KSEEB SSLC Class 10 Maths Solutions Karnataka State Syllabus

Expert Teachers at KSEEBSolutions.com has created Karnataka SSLC KSEEB Solutions for Class 10 Maths Pdf Free Download in English Medium and Kannada Medium of KTBS Karnataka State 10th Standard Maths Textbook Solutions Answers Guide, Textbook Questions and Answers, Notes Pdf, Model Question Papers with Answers, Study Material are part of KSEEB SSLC Class 10 Solutions.

Here we have given KSEEB Karnataka State Board Syllabus Class 10 Maths Textbook Solutions based on NCERT Syllabus. Students can also read Karnataka SSLC Maths Model Question Papers with Answers 2020-2021 Pdf.

Karnataka State Board Syllabus for Class 10 Maths Solutions

KSEEB Solutions for Class 10 Maths

10th Standard Maths Textbook Solutions Karnataka State Syllabus in English Medium

KSEEB Solutions For Class 10 Maths Chapter 1 Arithmetic Progressions

KSEEB 10th Maths Solutions Chapter 2 Triangles

SSLC Maths Solutions Class 10 KSEEB Chapter 3 Pair of Linear Equations in Two Variables

10th Standard Karnataka State Syllabus Maths Guide Chapter 4 Circles

Karnataka State Board 10th Maths Solutions Chapter 5 Areas Related to Circles

SSLC Maths Solutions KSEEB Chapter 6 Constructions

10th Maths In Kannada Solutions for Class 10 Chapter 7 Coordinate Geometry

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers

10th Maths Syllabus State Board Karnataka Solutions Chapter 9 Polynomials

SSLC Mathematics Solution Part 2 KSEEB Chapter 10 Quadratic Equations

Maths Guide For Class 10 State Syllabus Karnataka Chapter 11 Introduction to Trigonometry

Maths Solutions for Class 10 State Syllabus Karnataka Chapter 12 Some Applications of Trigonometry

Karnataka SSLC Maths Text Book Solutions Chapter 13 Statistics

Karnataka State Board Syllabus for Class 10 Maths textbook Solutions Chapter 14 Probability

Karnataka SSLC Maths Notes Pdf Chapter 15 Surface Areas and Volumes

Karnataka State Board Syllabus for Class 10 Maths Solutions in Kannada Medium

10th Standard Maths Textbook Solutions Karnataka State Syllabus in Kannada Medium

KSEEB Solutions For Class 10 Maths Chapter 1 Arithmetic Progressions

KSEEB 10th Maths Solutions Chapter 2 Triangles

SSLC Maths Solutions Class 10 KSEEB Chapter 3 Pair of Linear Equations in Two Variables

10th Standard Karnataka State Syllabus Maths Guide Chapter 4 Circles

Karnataka State Board 10th Maths Solutions Chapter 5 Areas Related to Circles

SSLC Maths Solutions KSEEB Chapter 6 Constructions

10th Maths In Kannada Solutions for Class 10 Chapter 7 Coordinate Geometry

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers

10th Maths Syllabus State Board Karnataka Solutions Chapter 9 Polynomials

SSLC Mathematics Solution Part 2 KSEEB Chapter 10 Quadratic Equations

Maths Guide For Class 10 State Syllabus Karnataka Chapter 11 Introduction to Trigonometry

Maths Solutions for Class 10 State Syllabus Karnataka Chapter 12 Some Applications of Trigonometry

Karnataka SSLC Maths Text Book Solutions Chapter 13 Statistics

Karnataka State Board Syllabus for Class 10 Maths textbook Solutions Chapter 14 Probability

Karnataka SSLC Maths Notes Pdf Chapter 15 Surface Areas and Volumes

We hope the given Karnataka SSLC KSEEB Solutions for Class 10 Maths Pdf Free Download in English Medium and Kannada Medium of KTBS Karnataka State 10th Standard Maths Textbook Solutions Answers Guide, Textbook Questions and Answers, akub koyyur 10th maths Notes Pdf, Model Question Papers with Answers, Study Material will help you.

If you have any queries regarding KSEEB Karnataka State Board Syllabus Class 10th Std Maths Text Book Solutions based on NCERT Syllabus, 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

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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 1

(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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 2

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 3

(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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 4
But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 5
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.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 7
Since, b2 – 4ac is less than 0.
∴ The quadratic equation (1) has no real roots.
Thus, the given situation is not possible.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 6
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

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 1
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 2

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
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 3
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
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 4

(iii) 336 and 54
a = 336, b = 54
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 5

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

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 an the nth term of the A.P.
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 7
Solution:
(i) a = 7, d = 3, n = 8, an =?
an = a + (n – 1) d
a8= 7 + (8 – 1) 3
= 7 + 7 × 3
= 7 + 21
∴ a8 = 28

(ii) a = -18, d =?, n = 10, an = 0
an = 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, an = -5
an = 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 =? an = 3.6
an = 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, an =?
an = a + (n – 1) d
= 3.5 + (105 – 1) (0)
= 3.5+ 104 × 0
= 3.5 +0
∴ an = 3.5

Question 2.
Choose the correct choice In the following and justify:
(i) 30th 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 =?
an = a + (n – 1) d
a30 = 10 + (30 – 1)(-3)
= 10 + 29(-3)
= 10 – 87
∴ a30 = 77
∴ Ans: (C) -77

(ii) 11th 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 =?
an = 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:
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 1
Solution:
(i) KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 2
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 =?
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 1

(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
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 2
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, an = 78, n =?
an = 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, an = 201. n =?
a + (n – 1) d = an
3 + (n – 1) 6 = 201
3 + 6n – 6 = 201
6n – 3 = 201
6n = 203
\(n=\frac{204}{6}\)
∴ n = 34

(ii) \(18,15 \frac{1}{2}, 13, \dots \dots 47\)
\(a=18, d=a_{2}-a_{1}=\frac{31-36}{2}=\frac{-5}{2}\)
an = -47, n =?
an = 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. an = -150.
a + (n – 1) d = an
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 31st term of an AP whose 11th term Is 38 and the 16th term is 73.
Solution:
a = 38, a16 = 83 a31 =?
an = 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.
∴ an = 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 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
n = 50, a3 = 12, an = 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).
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 3
Substituting the value of d.
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
∴ a = 12 – 4
a = 8.
∴ an = 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, an = 0, n =?
a3 = a + 2d = 4 ………….. (1)
a9 = a + 8d = -8 …………. (2)
From equation (1) – equation (2).
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 4
6d = 12
\({d}=\frac{-12}{6}\)
∴ d = -2
a + 2d = 4
a – 2(2) = 4
a – 4 =4
∴ a = 4 + 4
∴ a = 8
an = a + (n – 1) d
= 8 + (n – 1) (-2)
= 8 – 2n + 2
= 10 – 2n = 0 ∵ an = 0
\(n=\frac{10}{2}\)
∴ n = 5
∴ 5th term of this AP is Zero.

Question 10.
The 1 7th term of an AP exceeds its 17th 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 54th term?
Solution:
3, 15, 27, 39, ………… an =?, n =?
an = a54 + 132
a = 3, d = 15 – 3 = 12
an = a54 + 132
an = a + 53d + 132
3 + 53(12) + 132
= 3 + 636 + 132
∴ an = 771
an = 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.
∴ 65th 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
100th term of 1st set – 100th 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. an = 994.
n =?
a + (n – 1)d = an
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 nth terms of two APs: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?
Solution;
63, 65, 67,……….
a = 63. d = 65 – 63 =2. an =?
nth term of this is
an = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
an= 2n + 61 …………….. (i)
3, 10, 17, ………….
a = 3, d = 10 – 3 = 7, an =?
an = a + (n – 1)d
= 3 + (n – 1) 7
= 3 + 7n – 7
an = 7n — 4 ………….(ii)
Here. nth 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, an = 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
an = a + (n – 1) d
a20= 253 + (20 – 1) 5
= 253 + 19 × 5
= 253 + 95
∴ a20 = 348
∴ 20th 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)
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 5
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
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 6
a = 5000, d = 5200 – 5000 = 200,
an = 7000, n=?
a + (n – 1)d = an
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 ‘nth week is Rs. 20.75
∴ A.P. 5, 6, 75 , …………. , 20.75
a = 5, d = 6.75 – 5 = 1.75, an = 20.75.
n =?
a + (n – 1) d = an
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

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,
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 1
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 2
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 3

(ii) 2x2 + x – 4 = 0
Dividing all terms by 2,
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 4
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 5

(iii) 4x2 + \(4 \sqrt{3} x\) + 3 = 0
Dividing all terms by 4,
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 6

(iv) 2x2 + x + 4 = 0
Dividing all terms of the equation by 2,
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 7
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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 8

(ii) 2x2 + x – 4 = 0
Here, a = 2, b = 1, c = 4
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 9

(iii) 4x2 + \(4 \sqrt{3} x\) + 3 = 0
Here, a = 4, b = \(4 \sqrt{3} \), c = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 10
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 11

(iv) 2x2 + x + 4 = 0
Here, a = 2, b = 1, c = 4
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 12

Question 3.
Find the roots of the following equations:
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 13
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 14
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 15
(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)
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 16
(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:
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 18
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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 19
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.
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 20
∴ 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.
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 21
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 …………?
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 22
∴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. .
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 23
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

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
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 1
∴ 90 = 45 × 2 + 0
∴ HCF of 135 and 225 is 45.

(ii) 196 and 38220
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 2
∴ 38220 = 196 × 195 + 0
∴ HCF of 196 and 38220 is 196.

(iii) 867 and 2552
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 3
2 = 1 × 2 + 0
∴ HCF of 867 and 2552 is 1.

KSEEB Solutions

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
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1 6
∴ 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.

KSEEB Solutions

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 3 Pair of Linear Equations in Two Variables Ex 3.2

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Strength of the X Std. is 10
Number of boys be ‘y’, then
number of girls be ‘x’ .
x + y = 10 …….. (i)
x = y + 4
∴ x – y = 4 ………. (ii)
From equations (i) + (ii)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 1
\(\quad x=\frac{14}{2}=7\)
Substituting the value of ‘x’ in eqn. (i),
x + y = 10
7 + y = 10
y = 10 – 7
y = 4.
∴ Number of girls, x = 7
Number of boys, y = 4.

(ii) Cost of each pencil be Rs. ‘x’
Cost of pen be Rs. ‘y’
5x + 7y = 50 ………. (i)
7x + 5y = 46 ……….. (ii)
From equation (i) + equation (ii)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 2
∴ x + y = 8 …………… (iii)
from Eqn. (ii) – Eqn. (i),
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 3
∴ -x + y = 2 …………. (iv)
Eqn. (iii) + Eqn. (iv)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 4
∴ y = 5
Substituting the value of ‘y’ in eqn. (i)
x + y = 8
x + 5 = 8
∴ x = 8 – 5 x = 3
∴Cost of each pencil is Rs. 3.
Cost of each pen is Rs. 5.

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}\) find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
i) 5x – 4y + 8 = 0 a1= 5, b1 = -4, C1= 8
7x + 6y – 9 = 0 a2 = 7, b2= 6, c2= -9
\(\frac{a_{1}}{a_{2}}=\frac{5}{7} \quad \frac{b_{1}}{b_{2}}=\frac{-4}{6} \quad \frac{c_{1}}{c_{2}}=\frac{8}{-9}\)
Here, \(\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}\)
∴ Lines representing the pairs of linear equations intersect at a point.

ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
a1 =9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{9}{18}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{6}=\frac{1}{2}\)
\(\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}\)
Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)
∴ Representation of lines graphically are coincident.

iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Here, a1 = 6, b1 = -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
\(\frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1}\)
\(\frac{c_{1}}{c_{2}}=\frac{10}{9}\)
Here, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ Representation of lines graphically is parallel.

Question 3.
On compairing the ratios \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}, \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\) and \(\frac{c_{1}}{c_{2}}\) find out whether the following pair of linear equations are consistent, or inconsistent.
i) 3x + 2y = 5; 2x – 3y = 7
ii) 2x – 3y = 8; 4x-6y = 9
iii) \(\frac{3}{2} x+\frac{5}{3} y=7 : 9 x-10 y=14\)
iv) 5x – 3y = 11: -10x – 6y = -22
v) \(\frac{4}{3} x+2 y=8 ; 2 x+3 y=12\)
Solution:
i) 3x + 2y = 5 ⇒ 3x + 2y – 5 = 0
2x – 3y = 7 ⇒ 2x – 3y – 7 = 0
Here, a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7
\(\frac{a_{1}}{a_{2}}=\frac{3}{2} \quad \frac{b_{1}}{b_{2}}=-\frac{2}{3} \quad \frac{c_{1}}{c_{2}}=\frac{-5}{-7}=\frac{5}{7}\)
Here, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
∴ Graphical representation is intersecting lines.
∴ Pair of linear equations are consistent.

ii) 2x – 3y = 8 ⇒ 2x – 3y – 8 = 0
4x – 6y = 9 ⇒ 4x – 6y – 9 = 0
a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2}\)
\(\frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}\)
Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)
∴ Graphical representation is parallel lines.
∴ Equations are inconsistent.

iii) \(\frac{3}{2} x+\frac{5}{3} y=7 \quad \frac{3}{2} x+\frac{5}{3} y-7=0\)
9x – 10y = 14 ⇒ 9x – 10y – 14 = 0
\(a_{1}=\frac{3}{2}, \quad b_{1}=\frac{5}{3}, \quad c_{1}=-7\)
a2 = 9, b2 = -10, c2 = -17
\(\frac{a_{1}}{a_{2}}=\frac{3}{2} \times \frac{1}{9} \quad \frac{b_{1}}{b_{2}}=\frac{5}{3} \times \frac{-1}{6}\)
\(\frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{7}{14}=\frac{1}{2}\)
Here, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
∴ Pair of equations are consistent

iv) 5x – 3y = 11 ⇒ 5x – 3y – 11 = 0
-10x + 6y = – 22 ⇒ -10x + 6y + 22 = 0
a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = -22
\(\frac{a_{1}}{a_{2}}=\frac{5}{-10}=-\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{6}=-\frac{1}{2}\)
\(\frac{c_{1}}{c_{2}}=\frac{-11}{22}=-\frac{1}{2}\)
Here, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ Pair of equations are consistent
∴ Graphical representation is coninciding.

v) \(\frac{4}{3} x+2 y=8 \quad \frac{4}{3} x+26-8=0\)
2x + 3y = 12 ⇒ 2x + 3y – 12 = 0
\(a_{1}=\frac{4}{3}, \quad b_{1}=2, \quad c_{1}=-8\)
a2 = 2, b2 = 3, c2 = -12
\(\frac{a_{1}}{a_{2}}=\frac{4}{3} \times \frac{1}{2}=\frac{1}{6} \quad \frac{b_{1}}{b_{2}}=\frac{2}{3}\)
\(\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{2}{3}\)
Here, \(\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}\)
∴ Pair of equations are consistent

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent obtain the solution graphically
(i) x + y = 5, 2x + 2y = 10
(ii) x-y = 8 3x-3y= 16
(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0 4x – 3y – 5 = 0
Solution:
(i) x + y = 5 ⇒ x + y – 5 = 0
2x + 2y = 10 ⇒ 2x + 2y – 10 = 0
a1 = 1, b1 = 1, c1 = -5
a2 = 2, b2 = 2, c2 = -10
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{2} \quad \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{-5}{-10}=\frac{1}{2}\)
Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)
∴ Pair of equations are consistent
(i) x + y = 5
y = 5 – x

x 0 2 4
y = 5 – x 5 3 1

(ii) 2x + 2y = 10
x + y = 5
y = 5 – x

x 0 2 5
y = 5 – x 5 3 0

∴ We can give any value for ‘x’, i.e., solutions are infinite.
∴ Solution, P (5, 0) x = 5, y = 0
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 5

(ii) x – y = 8 ⇒ x – y – 8 = 0
3x – 3y = 16 ⇒ 3x – 3y – 16 = 0
Here, \(\frac{a_{1}}{a_{2}}=\frac{1}{3} \quad \frac{b_{1}}{b_{2}}=\frac{-1}{-3}=\frac{1}{3}\)
\(\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}\)
\(\quad \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)
∴ Linear equations are inconsistenttent.
∴ Algebraically it has no solution.
Graphical representation → Parallel Lines.
(i) x – y = 8
-y = 8 – x
y = -8 + x

x 8 10 9
y = -8+x 0 2 1

(ii) 3x – 3y = 16
-3y = 16 – 3x
3y = -16 + 3x
\(\quad y=\frac{-16+3 x}{3}\)

x 6 8
\(y=\frac{-16+3 x}{3}\) 0.8 2.6

No solution because it is inconsistent
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 6

(iii) 2x + y – 6 = 0
4x – 2y – 4 = 0
Here a1 = 2, b1 = 1, c1 = -6
a2 = 4, b2 = -2, c2 = -4
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{-2}\)
\(\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}\)
Here, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
Pair of equations are consistent. Algebraically both lines intersect.
Graphical Representation :
(i) 2x + y = 6
y = 6 – 2x

x 0 2
y = 6 – 2x 6 2

(ii) 4x – 2y – 4 = 0
4x – 2y = 4
-2y = 4 – 4x
2y = -4 + 4x
\(\quad y=\frac{-4+4 x}{2}\)

x 1 3
\(y=\frac{-4+4 x}{2}\) 0 4

Solution: intersecting point, P (2, 2) i.e., x = 2, y = 2
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 7

(iv) 2x – 2y – 2 = 0
4x – 3y – 5 = 0
a1 = 2, b1 = -2, c1 = -2
a2 = 4, b2 = -3, c2 = -5
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{-2}{-3}=\frac{2}{3}\)
\(\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}\)
Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\)
Pair of equations are consistent.
∴ Algebraically both lines intersect.
Graphical Representation :
(i) 2x – 2y – 2 =0
2x – 2y = 2
-2y = 2 – 2x
2y = -2 + 2x
\(\quad y=\frac{-2+2 x}{2}\)
∴ y = – 1 + x

x 2 4
y= -1+ x 1 3

(ii) 4x – 3y – 5 = 0
4x – 3y = 5
-3y = 5 – 4x
3y = -5 + 4x
\(\quad y=\frac{-5+4 x}{3}\)

x 2 5
\(y=\frac{-5+4 x}{3}\) 1 5

Solution: P(2, 1) i.e., x = 2, y = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 8

Question 5.
Half the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden.
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 9
Solution:
Length of rectangular garden be ‘x’ m.
Breadth of rectangular garden be ‘y’ m, then
Length of the garden is 4m more than its width.
x = y + 4 ……….. (i)
x – y = 4
Half of the circumference is 36 m.
\(\frac{2 x+2 y}{2}=36\)
2x + 2y = 72
x + y = 36 …………… (ii)
∴ x – y = 4 (i)
x + y = 36 (ii)
From eqn. (i) + eqn. (ii)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 10
\(\quad x=\frac{40}{2}\)
∴ x = 20 m.
Substituting the value of ’x’ in eqn. (i)
x – y = 4
20 – y = 4
-y = 4 – 20
-y = -16
y = 16 m.
∴ Length of the garden = 20 m.
Breadth of the garden = 16 m.

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:
Linear equation is 2x + 3y – 8 = 0.
(i) Linear equations for intersecting lines:
2x + 3y – 8 = 0
3x + 2y – 7 = 0
a1 = 2, b1 = 3, c1 = -8
a2 = 3, b2 = 2, c2 = -7
\(\frac{a_{1}}{a_{2}}=\frac{2}{3} \quad \frac{b_{1}}{b_{2}}=\frac{3}{2}\)
Here, when \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\) Geometrical representation is intersecting lines.

(ii) For Parallel lines :
2x + 3y – 8 = 0
2x + 3y – 12 = 0
a1 = 2, b1 = 3, c1= -8
a1 = 2, b1 = 3, c1 = -12
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{2}=\frac{1}{1} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{3}=\frac{1}{1}\)
\(\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{8}{12}=\frac{2}{3}\)
Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\) hence
Graphical representation is parallel lines.

(iii) For Intersecting lines :
2x + 3y – 8 = 0
4x + 6y – 16 = 0
a1 = 2, b1 = 3, c1 = -8
a1 = 4, b1 = 6, c1 = -16
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}\)
\(\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}\)
Here,
∴ Lines are intersecting.

Question 7.
Draw the graphs of the equations x – y + 1 =0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
x – y + 1 = 0 ………. (i)
3x – 2y – 12 = 0 ……….. (ii)
x – y + 1 = 0
-y = -x -1
y = x + 1

x 2 4
y = x + 1 3 5

3x + 2y – 12 = 0
2y = -3x + 12
\(\quad y=\frac{-3 x+12}{2}\)

x 0 2
\(y=\frac{-3 x+12}{2}\) 6 3

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 11
Graphs of these two equations intersect at A. Vertices formed for ∆ABC are,
A (2, 3), B (-1, 0), C (4, 0).

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2, 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

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 =
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 1

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 2
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
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 3
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
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 4
∴ 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
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 5

iv) Similarly, to find Area of the region representing Black region :
∴ A = πr2 – πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 6
∴ A = 2425.5 sq. cm.

v) Area of the region representing white:
∴ A = πr2 – πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 7
∴ 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.

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0 ,
(ii) 2x2 + x – 6 = 0
(iii) \(\sqrt{2}\)x2 + 7x + \(5 \sqrt{2}\) =0
(iv) 2x2 – x + \(\frac{1}{8}\) = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 1
x(x – 5) + 2 (x – 5) = 0
(x – 5) (x + 2) = 0
If x + 2 = 0, then x = -2
If x – 5 = 0, then x = 5
∴ x = -2 OR +5.

(ii) 2x2 + x – 6 = 0
2x2 + 4x – 3x – 6 = 0
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 2
2x(x + 2) – 3(x + 2) = 0
(x + 2) (2x – 3) = 0
If x + 2 = 0, then x = -2
If 2x – 3 = 0, then 2x = 3 x = \(\frac{3}{2}\)
∴ x = -2 OR \(\frac{3}{2}\)

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 3

iv) 2x2 – x + \(\frac{1}{8}\) = 0 Multiply by 8
8(2x2 -x + \(\frac{1}{8}\)) = 0
16x2 – 8x + \(\frac{8}{8}\) = 0
16x2 – 8x + 1 = 0
16x2 – 4x – 4x + 1 = 0
4x (4x – 1) – 1 (4x – 1) =
(4x – 1) (4x – 1) = 0
4x – 1 = 0 or 4x – 1 = 0
4x = 1 or 4x = 1
x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)
x = \(\frac{1}{4}\) are the roots of the equation 2x2 – x + \(\frac{1}{8}\) = 0

(v) 100x2 – 20x + 1 = 0
100x2 – 10x – 10x + 1 = 0
10x(10x – 1) – 1 (10x – 1) = 0
(10x – 1) (10x – 1) = 0
(10x – 1)2 = 0
If 10x – 1 = 0, then
10x = 1
∴ x = \(\frac{1}{10}\)
∴ Two roots are \(\frac{1}{10} , \frac{1}{10}\)

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 2.
Solve the problems:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let John had x marbles and Jivanti had (45 – x) marbles.
When both of them lost 5 marbles then equation becomes (x – 5) × (45 – x – 5) = 124
⇒ (x – 5) × (40 – x) = 124
⇒ x2 – 45x + 324 = 0
⇒ x2 – 9x – 36x + 324 = 0
⇒ x(x – 9) – 36(x – 9) = 0
⇒ (x – 9)(x – 36) = 0
Either x – 9 = 0 or x – 36 = 0
Thus, x = 9 or x = 36
∴ If John had 9 marbles, then Jivanti had 45 – 9 = 36 marbles.
If John had 36 marbles, then Jivanti had 45 – 36 = 9 marbles.

(ii) Let the number of toys produced in a day be x.
Then cost of 1 toy = \(\frac{750}{x}\)
⇒ \(\frac{750}{x}\) = 55 – x
⇒ 750 = 55x – x2
⇒ x2 – 55x + 750 = 0
⇒ x2 – 30x – 25x + 750 = 0
⇒ x(x – 30) – 25(x – 30) = 0
⇒ (x – 30)(x – 25) = 0
Either x – 30 = 0 or x – 25 = 0
x = 30 or x = 25

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 3.
Find two numbers whose sum is 27 and the product is 182.
Solution:
Sum of two numbers be 27
Let one number be ‘x’ other number be ’27 – x’
Product of two numbers = 182
x(27 – x)= 182
27x – x2 = 182
x2 – 27x + 182 = 0
x2 – 14x – 13x+ 182 = 0
x (x – 14) – 13 (x – 14) = 0
(x – 14) (x – 13) = 0
x + 14 = 0 (or) x – 13 = 0
x = – 14 or x = 13
x = 14, 13
∴ Two numbers are 14, 13 (or) 13, 14

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
In that numbers, let one of the numbers be ’x’.
It’s consecutive positive integer is (x + 1)
Sum of their squares is 365.
∴ (x)2 + (x + 1)2 = 365
x2 + x2 + 2x + 1 = 365
2x2 + 2x + 1 = 365
2x2 + 2x + 1 – 365 = 0
2x2 + 2x – 365 = 0
x2 + x – 182 = 0
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 5
x2 + 14x – 13x – 182 = 0
x(x + 14) – 13(x + 14) = 0
(x + 14) (x – 13) = 0
If x + 14 = 0, then x = -14
If x – 13 = 0, then x = 13
In these positive integer is 13.
∴ One number, x = 13
It consecutive number is, x + 1
= 13 + 1 = 14
∴ The Numbers are 14 and 13.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm., find the other two sides.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 6
In ⊥∆ABC,
∠ABC = 90°.
AB is altitude,
BC is base AC is Hypotenuse
Let Base, BC = x cm.
if so considered, :
Altitude AB = (x – 7) cm.
Hypotenuse AC = 13 cm.
As per Pythagoras theorem,
In a right angled ABC,
AB2 + BC2 = AC2
(x – 7)2 + (x)2 = (13)2
x2 – 14x + 49 + x2= 169
2x2 – 14x + 49 = 169
2x2 – 14x + 49 – 169 = 0
2x2 – 14x – 120 = 0
x2 – 7x – 60 = 0
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 7
x2 – 12x + 5x – 6 0 = 0
x(x – 12) + 5(x – 12) = o
(x – 12) (x + 5) =0
If x – 12 = 0, then x = 12
If x + 5 = 0, then x = -5
Positive value, x = 12
∴ Base, BC = x = 12 cm.
Altitude, AB = x – 7 = 12 – 7 = 5 cm.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
∴ Cost of production of each article = ₹ (2x + 3)
According to the condition,
Total cost = ₹ 90
⇒ x × (2x + 3) = 90
⇒ 2x2 + 3x = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 – 12x + 15x – 90 = 0
⇒ 2x(x – 6) + 15(x – 6) = 0
⇒ (x – 6)(2x + 15) = 0
Either x – 6 = 0 or 2x + 15 = 0
⇒ x= 6 or x = \(\frac{-15}{2}\)
But the number of articles produced can never be negative.
⇒ x = \(\frac{-15}{2}\) is rejected
∴ Cost of production of each article = ₹ (2 × 6 + 3) = ₹ 15
Thus, the required number of articles produced is 6 and the cost of each article is ₹ 15.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.
Find the area of the triangle whose vertices are:
i) (2, 3), (-1,0), (2,-4)
ii) (-5, -1), (3, -5), (5, 2)
Solution:
i) Let A (2, 3) = (x1, y1)
B (-1, 0) = (x2, y2)
C (2, -4) = (x3, y3).
Area of the triangle from the given data:
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 1
∴ Area of ∆ ABC = 10.5 sq. units

ii) Let A (-5, -1) = (x1, y1)
B (3, -5) = (x2, y2)
C (5, 2) = (x3, y3).
Area of the triangle from the given data :
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 2
∴ Area of ∆ ABC =32 sq. units

Question 2.
In each of the following find the value of ‘k’, for which the points are collinear.
i) (7, -2), (5, 1), (3, k)
ii) (8, 1), (k, -4), (2, -5)
Solution:
i) Let A (7,-2)= (x1, y1)
B (5, 1) = (x2, y2)
C (3, k) = (x3, y3).
Points are collinear, therefore the area of the triangle formed by these is zero (0).
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 3
-k + 4 = 0
-k = -4
∴ k = 4

ii) Let A (8, 1) = (x1, y1)
B (k, -4) = (x2, y2)
C (2, -5) = (x3, y3).
Area of Triangle ABC = 0
∴ ABC is a straight line.
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 4

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let the mid-points of ∆ABC are P, Q, R and also the mid-points of AB, BC and AC.
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 5
As per Mid-point formula,
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 6
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 7
\(=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
Area of ∆PQR : Area of ∆ABC
∴ 1 : 4

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 8
Solution:
Area of quadrilateral ABCD =?
In the quadrilateral ABCD, diagonal AC is drawn which divides ∆ABC, ∆ACD.
Sum of these triangles is equal to the Area of the quadrilateral.
i) Now, Area of ∆ABC :
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 9
= 10.5 sq. units

ii) Area of ∆ACD
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 10
∴ Area of quadrilateral:
= Area of ∆ABC + Area of ∆ACD = 10.5 + 17.5 = 28 sq. units.
∴ Area of quadrilateral ABCD = 28 sq.units.

Question 5.
You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 11
Solution:
AD Median is drawn to side BC which is the vertex of ∆ABC.
Median AD divides AABC into two triangles ∆ABD and ∆ADC which are equal in area.
Now Median AD bisects BC.
∴ BD = DC.
i) As per the Mid-Points formula,
Coordinates of D are
= \(\sqrt{(4)^{2},(0)^{2}}\)
= \(\sqrt{16,0}\)
= (4,0)
∴ Coordinates of D are (4, 0)

ii) Now, Area of ∆ABD:
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 12

iiii) Now, Area of ADC:
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 13
= \(\frac{1}{2}\) × -6
= – 3 sq. units.
∴ Area of ∆ADC = -3 sq. units.
∴ Area of ∆ABC :
= Area of ∆ABD + Area of ∆ADC
= (-3) +(-3)
= -6 sq. units.

iv) Now, Area of ∆ABC : (Direct Method)
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 14
KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 15

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3, drop a comment below and we will get back to you at the earliest

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