KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3

   

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Solutions Chapter 3 Lines and Angles Exercise 3.3.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.3

Question 1.
In Fig. 3.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 1
Answer:
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 2
Arms of the ∆PQR QP and RP are produced to S and T and ∠PQT = 110°, ∠SPR = 135°. ∠PRQ =?
Straight-line PR is on straight line SQ.
∠SPR and ∠RPQ are Adjacent angles.
∴ ∠SPR + ∠RPQ = 180°
135 + ∠RPQ = 180°
∴ ∠RPQ =180 – 135
∠RPQ = 45° (i)
Similarly QP straight line is on straight line TR.
∠TQP and ∠PQR are Adjacent angles.
∴ ∠RQP + ∠PQR = 180°
110 + ∠PQR = 180°
∠PQR = 180 – 110
∴ ∠PQR = 70°
Now, in ∆PQR,
∠QPR + ∠PQR + ∠PRQ = 180°
45 + 70 + ∠PRQ = 180°
115 + ∠PRQ = 180°
∠PRQ = 180 – 115
∴ ∠PRQ = 65°.

Reference Angle Calculator. The reference angle is defined as the smallest possible angle made by the terminal side of the given angle with the x-axis.

Question 2.
In Fig. 3.40, ∠X = 62°. ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 3
Answer:
In this figure, ∠X = 62°
∠XYZ = 54°
In ∆XYZ, YO and ZO are angular bisectors of ∠XYZ and ∠XZY.
Then, ∠OZY =?
∠OYZ =?
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 4
In ∆XYZ
∠X + ∠Y + ∠Z = 180°
62 + 54 + ∠Z= 180
116 + ∠Z= 180
∠Z= 180- 116
∴ ∠Z = 64°
YO is the angular bisector of ∠Y
∴ ∠OYZ = \(\frac{54}{2}\) = 27°
ZO is the angular bisector of ∠Z
∴ ∠OZY = \(\frac{64}{2}\) = 32°
∴ ∠OZY = 32°
Now, in ∆OYZ,
∠OYZ + ∠OZY + ∠YOZ = 180°
27 + 32 + ∠YOZ = 180
59 + ∠YOZ = 180
∠YOZ = 180 – 59
∴∠YOZ = 121°
∴ ∠OZY = 32°
∠YOZ = 121°

Question 3.
In Fig. 3.41, if AB||DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 5
Answer:
If AB || DE, ∠BAC = 35, ∠CDE = 53 then ∠DCE = ?
AB || DE, AE is the bisector.
∴∠BAC = ∠DEC = 35° (∵ Alternate angles)
∴∠DEC= 35°
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 6
Now in ∆CDE,
∠DCE + ∠CDE + ∠CED = 180°
∠DCE + 53 + 35 = 180
∠DCE + 88 = 180
∠DEC = 180 – 88
∴ ∠DCE = 92°.

Question 4.
In Fig. 3.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 7
Answer:
PQ and RS straight lines intersect at T.
If ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, then ∠SQT =?
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 8
In ∆PRT,
∠RPT + ∠PRT + ∠PTR = 180°
95 + 40 + ∠PTR = 180°
135 + ∠PTR = 180
∠PTR = 180 – 135
∴ ∠PTR = 45°
∠PTR = ∠STQ = 45° (∵ Vertically opposite angles)
In ∆TSQ,
∠STQ + ∠TSQ + ∠SQT =180
45 + 75 + ∠SQT = 180
120 + ∠SQT = 180
∴∠SQT = 180 – 120
∴ ∠SQT = 60°.

Question 5.
In Fig. 3.43, PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°. then find the value of x and y.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 9
Answer:
PQ ⊥ PS, PQ ⊥ SR, ∠SQR = 28°, ∠QRT = 65°, Then x = ?, y = ?
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 10
Solution: ∠QRT + ∠QRS = 180° (∵ Linear pairs)
65 + ∠QRS = 180
∠QRS = 180 – 65
∴∠QRS =115°
In ∆SRQ,
∠QRS + ∠SQR + ∠RSQ = 180°
115 + 28 + ∠RSQ = 180
∴∠RSQ =180 – 143
∴∠RSQ = 37
Now, ∠RSQ = ∠PQS
37° = x
∴x = 37
In ∆SPQ,
∠SPQ + ∠PSQ + ∠PQS = 180°
90 + y + 37 = 180
∴y = 180- 127
∴y = 53°.

Question 6.
In Fig. 3.44, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\)∠QPR.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 11
Answer:
Data: Arm QR of ∆PQR is produced upto S. Angular bisectors of ∠PQR and ∠PRS meet at T.
To Prove: ∠QTR = \(\frac{1}{2}\) ∠QPR
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 12
Proof: Let ∠PQR = 50°, and ∠PRS = 120°
∠PQT = ∠TQR = 25°
∠PRT = ∠TRS = 60°
QR arm of ∆ PQR is produced upto S.
∴Exterior angle ∠PRS = ∠PQR + ∠QPR
120 = 50 + ∠QPR
∴ ∠QPR = 120 – 50
∠QPR = 70°
∴ ∠PRQ = 60°
Now, in ∆TRQ,
∠TQR + ∠TRQ + ∠QTR = 180°
25 + 120 + ∠QTR = 180°
145 + ∠QTR = 180°
∠QTR = 180 – 145
∴ ∠QTR = 35°
Now, ∠QTR = 35° ∠QPR = 70°
∠QTR = \(\frac{70}{2}\)
∴ ∠QTR = \(\frac{10}{2}\) x ∠QPR.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2

   

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.2

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
i) x = 0
ii) x = -1
iii) x = 2
Answer:
i) f(x) = 5x – 4x2 + 3 x = 0 then,
f(0) = 5(0) – 4(0)2 + 3
=0 – 0 + 3
f(0) = 3

ii) f(x) = 5x – 4x2 + 3 x = -1 then,
f(-1) = 5(-1) – 4(-1)2 + 3
= -5 – 4(+1) + 3
= -5 – 4 + 3
= -9 + 3
f(-1) = -6

iii) f(x) = 5x – 4x2 + 3 x = 2 then,
f(2) = 5(2) – 4(2)2 + 3
= 5(2) – 4(4) + 3
= 10 – 16 + 3
= 13 – 16
f(2) = -3

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials :
i) p(y) = y2 – y + 1
ii) p(t) = 2 + t + 2t2 – t3
iii) p(x) = x3
iv) p(x) = (x – 1) (x + 1)
Answer:
i) (a) p(y) = y2 – y + 1
p(0) = (0)2 – 0 + 1
= 0 – 0 + 1
∴ p(0) = =1

(b) p(y) = y2 – y + 1
p(1) = (1)2 – 1 + 1
= 1 – 1 + 1
∴ p(1) = 1

(c) p(y) = y2 – y +
p(2) = (2)2 – 2 + 1
= 4 – 2 + 1
= 5 – 2
∴ p(2) = 3

ii) (a) p(t) = 2 + t + 2t2 – t3
p(0) = 2 + 0 + 22 – (0)3
= 2 + 0 + 0 – 0
∴ p(0) = 2

(b) p(t) = 2 + t + 2t2 – t3
p(1) = 2 + 1 + 2(1)2 – (1)3
= 2 + 1 + 2(1) – 1
= 2 + 1 + 2 – 1
∴ p(1) = 4

(c) p(t) – 2 + t + 2t2 – t3
p(2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 2(4) – 8
= 2 + 2 + 8 – 8
∴ p(2) = 4

iii) (a) p(x) = x3
p(0) = (0)3
p(0) = 0

(b) p(x) = x3
p(1) = (1)3
∴ p(1) = 1

(c) p(x) = x3
p(2) = (2)3
∴ p(2) = 8

iv) (a) p(x) = (x – 1)(x + 1)
p(x) = x2 – 1 [∵ (a + b)(a – b) = a2 – b]
p(0) = (0)2 – 1
= 0 – 1
∴ p(o) = -1

(b) p(x) = (x – 1)(x + 1)
p(x) = x2 – a [∵ (a + b)(a – b) = a2 – b2]
p(1) = (1)2 – 1
= 1 – 1
∴ p(1) = 0

(c) p(x) = (x – 1)(x + 1)
p(x) = x2 – 1 [∵ (a + b)(a – b) = a2 – b2]
p(2) = (2)2 – 1
= 4 – 1
∴ p(0) = 3

Question 3.
Verify whether the following are zeroes of the polynomial, induced against them.
i) p(x) = 3x + 1; \(x=-\frac{1}{3}\)
ii) p(x) = 5x – π; \(x=\frac{4}{5}\)
iii) p(x) = x2 – 1; x = 1, -1
iv) p(x) = (x + 1) (x – 2); x = -1, 2
v) p(x) = x2; x = 0
vi) p(x) = lx + m; \(x=-\frac{m}{l}\)
vii) p(x) = 3x2 – 1; \(x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)
Answer:
i) p(x) = 3x + 1; \(x=-\frac{1}{3}\)
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 1
Here value of polynomial is zero.
\(x=-\frac{1}{3}\) is not zero of the polynomial

ii) p(x) = 5x – π; \( x=\frac{4}{5}\)
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 2
Here value of polynomial is not zero.
\(x=\frac{4}{5}\) is not zero of the polynomial

iii) p(x) = x2 – 1; x = 1, -1
p(1) = (1)2 – 1
= 1 – 1
p(1) = 0
Here value of p(x) is zero.
hence its zero is 1.
p(x) = x2 – 1; x = -1
p(-1) = (-1)2 – 1
= 1 – 1
p(-1) = 0
Here value of p(x) is zero.
∴ -1 is zero.

iv) p(x) = (x – 1)(x – 2); x = -1, 2
p(x) = x2 – 2x + x – 2
p(x) = x2 – x + 2 x = -1
p(-1) = (-1)2 – (-1)2 + 2
= 1 + 1 + 2
p(-1) = 4
Here value of polynomila is not zero.
∴ -1 is not zero.
p(x) = x2 – x + 2 x = 2
p(2) = (2)2 – (2) + 2
= 4 – 2 + 2
p(-1) = 4
Here value of polynomial is not zero.
∴ 2 is not zero.

(v) p(x) = x2; x = 0
p(0) = (0)2
p(0) = 0
Here value of p(x) is zero.
∴ 0 is its zero.

vi) p(x) = lx + m; \(x=-\frac{m}{l}\)
\(\mathrm{p}\left(-\frac{\mathrm{m}}{l}\right)=l\left(-\frac{\mathrm{m}}{l}\right)+\mathrm{m}\)
= -m + m
= 0
Here p(x) is zero.
∴ \(-\frac{\mathrm{m}}{l}\) is its zero.

(vii) p(x) = 3x2 – 1; \(x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 3

viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 4
Here value of p(x) is not zero.
∴ \(\frac{1}{2}\) is not its zero.

Question 4.
Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Answer:
i) p(x) = x + 5
Let p(x) =0, then,
p(x) = x + 5 = 0
x = 0 – 5
∴ x = -5
-5 is zero of p(x).

ii) p(x) = x – 5
If p(x) = 0, then
p(x) = x – 5 = 0
x = 0 + 5
∴ x = 5
5 is the zero of p(x).

iii) p(x) = 2x + 5
If p(x)= 0, then
p(x) = 2x + 5 = 0
2x = – 5
∴ \(x=\frac{-5}{2}\)
\(\frac{-5}{2}\) is the zero of p(x).

iv) p(x) = 3x – 2
If p(x)= 0, then
p(x) = 3x – 2 = 0
3x = 2
∴ \(x=\frac{2}{3}\)
\(\frac{2}{3}\) is the zero of p(x).

v) p(x) = 3x
If p(x) = 0, then
p(x) = 3x = 0
∴ \(x=\frac{0}{3}\)
\(\frac{0}{3}\) is the zero of p(x)

vi) p(x) = ax, a ≠ 0
If p(x)= 0, then
p(x) = ax = 0
∴ \(x=\frac{0}{a}\) ∴ x = ∞(infinity)
∞ is the zero of p(x).

vii) p(x) = cx + d, c ≠ 0, c, d are real numbers
If p(x)= 0, then
p(x) = cx + d = 0
cx = 0 – d
cx = -d
∴ \(x=-\frac{d}{c}\)
\(-\frac{\mathrm{d}}{\mathrm{c}}\) is the zero of p(x).

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.1

   

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
i) 4x2 – 3x + 7
ii) y2 + \(\sqrt{2}\)
iii) \(3 \sqrt{t}+t \sqrt{2}\)
iv) \(\mathrm{y}+\frac{2}{\mathrm{y}}\)
v) x10 + y3 + t50
Answer:
i) 4x2 – 3x + 7
Here polynomial has one variable, i.e. x
ii) y2 + \(\sqrt{2}\)
Here polynomial has one variable, i.e. y
iii) \(3 \sqrt{t}+t \sqrt{2}\)
This is polynmomial with one variable, because T is only one variable.
iv) \(\mathrm{y}+\frac{2}{\mathrm{y}}\)
Here polynomial has one variable, ie. y.
v) x10 + y3 + t50
This polynomial is not having one variable because here 3 variables means ‘x’, y and ‘t’ are there.

Question 2.
Write the coefficients of x in each of the followng :
i) 2 + x2 + x
ii) 2 – x2 + x3
iii) \(\frac{\pi}{2}\)x2 + x
v) \(\sqrt{2} \mathrm{x}\) – 1
Answer:
i) 2 + x2 + x
Here, coefficient of x2 is 1.
ii) 2 – x2 + x3
Here coefficient of x2 is -1
iii) \(\frac{\pi}{2}\)x2 + x
Here coefficient of x2 is \(\frac{\pi}{2}\).
iv) \(\sqrt{2} \mathrm{x}\) – 1
Here coefficint of x2 is -1.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100
Answer:
i) A Bionomial of degree 35
E.g. f(x) = – x35 + 10
ii) A binomial of degree 100
E.g. f(y) = – y100.

Question 4.
Write the degree of each of the following polynomials :
i) 5x3 + 4x2 + 7x
ii) 4 – y2
iii) 5t – \(\sqrt{7}\)
iv) 3
Answer:
i) 5x3 + 4x2 + 7x Highest power (degree) 3
ii) 4 – y2 Highest power degree) 2
iii) 5t – \(\sqrt{7}\) Highest power (degree) 1
iv) 3 Highest power (degree) 0

Question 5.
Classify the folloiwng as linear, quadratic and cubic polynomials :
i) x2 + x
ii) x – x3
iii) y + y2 + 4
iv) 1 + x
iii) 3t
iv) r2
vii) 7x3
Answer:

Linear Polynomial Quadratic Polynomial Cubic Polynomial
iv) 1 + x i) x2 + x iii) y + y2 + 4
ii) x – x3
(v) 3t (vi) r2 (vii) 7x3

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1, drop a comment below and we will get back to you at the earliest.

Tili Kannada Text Book Class 9 Solutions Gadya Chapter 1 Avare Rajaratnam!

   

Students can Download Kannada Lesson 1 Avare Rajaratnam! Questions and Answers, Summary, Notes Pdf, Tili Kannada Text Book Class 9 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Tili Kannada Text Book Class 9 Solutions Gadya Bhaga Chapter 1 Avare Rajaratnam!

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KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

   

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.4.

Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches :
2, 3, 4, 5, 0, 1. 3, 3, 4, 3
Find the mean, median and mode of these scores.
Solution:
Number of goals scored by a team:
(i) 2, 3, 4, 5, 0, 1, 3, 3, 4, 3
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 1
= 2.8

(ii) If number of goals are arranged in ascending order,
0, 1, 2, 3, 3. 3, 3, 4, 4. 5
No. of observations =10 (even number)
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 1.1
= 5th observation.
= 3
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 1.2
= 5 + 1
= 6th observation ➝ 3
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 1.3
= 3

(iii) Mode is the value of the observation which occurs most frequently.
Here 3 is repeated 4 times.
∴ Mode = 3.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded : 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution:
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 2
Mode= 54.8

(ii) When marks are arranged in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations = 15 (odd number)
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 2.1
∴ Median = 52.

(iii) Mode: It is the repeated value.
Here 52 is repeated three times.
∴ Mode = 52.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63. find the value of x,
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Median = 63.
Number of observations = 10 (even number)
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 3
∴ x + 1 = 63 ∴ x = 63 – 1
∴ x = 62.

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Solution:
14, 25, 14, 28, 18, 17, 18, 14, 23, 22; 14, 18
Mode = ?
Arranging the scores in ascending order,
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Mode is the repeated value.
∴ Here 14 is repeated 4 times.
∴ Mode = 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 5
Solution:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Q 5.1
∴ The men salary of workers is Rs. 5083.33.

Question 6.
Give one example of a situation in which
(i) The mean is an appropriate measure of central tendency.
(ii) The mean is not an appropriate measure of central tendency but the median is appropriate of central tendency.
Solution:
(i) Each score is in lesser difference, it is easy to calculate the average than the median.
Ex: monthly salary of 5 persons, 10000, 10100, 10200, 10300, 10400
Mean of the scores is 10200.

(ii) If scores have more differences, it is easy to calculate the median than the mean.
Ex.: Marks obtained by 7 students in Mathematics :
2, 10, 20, 15, 4, 23, 3
Ascending Order : 2, 3, 4, 10, 15, 20, 23
Median =10 but the mean is 11.

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KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

   

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.3.

Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %).
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 10
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) Scale: x axis ➝ 1 horizontal axis = 1 cm.
y axis ➝ 5% = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 1.1
(ii) Reproductive health conditions.
(iii) Nerve and skin.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys In different sections of Indian society is given below.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 11
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
(i) Scale: x axis ➝ 1 horizontal = 1 cm.
y axis ➝ 100 girls = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 2.1
(ii) After verification we came to know that Strength of Girls is more in ST section and it is less in Urban section.

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 12
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats ?
(ii) A Political Party won the maximum number of seats.
Solution:
(i) Scale: x axis ➝ 1 horizontal(Party) = 1 cm.
y axis ➝ 10 Parties = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 3.1
(ii) ‘A’ political Party won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 13
(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data ?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long ? Why ?
Solution:
(i) Scale: x axis ➝ 1 Class interval = 1 cm.
y axis ➝ 2 leaves = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 4.1
(ii) Frequency Polygon.
(iii) No. Because from 145 to 153, length of leaves is 153.

Question 5.
The following table gives the life times of 400 neon lamps:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 14
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours ?
Solution:
(i) Scale: x axis ➝ 1 Class interval = 1 cm.
y axis ➝ 10 Bulbs = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 5.1
(ii) 184 lamps have a life time of more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 15
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
(i) Scale: x axis ➝ 5 marks = 1 cm.
y axis ➝ 2 frequency = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 6.1
Section B students scored less marks in great numbers.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 16
Represent the data of both the teams on the same graph by frequency polygons.
(Hint: First make the class intervals continuous.)
Solution:
Class interval is continuous,
1 – 6, 7 – 12 Here the difference is 1.
∴ \(\frac{1}{2}\) = 0.5 is lower limit is subtracted,
0.5 is taken upper limit, it becomes
0.5 – 6.5
6.5 – 12.5.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 7.1

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 17
Scale: x-axis ➝ 3.5 mark = 1 cm.
y – axis ➝ 1 frequency = 1 cm
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 7.3

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 18

Draw a histogram to represent the data above.
Solution:
Scale: x-axis ➝ 1 year = 1 cm.
y – axis ➝ 1 child = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 8.2

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 9
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie
Solution:
(i) Scale: x-axis ➝ 1 letter = 1 cm.
y – axis ➝ 4 surnames = 1 cm.
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 Q 9.1
(ii) Class interval which has maximum number of surnames is : 6 – 8.

We hope the KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6

   

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6

(Assume π = \(\frac{22}{7}\) unless stated otherwise)

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm3 = 1 l)
Solution:
The circumference of the base of cylindrical vessel, C = 132 cm.
height, h = 25 cm
C = 2πr = 132
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 1
r = 21 cm
∴ Volume of Cylinder, V = πr2h
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 1.1
= 34650 cm3
1000 cubic cm. = 1 litre
∴ 34650 cm3 = … ? …
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 1.2

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution:
Let the inner diameter of a cylindrical vessel be d1 and outer diameter be d2.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 2
height of pipe, h = 35 cm.
∴ volume of pipe, V = \(\pi r_{2}^{2} h-\pi r_{1}^{2} h\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 2.1
= 110 × 52
∴ V = 5720 cm3
Mass of pipe of 1 ccm is 0.6 gm.
∴ Mass of 5720 cm3. … ? …
= 5720 × 0.6
= 3432 Kg.

Question 3.
A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much ?
Solution:
Length of rectangular tin, l = 5 cm,
breadth, b = 4 cm,
height, h = 15 cm.
∴ Volume of rectangular tin, V = l × b × h
V = 5 × 4 × 15
= 300 cm3
Diameter of plastic cylinder, d = 7 cm,
∴ r = \(\frac{7}{2}\)
Volume of cylinder, v = \(\pi \mathrm{r}^{2} \mathrm{h}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 3
= 11 × 7 × 5
= 385 cm3.
∴ Plastic cylinder’s volume is great.
= Volume of cylindrical tin – Volume of rectangular tin.
= 385 – 300
= 85 cm3.
∴ Volume of plastic tin is greater than rectangular tin by 85 cm3.

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base,
(ii) its volume. (Use π = 3.14)
Solution:
Curved surface area of cylinder = 94.2 cm2
height, h = 5 cm.
r = ?
V = ?
(i) Curved Surface area of cylinder, L.S.A.
A= 2πrh
94.2 = 2 × 3.14 × 5 × r
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 4
= 3.01 cm.
= 3 cm.
(ii) Volume of cylinder, V = πr2h
= 3.14 × (3)2 × 5
= 3.14 × 9 × 5
= 141.3 cm3.

Question 5.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m2, find
(i) inner curved surface area of the vessel.
(ii) radius of the base,
(iii) capacity of the vessel.
Solution:
(i) Cost of painting for cylindrical vessel = Rs. 2200
Cost of painting is at the rate of Rs. 20 per m2.
∴ Curved Surface area of vessel = \(\frac{2200}{20}\)
= 110 m2

(ii) Depth of vessel, h = 10 m
r = ?
Curved surface area of vessel, V = 2πrh
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 5
(iii) Capacity of the Vessel = Volume of Vessel
V = πr2h
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 5.1
V = 96.25 m3.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Solution:
Volume of a cylindrical vessel = 15.4 litres
1000 c.cm. = 1 litre.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 6
height, h = 1 m,
radius, r = ?
Curved surface area =?
Volume of cylinder, V = πr2h
0. 0154 = \(\frac{22}{7}\) × r2 × 1
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 6.1
r = 0.7 m
Now, curved surface area of cylinder vessel,
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 6.2
= 0.44 × 1.07
A = 0.4708 m2

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7
Solution:
Diameter of pencil, d = 7 mm = 0.7 m.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7.1
length of pencil, h = 14 cm
diameter of pencil, d = 1 mm = 0.1 cm
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7.2
Volume of wood in pencil, V
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 7.3

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
Diameter of pencil, d = 7 cm
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6 Q 8
height of soup, h = 4 cm
Volume of soup given for 1 patient is 154 cm3
Volume of soup given for 250 patients .. ? ….
= 154 × 250 = 38500 cm3
For 1000 cm3, 1 litre
For 38500 cm3 ….. ? …
= \(\frac{38500}{1000}\)
= 38.5 litres soup.

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

   

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.2.

Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The blood groups of 30 students of Class VIII are recorded as follows :
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B. O
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
Answer:
The blood groups of 30 students of class VIII are as follows
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 1
i) Out of 30 students only three students have blood group AB. This is rarest blood group.
ii) Most common blood group students are 12. Hence ‘O’ is the maximum blood group.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 2.1
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 – 5 (5 not included). What main features do you observe from this tabular representation ?
Solution:
Size of the class interval = 5
First class interval is 0 – 5 (5 is not included).
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 2
More number of people, i.e. 5 to 15 Km come.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 3
(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about ?
(iii) What is the range of this data ?
Solution:
Size of class interval = 2
(i) Frequency distribution table :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 3.1
(ii) Relative humidity is more, these data are taken during rainy season.
(iii) Maximum humidity = 99.2
Minimum humidity = 84.0
∴ Range = Maximum – Minimum
= 99.2 – 84.9
= 14.3

Question 4.
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 4
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table ?
Solution:
Maximum height : 150 cm.
Minimum height : 173 cm.
Size of class interval = 5
(i) Frequence distribution table for grouped data:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 4.1
(ii) Referring to above table, 50% students have height less than 165 cm.
Height of majority students is 160 – 165 cm.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 5
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per millon ?
Solution:
(i) Grouped frequency distribution table:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 5.1
(ii) In 8 days the concentration of sulphur dioxide is more than 0.11 parts per million.

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 6
Prepare a frequency distribution table for the data given above.
Solution:
Frequency distribution table :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 6.1

Question 7.
The value of n upto 50 decimal places is given below :
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits ?
Solution:
Frequency distribution table :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 7
(ii) Digits which have more frequency are 3 and 9.
‘0’ is the digit which has less frequency.

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 8
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
(ii) How many children watched television for 15 or more hours a week ?
Solution:
(i) Size of Class Interval = 5
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 8.1
(ii) Number of children watched TV for 15 days or more hours a week is 2.

Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows :
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 9
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution:
Size of the class interval = 0.5
Class interval : 2 – 2.5, 2.5 – 3 …….
Grouped Frequency distribution table:
KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 Q 9.1

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Statistics Ex 14.2 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Statistics Exercise 14.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5

   

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Solution:
Match box is in the form of cuboid.
Its length, l = 4 cm.
breadth, b = 2.5 cm.
height, h = 1.5 cm.
∴ Volume of Cuboid, V = l × b × h
= 4 × 2.5 × 1.5
V = 15 cm3.
Volume of 1 match box is 15 cm3.
Volume of 12 match boxes …?… .
= 15 × 12
= 180 cm3.

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold ? (1 m3 = 1000 l)
Solution:
Length of cuboidal water tank, l = 6 m
breadth b = 5 m
height, h =4.5 m
∴ Volume of cuboidal water tank, V= l × b × h
= 6 × 5 × 4.5
V= 135 m3.
1 m3 = 1000 l.
135 m3 = ? = 135 × 1000
= 135000 lit.
∴ Number of litres of water tank = 1350000 litres.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid ?
Solution:
Volume of cuboidal vessel, V = 380 m3.
length, l = 10 m
breadth, b = 8 m height, h = ?
Volume of vessel, V= l × b × h
380 = 10 × 8 × h
380 = 80 h
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 3
∴ h = 4.75 m.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3.
Solution:
Length of cuboidal pit, l = 8 m
breadth, b = 6 m height, h = 3 m
∴ Volume of pit, V = l × b × h
= 8 × 6 × 3
= 144 m3.
∴ Cost of digging a cuboidal 1 m3 is Rs. 30
Cost of digging a cuboidal pit for 144 m3 … ? …
= 144 × 30
= Rs. 4320.

Question 5.
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
Volume of cuboidal tank, V = 50000 lit.
1 c.c = 1000 litres
∴ Volume of tank = \(\frac{50000}{1000}\) = 50 c.c
length of tank, l = 2.5 m
height, h = 10m
breadth,b = ?
Volume of cuboid, V = l × b × h
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 5
∴ b = 2m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last ?
Solution:
Length of cuboid water tank, l = 20 m
breadth, b = 15 m
height, h = 6 m
Volume, V = ?
Volume of water tank, V = l × b × h
= 20 × 15 × 6
= 1800 m3.
1m3 = 1000 litres
∴ 1800m3 = 1800 x 1000
= 1800000 litres.
Water required daily per man= 150 litres
∴ Quantity of water required for 4000 men?
= 4000 × 150
= 600000 litres.
Quantity of water for 1 day = 600000
Number of days for consuming 1800000 litres … ? …
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 6
= 3 Days.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown
Solution:
Length of rectangular godown, l1 = 40 m
breadth, b1 = 25 m
height, h1 = 10 m
∴ Volume of godown, V= l1 × b1 × h1
= 40 × 25 × 10
= 10000 m3.
Length of wooden crate, l2 = 1.5 m
breadth, b2 = 1.25 m
height, h2 = 0.5 m
∴ Volume of wooden crate. V = l2 × b2 x h2
= 1.5 × 1.25 × 0.5
= 0.9375 m3.
For 0.9375 m3, 1 crate
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 7
= 10666.66

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Each side of a cube, a = 12 cm.
∴ Volume of cube, V = a3
= (12)3
= 1728 cm3.
Solid cube is cut into 8 cubes of equal volume.
Volume of each small cube,
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 8
∴ l3 = 216
∴ l = 6 cm.
∴ Each side of small cube is 6 cm.
Ratio of outer area :
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 8.1

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
Depth of river, h = 3m
breadth, b = 40 m
1 Km. = 1000 m
∴ 2 Km. = 2000 m.
Speed of the river per hour is 2 Km.
For 60 minutes, 2000m
For 1 minute …?… = \(\frac{2000}{60}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5 Q 9
∴ Depth of river in 1 minute, l = \(\frac{100}{3}\) m.
Volume of water flows in 1 minute,
V = l × b × h
= \(\frac{100}{3}\) × 40 × 3
= 4000 m3
∴ In One minute, 4000 m3 water reaches the sea.

KSEEB Solutions for Class 9 Maths

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4

   

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4

(Assume π = \(\frac{22}{7}\), unless stated otherwise)

Question 1.
Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm.
Solution:
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 1
Surface area of the sphere = \(4 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 1.1
= 22 × 3 × 21
= 1386 cm2.

(ii) r = 5.6 cm. = \(\frac{56}{10}\)cm.
Surface area of the sphere = \(4 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 1.2
= 394.24 cm2.

(iii) r = 14 cm.
Surface area of the sphere = \(4 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 1.3
= 88 × 28
= 2464 cm2.

Question 2.
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m.
Solution:
(i) Diameter, d = 14 cm.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 2
Surface area of a sphere = \(4 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 2.1
= 616 cm2.
(ii) Diameter, d = 21 cm.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 2.2
Surface area of a sphere = \(4 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 2.3
= 1386 cm2.
(iii) Diameter, d = 3.5 m. = \(\frac{7}{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 2.4
Surface area of a sphere = \(4 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 2.5
= 38.5 m2.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm.
(Use π = 3.14 )
Solution:
r = 10 cm.
Total surface area of a hemisphere = \(3 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 3
= 942 cm2.

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Radius of spherical balloons be r1 and r2, r1 = 7 cm. and r2 = 14 cm.
Ratio of surface area
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 4
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 4.1
∴ Ratio of surface areas of the balloons = 1 : 4.

Question 5.
A hemispherical bowl made of brass has an inner diameter of 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2.
Solution:
Inner diameter of hemispherical bowl = 10.5 cm.
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 5
Curved surface area of the bowl = \(2 \pi r^{2}\)
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 5.1
= 173.25 cm2.
Cost of tin-plating for 100 cm2 is Rs. 16.
Cost of tin-plating for 173.25 sq.cm. …?…
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 5.2
= Rs. 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm2.
Solution:
Total surface area of a sphere= 154 cm2
Radius, r =?
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 6
∴ r = 3.5 cm

Question 7.
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the diameter of earth be d’ unit
∴ Diameter of Moon = \(\frac{d}{4}\)
radius of the earth = \(\frac{\mathrm{d}}{2}\)
radius of the moon = \(\frac{1}{2} \times \frac{\mathrm{d}}{4}=\frac{\mathrm{d}}{8}\)
Ratio of surface area,
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 7
= 1 : 16

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Thickness of the hemispherical bowl = 0.25 cm.
inner radius, r = 5 cm.
∴ Outer Surfce area= 5 + 0.25 = 5.25 cm.
Outer Curved Surface area = 2πr2
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 8
= 173.25 cm2.

Question 9.
A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder
(iii) ratio of the areas obtained in (i) & (ii).

Solution:
(i) If radius of a sphere is ‘r’ cm, its Surface area = 4πr2
(ii) height of cylinder, h = diameter of sphere
h = r + r
∴ h = 2r
∴ Curved surface area of cylinder = 2πrh
= 2πr × h
= 2πr × 2r
= 4πr2
KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4 Q 9.1
= 1 : 1

KSEEB Solutions for Class 9 Maths
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