KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex 4.2

   

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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.

KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex 13.3

   

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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 = \overline{X}=\frac{\Sigma x}{N}=\frac{421}{10}=42.1

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

Weight (kg) 29 30 31 32 33
No. of children 02 01 04 03 05

Answer:

Weigh t(kg) No. of children(x) ‘ f
29 02 58
30 01 30
31 04. 124
32 03 96
33 05 164
N= 15 Zf = 473

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

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

Mark 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 3 7 10 6 8 2 4

Answer:

Marks Frequency Midpoint fx
10-20 3 15 45
20-30 7 25 175
30-40 10 35 350
40-50 6 45 270
50-60 8 55 440
60-70 2 65 130
70-80 4 75 300
N = 40 Efx = 1710

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

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

Mark 15-19 20-24 25-29 30-34 35-39 40-44
Frequency 6 5 9 12 6 2

Answer:

Marks. Frequency Midpoint f(X)
15-19 6 17 102
20-24 5 22 *110
25-29 9 27 243
30-34 12 32 384
35-39 6 37 222
40-44 2 42 84
N = 40 Efx = 1145

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,
\frac{N+1}{2}+\frac{9+1}{2}=\frac{10}{2}=5^{\text { th }}
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 = \frac{34+41}{2}=\frac{75}{2}=37.5

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

Class interval 110-119 120-129 130-139 140-149 150-159 160-169
Frequency 6 8 15 10 6 5

Answer:

Class interval Frequency (f) Cumulative frequency (fc)
110-119 6 6
120-129 8 14
130-139 15 29
140-149 10 45
160-169 5 50
 N = 50

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
KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex. 13.3 1

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

Class interval 0-5 5-10 10-15 15-20 20-25 25-30
Frequency 5 3 9 10 8 5

Answer:

Class interval Frequency (t) Cumulative frequency (fc)
0-5 5 5
5-10 3 8
10-15 9 , 17
15-20 10 27
20-25 8 35
25-30 5 40
N = 40

N = 40
\(\frac{N}{2}=\frac{40}{2}=20\)
∴ Median class is 15 – 20
LRL = 15
Fc = 17
Fm = 9
i = 5
KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex. 13.3 2

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

X 5 10 12 15 20 30 40
f 4 8 11 13 16 12 9

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

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

   

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

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

   

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

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KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2

   

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

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex 4.1

   

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.
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex. 4.1 1

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.
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex. 4.1 2
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex. 4.1 3

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)

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.1

   

Students can Download Maths Chapter 9 Commercial Arithmetic Ex 9.1 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 9 Commercial Arithmetic Ex 9.1

Question 1.
In a school 30% students play chess 60% play carrom and the rest play other games. If the total number of students in the school is 900 find the exact number of students who play each game.
Answer:
30% Students play chess. Total number of students is 900.
The number of players who play chess = 30% of 900.
\frac { 300 }{ 100 } × 900 = 270
The number of players who play carrom board = 60% of 900
\frac { 60}{ 100 } × 900 = 540
The number of players who play other games = 10% of 900
\frac { 10 }{ 100 } × 900 = 90

Question 2.
In a school function Rs. 360 remained after spending 82% of the money. How much money was there in the beginning? verify your answer.
Answer:
Let the money, in the beginning, be ₹ 100 The money spent is ₹ 82.
∴ The amount remaining is 100 – 82 = ₹ 18.
If the remaining money is ₹ 18 the money, in the beginning, is ₹ 100.
If the remaining money is ₹ 360 the money in the beginning is
\frac { 100 }{ 18 } × 360 = 2000
Verification :
\frac { 82}{ 100 } × 2000 = 1640
Money remained = 2000 – 1640 = ₹ 360.

Question 3.
Akshay’s income is 20% less than that of Ajay what percent is Ajay’s income more than that of Akshay?
Answer:
Let Ajay’s income be Rs. 100.
Then Akshay’s income is (100 – 20) = Rs. 80.
If Akshay’s income is Rs. 80.
Ajay’s income is Rs. 100.
If Akshay’s income is Rs. 1 Ajay’s income is
\(\frac{100}{80}\) × 1 = Rs. \(\frac { 100 }{ 80 }\)
Changing the scale to 100.
Ajay’s income is Rs. 100
Akshay’s income is
= \(\frac { 100 }{ 80 }\) × 100 = 125
∴ Ajay’s income is 125 – 100 = 25% more than that of Akshay.

Question 4.
A daily wage employee spends 84% of his weekly earning. If he saves Rs. 384 find his weekly earning.
Answer:
Let his weekly earning be ₹ 100.
His expenditure is ₹ 84.
Therefore his savings is (100 – 84) = ₹ 16
If the savings is ₹ 16 then the income is ₹ 100.
If the savings is ₹ 384 then the income is
\frac { 100 }{ 16 } × 384 = ₹  2400
∴ His weekly earning is ₹ 2400.

Question 5.
A factory announces a bonus of 10% its employees. If an employee gets Rs. 10,780 find his actual salary.
Answer:
Let the salary of the employee be ₹ 100.
He gets a bonus of ₹ 10.
He gets 100 + 10 = ₹ 110 including bonus.
∴ If the employee gets ₹ 10.780 his salary is
\frac { 100 }{ 110 } × 10780 = 9800
∴ The salary of the employee is ₹ 9800

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex 1.3

   

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Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.3

Question 1.
Find the quotient and remainder when each of the following number is divided by 13 : 8, 31, 44, 85, 1220.
Answer:
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex. 1.3 1

Question 2.
Find the quotient and the remainder when each of the following numbers are divided by 304.
128, 636, 785, 1038, 2236, 8858
Answer:
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex. 1.3 2
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex. 1.3 3

Question 3.
Find the least natural number larger than 100 which leaves the remainder 12 when divided by 19.
Answer:
The least number greater than 100 divisible by 19 is 114.
Adding 12 to this number we get 114 + 12 = 126.
126 leaves 12 as the remainder when divided by 19.
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex. 1.3 4

Question 4.
What is the least natural number you have to add to 1024 to get a multiple of 181?
Answer:
The Multiples of 181 are 181, 362, 543, 724, 905, 1086.
The nearest multiple of 181 to 1024 is 1.086.
Hence 62 (1086 – 1024) must be added to 1024 to get a multiple of 181.

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Ex 7.2

   

Students can Download Maths Chapter 7 Rational Numbers Ex 7.2 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 7 Rational Numbers Ex 7.2

Question 1.
Write down ten rational numbers which are equivalent to \(\frac { 5 }{ 7 }\) and the denominator not exceeding 80.
Answer:
Multiply both numerator and denominator by 2, 3, 4………
\(\frac{10}{14}, \frac{15}{21}, \frac{20}{28}, \frac{35}{35}, \frac{30}{42}, \frac{35}{49}, \frac{40}{56}, \frac{45}{63}, \frac{50}{70}, \frac{55}{77}\)

Question 2.
Write down 15 rational numbers which are equivalent to \(\frac { 11 }{ 5 }\) and the numerator not exceeding 180.
Answer:
\(\begin{array}{l}{\frac{22}{10}, \frac{33}{15}, \frac{44}{20}, \frac{55}{25}, \frac{66}{30}, \frac{77}{35}, \frac{88}{40}, \frac{99}{45}} \\ {\frac{110}{50}, \frac{121}{55}, \frac{132}{60}, \frac{143}{65}, \frac{154}{70}, \frac{165}{75}, \frac{176}{80}}\end{array}\)

Question 3.
Write down 10 positive rational numbers such that the sum of the numerator and the denominator of each is 11. Write them in decreasing order.
Answer:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Ex 7.2 1

Question 4.
Write down ten positive rational numbers such that numerator – denominator for each of them is -2. Write to them in increasing order.
Answer:
Numerator – denominator = – 2
therefore the denominator is greater than the numerator by 2.
\(\frac{1}{3}, \frac{2}{4}, \frac{3}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{9}, \frac{7}{9}, \frac{8}{10}, \frac{9}{11}, \frac{10}{12}\)

Question 5.
Is \(\frac { 3 }{ -2 }\) a rational number? If so, how do you write it in the form conforming to the definition of a rational number (that is, the denominator as positive integer)?
Answer:
\(\frac{3}{-2}\) is a rational number because the denominator is negative.
It can be written as \(\frac{3}{-2}\) since \(\frac{3}{-2}\) is same as \(\frac{3}{-2}\)

Question 6.
Earlier you have studied decimals 0.9, 0.8, can you’ write these as rational numbers?
Answer:
\(0.9=\frac{9}{10} \text { and } 0.8=\frac{8}{10}=\frac{4}{5}\)

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Ex 7.1

   

Students can Download Maths Chapter 7 Rational Numbers Ex 7.1 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 7 Rational Numbers Ex 7.1

1. Identify the property in the following statements.

Question (i)
2 + (3 + 4) = (2 + 3) + 4
Answer:
Associative property of addition.

Question (ii)
2 × 8 = 8 × 2
Answer:
Commutative property of multiplication

Question (iii)
8 × (6 + 5) = (8 × 6) + (8 × 5)
Answer:
Distributive property of multiplication over addition in integers.

Question 2.
Find the additive inverse of the following integers.
Answer:
Integer                    Additive inverse
6                                      -6
9                                      -9
123                                -123
-76                                   76
-85                                   85
1000                             -1000

3. Find the integer m in the following:

Question (i)
m + 6 = 8
Answer:
m = 8 – 6

Question (ii)
m + 25 = 15
Answer:
m =15 – 25
m = -10

Question (iii)
m – 40 = -26
Answer:
m = – 26 + 40
m = 14

Question (iv)
m + 28 = – 49
Answer:
m = – 49 – 28
m = – 77

Question 4.
Write the following in increasing order: 21, -8, -26, 85, 33, -333, -210, 0, 2011
Answer:
-333 < -210 < -26 < -8 < 0 < 21 < 33 < 85 < 2011.
-333, -210, -26, -8, 0, 21, 33, 85, 2011

Question 5.
Write the following in decreasing order: 85, 210, -58, 2011, -1024, 528, 364, -10000,12
Answer:
2011 > 528 > 364 > 210 > 85 > 12 > – 58 > -1024 < -10,000
2011, 528, 364, 210, 85, 12, -58, -1024, -10,000.

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