Category: Class 11

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.3 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 3.3

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.3 Text Book Back Questions and Answers

Question 1.
If the equation ax² + 5xy – 6y² + 12x + 5y + c = 0 represents a pair of perpendicular straight lines, find a and c.
Solution:
Comparing ax² + 5xy – 6y² + 12x + 5y + c = 0 with ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = a, 2h = 5, (or) h = \(\frac{5}{2}\), b = -6, 2g = 12 (or) g = 6, 2f = 5 (or) f = \(\frac{5}{2}\), c = c
Condition for pair of straight lines to be perpendicular is a + b = 0
a + (-6) = 0
a = 6
Next to find c. Condition for the given equation to represent a pair of straight lines is

R₁ → R₁ – R₃
Expanding along first row we get 0 – 0 + (6 – c) [\(\frac{25}{4}\) + 36] = 0
(6-c) [\(\frac{25}{4}\) + 36] = 0
6 – c = 0
6 = c (or) c = 6

Question 2.
Show that the equation 12x² – 10xy + 2y² + 14x – 5y + 2 = 0 represents a pair of straight lines and also find the separate equations of the straight lines.
Solution:
Comparing 12x² – 10xy + 2y² + 14x – 5y + 2 = 0 with ax² + 2hxy + by² + 2gh + 2fy + c = 0
We get a = 12, 2h = -10, (or) h = -5, b = 2, 2g = 14 (or) g = 7, 2f = -5 (or) f = \(-\frac{5}{2}\), c = 2
Condition for the given equation to represent a pair of straight lines is \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=0\)
\(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
12 & -5 & 7 \\
-5 & 2 & \frac{-5}{2} \\
7 & \frac{-5}{2} & 2
\end{array}\right|\)

= \(\frac{1}{4}\) [12(16 – 25) + 5(-40 + 70) + 7(50 – 56)]
= \(\frac{1}{4}\) [12(-9) + 5(30) + 7(-6)]
= \(\frac{1}{4}\) [-108 + 150 – 42]
= \(\frac{1}{4}\) [0]
= 0
∴ The given equation represents a pair of straight lines.
Consider 12x² – 10xy + 2y² = 2[6x² – 5xy + y²] = 2[(3x – y)(2x – y)] = (6x – 2y)(2x – y)
Let the separate equations be 6x – 2y + l = 0, 2x – y + m = 0
To find l, m
Let 12x² – 10xy + 2y² + 14x – 5y + 2 = (6x – 2y + l) (2x – y + m) ……. (1)
Equating coefficient of y on both sides of (1) we get
2l + 6m = 14 (or) l + 3m = 7 ………… (2)
Equating coefficient of x on both sides of (1) we get
-l – 2m = -5 ……… (3)
(2) + (3) ⇒ m = 2
Using m = 2 in (2) we get
l + 3(2) = 7
l = 7 – 6
l = 1
∴ The separate equations are 6x – 2y + 1 = 0, 2x – y + 2 = 0.

Question 3.
Show that the pair of straight lines 4x² + 12xy + 9y² – 6x – 9y + 2 = 0 represents two parallel straight lines and also find the separate equations of the straight lines.
Solution:
The given equation is 4x² + 12xy + 9y² – 6x – 9y + 2 = 0
Here a = 4, 2h = 12, (or) h = 6 and b = 9
h² – ab = 6² – 4 × 9 = 36 – 36 = 0
∴ The given equation represents a pair of parallel straight lines
Consider 4x² + 12xy + 9y² = (2x)² + 12xy + (3y)²
= (2x)² + 2(2x)(3y) + (3y)²
= (2x + 3y)²
Here we have repeated factors.
Now consider, 4x² + 12xy + 9y² – 6x – 9y + 2 = 0
(2x + 3y)² – 3(2x + 3y) + 2 = 0
t² – 3t + 2 = 0 where t = 2x + 3y
(t – 1)(t – 2) = 0
(2x + 3y – 1) (2x + 3y – 2) = 0
∴ Separate equations are 2x + 3y – 1 = 0, 2x + 3y – 2 = 0

Question 4.
Find the angle between the pair of straight lines 3x² – 5xy – 2y² + 17x + y + 10 = 0.
Solution:
The given equation is 3x² – 5xy – 2y² + 17x + y + 10 = 0
Here a = 3, 2h = -5, b = -2
If θ is the angle between the given straight lines then

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Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.2 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 3 Analytical Geometry Ex 3.2

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.2 Text Book Back Questions and Answers

Question 1.
Find the angle between the lines whose slopes are \(\frac{1}{2}\) and 3.
Solution:
Given that m₁ = \(\frac{1}{2}\) and m₂ = 3.
Let θ be the angle between the lines then

tan θ = 1
tan θ = tan 45°
∴ θ = 45°

Question 2.
Find the distance of the point (4, 1) from the line 3x – 4y + 12 = 0.
Solution:
The length of perpendicular from a point (x₁, y₁) to the line ax + by + c = 0 is d = \(\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)
∴ The distance of the point (4, 1) to the line 3x – 4y + 12 = 0 is
[Here (x₁, y₁) = (4, 1), a = 3, b = -4, c = 12]

Question 3.
Show that the straight lines x + y – 4 = 0, 3x + 2 = 0 and 3x – 3y + 16 = 0 are concurrent.
Solution:
The lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 are concurrent if
\(\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|=0\)
The given lines x + y – 4 = 0, 3x + 0y + 2 = 0, 3x – 3y + 16 = 0
\(\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|=\left|\begin{array}{rrr}
1 & 1 & -4 \\
3 & 0 & 2 \\
3 & -3 & 16
\end{array}\right|\)
= 1(0 + 6) – 1(48 – 6) – 4(-9 – 0)
= 6 – (42) + 36
= 42 – 42
= 0
The given lines are concurrent.

Question 4.
Find the value of ‘a’ for which the straight lines 3x + 4y = 13; 2x – 7y = -1 and ax – y – 14 = 0 are concurrent.
Solution:
The lines 3x + 4y = 13, 2x – 7y = -1 and ax – y – 14 = 0 are concurrent.
\(\left|\begin{array}{rrr}
3 & 4 & -13 \\
2 & -7 & 1 \\
a & -1 & -14
\end{array}\right|=0\)
⇒ 3(98 + 1) – 4(-28 – a) – 13(-2 + 7a) = 0
⇒ 3(99) + 112 + 4a + 26 – 91a = 0
⇒ 297 + 112 + 26 + 4a – 91a = 0
⇒ 435 – 87a = 0
⇒ -87a = -435
⇒ a = \(\frac{-435}{-87}\) = 5

Question 5.
A manufacturer produces 80 TV sets at a cost of ₹ 2,20,000 and 125 TV sets at a cost of ₹ 2,87,500. Assuming the cost curve to be linear, find the linear expression of the given information. Also, estimate the cost of 95 TV sets.
Solution:
Let x represent the TV sets, andy represent the cost.

The equation of straight line expressing the given information as a linear equation in x and y is

1(y – 2,20,000) = (x – 80)1500
y – 2,20,000 = 1500x – 80 × 1500
y = 1500x – 1,20,000 + 2,20,000
y = 1500x + 1,00,000 which is the required linear expression.
When x = 95,
y = 1,500 × 95 + 1,00,000
= 1,42,500 + 1,00,000
= 2,42,500
∴ The cost of 95 TV sets is ₹ 2,42,500.

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.7 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 2.7

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 2.6 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 2 Algebra Ex 2.6

Samacheer Kalvi 11th Business Maths Algebra Ex 2.6 Text Book Back Questions and Answers

Question 1.
Expand the following by using binomial theorem:
(i) (2a – 3b)⁴
(ii) \(\left(x+\frac{1}{y}\right)^{7}\)
(iii) \(\left(x+\frac{1}{x^{2}}\right)^{6}\)
Solution:

Question 2.
Evaluate the following using binomial theorem:
(i) (101)⁴
(ii) (999)⁵
Solution:
(i) (x + a)ⁿ = nC₀ xⁿ a⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + ……… + nC_r x^n-r a^r + …… + nC_n aⁿ
(101)⁴ = (100 + 1)⁴ = 4C₀ (100)⁴ + 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² + 4C₃ (100)¹ (1)³ + 4C₄ (1)⁴
= 1 × (100000000) + 4 × (1000000) + 6 × (10000) + 4 × 100 + 1 × 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 10,40,60,401

(ii) (x + a)ⁿ = nC₀ xⁿ a⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + ……… + nC_r x^n-r a^r + …… + nC_n aⁿ
(999)⁵ = (1000 – 1)⁵ = 5C₀ (1000)⁵ – 5C₁ (1000)⁴ (1)¹ + 5C₂ (1000)³ (1)² – 5C₃ (1000)² (1)³ + 5C₄ (1000)⁵ (1)⁴ – 5C₅ (1)⁵
= 1(1000)⁵ – 5(1000)⁴ – 10(1000)³ – 10(1000)² + 5(1000) – 1
= 1000000000000000 – 5000000000000 + 10000000000 – 10000000 + 5000 – 1
= 995009990004999

Question 3.
Find the 5th term in the expansion of (x – 2y)¹³.
Solution:
General term is t_r+1 = nC_r x^n-r a^r
(x – 2y)¹³ = (x + (-2y))¹³
Here x is x, a is (-2y) and n = 13
5th term = t₅ = t₄₊₁ = 13C₄ x^13-4 (-2y)⁴
= 13C₄ x⁹ 2⁴ y⁴
= \(\frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\) × 2 × 2 × 2 × 2× x⁹y⁴
= 13 × 11 × 10 × 8x⁹y⁴
= 13 × 880x⁹y⁴
= 11440x⁹y⁴

Question 4.
Find the middle terms in the expansion of
(i) \(\left(x+\frac{1}{x}\right)^{11}\)
(ii) \(\left(3 x+\frac{x^{2}}{2}\right)^{8}\)
(iii) \(\left(2 x^{2}-\frac{3}{x^{3}}\right)^{10}\)
Solution:
(i) General term is t_r+1 = nC_r x^n-r a^r
Here x is x, a is \(\frac{1}{x}\) and n = 11, which is odd.
So the middle terms are \(\frac{t_{n+1}}{2}=\frac{t_{11+1}}{2}, \frac{t_{n+3}}{2}=\frac{t_{11+3}}{2}\)
i.e. the middle terms are t₆, t₇

(ii) Here x is 3x, a is \(\frac{x^{2}}{2}\), n = 8, which is even.
∴ The only one middle term = \(\frac{t_{n+1}}{2}=\frac{t_{8+1}}{2}\) = t₅
General term t_r+1 = nC_r x^n-r a^r

(iii) \(\left(2 x^{2}-\frac{3}{x^{3}}\right)^{10}=\left(2 x^{2}+\frac{-3}{x^{3}}\right)^{10}\) compare with the (x + a)ⁿ
Here x is 2x², a is \(\frac{-3}{x^{3}}\), n = 10, which is even.
So the only middle term is \(\frac{t_{n+1}}{2}=\frac{t_{10}}{2}+1\) = t₆
General term t_r+1 = nC_r x^n-r a^r
t₆ = t₅₊₁ = t_r+1

Question 5.
Find the term in dependent of x in the expansion of
(i) \(\left(x^{2}-\frac{2}{3 x}\right)^{9}\)
(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)
(iii) \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\)
Solution:
(i) Let the independent form of x occurs in the general term, t_r+1 = nC_r x^n-r a^r
Here x is x², a is \(\frac{-2}{3 x}\) and n = 9

Independent term occurs only when x power is zero.
18 – 3r = 0
⇒ 18 = 3r
⇒ r = 6
Put r = 6 in (1) we get the independent term as 9C₆ x⁰ \(\frac{(-2)^{6}}{3^{6}}\) = 9C₃ \(\left(\frac{2}{3}\right)^{6}\)
[∵ 9C₆ = 9C_9-6 = 9C₃]

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}=\left(x+\frac{-2}{x^{2}}\right)^{15}\) compare with the (x + a)ⁿ
Here x is x, a is \(\frac{-2}{x^{2}}\), n = 15.
Let the independent term of x occurs in the general term

Independent term occurs only when x power is zero.
15 – 3r = 0
15 = 3r
r = 5
Using r = 5 in (1) we get the independent term
= 15C₅ x⁰ (-2)⁵ [∵ (-2)⁵ = (-1)⁵ 2⁵ = -2⁵]
= -32(15C₅)

(iii) \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\) Compare with the (x + a)ⁿ.
Here x is 2x², a is \(\frac{1}{x}\), n = 12.
Let the independent term of x occurs in the general term.


Independent term occurs only when x power is zero
24 – 3r = 0
24 = 3r
r = 8
Put r = 8 in (1) we get the independent term as
= 12C₈ 2^12-8 x⁰
= 12C₄ × 2⁴ × 1
= 7920

Question 6.
Prove that the term independent of x in the expansion of \(\left(x+\frac{1}{x}\right)^{2 n}\) is \(\frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot(2 n-1) 2^{n}}{n !}\)
Solution:
There are (2n + 1) terms in expansion.
∴ t_n+1 is the middle term.

Question 7.
Show that the middle term in the expansion of is (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1) 2^{n} x^{n}}{n !}\)
Solution:
There are 2n + 1 terms in expansion of (1 + x)²ⁿ.
∴ The middle term is t_n+1.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.5

Samacheer Kalvi 11th Business Maths Algebra Ex 2.5 Text Book Back Questions and Answers

By the principle of mathematical induction, prove the following:

Question 1.
1³ + 2³ + 3³ + ….. + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\) for all x ∈ N.
Solution:
Let P(n) be the statement 1³ + 2³ + …… + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\) for all n ∈ N.
i.e., p(n) = 1³ + 2³ + …… + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\), for all n ∈ N
Put n = 1
LHS = 1³ = 1
RHS = \(\frac{1^{2}(1+1)^{2}}{4}\)
= \(\frac{1 \times 2^{2}}{4}\)
= \(\frac{4}{4}\)
= 1
∴ P(1) is true.
Assume that P(n) is true n = k
P(k): 1³ + 2³ + …… + k³ = \(\frac{k^{2}(k+1)^{2}}{4}\)
To prove P(k + 1) is true.
i.e., to prove 1³ + 2³ + ……. + k³ + (k + 1)³ = \(\frac{(k+1)^{2}((k+1)+1)^{2}}{4}=\frac{(k+1)^{2}(k+2)^{2}}{4}\)
Consider 1³ + 2³ + …… + k³ + (k + 1)³ = \(\frac{k^{2}(k+1)^{2}}{4}\) + (k + 1)³
= (k + 1)² [\(\frac{k^{2}}{4}\) + (k + 1)]
= (k + 1)² \(\left[\frac{k^{2}+4(k+1)}{4}\right]\)
= \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)
⇒ P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all n ∈ N.

Question 2.
1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N.
Solution:
Let P(n) denote the statement
1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)
Put n = 1
LHS = 1(1 + 1) = 2
RHS = \(\frac{1(1+1)(1+2)}{3}=\frac{1(2)(3)}{3}\) = 2
∴ P(1) is true.
Now assume that the statement be true for n = k
(i.e.,) assume P(k) be true
(i.e.,) assume 1.2 + 2.3 + 3.4 + …… + k(k + 1) = \(\frac{k(k+1)(k+2)}{3}\) be true
To prove: P(k + 1) is true
(i.e.,) to prove: 1.2 + 2.3 + 3.4 + …… + k(k + 1) + (k + 1) (k + 2) = \(\frac{(k+1)(k+2)(k+3)}{3}\)
Consider 1.2 + 2.3 + 3.4 + ……. + k(k + 1) + (k + 1) (k + 2)
= [1.2 + 23 + …… + k(k + 1)] + (k + 1) (k + 2)
= \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2)
= \(\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\)
= \(\frac{(k+1)(k+2)(k+3)}{3}\)
∴ P(k + 1) is true.
Thus if P(k) is true, P(k + 1) is true.
By the principle of Mathematical ‘induction, P(n) is true for all n ∈ N.
1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

Question 3.
4 + 8 + 12 + ……. + 4n = 2n(n + 1), for all n ∈ N.
Solution:
Let P(n) denote the statement 4 + 8 + …….. + 4n = 2n(n + 1)
i.e., P(n) : 4 + 8 + 12 + … + 4n = 2n(n + 1)
Put n = 1,
P(1): LHS = 4
RHS = 2 (1)(1 + 1) = 4
P(1) is true.
Assume that P(n) is true for n = k
P(k): 4 + 8 + 12 + ……. + 4k = 2k(k + 1)
To prove P(k + 1)
i.e., to prove 4 + 8 + 12 + ……. + 4k + 4(k + 1) = 2(k + 1) (k + 1 + 1)
4 + 8 + 12 + …… + 4k + (4k + 4) = 2(k + 1) (k + 2)
Consider, 4 + 8 + 12 + …….. + 4k + (4k + 4) = 2k(k + 1) + (4k + 4)
= 2k(k + 1) + 4(k + 1)
= 2k² + 2k + 4k + 4
= 2k² + 6k + 4
= 2(k + 1)(k + 2)
P(k + 1) is also true.
∴ By Mathematical Induction, P(n) for all value n ∈ N.

Question 4.
1 + 4 + 7 + ……. + (3n – 2) = \(\frac{n(3 n-1)}{2}\) for all n ∈ N.
Solution:
Let P(n) : 1 + 4 + 7 + ……. + (3n – 2) = \(\frac{n(3 n-1)}{2}\)
Put n = 1,
LHS = 1
RHS = \(\frac{1(3-1)}{2}\) = 1
∴ P(1) is true.
Assume P(k) is true for n = k
P(k): 1 + 4 + 7 + ……. + (3k – 2) = \(\frac{k(3 k-1)}{2}\)
To prove P(k + 1) is true, i.e., to prove
1 + 4 + 7 + ……. + (3k – 2) + (3(k + 1) – 2) = \(\frac{(k+1)(3(k+1)-1)}{2}\)
1 + 4 + 7 + ……. + (3k – 2) + (3k + 3 – 2) = \(\frac{(k+1)(3 k+2)}{2}\)
1 + 4 + 7 + …… + (3k + 1) = \(\frac{(k+1)(3 k+2)}{2}\)
1 + 4 + 7 + …… + (3k – 2) + (3k + 1) = \(\frac{k(3 k-1)}{2}\) + (3k + 1)
= \(\frac{k(3 k-1)+2(3 k+1)}{2}\)
= \(\frac{3 k^{2}-k+6 k+2}{2}\)
= \(\frac{3 k^{2}+5 k+2}{2}\)
= \(\frac{(k+1)(3 k+2)}{2}\)
∴ P(k + 1) is true whenever P(k) is true.
∴ By the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 5
3²ⁿ – 1 is divisible by 8, for all n ∈ N.
Solution:
Let P(n) denote the statement 3²ⁿ – 1 is divisible by 8 for all n ∈ N
Put n = 1
P(1) is the statement 3²⁽¹⁾ – 1 = 3² – 1 = 9 – 1 = 8, which is divisible by 8
∴ P(1) is true.
Assume that P(k) is true for n = k.
i.e., 3^2k – 1 is divisible by 8 to be true.
Let 3^2k – 1 = 8m
To prove P(k + 1) is true.
i.e., to prove 3^2(k+1) – 1 is divisible by 8
Consider 3^2(k+1) – 1 = 3^2k+2 – 1
= 3^2k . 3² – 1
= 3^2k (9) – 1
= 3^2k (8 + 1) – 1
= 3^2k × 8 + 3^2k × 1 – 1
= 3^2k (8) + 3^2k – 1
= 3^2k (8) + 8m (∵ 3^2k – 1 = 8m)
= 8(3^2k + m), which is divisible by 8.
∴ P(k + 1) is true wherever P(k) is true.
∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 6.
aⁿ – bⁿ is divisible by a – b, for all n ∈ N.
Solution:
Let P(n) denote the statement aⁿ – bⁿ is divisible by a – b.
Put n = 1. Then P(1) is the statement: a¹ – b¹ = a – b is divisible by a – b
∴ P(1) is true. Now assume that the statement be true for n = k
(i.e.,) assume P(k) be true, (i.e.,) a^k – b^k is divisible by (a – b) be true.
⇒ \(\frac{a^{k}-b^{k}}{a-b}\) = m (say) where m ∈ N
⇒ a^k – b^k = m(a – b)
⇒ a^k = b^k + m(a – b) ……. (1)
Now to prove P(k + 1) is true, (i.e.,) to prove: a^k+1 – b^k+1 is divisible by a – b
Consider a^k+1 – b^k+1 = a^k . a – b^k . b
= [b^k + m(a – b)] a – b^k . b [∵ a^k = bm + k(a – b)]
= b^k . a + am(a – b) – b^k . b
= b^k . a – b^k . b + am(a – b)
= b^k(a – b) + am(a – b)
= (a – b) (b^k + am) is divisible by (a – b)
∴ P(k + 1) is true.
By the principle of Mathematical induction. P(n) is true for all n ∈ N.
∴ aⁿ – bⁿ is divisible by a – b for n ∈ N.

Question 7.
5²ⁿ – 1 is divisible by 24, for all n ∈ N.
Solution:
Let P(n) be the proposition that 5²ⁿ – 1 is divisible by 24.
For n = 1, P(1) is: 5² – 1 = 25 – 1 = 24, 24 is divisible by 24.
Assume that P(k) is true.
i.e., 5^2k – 1 is divisible by 24
Let 5^2k – 1 = 24m
To prove P(k + 1) is true.
i.e., to prove 5^2(k+1) – 1 is divisible by 24.
P(k): 5^2k – 1 is divisible by 24.
P(k + 1) = 5^2(k+1) – 1
= 5^2k . 5² – 1
= 5^2k (25) – 1
= 5^2k (24 + 1) – 1
= 24 . 5^2k + 5^2k – 1
= 24 . 5^2k + 24m
= 24 [5^2k + 24]
which is divisible by 24 ⇒ P(k + 1) is also true.
Hence by mathematical induction, P(n) is true for all values n ∈ N.

Question 8.
n(n + 1) (n + 2) is divisible by 6, for all n ∈ N.
Solution:
P(n): n(n + 1) (n + 2) is divisible by 6.
P(1): 1 (2) (3) = 6 is divisible by 6
∴ P(1) is true.
Let us assume that P(k) is true for n = k
That is, k (k + 1) (k + 2) = 6m for some m
To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2)(k + 3) is divisible by 6.
P(k + 1) = (k + 1) (k + 2) (k + 3)
= (k + 1)(k + 2)k + 3(k + 1)(k + 2)
= 6m + 3(k + 1)(k + 2)
In the second term either k + 1 or k + 2 will be even, whatever be the value of k.
Hence second term is also divisible by 6.
∴ P (k + 1) is also true whenever P(k) is true.
By Mathematical Induction P (n) is true for all values of n.

Question 9.
2ⁿ > n, for all n ∈ N.
Solution:
Let P(n) denote the statement 2ⁿ > n for all n ∈ N
i.e., P(n): 2ⁿ > n for n ≥ 1
Put n = 1, P(1): 2¹ > 1 which is true.
Assume that P(k) is true for n = k
i.e., 2^k > k for k ≥ 1
To prove P(k + 1) is true.
i.e., to prove 2^k+1 > k + 1 for k ≥ 1
Since 2^k > k
Multiply both sides by 2
2 . 2^k > 2k
2^k+1 > k + k
i.e., 2^k+1 > k + 1 (∵ k ≥ 1)
∴ P(k + 1) is true whenever P(k) is true.
∴ By principal of mathematical induction P(n) is true for all n ∈ N.

Students can download 11th Business Maths Chapter 1 Matrices and Determinants Ex 1.4 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 1.4

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.

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.4 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 2.4

Samacheer Kalvi 11th Business Maths Algebra Ex 2.4 Text Book Back Questions and Answers

Question 1.
If ⁿP_r = 1680 and ⁿC_r = 70, find n and r.
Solution:
Given that ⁿP_r = 1680, ⁿC_r = 70
We know that ⁿC_r = \(\frac{n \mathrm{P}_{r}}{r !}\)
70 = \(\frac{1680}{r !}\)
r! = \(\frac{1680}{70}\) = 24
r! = 4 × 3 × 2 × 1 = 4!
∴ r = 4

Question 2.
Verify that ⁸C₄ + ⁸C₃ = ⁹C₄.
Solution:
LHS = ⁸C₄ + ⁸C₃
= \(\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}\)
= 7 × 2 × 5 + 8 × 7
= 70 + 56
= 126
RHS = ⁹C₄
= \(\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)
= 9 × 7 × 2
= 126
∴ LHS = RHS
Hence verified.

Question 3.
How many chords can be drawn through 21 points on a circle?
Solution:
To draw a chord we need two points on a circle.
∴ Number chords through 21 points on a circle = ²¹C₂ = \(\frac{21 \times 20}{2 \times 1}\) = 210.

Question 4.
How many triangles can be formed by joining the vertices of a hexagon?
Solution:
A hexagon has six vertices. By joining any three vertices of a hexagon we get a triangle.
∴ Number of triangles formed by joining the vertices of a hexagon = ⁶C₃ = \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\) = 20.

Question 5.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Solution:
In this problem first, we have to select consonants and vowels.
Then we arrange a five-letter word using 3 consonants and 2 vowels.
Therefore here both combination and permutation involved.
The number of ways of selecting 3 consonants from 7 is ⁷C₃.
The number of ways of selecting 2 vowels from 4 is ⁴C₃.
The number of ways selecting 3 consonants from 7 and 2 vowels from 4 is ⁷C₃ × ⁴C₂.
Now with every selection number of ways of arranging 5 letter word
= 5! × ⁷C₃ × ⁴C₂
= 120 × \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1}\)
= 25200

Question 6.
If four dice are rolled, find the number of possible outcomes in which atleast one die shows 2.
Solution:
When a die is rolled number of possible outcomes is selecting an event from 6 events = ⁶C₁
When four dice are rolled number of possible outcomes = ⁶C₁ × ⁶C₁ × ⁶C₁ × ⁶C₁
When a die is rolled number of possible outcomes in which ‘2’ does not appear is selecting an event from 5 events = ⁵C₁
When four dice are rolled number of possible outcomes in which 2 does not appear = ⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C₁
Therefore the number of possible outcomes in which atleast one die shows 2
= ⁶C₁ × ⁶C₁ × ⁶C₁ × ⁶C₁ – ⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C₁
= 6 × 6 × 6 × 6 – 5 × 5 × 5 × 5
= 1296 – 625
= 671
Note: when two dice are rolled number of possible outcomes is 36 and the number of possible outcomes in which 2 doesn’t appear = 25. When two dice are rolled the number of possible outcomes in which atleast one die shows 2 = 36 – 25 = 11. Use the sample space, S = {(1, 1), (1, 2),… (6, 6)}.

Question 7.
There are 18 guests at a dinner party. They have to sit 9 guests on either side of a long table, three particular persons decide to sit on one side and two others on the other side. In how many ways can the guests to be seated?
Solution:
Let A and B be two sides of the table 9 guests sit on either side of the table in 9! × 9! ways.
Out of 18 guests, three particular persons decide to sit namely inside A and two on the other side B. remaining guest = 18 – 3 – 2 = 13.
From 13 guests we can select 6 more guests for side A and 7 for the side.
Selecting 6 guests from 13 can be done in ¹³C₆ ways.
Therefore total number of ways the guest to be seated = ¹³C₆ × 9! × 9!
= \(\frac{13 !}{6 !(13-6) !} \times 9 ! \times 9 !\)
= \(\frac{13 !}{6 ! \times 7 !} \times 9 ! \times 9 !\)

Question 8.
If a polygon has 44 diagonals, find the number of its sides.
Solution:
A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon.
A number of line segments obtained by joining the vertices of a n sided polygon taken two at a time = Number of ways of selecting 2 out of n.
= ⁿC₂
= \(\frac{n(n-1)}{2}\)
Out of these lines, n lines are the sides of the polygon, Sides can’t be diagonals.
∴ Number of diagonals of the polygon = \(\frac{n(n-1)}{2}\) – n = \(\frac{n(n-3)}{2}\)
Given that a polygon has 44 diagonals.
Let n be the number of sides of the polygon.
\(\frac{n(n-3)}{2}\) = 44
⇒ n(n – 3) = 88
⇒ n² – 3n – 88 = 0
⇒ (n + 8) (n – 11)
⇒ n = -8 (or) n = 11
n cannot be negative.
∴ n = 11 is number of sides of polygon is 11.

Question 9.
In how many ways can a cricket team of 11 players be chosen out of a batch of 15 players?
(i) There is no restriction on the selection.
(ii) A particular player is always chosen.
(iii) A particular player is never chosen.
Solution:
(i) Number of ways choosing 11 players from 15 is ¹⁵C₁₁ = ¹⁵C₄
= \(\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}\)
= 15 × 7 × 13
= 1365.

(ii) If a particular is always chosen there will be only 14 players left put, in which 10 are to selected in ¹⁴C₁₀ ways.
¹⁴C₁₀ = ¹⁴C₄
= \(\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}\)
= \(\frac{14 \times 13 \times 11}{2}\)
= 91 × 11
= 1001 ways

(iii) If a particular player is never chosen we have to select 11 players out of remaining 14 players in ¹⁴C₁₁ ways.
i.e., ¹⁴C₃ ways = \(\frac{14 \times 13 \times 12}{3 \times 2 \times 1}\) = 364 ways.

Question 10.
A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways this can be done when
(i) atleast two ladies are included.
(ii) atmost two ladies are included.
Solution:
(i) A committee of 5 is to be formed.

(ii) Almost two ladies are included means maximum of two ladies are included.

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.3 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 2.3

Samacheer Kalvi 11th Business Maths Algebra Ex 2.3 Text Book Back Questions and Answers

Question 1.
If ⁿP₄ = 12(ⁿP₂), find n.
Solution:
Given that ⁿP₄ = 12(ⁿP₂)
n(n – 1) (n – 2) (n – 3) = 12n(n – 1)
Cancelling n(n – 1) on both sides we get
(n – 2) (n – 3) = 4 × 3
We have product of consecutive number on both sides with decreasing order.
n – 2 = 4
∴ n = 6

Question 2.
In how many ways 5 boys and 3 girls can be seated in a row so that no two girls are together?
Solution:
5 boys can be seated among themselves in ⁵P₅ = 5! Ways. After this arrangement, we have to arrange the three girls in such a way that in between two girls there atleast one boy. So the possible places girls can be placed with the × symbol given below.
× B × B × B × B × B ×
∴ There are 6 places to seated by 3 girls which can be done 6P3 ways.
∴ Total number of ways = 5! × ⁶P₃
= 120 × (6 × 5 × 4)
= 120 × 120
= 14400

Question 3.
How many 6-digit telephone numbers can be constructed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?
Solution:
Given that each number starts with 35. We need a 6 digit number. So we have to fill only one’s place, 10’s place, 100th place, and 1000th places. We have to use 10 digits.

In these digits, 3 and 5 should not be used as a repetition of digits is not allowed. Except for these two digits, we have to use 8 digits. One’s place can be filled by any of the 8 digits in different ways, 10’s place can be filled by the remaining 7 digits in 7 different ways.

100th place can be filled by the remaining 6 different ways and 1000th place can be filled by the remaining 5 digits in 5 different ways.

∴ Number of 6 digit telephone numbers = 8 × 7 × 6 × 5 = 1680

Question 4.
Find the number of arrangements that can be made out of the letters of the word “ASSASSINATION”.
Solution:
The number of letters of the word “ASSASSINATION” is 13.
The letter A occurs 3 times
The letter S occurs 4 times
The letter I occur 2 times
The letter N occurs 2 times
The letter T occurs 1 time
The letter O occurs 1 time
∴ Number of arrangements = \(\frac{13 !}{3 ! 4 ! 2 ! 2 ! 1 ! 1 !}=\frac{13 !}{3 ! 4 ! 2 ! 2 !}\)

Question 5.
(a) In how many ways can 8 identical beads be strung on a necklace?
(b) In how many ways can 8 boys form a ring?
Solution:
(a) Number of ways 8 identical beads can be stringed by \(\frac{(8-1) !}{2}=\frac{7 !}{2}\)
(b) Number of ways 8 boys form a ring = (8 – 1)! = 7!

Question 6.
Find the rank of the word ‘CHAT’ in the dictionary.
Solution:
The letters of the word CHAT in alphabetical order are A, C, H, T. To arrive the word CHAT, first, we have to go through the word that begins with A. If A is fixed as the first letter remaining three letters C, H, T can be arranged among themselves in 3! ways. Next, we select C as the first letter and start arranging the remaining letters in alphabetical order. Now C and A is fixed remaining two letters can be arranged in 2! ways. Next, we move on H with C, A, and H is fixed the letter T can be arranged in 1! ways.
∴ Rank of the word CHAT = 3! + 2! + 1! = 6 + 2 + 1 = 9

Note: The rank of a given word is basically finding out the position of the word when possible words have been formed using all the letters of the given word exactly once and arranged in alphabetical order as in the case of dictionary. The possible arrangement of the word CHAT are (i) ACHT, (ii) ACTH, (iii) AHCT, (iv) AHTC, (v) ATCH, (vi) ATHC, (vii) CAHT, (viii) CATH, (ix) CHAT. So the rank of the word occurs in the ninth position.
∴ The rank of the word CHAT is 9.

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.