Class 8

Siri Kannada Text Book Class 8 Answers Solutions Guide Notes Pdf free download is part of KSEEB Solutions for Class 8. Here we have given Karnataka State Board Syllabus 8th Standard 1st Language Siri Kannada Textbook NCERT Solutions. Students can also read Tili Kannada Text Book Class 8 Solutions of 2nd language.

Siri Kannada Text Book Class 8 Answers Solutions Guide (1st Language)

Class 8 Siri Kannada Gadya Bhaga​ Karnataka State Board Solutions

• Chapter 1 Maggada​ Saheba
• Chapter 2 Niru Kodada​ Nadinalli
• Chapter 3 Talakadina​ Vaibhava
• Chapter 4 Sarthaka​ Badukina​ Sadhaka​
• Chapter 5 Huvada​ Hudugi
• Chapter 6 Yashodhare
• Chapter 7 Amma
• ​Chapter 8 Saptakshari Mantra​

Class 8 Siri Kannada Padya Bhaga Karnataka State Board Solutions

• Chapter 1 Kannadigara​ Tayi
• Chapter 2 Sanna Sangati
• Chapter 3 Geletana
• Chapter 4 Bharavase
• Chapter 5 Vachanamrut
• Chapter 6 Someshwara Shataka
• Chapter 7 Jeevana Darshana
• Chapter 8 Ramadhanya Charite

Class 8 Siri Kannada Pathya Puraka Adhyayana Karnataka State Board Solutions

• Chapter 1 Kattuvevu Naavu
• Chapter 2 Sarthaka
• Chapter 3 Aahuti
• Chapter 4 Magalige Bareda Patra
• Chapter 5 Aatoriksada Rasaprasangagalu

Tili Kannada Text Book Class 8 Answers Solutions Guide (2nd Language)

Class 8 Tili Kannada Gadya Bhaga​ Karnataka State Board Solutions

• Chapter 1 Buddhana Salahe (Nirupama)
• Chapter 2 Kanasu Mattu Sandesha (A.P.J. Abdul Kalam)
• Chapter 3 Gandhijiya Balya (Goruru Ramaswami Ayyangar)
• Chapter 4 Sukri Bommana Gowda (Santi Nayaka)
• Chapter 5 Blood Group (Vijayamala Ranganath)
• Chapter 6 Parivartan (G. S. Basavarajasastri)
• Chapter 7 Ondu Marada Bele (Ha. Ma. Nayaka)
• Chapter 8 Asanada Mele Asana (Rasi)

Class 8 Tili Kannada Padya Bhaga Karnataka State Board Solutions

• Chapter 1 Anveshane (G. S. Sivarudrappa)
• Chapter 2 Harida Hakkigalu (R.V. Bhandari)
• Chapter 3 Jyotiye Agu Jagakella (Janapadageete)
• Chapter 4 Nanna Hageye (Su.Ram. Ekkundi)
• Chapter 5 Tugi Tugi Maragale (N. S. Lakshminarayanabhatta)
• Chapter 6 Male Barali (Savita Nagabhushana)
• Chapter 7 Hakkigalu (Siddhalingayya)
• Chapter 8 Gauravisu Jeevanava (D.V.G)

Class 8 Tili Kannada Saiddhantika Vyakarana

• Kannada Varnamale
• Gunataksharagalu – Samyuktaksharagalu
• Kannada Sandhigalu
• Namapada – Vibhakti Pratyaya Galu
• Tatsama – Tadbhava Galu
• Vachanagalu – Vakya Rachane
• Lingagalu
• Dvirukti – Jodi Nudi Nudigattugalu
• Patra Lekhana
• Prabandhagalu
• Gadegala Vistarane

We hope the given Siri Kannada Text Book Class 8 Solutions Answers Guide Notes Pdf free download will help you. If you have any queries regarding Karnataka State Board Syllabus 8th Standard 1st Language Siri Kannada Textbook Answers, drop a comment below and we will get back to you at the earliest.

Tili Kannada Text Book Class 8 Answers Solutions Guide Notes Pdf free download is part of KSEEB Solutions for Class 8. Here we have given Karnataka State Board Syllabus 8th Standard 2nd Language Tili Kannada Textbook Solutions.

Tili Kannada Text Book Class 8 Answers Solutions Guide (2nd Language)

Class 8 Tili Kannada Gadya Bhaga​ Karnataka State Board Solutions

• Chapter 1 Buddhana Salahe (Nirupama)
• Chapter 2 Kanasu Mattu Sandesha (A.P.J. Abdul Kalam)
• Chapter 3 Gandhijiya Balya (Goruru Ramaswami Ayyangar)
• Chapter 4 Sukri Bommana Gowda (Santi Nayaka)
• Chapter 5 Blood Group (Vijayamala Ranganath)
• Chapter 6 Parivartan (G. S. Basavarajasastri)
• Chapter 7 Ondu Marada Bele (Ha. Ma. Nayaka)
• Chapter 8 Asanada Mele Asana (Rasi)

Class 8 Tili Kannada Padya Bhaga Karnataka State Board Solutions

• Chapter 1 Anveshane (G. S. Sivarudrappa)
• Chapter 2 Harida Hakkigalu (R.V. Bhandari)
• Chapter 3 Jyotiye Agu Jagakella (Janapadageete)
• Chapter 4 Nanna Hageye (Su.Ram. Ekkundi)
• Chapter 5 Tugi Tugi Maragale (N. S. Lakshminarayanabhatta)
• Chapter 6 Male Barali (Savita Nagabhushana)
• Chapter 7 Hakkigalu (Siddhalingayya)
• Chapter 8 Gauravisu Jeevanava (D.V.G)

Class 8 Tili Kannada Saiddhantika Vyakarana

• Kannada Varnamale
• Gunataksharagalu – Samyuktaksharagalu
• Kannada Sandhigalu
• Namapada – Vibhakti Pratyaya Galu
• Tatsama – Tadbhava Galu
• Vachanagalu – Vakya Rachane
• Lingagalu
• Dvirukti – Jodi Nudi Nudigattugalu

Class 8 Tili Kannada Rachana Bhaga

• Patra Lekhana
• Prabandhagalu
• Gadegala Vistarane

We hope the given Tili Kannada Text Book Class 8 Solutions Answers Guide Notes Pdf free download will help you. If you have any queries regarding Karnataka State Board Syllabus 8th Standard 2nd Language Tili Kannada Textbook Answers, drop a comment below and we will get back to you at the earliest.

Students can Download Maths Chapter 12 Tribhujagala Rachane Ex 12.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 12 Tribhujagala Rachane Ex 12.4

Students can Download Maths Chapter 4 Factorisation Ex 4.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 4 Factorisation Ex 4.2

1. In the following, you are given the product pq and the sum p + q. Determine p and q.

Question i.
pq = 18, p + q = 11
Answer:
p = 9, q = 2

Question ii.
pq = 32 and p + q = – 12
Ans.
p = -8, q = – 4

Question iii.
pq = -24 and p + q = 2
Answer:
p = 6, q = – 4

Question iv.
pq = -12 and p + q = 11
Answer:
p = 12, q = – 1

Question v.
pq = – 4 and p + q = – 5
Answer:
p = – 6 and q = 1

Question vi.
pq = – 44 and p + q = – 7
Answer:
p = -11, q = 4

2. Factorise.

Question i.
x² + 6x + 8
Answer:
x² + 4x + 2x + 8
x(x + 4) + 2(x + 4)
(x + 4)(x + 2)

Question ii.
x² + 4x + 3
Answer:
x² + 3x + x + 3
x(x + 3) + 1 (x + 3)
(x + 3)(x + 1)

Question iii.
a² + 5a + 6
Answer:
a² + 3a + 2a + 6
a(a + 3) + 2(a + 3)
(a + 3)(a + 2)

Question iv.
a² – 5a + 6
Answer:
a² – 3a – 2a + 6
a(a – 3) – 2(a – 3)
(a – 3)(a – 2)

Question v.
a² – 3a – 40
Answer:
a² – 8a + 5a – 40
a(a – 8) + 5(a – 8)
(a – 8)(a + 5)

Question vi.
x² – x – 72
Answer:
x² – 9x + 8x – 72
x(x – 9) + 8(x – 9)
(x – 9)(x + 8)

3. Factorise :

Question i.
x² + 14x + 49
Answer:
x² + 14x + 49
Identity a² + 2ab + b² = (a + b)²
= x² + 2 × x × 7 + 7²
= (x + 7 )²

Question ii.
4x² + 4x + 1
Answer:
4x² + 4x + 1
Identity a² +2ab + b² =(a + b)²
= (2x)² + (2)(x)(l) +1²
= (2x +1)²

Question iii.
a²-10a+ 25
Answer:
a² – 10a + 25
Identity a² – 2ab + b² = (a – b)²
= a² – (2)(a)(5) + 5²
= (a-5)²

Question iv.
2×2 – 24x + 72
Answer:
2x² – 24x + 72
= 2(x² – 12x + 36)
Identity a² – 2ab + b² = (a – b)²
= 2(X² – (2)(x)(6) + 6²)
= 2(x-6)²

Question v.
p² – 24p + 144
Answer:
p2 – 24p + 144
Identity a² – 2ab + b² = (a – b)²
= p² – (2)(p)(12²) + 12²
= (p – 12)²

Question vi.
x3-12×2 + 36x
Answer:
x3 – 12×2 + 36x
Identify a² – 2ab + b2 =(a – b)²
=x [x² -12.x + 36]
= x[x² – (2)(x)(6) + 6²]
= x(x – 6)²

We get factors of 144 are 1, 4, 2, 3, 6, 8, 16, 9, 12, 18, 24, 36, 48, 72, 144.

Students can Download Maths Chapter 13 Statistics Ex 13.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 13 Statistics Ex 13.3

Question 1.
Runs scored by 10 batsmen in a one-day cricket match are given. Find the average run scored. 23, 54, 08, 94, 60, 18, 29, 44, 05, 86
Answer:
Σ x = 23 + 54 + 08 + 94 + 60 + 18 + 29 + 44 + 05 + 86 = 421
N = 10
Average = Mean = 

Question 2.
Find the mean weight form the following table:

Answer:

Mean = \(\frac{\Sigma f_{X}}{N}=\frac{473}{15}=31.53\)

Question 3.
Calculate the mean for the following frequency distribution.

Answer:

Mean = \(\frac{\Sigma f_{x}}{N}=\frac{1710}{40}=42.75\)

Question 4.
Calculate the mean for the following frequency distribution.

Answer:

Mean = \(\overline{X}=\frac{\Sigma f_{X}}{N}=\frac{1145}{40}=28.625\)

Question 5.
Find the median of the data 15,22, 9,20, 6,18,11,25,14.
Answer:
6, 9, 11, 14,(15), 18, 20, 22,25 (Ascending order)
N = 9,

Median = 15.

Question 6.
Find the median of the data 22,28,34, 49, 44, 57,18,10,33, 41, 66, 59.
Answer:
10, 18, 22, 28, 33, 34, 41, 44, 49, 57, 59, 66 (Ascending order)
N = 10
∴ Median = 

Question 7.
Find the median for the following frequency distribution table.

Answer:

N = 50
\(\frac{\mathrm{N}}{2}=\frac{50^{25}}{\not 2}=25\)
∴ Median class is 130 – 139
LRL = 129.5
Fc = 14
Fm = 15
i = 10

Question 8.
Find the median for the following frequency distribution table.

Answer:

N = 40
\(\frac{N}{2}=\frac{40}{2}=20\)
∴ Median class is 15 – 20
LRL = 15
Fc = 17
Fm = 9
i = 5

Question 9.
Find the mode for the following data.
(i) 4,3,1,5,3, 7, 9,6 Answer: Mode = 3
(ii) 22,36,18,22,20,34,22, 42, 46,42
Answer:
Mode = 22

Question 10.
Find the mode for the following data

Answer:
Mode = 20 (It has the highest frequency)

Students can Download Kannada Lesson 1 Buddhana Salahe Questions and Answers, Summary, Notes Pdf, Tili Kannada Text Book Class 8 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Tili Kannada Text Book Class 8 Solutions Gadya Bhaga Chapter 1 Buddhana Salahe

Buddhana Salahe Questions and Answers, Summary, Notes

Buddhana Salahe Summary in Kannada

Students can Download Kannada Poem 1 Kannadigara​ Tayi​​​​ Questions and Answers, Summary, Notes Pdf, Siri Kannada Text Book Class 8 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Siri Kannada Text Book Class 8 Solutions Padya Bhaga Chapter 1 Kannadigara​ Tayi

Kannadigara​ Tayi​​ Questions and Answers, Summary, Notes

Kannadigara​ Tayi​​ Summary in Kannada

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2

Question 1.
If 4 is added to a number and the sum is multiplied 3, the result is 30. Find the number.
Answer:
Let the number be ‘x’.
If 4 is added to it will be x + 4.
Sum is multiplied by 3 the result is 30.
∴ (x + 4)3 = 30
3x +12 = 30
3x = 30 – 12
3x = 18
x = \(\frac { 18 }{ 3 }\)
x = 6
∴The number is 6

Question 2.
Find three consecutive odd numbers whose sum is 219.
Answer:
Let the odd number be ‘x’.
The next two consecutive numbers are x + 2 and x + 4
x + (x + 2) + (x + 4) = 219
3x + 6 = 219
3x = 219 – 6
3x = 213
x = \(\frac { 213 }{ 3 }\)
x = 71
x + 2 = 71 + 2 = 73
x + 4 = 71 + 4 = 75
Three consecutive odd numbers are 71, 73, 75

Question 3.
A number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.
Answer:
Let the number be x
Number subtracted by 30 = 30 – x
14 subtracted by 3 times the number 3x – 14
∴30 – x = 3x – 14
30+ 14 = 3x + x
44 = 4x
x = \(\frac { 44 }{ 4 }\)
x =11
∴ The number is 11

Question 4.
If 5 is subtracted from three times a number the result is 16. Find the number.
Answer:
Let the number be x, 5 is subtracted from 3 times the number the result is 16.
3x – 5 = 16
3x= 16 + 5
3x = 21
x = \(\frac { 21 }{ 3 }\)
x = 7
∴ The number is 7

Question 5.
Find two numbers such that one of them exceeds the other by 9 and their sum is 81.
Answer:
Let the number is x. The other number is x + 9.
Their sum is 81
∴ x + (x + 9) = 81
2x = 81 – 9
2x = 72
x = \(\frac{72}{2}\)
x = 36
x + 9 = 36 + 9 = 45
∴ The number are 36 and 45

Question 6.
Prakruthi’s age is 6 time Sahil’s age. After 15 years prakruthi will be 3 times as old as Sahil. Find their age.
Answer:
Let Sahil’s present age be x. Prakruthi’s present age is 6x, 15 years later Sahil age will be (x + 15) years and Prakruthi age will be (6x + 15) years.
Given that the Prakruthi age will be 3 times as old as Sahil.
∴6x +15 = 3(x + 15)
6x + 15 = 3x + 45
6x – 3x = 45 – 15
3x = 30
x = \(\frac{30}{3}\) = 10
Sahils age = x = 10 years Prakruthis age = 6x = 6 x 10 = 60 years

Question 7.
Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will be twice that of his son. Find their present age.
Answer:
Ahmed’s presents age be x
Fathers present age = 3x
12 years later Ahmed’s age = x + 12
and father’s age = 3x + 12
Given 3x + 12 = 2(x + 12)
3x+ 12 = 2x + 24
3x – 2x = 24 – 12
x = 12 years
∴ Ahmed’s age = 12 years
Fathers age = 3x = 3 × 12 = 36 years.

Question 8.
Sanju is 6 years older than his brother Nishu. If the sum of their ages is 28 years what are their present ages.
Answer:
Let Nishu’s age be ‘x’ Sanju’s age = x + 6
Sum of their ages = 28 x + (x + 6) = 28
2x + 6 = 28
2x = 28 – 6
2x = 22
x = \(\frac{22}{2}\)
x = 11
Nishu’sage = x = 11 years
Sanju’ s age = x + 6 = 11 + 6 = 17 years

Question 9.
Viji is twice as old as his brother Deepu. If the difference of their ages is 11 years, find their present age.
Answer:
Let Deep’s age be ‘x’, Viji’s age is 2x
Difference of their age = 11
2x – x = 11
x = 11
∴ Deepu’s age = x = 11 years
Viji’s age = 2x = 2 x 11 = 22 years

Question 10.
Mrs. Joseph is 27 years older than her daughter Bindu. After 8 years she will be twice as old as Bindu. Find their present ages.
Answer:
Let Bindu’s present age be ‘x’ years.
Mrs. Joseph’s present age = x + 27 years.
After 8 years Bindu’s age = x + 8 and Mrs.
Josephs age = x + 27 + 8
= x + 35 years.
Given that x + 35 = 2(x + 8)
x + 35 = 2x + 16
35 – 16 = 2x – x
19 = x
x= 19
Bindu’s present age = 19 years
Mrs.Joseph’s age = x + 27 = 19 + 27 = 46 years

Question 11.
After 16 years Leena will be three times as old as she is now. Find her present age.
Answer:
Let Leena’s present age be ‘x’ years
16 years later she will be (x + 16) years
Given x + 16 = 3x
16 = 2x
x = \(\frac { 16 }{ 2 }\)
x = 8 years
Leena’ s present age = 8 years

Question 12.
A rectangle has a length which is 5 cm less than twice its breadth. If the length is decreased by 5 cm and breadth is increased by 2 cm the perimeter of the resulting rectangle will be 74 cm. Find the length and breadth of the original rectangle.
Answer:
Let the breadth of the original rectangle be ‘b’ twice the breadth is 2b.
The length of the rectangle is 5 cm less than twice the breadth.
∴ Length = 2b – 5
If the length is decreased by 5 then length
is 2b – 5 – 5 = 2b – 10
If the breadth is increased by 2cm then breadth is b + 2 cm.
Perimeter of new rectangle = 2(length + breadth)
74 = 2(2b – 10 + b + 2)
74 = 2(3b – 8)
74 + 16 = 6b 90 = 6b
b = \(\frac{90}{6}\)
b = 15cm.
Breadth of the original rectangle = 15 cm
Length of the original rectangle = 2b – 5
= 2 × 15 – 5
= 30 – 5
= 25 cm

Question 13.
The length of a rectangular field is twice its breadth. If the perimeter of the field is 288m. Find the dimensions of the field.
Answer:
Let breadth of the rectangular field bee ‘x’m
∴ Its length = 2x
Its perimeter = 288
2(1 + b) = 288
2(2x + x) = 288
2(3x) = 288
6x = 288
x = \(\frac { 288 }{ 6 }\) = 48
Its length = 2x = 2 × 48 = 96m
breadth = x = 48m

Question 14.
Srishti’s salary is the same as 4 times Azar’s salary. If together they earn Rs 3750 a month find their individual salaries.
Answer:
Let Axar’s salary be x. Sristi’s salary is 4x together they earn Rs. 3750
∴ x + 4x = 3750
5x = 3750
x = \(\frac{3750}{5}\)
x = 750
4x = 4 × 750 = 3000
∴ Azar’s salary is Rs.750 and Sristi’s salary is Rs.3000

Students can Download Maths Chapter 4 Factorisation Ex 4.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 4 Factorisation Ex 4.1

1. Resolve into factors.

Question i.
x² + xy
Answer:
x² + xy = x(x + y)

Question ii.
3x² – 6x
Answer:
3x² – 6x = 3x (x – 2)

Question iii.
(1.6)a² – (0.8)a
Answer:
(1.6)² – (0.8)a
= (0.8 x 2a²) – (0.8)a
= 0. 8a(2a- 1)

Question iv.
5 – 10m – 20n
Answer:
5 – 10m -20n = 5(1 – 2m – 4n)

2. Froctorise:

Question i.
a² + ax + ab + bx
Answer:
a² + ax + ab + bx a(a + x) + b(a + x)
(a + x)(a + b)

Question ii.
3ac + 7bc – 3ad – 7bd
Answer:
3ac + 7bc – 3ad – 7bd
c(3a + 7b – d(3a + 7b) (3a + 7b) (c-d)

Question iii.
3xy – 6zy – 3xt + 6zt
Answer:
3y (x – 2z) – 3t(x – 2z)
(x – 2z) (3y – 3t)

Question iv.
y3 + 3y² + 2y – 6 – xy + 3x
Answer:
y² (y – 3) + 2(y – 3) – x(y – 3)
(y- 3) (y² + 2 – x)

3. Factorise:

Question i.
4a² – 25
Answer:
4a² – 25
= (2a)² – 52
[a² – b²=(a+b)(a-b)]
= (2a + 5)(2a – 5)

Question ii.
\(x^{2}-\frac{9}{16}\)
Ans.

Question iii.
x⁴ – y4
Answer:
x⁴ – y⁴
= (x²)² – (y²)²
Identity a² – b² = (a + b)(a – b)
= (x² + y²)(x² – y²)
= (x²+ y²)(x + y)(x – y)

Question iv.
\(\left(7 \frac{3}{10}\right)^{2}-\left(2 \frac{1}{10}\right)^{2}\)
Ans.

Question v.
(0.7)² – (0.3)²
Answer:
(0.7)² – (0.3)²
Identity a² – b² = (a + b)(a – b)
= (0.7 + 0.3)(0.7 – 0.3)
= (1.0)(0.4)
= 0.4

Question vi.
(5a-2b)²-(2a-b)²
Answer:
(5a-2b)² – (2a-b)²
Identity a² – b² = (a + b)(a – b)
= [5a – 2b + 2a – b]
= [5a – 2b – (2a – b)]
= (7a – 3b)[5a – 2b – 2a + b]
= (7a – 3b)(3a – b)