b) What are the advantages of Railway transport?
Answer:

1. Railways are well suited for carrying heavy and bulky goods over long distances.
2. It provides long distance travel throughout the day and night service.
3. It can provide better production and safety to the goods than motor transport.
4. Though initial investment is large, in the long run the operating expenses will be very low in railways and it will prove a cheaper mode of transport.
5. It has regular schedule of timing and is available throughout the year.
6. It provides unaffected services whether rainy or sunny weather conditions.