KSEEB Solutions

Expert Teachers at KSEEBSolutions.com has created KSEEB Solutions for Class 3 all subjects Pdf Free Download in English Medium and Kannada Medium of 3rd Standard Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material, are part of KSEEB Solutions. Here we have given KTBS Karnataka State Board Syllabus for Class 3 Textbook Solutions.

Karnataka State Board Syllabus for Class 3 Textbooks Solutions

• KSEEB Solutions for Class 3 Maths
• KSEEB Solutions for Class 3 EVS
• KSEEB Solutions for Class 3 English
• Savi Kannada Text Book Class 3 Solutions

We hope the given KSEEB Solutions for Class 3 all subjects Pdf Free Download in English Medium and Kannada Medium of 3rd Std Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material will help you. If you have any queries regarding KTBS Karnataka State Board Syllabus for Class 3 Textbooks Solutions, drop a comment below and we will get back to you at the earliest.

Expert Teachers at KSEEBSolutions.com has created KSEEB Solutions for Class 4 all subjects Pdf Free Download in English Medium and Kannada Medium of 4th Standard Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material, are part of KSEEB Solutions. Here we have given KTBS Karnataka State Board Syllabus for Class 4 Textbook Solutions.

Karnataka State Board Syllabus for Class 4 Textbooks Solutions

• KSEEB Solutions for Class 4 Maths
• KSEEB Solutions for Class 4 EVS
• KSEEB Solutions for Class 4 English
• Savi Kannada Text Book Class 4 Solutions

We hope the given KSEEB Solutions for Class 4 all subjects Pdf Free Download in English Medium and Kannada Medium of 4th Std Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material will help you. If you have any queries regarding KTBS Karnataka State Board Syllabus for Class 4 Textbooks Solutions, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.
iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
Solution:
i) Rent for            Rent for            Rent for
1st km                  2nd km             3rd km
Rs. 15                     Rs. 8                 Rs. 8
d = a₂ – a₁ = 8 – 15 = -7
d = a₃ – a₂ = 8 – 8 = 0
Here ’d’ is not constant.
Hence it does not form an AP.

ii) Let the amount of air present in a cylinder be ‘x’.

1. \(\frac{1}{4} \text { of } 1=\frac{1}{4}\) Amount of air goes out
   \(1-\frac{1}{4}=\frac{3}{4}\) Air remaining
2. \(\frac{1}{4} \text { of } \frac{3}{4}=\frac{3}{16}\) Amount of air goes out
   \(\frac{3}{4}-\frac{3}{16}=\frac{9}{16}\) Air remaining
3. \(\frac{1}{4} \text { of } \frac{9}{16}=\frac{9}{64}\) Amount of air goes out
   \(\frac{9}{16}-\frac{9}{64}=\frac{27}{64}\) Air remaining

Set \(\frac{3}{4}, \frac{9}{16}, \frac{27}{64}, \ldots .\)
\(\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{9}{16}-\frac{3}{4} \quad \mathrm{d}=\frac{-3}{16}\)
\(\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\frac{27}{64}-\frac{9}{16} \quad \mathrm{d}=\frac{-9}{64}\)
Here the value of ‘d’ is not constant.
∴ The given list of numbers does not form an A.P.

(iii) Cost of Digging for the

d = a₂ – a₁ = 200 – 150 = 50
d = a₃ – a₂ = 250 – 200 = 50
Here value of ‘d’ is constant.
This forms an AP.

iv) Amount of money deposited in the
d = a₂ – a₁ = 11,664 – 10.800 = Rs. 864
d = a₃ – a₂ = 12,597,12 – 11.664 = Rs. 933.12
Here value of ‘d’ is not constant.
∴ The given list of numbers does not form an AP.

Question 2.
Write first four terms of the AP, when the first term ‘a’ and the common difference’d’ are given as follows:
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = -3
iv) a = -1, d = \(\frac{1}{2}\)
v) a = -1.25, d = 0.25
Solution:
i) a = 10, d = 10
First four terms are,
a, a + d, a + 2d, a + 3d
10, 10 + 10, 10 + 20, 10 + 30
10. 20, 30, 40.

ii) a = -2, d = 0
First four terms are,
a, a + d, a + 2d, a + 3d
-2. -2 + 0, -2 + 0, -2 + 0
-2. -2. -2, -2

iii) a = 4, d = -3
First four terms are,
a, a + d, a + 2d, a + 3d
4, 4 – 3, 4 – 2 × 3, 4 – 3 ×3
4, 1, -2, -5

iv) a = -1, d = \(\frac{1}{2}\)
First four terms are,
a, a + d, a + 2d, a + 3d
\(-1,-1+\frac{1}{2},-1+2 \times \frac{1}{2},-1+3 \times \frac{1}{2}\)
\(-1,-\frac{1}{2}, 0, \frac{1}{2}\)

v) a = -1.25, d = 0.25
First four terms are,
a, a + d, a + 2d, a + 3d
-1.25, -1.25 + 0.25, -1.25 + 2 × 0.25, -1.25 + 3 ×0.25
-1.25, -1, -0.75, -0.50

Question 3.
For the following APs, write the first term and the common difference :
(i) 3, 1,-1,-3, ………..
(ii) -5, -1, , 7, …………
(iii) \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \ldots \ldots\)
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
(i) 3, 1,-1,-3
First term, a = 3,
Common difference, d = a₂ – a₁ = 1 – 3
d = -2.

(ii) -5, -1, 3, 7, ……….
First term, a = -5,
Common difference, d = a₂ – a₁ = -1 – (-5)
d = -1 + 5
d = 4.

(iii) \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \ldots \ldots\)
First term, \(\mathrm{a}=\frac{1}{2}\)
Common difference, d = a₂ – a₁
\(=\frac{1}{2}-\frac{1}{2}\)
d= 0

iv) 0.6, 1.7, 2.8, 3.9
First term, a = 0.6,
Common difference, d = a₂ – a₁ = 1.7 – 0.6
d= 1.1

Question 4.
Which of the following are APs? If they form an AP, find the common difference ’d’ and write three more terms.
i) 2, 4, 8, 16, …….
ii) \(2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \ldots\)
iii) -1.2, -3.2, -5.2, -7.2
iv) -10, -6, -2, 2, …….
v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2} \ldots \ldots\)
vi) 0.2, 0.22, 0.222, 0.2222, ………
vii) 0, -4, -8, -12, ………
viii) \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \ldots\)
ix) 1, 3, 9, 27, ……..
x) a, 2a, 3a, 4a, ………..
xi) a, a², a³, a⁴, ………..
xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \ldots\)
xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}\)
xiv) 1¹, 3², 5², 7²,
xv) 1¹, 5², 7², 73, ……….
Solution:
i) 2, 4, 8, 16…………
d = a₂ – a₁= 4 – 2 = 2
d = a₂ – a₁ = 8 – 4 = 4
Here ‘d’ is not constant.
∴ the given List of numbers does not form an AP.

ii) \(2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \ldots\)
\(\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{5}{2}-\frac{2}{1}=\frac{5-4}{2}=\frac{1}{2}\)
\(\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\frac{3}{1}-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\)
Here ‘d’ is constant.
∴ the given List of numbers form an AP.
Next three terms are: \(4,\frac{9}{2}, 5\)

iii) -1.2, -3.2. -5.2, -7.2
d = a₂ – a₁ = -3.2 – (-1.2) = -3.2+ 1.2
d = -2
d= a₃ – a₂ = -5.2 – (-3.2) = -5.2 + 3.2
d = -2
Here d is constant.
∴ This is an Arithmetic Progression.
Next three terms are:
-7.2 – 2 = -9.2
-9.2 – 2 = -11.2
-11.2 – 2 = -13.2
-11.2-2 = -13.2

iv) -10. -6. -2, 2,..,…
d = a₂ – a₁ = -6 – (- 10) = -6 + 10
d = 4
d = a₃ – a₂= -2 – (-6) = -2 + 6
d = 4
Here d is constant.
∴ Given set of numbers form an A.P.
Further three terms are
2 + 4 = 6
4 + 4 = 8
8 + 4 = 12

v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2} \ldots \ldots\)
d = a₂ – a₁ = \(=3+\sqrt{2}-3\)
\(\mathrm{d}=\sqrt{2}\)
d = a₃ – a₂ = \(=3+2 \sqrt{2}-(3+\sqrt{2})\)
\(=3+2 \sqrt{2}-3+\sqrt{2}\)
\(\mathrm{d}=\sqrt{2}\)
Here ‘d’ is constant
∴ Given set of numbers forms an A.P.
Further three terms are:
\(3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}\)
\(3+4 \sqrt{2}+\sqrt{2}=3+5 \sqrt{2}\)
\(3+5 \sqrt{2}+\sqrt{2}=3+6 \sqrt{2}\)

vi) 0.2. 0.22. 0.222. 0.2222
d = a₂ – a₁ = 0.22 – 0.2
d = 0.02
d = a₃ – a₂ = 0.222 – 0.22
d = 0.002
Here d is not constant,
∴ Given set of numbers do not form an A.P.

vii) 0,-4, -8, -12, …………
a₁ = 0, a₂ = -4
d = a₂ – a₁ = -4 – 0
d = -4
d = a₃ – a₂ = -8 – (-4) = -8 + 4
d = -4
Here d is constant.
∴ Given set of numbers form an AP.
Further three terms are:
-12 – 4 = -16
-16 – 4 = -20
-20 – 4 = -24
∴ Next three terms are: -16. -20, -24.

viii) \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \ldots\)
\(a_{1}=-\frac{1}{2}, a_{2}=-\frac{1}{2}\)
\(\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=-\frac{1}{2}-\left(-\frac{1}{2}\right)\)
\(d=-\frac{1}{2}+\frac{1}{2}=0\)
\(\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=-\frac{1}{2}-\left(-\frac{1}{2}\right)\)
\(\mathrm{d}=-\frac{1}{2}+\frac{1}{2}=0\)
Here d is constant.
∴ Given set of numbers form an AP.
Next three terms are:
\(-\frac{1}{2}+0=-\frac{1}{2}\)
\(-\frac{1}{2}+0=-\frac{1}{2}\)
\(-\frac{1}{2}+0=-\frac{1}{2}\)
∴ Next three terms are: \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)

ix) 1, 3, 9, 27
d = a₂ – a₁ = 3—1
d = 2
d = a₃ – a₂
d = 6
Here ‘d’ not is constant
∴ Given set of numbers do not form an A.P.

x) a, 2a, 3a,4a
a₁= a a₂ = 2a
d = a₂ – a₁ = 2a-a
d = a
d = a₃ – a₂ = 3a-2a
d = a
Here d’ is constant.
∴ Given set of numbers form an A.P.
Further three terms are :
4a + a = 5a
5a + a = 6a
6a + a = 7a
∴ Next three terms are: 5a, 6a, 7a.

xi) a, a², a³, a⁴
a₁ = a. a₂ = a²
d = a₂ – a₁ = a² – a
d= a(a – 1)
d = a₃ – a₂ = a³ – a²
d = a² (a – 1)
Here ‘d is not constant.
∴ Given set of numbers do not form an A.P.

xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \ldots\)
\(a_{1}=\sqrt{2}, a_{2}=\sqrt{8}\)
\(\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\sqrt{8}-\sqrt{2}\)
\(\mathrm{d}=2 \sqrt{2}-\sqrt{2}=\sqrt{2}\)
\(\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\sqrt{18}-\sqrt{8}\)
\(\mathrm{d}=3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}\)
Here d’ is constant.
∴ Given set of numbers form an A.P.
Succeding three terms are :
\(\sqrt{32}+\sqrt{2}=4 \sqrt{2}+\sqrt{2}=5 \sqrt{2}\)
\(5 \sqrt{2}+\sqrt{2}=6 \sqrt{2}\)
\(6 \sqrt{2}+\sqrt{2}=7 \sqrt{2}\)
\(5 \sqrt{2}, 6 \sqrt{2}, 7 \sqrt{2}\)
∴ Next three terms are: \(\sqrt{50}, \sqrt{72}, \sqrt{98}\)

xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}\)
\(a_{1}=\sqrt{3}, a_{2}=\sqrt{6}\)
\(\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\sqrt{6}-\sqrt{3}\)
\(\mathrm{d}=\mathrm{a}_{3}-\mathrm{a}_{2}=\sqrt{9}-\sqrt{6}\)
\(\mathrm{d}=3-\sqrt{6}\)
Here ‘d is not constant.
∴ Given set of numbers do not form an A.P.

xiv) 1¹,3²,5², 7², …………
1, 9, 25, 49, ………….
a₁ = 1, a₂ = 9
d = a₂ – a₁ = 9 – 1
d = 8
d = a₃ – a₂ = 25 – 9
d = 16
Here ‘d’ is not constant,
∴ Given set of numbers do not form an A.P.

xv) 1¹, 5², 7², 73, ……….
1, 25. 49, 73, ……….
a₁ = 1. a₂ = 25, a₃ = 49, a₄ = 73
d = a₂ – a₁ = 25 – 1
d = 24
d = a₃ – a₂ = 49 – 25
d = 24
d = a₄ – a₃ = 73 – 49
d = 24
Here ‘d’ is constant.
∴ Given set of numbers form an A.P.
Succeeding three terms are:
73 + 24 = 97
97 + 24 = 121
121 + 24 = 145
∴ Next three terms are: 97, 121, 145.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
(i) 2x² – 3x + 5 = 0
(ii) 3x² – \(4 \sqrt{3} x\) + 4 = 0
(iii) 2x² – 6x + 3 = 0
Solution:
i) 2x² – 3x + 5 = 0
It is in the form of ax² + bx + c = 0
a = 2, b = – 3 and c = 5
Nature of roots = b² – 4ac
Discriminant = (- 3)² – 4 × (2)(5)
= 9 – 40
= – 31
Discriminant = – 31 < 0
The given Quadratic equation has no real roots.

(ii) 3x² – \(4 \sqrt{3} x\) + 4 = 0
Here, a = 3, b = \(4 \sqrt{3}\), c = 4

(iii) 2x² – 6x + 3 = 0
Here, a = 2, b = -6, c = 3
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24 = 12
∴ b² – 4ac = 12 > 0.
∴ It has two distinct roots.
2x² – 6x + 3 = 0
Here, a = 2, b = -3, c = 3

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots
(i) 2x² + kx+3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) Comparing the given quadratic equation with ax2+ bx + c = 0, we get a = 2,b = k,c = 3
∴ b² – 4ac = (k)² – 4(2)(3) = k² – 24
∵ For a quadratic equation to have equal roots, b² – 4ac = 0

(ii) kx(x – 2) + 6 = 0 ⇒ kx² – 2kx + 6 = 0
Comparing kx² – 2kx + 6 = 0 with ax² + bx + c = 0, we get
a = k, b = -2k, c = 6
∴  b² – 4ac = (-2k)² – 4(k)(6) = 4k² – 24k
Since, the roots are equal

But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth.

Solution:
Let the breadth of rectangular mango grove is ‘x’ and length of rectangular mango grove is 2x
Area of rectangular mango grove = 800 m²
length × breadth = 800
x (2x)= 800
2x² = 800
x² = \(\frac{800}{2}\) = 400
x = ± \(\sqrt{400}\)
x = ± 20m
∴ breadth of mango grove x = 20 m
length of mango grove = 2x = 2 × 20 = 40 m
It is possible to construct a rectangular mango grove of length 40m and breadth 20m.

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the age of one friend = x years
∴ Age of other friend = (20 – x) years
Four years ago,
Age of one friend = (x – 4) years
Age of other friend = (20 – x – 4) years
= (16 -x) years
According to the condition,
(x – 4) × (16 – x) = 48
⇒ 16x – 64 – x² + 4x = 48
⇒ – x² + 20x – 64 – 48 = 0
⇒ -x² + 20x – 112 = 0
⇒ x² – 20x + 112 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get a = 1, b = – 20 and c = 112

Since, b² – 4ac is less than 0.
∴ The quadratic equation (1) has no real roots.
Thus, the given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m. and area 400 m²? If so, find its length and breadth.
Solution:

Let the length of rectangular park is ‘x’
Perimeter of the rectangular park = 80m
2 (length + breadth) = 80
length + breadth = \(\frac{80}{2}\) = 40
breadth = (40 – x)
area of rectangle = l × b
x (40 – x) = 400
40x – x² = 400
x² – 40x + 400 = 0
x² – 20x – 20x + 400 = 0
x (x – 20) – 20 (x – 20) = 0
(x – 20) (x – 20) = 0
x – 20 (or) x – 20 = 0
x = 20 m
length of rectangular park = 20 m and breadth of rectangular park = (40 – x)
= 40 – 20 = 20m
∴ It is possible to design rectangular park of length 20 m and breadth is 20 m.
∴ The park is a square having 20m side.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2

Question 1.
Express each number as a product of its prime factors :
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
a = 26, b = 91

26 = 2 × 13
91 = 7 × 13
LCM of (26, 91) = 2 × 13 × 7 = 182
HCF of (26, 91) = 13
Verification
LCM of (26, 91) × H C F (26, 91) = 182 × 13 = 2366
product of two numbers = 26 × 91 = 2366
Hence LCM × HCF = Product of two numbers.

(ii) 510 and 92
a = 510, b = 92

(iii) 336 and 54
a = 336, b = 54

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) 12, 15 and 21
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
∴ H.C.F. = 3
L.C.M. = 3 × 2 × 2 × 5 × 7 = 420
∴ HCF × LCM = Product of three numbers
3 × 420 = 12 × 15 × 21
1260 = 1260

ii) 17, 23 and 29
17, 23 and 29 are prime numbers
LCM of (17, 23, 29) = 17 × 23 × 29
= 391 × 29
= 11339
HCF of (17, 23, 29) = 1
(∵ 17, 23 and 29 have no common Factor)

(iii) 8, 9 and 25
8 = 2 × 2 × 2= 23
9 = 3 × 3 = 32
24 = 5 × 5 = 52
∴ H.C.F. = 1
L.C.M. = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
∴ HCF × LCM = Product of three numbers
1 × 1800= 8 × 9 × 25
1800 = 1800

Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
H.C.F × L.C.M = a × b
(a, b) × (a, b) = a × b
9 × x = 306 × 657
x = \(\frac{306 \times 657}{9}\)
∴ x = 22338
∴ L.C.M of (306, 657) = 22338

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number ‘n’.
Solution:
If the number ends with digit 0, then the prime factorization of the number must contain the factors of 10 as 2 × 5.
Now the prime factorization of 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
Since it does not contain the factors of 10 as 2 × 5, thus ⁿ cannot end with the digit 0, for any natural number n.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
i) 7 × 11 × 13 + 13 = 13(77 + 1)
= 13 × 78
∴ It is a composite number
[It is having more than two factors]

ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
=5[7 × 6 × 4 × 3 × 2 × 1 + 1]
= 5 (1008 + 1)
= 5 × 1009 = 5045
∴ It is a composite number.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting
Solution:
Sonia takes 18 minutes to drive one round of the field.
Ravi takes 12 minutes for the same.
Suppose they both start at the same point and at the same time.
Time to meet both again at the starting point,
For this, we have to find out HCF of 18 and 12.
18 = 2 × 3 × 3 = 2 × 3²
12 = 2 × 2 × 3 = 2² × 3
∴ H.C.F. = 2 × 3 = 6
∴ After 6 minutes, they both meet at the same point.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2

Question 1.
Fill in the blanks In the following table,
given that ‘a’ is the first term, ‘d’ is the common difference and a_n the n^th term of the A.P.

Solution:
(i) a = 7, d = 3, n = 8, a_n =?
a_n = a + (n – 1) d
a₈= 7 + (8 – 1) 3
= 7 + 7 × 3
= 7 + 21
∴ a₈ = 28

(ii) a = -18, d =?, n = 10, a_n = 0
a_n = a + (n – 1) d
0 = -18 + (10 – 1) d
0 = -18 + 9d
18 = 9d
9d = 18
\(\mathrm{d}=\frac{18}{9}=2\)

(iii) a =?, d = -3, n = 18, a_n = -5
a_n = a + (n – 1) d
-5 = a + (18 – 1) (-3)
= a + 17(-3)
-6 = 1 – 51
∴ a = -5 + 51 = 46

(iv) a = -18.9, d = 2.5, n =? a_n = 3.6
a_n = a + (n – 1) d
3.6= -18.9 + (n – 1) (2.5)
3.6= -18.9 + 2.5n – 2.5
3.6 = 2.5n – 21.4
2.5n = 3.6 + 21.4
2.5n = 25
\(n=\frac{25}{2.5}=\frac{250}{25}\)
∴ n = 10.

(v) a = 3.5, d = 0, n = 105, a_n =?
a_n = a + (n – 1) d
= 3.5 + (105 – 1) (0)
= 3.5+ 104 × 0
= 3.5 +0
∴ a_n = 3.5

Question 2.
Choose the correct choice In the following and justify:
(i) 30^th term of the AP: 10, 7, 4, ……….. is
A) 97
B) 77
C) -77
D) -87
Solution:
a = 10. d = 7 – 10 = -3, n = 30, a₃₀ =?
a_n = a + (n – 1) d
a₃₀ = 10 + (30 – 1)(-3)
= 10 + 29(-3)
= 10 – 87
∴ a₃₀ = 77
∴ Ans: (C) -77

(ii) 11^th term of the A.P.
\(-3,-\frac{1}{2}, 2, \dots \ldots,\) is
A) 28
B) 22
C) -38
D) \(-48 \frac{1}{2}\)
Solution:
\(a=-3, \quad d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=2 \frac{1}{2}\)
n = 11, a₁₁ =?
a_n = a + (n – 1) d
\(a_{11}=-3+(11-1)\left(2 \frac{1}{2}\right)\)
\(=-3+10\left(\frac{5}{2}\right)\)
= -3 +25
∴ a₁₁ = 22
∴ Ans: (B) 22

Question 3.
In the following APs find the missing terms in the boxes:

Solution:
(i)
a = 2, a + d =?, a + 2d = 26
a + 26 = 26
2 + 2d = 26
2d = 24
∴ d=12 . .
∴ a + d = 2 + 12 = 14
∴ Ans: 14

(ii) Here, a =?, a + d = 13, a + 2d=?,
a + 3d = 3
a + d + 2d = 3
13 + 2d = 3
2d = 3 – 13
2d = -10
∴ d = -5
a + d = 13
a + (-5) = 13
a – 5 = 13
∴ a – 13 + 5 = 18
∴ a = 18
a + 2d =?
= 18 + 2(-5)
= 18 – 10
a + 2d = 8
∴ Ans: 18, 8

(iii) a = 5, a + d =?, a + 2d =?

(iv) a = -4, a +d=? a + 2d =?
a+3d = ? a + 4d =? a + 5d = 6
a + 5d = 6
-4 + 5d = 6
5d = 6 + 4
5d = 10
\(\mathrm{d}=\frac{10}{5}\)
∴ d=2.
a + d = -4 + 2 = -2
a + 2d = -4 + 2 (2) = -4 + 4 = 0
a + 3d = -4 + 3 (2) = -4 + 6 = 2
a + 4d = -4 + 4 (2) = -4 + 8 = 4
∴ Ans: -2, 0, 2, 4

(v) a =? a + d = 38, a + 2d =?
a + 3d =?, a + 4d =?a + 5d = 22

4d = -60
\(\mathrm{d}=-\frac{-60}{4} \quad=-15\)
a + d = 38
a – 15=38
a = 38 + 15 = 53
a + 2d = 53 + 2(-15) = 53 – 30 = 23
a + 3d = 53 + 3(-15) = 53 – 45 = 8
a + 4d =53 + 4(-15) = 53 – 60 = 7
∴ 53, 23, 8, -7.

Question 4.
Which term of the AP : 3, 8, 13. 18, ………. is 78?
Solution:
a = 3, d = 8 – 3 = 5, a_n = 78, n =?
a_n = a + (n – 1) d
78 = 3+(n—1)(5)
78 = 3 + 5n – 5
78 = 5n – 2
5n = 78 + 2
\(n=\frac{80}{5}\)
∴ n = 16

Question 5.
Find the number of terms In each of the following APs:
(i) 7, 13, 19, ………… 201
(ii) \(18,15 \frac{1}{2}, 13, \dots \dots 47\)
Solution:
(i) 7, 13, 19, ………. 201
a = 3, d = 13 – 7 = 6, a_n = 201. n =?
a + (n – 1) d = a_n
3 + (n – 1) 6 = 201
3 + 6n – 6 = 201
6n – 3 = 201
6n = 203
\(n=\frac{204}{6}\)
∴ n = 34

(ii) \(18,15 \frac{1}{2}, 13, \dots \dots 47\)
\(a=18, d=a_{2}-a_{1}=\frac{31-36}{2}=\frac{-5}{2}\)
a_n = -47, n =?
a_n = a + (n – 1) d
\(-47=18+(n-1)\left(\frac{-5}{2}\right)\)
\(-47-18=(n-1)\left(\frac{-5}{2}\right)\)
\((n-1)\left(\frac{-5}{2}\right)=-65\)
\(n-1=-65 \times \frac{-2}{5}\)
n – 1 = -13 × -2
n – 1 = + 26
∴ n = 26 + 1
∴ n = 27

Question 6.
Check whether -150 is a term of the A.P: 11, 8, 5, 2, ………..
Solution:
11, 8, 5, 2, ……….. -150
a = 11, d = 8 11 = -3. a_n = -150.
a + (n – 1) d = a_n
11 + (n – 1) (-3) = -150
11 – 3n + 3 = -150
-3n + 14 = -150
-3n = -150 – 14
-3n = -164
3n = 164
\(n=\frac{164}{3}\)
Here value of ‘n’ is not perfect. Hence -150 is not a term of the A.P.

Question 7.
Find the 31^st term of an AP whose 11^th term Is 38 and the 16^th term is 73.
Solution:
a = 38, a₁₆ = 83 a₃₁ =?
a_n = a + (n – 1) d
a₁₆ = a + (16 – 1) d
a + 15d = 83
38 + 15d = 83
15d = 83 – 38
15d = 45
\(d=\frac{45}{15}\)
∴ d=3.
∴ a_n = a + (n – 1)d
a₃₁ = 38 + (31 – 1) 3
= 38 + 30 × 3
= 38 + 90
∴ a = 128.

Question 8.
An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term.
Solution:
n = 50, a₃ = 12, a_n = 106, a₂₉ =?
a₁₁ = a + (n – 1) d
a₅₀ = a + (50 – 1) d = 106
∴ a + 49d = 106 ……………… (1)
a₃ = a + 2d = 12 ………………. (2)
Subtracting equation (2) in equation (1).

Substituting the value of d.
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
∴ a = 12 – 4
a = 8.
∴ a_n = a + (n – 1) d
a₂₉ = 8 + (29 – 1) 2
= 8 + 28 × 2
= 8 + 56
∴ a₂₉ = 64

Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
a₃ = 4, a₉ = -8, a_n = 0, n =?
a₃ = a + 2d = 4 ………….. (1)
a₉ = a + 8d = -8 …………. (2)
From equation (1) – equation (2).

6d = 12
\({d}=\frac{-12}{6}\)
∴ d = -2
a + 2d = 4
a – 2(2) = 4
a – 4 =4
∴ a = 4 + 4
∴ a = 8
a_n = a + (n – 1) d
= 8 + (n – 1) (-2)
= 8 – 2n + 2
= 10 – 2n = 0 ∵ a_n = 0
\(n=\frac{10}{2}\)
∴ n = 5
∴ 5^th term of this AP is Zero.

Question 10.
The 1 7^th term of an AP exceeds its 17^th term by 7. Find the common difference.
Solution:
a₁₇ = a₁₀ + 7, d =?
a + 16d = a + 9d + 7
a +1 6d – a – 9d = 7
7d = 7
\(\mathrm{d}=\frac{7}{7}\)
∴ = 1

Question 11.
Which term of the AP: 3, 15, 27. 39, … will be 132 more than its 54^th term?
Solution:
3, 15, 27, 39, ………… a_n =?, n =?
a_n = a₅₄ + 132
a = 3, d = 15 – 3 = 12
a_n = a₅₄ + 132
a_n = a + 53d + 132
3 + 53(12) + 132
= 3 + 636 + 132
∴ a_n = 771
a_n = a + (n – 1) d = 771
= 3 + (n – 1)12 = 771
3 + 12n — 12 = 771
12n – 9 = 771
12n = 771 + 9
12n = 780
\(n=\frac{780}{12}\)
∴ n = 65.
∴ 65^th term is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100. what is the difference between their 1 000th terms?
Solution:
Having common difference ‘d’, the APs
1st set a, a+d, a+2d
2nd set b, b + d, b + 2d
100^th term of 1st set – 100^th term of 2nd set = 100
∴ a + 99d – (b + 99d) = 100
a + 99d – b – 99d = 100
a – b= 100
Similarly.
1000th term of 1st set = 1 + 999d
1000 th term of 2nd set = b + 999d
Their difference
= a + 999d – (b + 999d)
= a + 999d – b – 999d
= a – b.

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
First three-digit number divisible by 7
105 and the List number Is 994.
∴ AP is 105, 112. 119 994.
a = 105, d = 112 – 105 = 7. a_n = 994.
n =?
a + (n – 1)d = a_n
105 + (n – 1) 7 = 994
105 + 7n – 7 = 994
7n + 98 = 994
7n = 994 – 98
7n = 896
\(n=\frac{896}{7}\)
∴ n = 128.
∴ Numbers with 3 digits divisible by 7 are 128.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4. after 10 are 12, 16, 20….
Multiples of 4 upto 250 is 248
∴ A.P. is 12, 16, 20, …….. 248
a = 12, d = 16 – 12 = 4
n =?
a = a + (n – 1) d = 248
12 + 4n – 4 = 248
4n + 8 = 248
4n = 248 – 8
4n = 240
\(n=\frac{240}{4}\)
∴ n = 60
∴ Multiples of 4 lie between 10 and 250 is 60.

Question 15.
For what value of n’. are the n^th terms of two APs: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?
Solution;
63, 65, 67,……….
a = 63. d = 65 – 63 =2. a_n =?
n^th term of this is
a_n = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
a_n= 2n + 61 …………….. (i)
3, 10, 17, ………….
a = 3, d = 10 – 3 = 7, a_n =?
a_n = a + (n – 1)d
= 3 + (n – 1) 7
= 3 + 7n – 7
a_n = 7n — 4 ………….(ii)
Here. n^th terms of second AP are equal.
∴ equation (i) = equation (ii)
2n + 61 = 7n – 4
2n – 7n = -4 – 61
5n = 65
5n =65
\(n=\frac{65}{5}\)
∴ n = 13
∴13th terms of the two given APs are equal.

Question 16.
DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution:
a = 16, a₇ = a₅ + 12, A.P =?
a₇ = a₅ + 12
a + 6d = a + 4d + 12
a + 6d – a – 4d = 12
2d = 12 ∴ d = 6.
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
∴ a = 16 – 12 ∴ a = 4
a = 4, d = 6.
∴ A.P.:
a, a + d, a + 2d, …………………
4, 4 + 6, 4 + 12, …………….
4, 10, 16, ………….

Question 17
Find the 20th term from the Last term of theAP: 3, 8, 13, …………., 253.
Solution:
3, 8, 13, ………….., 253
a = 3. d = 8 – 3 = 5, a_n = 253
20th term from the last term of the AP starting from 253 =?
253, 258, 263, ………… a₂₀ =?
a = 253, d = 258 – 253 = 5, n = 20
a_n = a + (n – 1) d
a₂₀= 253 + (20 – 1) 5
= 253 + 19 × 5
= 253 + 95
∴ a₂₀ = 348
∴ 20^th term from the last term of the AP is 348.

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. FInd the first three terms of the AP.
Solution:
a₄ + a₈ = 24 …………. (1)
a₆ + a₁₀ = 44 ………… (2)
But A.P is a, a + d, a + 2d
from equation (1).
a₄ + a₈ = 24
a + 3d + a + 7d = 24
2a + 10d = 24 ………….. (3)
from equation (2).
a₆ + a₁₀ = 44
a + 5d + a + 9d = 44
2a + 14d = 44 ………… (4)
Subtracting eqn. (4) from equation (3)

4d = 20
\(d=20 / 4\)
∴ d = 5
Substituting the value of d in equation (3)
2a + 10d = 24
2a + 10(5) = 24
2a + 50 = 24
2a = 24 – 50
2a = -26
a = 26/2 ,
∴ a = -13 .
∴ AP: a, a + d, a + 2d, ………..
-13, -13 + 5, -13 + 2(5), ………
-14, -8, -3, …………

Question 19.
Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an Increment of Rs. 200 each year. in which year did his income reach Rs. 7000?
Solution:
Payment of Subba Rao in the year 1995 = Rs. 5000
increment = Rs. 200
∴ The payment he received in the year 1996 is Rs. 5.200

a = 5000, d = 5200 – 5000 = 200,
a_n = 7000, n=?
a + (n – 1)d = a_n
5000 + (n – 1) 200 = 7000
5000 + 200n – 200 = 7000
200n + 4800 = 7000
200n = 7000 – 4800
200n = 2200
\(n=\frac{2200}{200}\)
∴ n = 11
Value of ‘n’ is 11
∴ From 1995 to 10 years means 2005. his salary becomes Rs. 7,000.

Question 20.
Ramkall saved Rs. 5 In the first week of a year and then Increased her weekly savings by Rs. 1.75. If in the nüI week. her weekly savings become Rs. 20.75. find ‘n’.
Solution:
Savings in the First week = Rs. 5.
Savings in the Second week 5 + 1.75 = Rs. 6.75
Savings in ‘n^th week is Rs. 20.75
∴ A.P. 5, 6, 75 , …………. , 20.75
a = 5, d = 6.75 – 5 = 1.75, a_n = 20.75.
n =?
a + (n – 1) d = a_n
5 + (n – 1) (1.75) = 20.75
5 + 1.75n – 1.75 = 20.75
1 .75n + 3.25 = 20.75
1.75n = 20.75 – 3.25
1.75n = 17.5
\(n=\frac{17.5}{1.75}\)
\(n=\frac{1750}{175}\)
∴ n = 10
∴ In the 10th week, her savings becomes Rs. 20.75.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² – 7x + 3 = 0.
(ii) 2x² + x – 4 = 0
(iii) 4x² + \(4 \sqrt{3} x\) + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) 2x² – 7x + 3 = 0
Dividing all terms by 2,

(ii) 2x² + x – 4 = 0
Dividing all terms by 2,

(iii) 4x² + \(4 \sqrt{3} x\) + 3 = 0
Dividing all terms by 4,

(iv) 2x² + x + 4 = 0
Dividing all terms of the equation by 2,

Here ‘x’ has no fixed value because \(\sqrt{-31}\) is not a square root.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0.
(ii) 2x² + x – 4 = 0
(iii) 4x² + \(4 \sqrt{3} x\) + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) 2x² – 7x + 3 = 0.
Here, a = 2, b = -7, c = 3

(ii) 2x² + x – 4 = 0
Here, a = 2, b = 1, c = 4

(iii) 4x² + \(4 \sqrt{3} x\) + 3 = 0
Here, a = 4, b = \(4 \sqrt{3} \), c = 3

(iv) 2x² + x + 4 = 0
Here, a = 2, b = 1, c = 4

Question 3.
Find the roots of the following equations:

Solution:


(x – 2) (x – 1) = 0
If x – 2 = 0, then x = 2
If x – 1 = 0, then x = 1
∴ x = 2 OR x = 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be ‘x’ years.
After 5 years his age will be (x + 5)
3 years back his age was (x – 3)

(x – 7) (x + 3) = 0
If x – 7 = 0, then x = 7
If x + 3 = 0, then x = -3
∴ x = 7 OR x = -3
∴ Present age of Rehman is 7 years.

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in mathematics is Y and marks in English is (30 – x)
Shefali’s gets 2 more marks in mathematics then x + 2 and 3 marks less in English is (30 – x – 3)
Product of Shefali’s marks = 210
(x + 2) (30 – x – 3) = 210
(x + 2) (27 – x) = 210
27x + 54 – x² – 2x = 210
– x² + 25x + 54 = 210
x² – 25x + 210 – 54 = 0
x² – 25x + 156 = 0
x² – 13x – 12x + 156 = 0
x(x – 13) – 12 (x – 13) = 0
(x – 13) (x – 12) = 0
x – 13 = 0 (or) x – 12 = 0
x = 13 (or) x = 12
Therefore, Shefali’s marks in Mathematics =13
Marks in English = 30 – 13 = 17 (or) Shefali’s marks in Mathematics = 12 then.
Marks in English = 30 – 12 = 18

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:

Let shorter side of rectangle ABCD be x metre,
Then, longer side of rectangle ABCD,
AB = (x + 30) m.
Diagonal AC = (x + 60) m.
In ⊥∆ABC, ∠B = 90°.
As per Pythagoras theorem,
AB² + BC² = AC²
(x + 30)² + (x)² = (x + 60)²
x² + 60x + 900 + x² = x² + 120x + 3600
2x² + 60x + 900 = x² + 120x + 3600
2x² – x² + 60x – 120x + 900 – 3600 = 0
x² – 60x – 2700 = 0

x² – 90x + 30x – 2700 = 0
x(x – 90) + 30 (x – 90) = 0
(x – 90) (x + 30) = 0
If x – 90 = 0, then x = 90
If x + 30 = 0, then x = -30
Shorter side of rectangle, BC = x = 90 m.
Longer side of rectangle, AB = x + 30 = 90 + 30= 120 m.
Diagonal of rectangle, AC = x + 60 = 90 + 60 = 150 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smellier number is 8 times the larger number. Find the two numbers.
Solution:
Let the small number be ‘x’.
The Square of the smaller number is 8 times the larger number.
∴ x² = 8 × Larger number.
∴ Larger number = \(\frac{1}{8}\) x².
The difference between these numbers is 180.

∴ Smaller number is 12,
Larger number is 18.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let initial speed of the train be x km/h,
Then, time taken to cover 360 km
If the speed had been 5 km/h more,
\(\frac{360}{x}\) Hr.
then time taken to cover is \(\frac{360}{x+5}\) Hr.

x(x + 365) = 360 (x + 5)
x² + 365x = 360x + 1800
x² + 365x – 360x – 1800 = 0
x² + 5x – 1800 = 0
x² + 45x – 40x – 1800 = 0
x(x + 45) – 40 (x + 45) = 0
(x + 45) (x – 40) = 0
If x + 45 = 0, then x = – 45
If x – 40 = 0, then x = 40
∴ Initial speed of a train is 40 km/hr.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. .
Solution:
Let the time required to fill up the tank for tap having a larger diameter be ‘x’ Hr. For ‘x’ hour part of the tank filling is 1.
For \(9 \frac{3}{8}\) hr. part of the tank filling ………..?
i.e., \(\frac{75 \times 1}{8 x}\) part
Time required for smaller diameter is = (x – 10)Hr.
For (x – 10)Hr part of the tank, filling is 1
For \(9 \frac{3}{8}\) Hr. part of the tank is …………?

∴Time required for larger diameter, x = 25 Hr.
∴ Time required for smaller diameter = x – 10 = 15 Hr.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let Average speed of passenger train be = x km/h.
. But Average speed of Express train be = (x + 11) km/h.
Time required for Passenger:
For travelling x km time required is 1 Hr
For travelling 132 km time required …….?
\(\frac{132}{x}\) Hr.
Time required for Express train :
For travelling (x + 11) km is 1 Hr.
For travelling 132 km …………?
i.e., \(\frac{132}{(x+11)}\) Hr
Express train required 1 Hr. less comparing to Passenger train. .

x(x + 143) = 132 (x+ 11)
x² + 143x = 132x + 1452
x² + 143x – 132 x – 1452 = 0
x² + 11x – 1452 = 0
x² + 44x – 33x – 1452 = 0
x(x + 44) – 33(x + 44) = 0
(x + 44) (x – 33) = 0
If x + 44 = 0, then x = -44
If x – 33 = 0, then x = 33
∴ Average speed of passenger train is 33 km/hr.
∴ Average speed of Express train = x + 11 = 33 + 11 = 44 km/hr.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1

Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225

∴ 90 = 45 × 2 + 0
∴ HCF of 135 and 225 is 45.

(ii) 196 and 38220

∴ 38220 = 196 × 195 + 0
∴ HCF of 196 and 38220 is 196.

(iii) 867 and 2552

2 = 1 × 2 + 0
∴ HCF of 867 and 2552 is 1.

Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, let q be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we get a = 6q + r
where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,
a = 6q + 0 = 6q or a = 6q + 1
or a = 6q + 2 or a = 6q + 3
or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ‘a’.
[∵ 6q = 2(3q) = 2m₁ 6q + 2 = 2(3q + 1) = 2m₂,
6q + 4 = 2(3 q + 2) = 2m₃]
But ‘a’ being an odd integer, we have :
a = 6q + 1, or a = 6q + 3, or a = 6q + 5

Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?
Solution:
HCF of 32 and 616

∴ 32 = 8 × 4 + 0
∴ HCF of 32 and 616 is 8.
∴ Both groups can march in 8 columns.

Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3g, 3q + 1 or 3g + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]
Solution:
Let us consider an arbitrary positive integer as ‘x’ such that it is of the form
3q, (3q + 1) or (3q + 2)
For x = 3q, we have x² = (3q)²
⇒ x² = 9q² = 3(3q²) = 3m ………. (1)
Putting 3q² = m, where m is an integer.
For x = 3q + 1,
x² = (3q + 1)² = 9q² + 6q + 1
= 3(3q² + 2q) + 1 = 3m + 1 ………… (2)
Putting 3q² + 2q = m, where m is an integer.
For x = 3q + 2,
x² = (3q + 2)²
= 9q² + 12q + 4 = (9q² + 12q + 3) + 1
= 3(3q² + 4q + 1) + 1 = 3m + 1 ……….. (3)
Putting 3q² + 4q +1 = m, where m is an integer.
From (1), (2) and (3),
x² = 3m or 3m + 1
Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let a and b be two positive integers and a > b
a = (b × q) + r where q and r are positive integers and 0 ≤ r < b.
Let b = 3 (if 9 is multiplied by 3 a perfect cube number is obtained)
a = 3q + r where 0 ≤ r < 3
(i) if r = 0, a = 3q
(ii) if r = 1, a = 3q + 1
(iii) if r = 2 a = 3q + 2
Consider cubes of these,
case i) a = 3q
a³ = (3q)³ = 27q³ = 9(3q³)
a³ = 9m
where m= 3q³ & m is an integer

case ii) a = 3q + 1
a³ = (3q + 1)³
(a + b)³ = a³ + b³ + 3a²b + 3ab²
a³ = 27q³ + 1 + 27q³ + 9q
a³ = 27 q³ + 27 q² + 9q + 1
a³ = 9 (3q³ + 3q² + q) + 1
a³ = 9m + 1
where m = 3q³ + 3q² + q & m is an integer

case iii) a = 3q + 2
a³ = (3q + 2)³ = 27q³ + 8 + 54q² + 36q
a³ = 27q³ + 54q² + 36q + 8
= 9 (3q³ + 6q² + 4q) + 8
a³ = 9m + 8
where m = 3q³ + 6q² + 4q and m is an integer.
∴ Cube of any positive integer is either of the form 9m, 9m + 1 (or) 9m + 8 for some integer m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:
Radius of 1st Circle, r₁ = 19 cm.
Circumference of 1st circle, C₁ = 2πr₁
\(=2 \times \frac{22}{7} \times 19\)
Radius of 2nd circle, r₂ = 9 cm.
Circumference of 2nd circle, C₂ = 2πr₂
\(=2 \times \frac{22}{7} \times 9\)
Sum of the circumferences of the two circles
\(=2 \times \frac{22}{7} \times 19+2 \times \frac{22}{7} \times 9\)
\(C_{3}=2 \times \frac{22}{7}(19+9)\)
\(=2 \times \frac{22}{7} \times 28\)
= 2 × 22 × 4
∴ C₃ = 176 cm.
Circumference of 3rd circle, C₃ = 176 cm. then, r₃ = ?
2πr₃ = C₃
\(2 \times \frac{22}{7} \times r_{3}=176\)
\(r_{3}=176 \times \frac{7}{22} \times \frac{1}{2}\)
∴ r₃ = 28 cm

Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
Radius of 1st Circle, r₁ = 8 cm.
Area of 1st Circle, A₁ = π r₁²
\(=\frac{22}{7} \times(8)^{2}\)
\(A_{1}=\frac{22}{7} \times 64\)
Radius of 2nd Circle, r₂ = 6 cm.
Area of 2nd Circle, A₂ = π r₂²
\(=\frac{22}{7} \times(6)^{2}\)
\(A_{2}=\frac{22}{7} \times 36\)
Total area of two circles =

Question 3.
The figure given below depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black, and White. The diameter of the region representing the Gold score is 21 cm. and each of the other bands is 10.5 cm. wide. Find the area of each of the five scoring regions.

Solution:
i) The diameter of the region representing Gold Score :
21
d = 21 cm. ∴ Radius, \(r=\frac{21}{2} \mathrm{cm}\)
∴ Area of the region representing Gold score
= πr²

A = 346.5 sq. cm.

ii) Radius of the region representing Gold score and Red score:
r = 10.5 + 10.5 = 21 cm.
∴ Area of the region representing Red score: Area of the region representing Gold and Red Score — Area of the region representing Gold score.
= πr² – πr²

∴ A = 1039.5 sq.cm.

iii) Total radii of the region representing Gold, Red and Blue region :
r =10.5 + 10.5 + 10.5 = 31.5 cm
\(=\frac{63}{2} \mathrm{cm}\)
∴ Area of the region representing Blue region:
[ Area of the region representing Gold, Red and Blue regions ] — Area of the region representing Gold and Red region
∴ A = πr² – πr²

iv) Similarly, to find Area of the region representing Black region :
∴ A = πr² – πr²

∴ A = 2425.5 sq. cm.

v) Area of the region representing white:
∴ A = πr² – πr²

∴ A = 3118.5 sq. cm

Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution:
Diameter of the Wheel of the car = 80 cm.
Radius of the wheel = 40 cm.
Circumference of the Wheel = 2πr
\(=2 \times \frac{22}{7} \times 40\)
\(=\frac{44 \times 40}{7}\)
= 251.4 cm
Speed of the car = 66 km/hr.
\(=\frac{66 \times 100000}{60} \mathrm{cm} . / \mathrm{min}\)
= 110000 cm/min.
Distance travelled by car in 10 minutes
= 110000 × 10
= 1100000 cm.
Let the number of revolutions of the wheel be ’n’, then
n × 251.4 = 1100000
∴ \(\mathrm{n}=\frac{1100000}{251.4}\)
n = 4375
∴ Car can complete 4,375 revolutions in 10 minutes.

Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
A) 2 units
B) n units
C) 4 units
D) 7 units
Solution:
A) 2 units
Circumference of a circle = Area of the circle.
2πr = πr²
2 × π × r = π × r × r
∴ 2 = r
∴ r = 2 units.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1, drop a comment below and we will get back to you at the earliest.