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.

KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1

   

KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.1

Question 1.
Complete the following statements :

  1. Probability of an event E + Probability of the event ‘not E’ = _____
  2. The Probability of an event that cannot happen is _____. Such an event is called ____.
  3. The probability of an event that is certain to happen is ____. Such an event is called ____.
  4. The sum of the probabilities of all the elementary events of an experiment is ____.
  5. The probability of an event is greater than or equal to ____ and less than or equal to ____.

Answers:

  1. 1 : Probability of an event E + Probability of the event ‘not E’ = 1.
  2. 0, impossible: The probability of an event that cannot happen is 0. Such an event is called impossible event.
  3. 1, certain: The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
  4. 1: The sum of the probabilities of all the elementary events of an experiment is 1.
  5. 0, 1: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) The outcome is not equally likely because the car starts normally only when there is some defect, the car does not start.
(ii) The outcome is not equally likely because the outcome depends on the training of the player.
(iii) The outcome in the trial of true-false question is, either true or false. Hence, the two outcomes are equally likely.
(iv) A baby can be either a boy or a girl and both the outcomes have equally likely chances.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer:
Because Tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game. By this coin may be head or tail (only one). It has an equally likely outcome.

Question 4.
Which of the following cannot be the probability of an event?
A) \(\frac{2}{3}\)
B) -1.5
C) 15%
D) 0.7.
Answer:
B) -1.5; This cannot be the probability of an event because possibility of event should not be less than 0 and more than 1. Hence -1.5 is lesser than 0.

Question 5.
If P(E) = 0.05, what is the probability of ‘not E’ ?
Solution:
IfP(E) = 0.05, then P(\(\overline{\mathrm{E}}\))=?
But, P(E) + P(\(\overline{\mathrm{E}}\)) = 1 .
∴ P(\(\overline{\mathrm{E}}\)) = 1 – 0.05
∴ P(\(\overline{\mathrm{E}}\)) = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag what is the probability that she takes out.
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Bag has only lemon flavoured candies. It has no orange candies.
∴ Possibility, P(E) = 0.
(ii) A lemon flavoured candy is possible. Because Bag contains all lemon flavoured candies.
∴ Possibility, P(F) = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let P(E) be an event 2 students have the same birthday i.e P(E) =?
P(\((\bar{E})\)) is an event 2 students not have same birth day P(\((\bar{E})\)) = 0.992
P(E) + P(\((\bar{E})\)) = 1
P(E) = 1 – P(\((\bar{E})\)) = 1.000 – 0.992 = 0.008
The probability of 2 students having the same birthday is 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red ?
Solution:
A bag contains 3 red balls, 5 black balls. Totally there are 8 balls.
∴ n(S) = 8
i) Possibility that the red ball drawn,
n(E) = 3
∴ Probability, P(E) = \(\frac{n(E)}{n(S)}=\frac{3}{8}\)
ii) Possibility that the 1 black ball drwn is
n(F) = 5
∴ Probability, P(F) = \(\frac{n(F)}{n(S)}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ?
(ii) white ?
(iii) not green ?
Solution:
A box contains 5 red marbles, 8 white marbles and 4 green marbles.
∴ Total number of marbles, n(S)
= 5 + 8 + 4
= 17
(i) Probability that the 1 red marble drawn is n(A) = 5
∴ Probability, P(A) = \(\frac{n(A)}{n(S)}=\frac{5}{17}\)
(ii) Possibility that 1 white marble drawn, n(B) = 8
∴ Probability, P(B) = \(\frac{n(B)}{n(S)}=\frac{8}{17}\)
(iii) Possibility that 1 not green marble ?
P(C) = 17 – 4 = 13 (∵ Except 4 green marbles)
∴ Probability, P(C) = \(\frac{n(C)}{n(S)}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin ?
(ii) will not be a Rs. 5 coin ?
Solution:
Number of 50 ps coins = 100
Number of Re. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
∴ Total number of coins, n(S) = 180
(i) Possibility of one 50 ps coin :
n(A) = 100
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14 Q 10
(ii) Possibility of one Rs. 5 coin:
n(B) = 180 – 10 = 170
(∵ 10 coins are Rs. 5)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14 Q 10.1

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper taks out one fish at random from a tank containing 5 male fish and 8 female fish (see figure given)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 11
What is the probability that the fish taken out is a male fish ?
Solution:
Number of male fish = 5
Number of female fish = 8
∴ Total number of fish, n(S) = 5 + 8 = 13
Probability that the fish taken out is male,
n(E) = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 11.1

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see the Figure) and these are equally likely outcomes.
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12
What is the probability that it will point at
(i) 8?
(ii) an odd number
(iii) a number greater than 2 ?
(iv) a number less than 9 ?
Solution:
Total numbers in the spinning = 8
∴ n(A) = 1
(i) Possibility that points one number is 8 : n(A) = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.1
(ii) Possibility that points one odd number : 1, 3, 5, 7
∴ n(B) = 4
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.2
(iii) A number greater than 2 :
3, 4, 5, 6, 7, 8
∴ n(C) = 6
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.3
(iv) A number less than 9 :
1, 2, 3, 4, 5, 6, 7, 8
∴ n(D) = 8
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.4

Question 13.
A die thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Possible event when a die is thrown:
1, 2, 3, 4, 5, 6
∴ n(S) = 6
(i) Probability of getting a Prime number : 2, 3, 5
∴ n(A) = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 13
(ii) A number lying between 2 and 6 : 3, 4, 5
∴ n(B) = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 13.1
(iii) Possibility of odd number : 1, 3, 5
∴ n(C) = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 13.2

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour,
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Solution:
Total number of cards = 52
(i) P(One red king) = \(\frac{2}{52}=\frac{1}{26}\)
(ii) Number of a face card :
4 king, 4 queen, 4 jack = Total 12 cards
∴ P(1 face card) = \(\frac{12}{52}=\frac{3}{13}\)
(iii) One red colour face card = 6
∴ P(1 red face card) = \(\frac{6}{52}=\frac{3}{26}\)
(iv) P(Heart Jack) = \(\frac{1}{52}\)
(v) P(1 spade) = \(\frac{13}{52}=\frac{1}{4}\)
(vi) P(Diamond card) = \(\frac{1}{52}\)

Question 15.
Five cards— the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is
(a) an ace?
(b) a queen ?
Solution:
Five cards— the ten, jack, queen, king and ace of diamonds.
(i) Probability of getting queen = \(\frac{1}{5}\)
(ii) If the queen is drawn and put aside, the probability of getting other cards = 4.
(∵ 5 – 1 = 4)
∴ a) Probability that the card is ace = \(\frac{1}{4}\)
b) Probability that the card is queen = \(=\frac{0}{4}=0\)

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens =12
Number of good pens =132
Total Number of pens = 12 + 132 = 144
The probability that the pen taken out is good one,
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 16

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that the bulb is defective ?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution:
Out of 20 bulbs, contain 4 defective ones.
(i) ∴ Probability of defective bulb = \(\frac{4}{20}\)
(ii) Out of 20, one bulb is drawn at random from the lot then remaining bulbs = 19.
Out of 19 bulbs, the bulbs which arc not defective = 19 – 4 = 15
∴ The probability that the bulb is not defective, P(E) = \(\frac{15}{19}\)

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
Number of discs which are numbered from 1 to 30,
n(S) = 90
(i) A two-digit number :
Out of 90, one digit number = 9
∴ 2-digit numbers = 90 – 9 = 81
∴ 2-digit numbers, n(E) = 81
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 18
(ii) A perfect square number :
1, 4, 9, 16, 25, 36, 49, 64, 81
∴ n(E) = 9
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 18.1
(iii) A number divisible by 5 :
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 .
∴ n(E) = 18
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 18.2

Question 19.
A child has a die whose six faces show the letters as given below :
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 19
The die is thrown once. What is the probbility of getting (i) A ?, (ii) D ?
Solution:
Number of faces in a die is 6.
∴ n(S) = 6
(i) The probability of getting face ‘A’:
∵ n(E) = 2
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 19.1
(ii) The probbility of getting face ‘D’ is 1.
n(E) = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 19.2

Question 20.
Suppose you drop a die at random on the rectangular region as shown in the following figure. What is the probability that it will land inside the circle with diameter 1 m ?
(* Not from the examination point of view)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 20
Solution:
(i) Length of a rectnagular region is 3m.
Breadth of a rectangular region is 2m.
∴ Area of rectangle = length × breadth
= 3 × 2 = 6 sq.m.
(ii) A circle with diameter 1 m.
∴ Radius, r = \(\frac{1}{2}\) m
Area of Circle = \(\pi r^{2}\)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 20.1
∴ The probability that die will land inside the circle
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 20.2

Question 21.
A lot consists of 144 ball pens of which 20 are defective arid the others are good. Nuri will buy a pen if it is good, but will not buy If It is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy It?
Solution:
Total number of pens, n(S) = 144.
Number of defective pens = 20
∴ Number of good pens = 144 – 20
= 124
(i) Number of good pens Nun buys = 124
∴ n(E) = 124
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 21
(ii) Number of pens which Nuri do not buy = 20
∴ n(E) = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 21.1

Question 22.
Refer to Example 13.
(i) Complete the following table:
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 22
(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore each of them has a probability \(\frac{1}{11}\). Do you agree with this argument ? Justify your answer.
Solution:

∵ The two dice are thrown together.
∴ Following are the possible outcomes :
(1, 1) ; (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1) ; (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1) ; (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6,.2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ Total number of possible outcomes is 6 × 6 = 36
(i) (a) The sum on two dice is 3 for (1, 2) and (2, 1)
∴ Number of favourable outcomes = 2
⇒ P(3) = \(\frac{2}{36}\)

(b) The sum on two dice is 4 for (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3
⇒ P(4) = \(\frac{3}{36}\)

(c) The sum on two dice is 5 for (1, 4), (2, 3), (3, 2) and (4,1)
∴ Number of favourable outcomes = 4
⇒ P(5) = \(\frac{5}{36}\)

(d) The sum on two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2) and (5,1)
∴ Number of favourable outcomes = 5
⇒ P(6) = \(\frac{5}{36}\)

(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1)
∴ Number of favourable outcomes = 6
⇒ P(7) = \(\frac{62}{36}\)

(f) The sum on two dice is 9 for (3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcomes = 4
⇒ P(9) = \(\frac{4}{36}\)

(g) The sum on two dice is 10 for (4, 6), (5, 5), (6,4)
∴ Number of favourable outcomes = 3
⇒ P(10) = \(\frac{3}{36}\)

(h) The sum on two dice is 11 for (5, 6) and (6,5)
∴ Number of favourable outcomes = 2
⇒ P(11) = \(\frac{2}{36}\)

Thus, the complete table is as follows:
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 22.9

(ii) No. The number of all possible outcomes is 36 not 11.
∴ The argument is not correct.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Events when one rupee coin is tossed 3 times :
HHH, TTT, HHT, HTH, HTT, THT, TTH
∴ n(S) = 8
Out of these Event except HHH, TTT
Hanif loosing the game, n(E) = 6
(∵ 8 – 2 = 6)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 23

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either Urne?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.)
Solution:
The possibility that one die is thrown:
n(S) = 6 × 6 = 36
(i) 5 will come up either time:
(1, 5), (2. 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3) (5, 4), (5, 6)
∴ Possible events = 11.
∴ Probability that 5 will not come up either time : 36 – 11 = 25
n(E) = 25
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 24
(ii) Possibility that 5 will come up at least ones: 11
∴ n(E) = 11
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 24.1

Question 25.
Which of the following arguments are correct and which are not correct ? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
(i) Given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So total outcomes is 4.
∴ The required probability = \(\frac{1}{4}\).
(ii) Given argument is correct.
Since, total numebr of possible outcomes = 6
Odd numbers = 3 and even numbers = 3
So, favourable outcomes = 3 (in both the cases even or odd).
∴ Probability = \(\frac{3}{6}=\frac{1}{2}\)

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2

   

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2

(Unless stated otherwise, take n \(=\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of ‘n’.
Solution:
Radius of hemisphere, r = 1 cm.
Volume of hemisphere, V \(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2 \pi}{3} \times(1)^{3}\)
Radius of Cone, r = 1 cm,
Height of Cone, h = 1 cm.
∴ Volume of a cone, V \(=\frac{1}{3} \pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 1
∴ Volume of Total cube \(=\frac{2 \pi}{3}+\frac{\pi}{3} \mathrm{cm}^{3}\)
= πcm3

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 2
solution:
Volume of cylinder, V = πr2h
= π × (1.5)2 × 8
= 18π cm3.
Volume of Cylinder V \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \pi \times(1.5)^{2} \times 2\)
\(=\frac{3}{2} \pi \mathrm{cm}^{3}\)
∴ The volume of air contained in the model that Rachel made.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 2.1
= 66 cm3.

Question 3.
A Gulab Jamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see figure given aside).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3.1
Solution:
Length of cylindrical Gulab jamun,
l = 5 – (1.4 + 1.4) = 2.2 cm.
Radius, r = 1.4 cm.
∴ Volume of Cylinder (1 Gulab Jamun), = Left hemisphere + Volume of Cylinder + Right hemisphere
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3.2
∴ Total volume of such 45 Gulab jamun,
\(=45 \times \frac{22}{3} \times 0.28 \times 12.2\)
= 22 × 0.28 × 183
= 1127.28 cm3.
∴ Gulab Jamun contains sugar syrup upto about 30% of its volume.
∴ \(1127.28 \times \frac{30}{100}\)
= 338 cm3.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see
Figure)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 4
Solution:
Radius of conical depression, r = 0.5 cm.
(Depth) height, h = 1.4 cm.
Volume of 1 depression which is conical, \(=\frac{1}{3} \pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 4.1
Volume of such 4 depressions
\(=4 \times \frac{11}{30} \mathrm{cm}^{3}\)
\(=\frac{44}{30} \mathrm{cm}^{3}\)
∴ Total volume of wooden pen-stand, \(=(15 \times 10 \times 3.5)-\frac{44}{30}\)
= 525 – 1.47
= 523.53 cm3

Question 5
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Volume of water in a Cone,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 5
Volume of lead shot \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \pi \times(0.5)^{3}\)
\(=\frac{\pi}{6} \mathrm{cm}^{3}\)
Let the number of lead balls kept in vessel \(\frac{1}{4}\) of the water flows out,
\(\Rightarrow n \times \frac{\pi}{6}=\frac{1}{4} \times \frac{200 \pi}{3}\)
\(\Rightarrow n \times \frac{\pi}{6}=\frac{100}{6} \pi\)
∴ n = 100
∴ Number of lead balls =100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. (Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14).
Solution:
Height of solid iron pole, h = 220 cm.
Base diameter of pole, d = 24 cm.
∴ Radius, r = 12 cm.
Height of cylinder, h = 60 cm.
Radius of cylinder, r= 8 cm.
Mass of the pole = ?
Volume of 1st Cylinder, V = πr2h
= π × (12)2 × 220
Volume of 2nd cylinder, V = πr2h
= 77 × (8)2 × 60
∴ Total Volume = π × (12)2 × 220 + π × (8)2 × 60
= {144 × 220 + 64 × 60}π
= 35520π
= 111532.8 cm3.
Mass of iron about 1 ccm is 8 gm.
∴ Mass of iron 111532.8 ……. ??
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 6
= 111.5328 × 8 kg.
= 892.26 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm. and its height is 180 cm.
Solution:
(i) Radius of cylinder GHEF, r = 60 cm.
Height, h = 120 + 60
= 180 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7
Volume of cylinderical vessel \(=\pi r^{2} h\)
= π × (60)2 × 180
(ii) Volume of hemisphere + Volume of cone
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7.1
(iii) Quality of water remained in cylinder.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7.2

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and n = 3.14.
Solution:
Diameter of spherical glass vessel = 8.5 cm.
∴ r = 4.25 cm.
Length of cylindrical neck = 8 cm.
Diameter = 2 cm.
∴ Radius, r = 1 cm.
Volume of water she found = 345 cm3.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8
(i) Volume of Cylinder neck, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8.1
(ii) Volume of cylindrical vessel \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \pi \times(4.25)^{3}\)
∴ Total volume of water, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8.2
∴ Her answer is not correct.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4

   

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4

Question 1.
The following distribution gives the dialy income of 50 workers of a factory.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1.1
n = 50, ∴ \(\frac{n}{2}\) = 25
On a graph paper mark the following points :
(120, 12), (140, 26), (160, 34), (180, 40), (200, 50).
For the Ogive graph,
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1.2

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2.1
To draw ogive of the less than type.
we have to join the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
From the graph, Median is 46.5 kg
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2.2
n = 35 ∴ \(\frac{\mathrm{n}}{2}\) = 17.5
Class interval which has median is = (46 – 48)
l = 46, n = 35, f = 14, cf = 14, h = 2
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2.3
= 46.5
∴ Median = 46.5 kg.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3.1
We can draw Ogive graph by plotting ordered pairs:
(50, 100), (55, 98), (60, 90), (65, 78). (70, 54), (75, 16).
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3.2

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3

   

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.1
(i) n = 68, ∴ \(\frac{n}{2}\) = 34
The median is in class interval (125 – 145)
l = 125, n = 68, f = 20, cf = 22, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.2
= 125 + 12
∴ Median = 137 units.

(ii) To find out Mode:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.3
Here, Mode is in class interval (125 – 145)
Maximum frequency, l = 125, f<sub>1</sub> = 20, f<sub>0</sub> = 13, f<sub>2/sub> = 14, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.4
= 125 + 10.76
= 135.76
∴ Mode = 135.76 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of ‘x’ and ‘y’
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 2
Solution:
Median is 28.5
Class interval which has median is = (20 – 30)
l = 20, n = 60, f = 20, cf = 5 + a × h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 2.1
17 = 25 – x
∴ x = 8
5 + x + 20 + 15 + y + 5 = 60
x + y – 45 = 60
x + y = 15
y = 15 – x
y = 15 – 8
∴ y = 7
∴ x = 8, y = 7

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.1
n = 100 ∴ \(\frac{n}{2}\) = 50
Median in which C.I. is (35 – 40)
l = 35, n = 100, f = 33, cf = 45, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.2
= 35 + 0.76
∴ Median = 35.76 years

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, 171.5 – 180.5)
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4.1
n = 40 ∴ \(\frac{n}{2}\) = 20
C.I. which has median is (144.5 – 153.5)
l = 144.5, n = 40, f= 12, cf = 17, h = 9
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4.2
= 144.5 + 2.25
∴ Median = 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5
Find the median life time of a lamp.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.1
n = 400 ∴ \(\frac{n}{2}=\frac{400}{2}=200\)
Class interval having median is = (3000 – 3500)
l = 3000, n = 400, f = 86, cf = 130, h = 500
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.2
= 3000 + 406.98
∴ Median = 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.1
(i) n = 100, ∴ \(\frac{n}{2}\) = 50
Class interval having median is (7 – 10)
l = 7, n = 100, f = 40, cf= 36, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.2
= 7 + 1.05
∴ Median = 8.05 Letters

(ii) Clall interval which has mode is (7 – 10)
Maximum frequency, l = 7, f1 = 40, f0 = 30, f2 = 16, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.3
= 7 + 0.88
∴ Mode = 7.88

(iii) Mean(\(\overline{X}\)) : Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.4
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.5
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32
∴ (i) Median = 8.05 letters
(ii) Mode = 7.88
(iii) Mean = 8.32.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7.1
(i) n = 30, ∴ \(\frac{\mathbf{n}}{2}\) = 15
Class interval having median is (55 – 60)
l = 55, n = 30, f = 6, cf = 13, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7.2
= 55 + 1.67
∴ Median = 56.67 kg.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2

   

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 1
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency ?
Solution:
(i) Here the maximum class interval is 35 – 45. This class interval has mode
l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 1.1
∴ Mode = 36.8 years

(ii) Mean (\(\overline{X}\)) : Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 1.2
Here, a = 30
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 1.3
= 30 + 5.37
= 35.37
∴ Mode = 36.8
∴ Mean = 35.37
Hence Age group 36.8 patients are admitted more in number.
Mean age group patients are 35.57

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 2
Determine the modal lifetimes of the components:
Solution:
Class interval which has mode = 60 – 80
Here, l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 2.1
∴ Mode = 65.63 Hours

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also. And the mean monthly expenditure :
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3
Solution:
(i) Mode is in class interval 1500 – 2000
Here, l = 1500, f1 = 40, f0 = 24, f2 = 33, h = 500
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.1

(ii) Mean (\(\overline{X}\)): Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.2
Here, a = 2750
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.4
= 2750 – 87.50
= 2662.5
∴ Modal monthly expenditure = 1847.8
∴ Mean Monthly expenditure = Rs. 2662.5

Question 4.
The following distribution gives the state- wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 4
Solution:
(i) Class interval which has mode 30 – 35
Here, l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 4.1
∴ Mode = 30.6
(ii) Mean (\(\overline{X}\)) : Step Deviation Method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 7
By Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 4.3
= 32.5 – 3.3
= 29.2
∴ Mode = 30.6
∴ Mean = 29.2
We conclude that more number of states has teacher-student ratio about 30.6.
Mean of Age ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 5
Find the mode of the data.
Solution:
Class interval which has mode is = 4000 – 5000
Maximum frequency is
Here, l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 5.1
= 4000 + 608.7
= 4608.7
∴ Mode = 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 6
Solution:
Class interval which has mode is 40- 50
Maximum frequency is 20.
Here, l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 6.1
= 40 + 4.7
= 44.7
∴ Mode = 44.7 cars.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1

   

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1

(Unless stated otherwise, take π \(=\frac{22}{7}\))

Question 1.
2 cubes each of volume 64cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
If measurement of each side of each cube be l cm, then
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 1
(l)3 \(=\sqrt[3]{64}\)
∴ l = 4 cm.
Length of cubes if both joined together,
Length, l = 4 + 4 = 8 cm.
breadth, b = 4 cm.
height, h = 4 cm.
Surface area of the cuboid,
= 2lb + 2lh + 2bh
= 2(8 × 4) + 2(8 × 4) + 2(4 × 4)
= 2 × 32 + 2 × 32 + 2 × 16 = 64 + 64 + 32
= 160 sq.cm.

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
(i) Diameter of a hollo hemisphere = 14 cm.
∴ Radius of a hollow hemisphere, r
r = 7 cm.
Inner curved surface of a hemisphere = 2π2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 2.1
= 14 × 22
= 308 sq. cm.
(ii) Radius of hollow cylinder, r = 7 cm.
Height, h = 13 – 7 = 6 cm.
∴ Inner Surface area of the Cylinder,
= 2πrh
\(=2 \times \frac{22}{7} \times 7 \times 6\)
= 44 × 6
= 264 sq.cm.
∴ Inner surface area of the vessel = Area of hemisphere + Area of cylinder
= 308 + 264
= 572 sq. cm.

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
(i) Radius of base of circular toy,
r = 3.5 cm.
∴ Surface Area of hemisphere, = 2πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.1
= 77 sq.cm
(ii) Let slant height of cone is l cm. then
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.2
∴ Curved surface area of a cone = πrl
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.4
= 137.5 sq.cm.
∴ Total surface area of a toy = Surface area of hemisphere + Curved surface area of a cone
= 77.0 + 137.5
= 214.5 sq.cm.

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Solution:
(i) A cubical block of side 7 cm is surmounted by a hemisphere.
∴ Diameter of hemisphere is = 7 cm.
Radius of hemisphere, r = 3.5 cm.
∴ Surface area of hemisphere = Curved surface area – Area of base of hemisphere
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 4
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 4.1
= 38.5
(ii) The curved surface area of square,
=6 × l2
= 6 × (7)2
= 6 × 49
= 294 sq. cm.
∴ Total surface area of cubical block, = Curved surface area of cubical block + Surface area of hemisphere.
= 294 + 38.5
=332.5sq.cm.

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the cube. Determine the surface area of the remaining solid.
Solution:
Let, Length of each edge of cubical block = Diameter of hemisphere = l
∴ Radius of hemisphere, \(r=\frac{l}{2} unit\).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 5
Total surface area of newly formed cube = Curved surface area of cube – Upper part of Cube + Area of hemisphere which is depressed
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 5.1

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Figure given below). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 6
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 6.1
(i) Surface area of cylindrical capsule, = 2πrh
\(=2 \pi \times\left(\frac{5}{2}\right) \times 9\)
= 45π mm2,
(ii) Surface area of both hemispheres,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 6.2
= 25π mm2
∴ Total surface area of capsule, = 45π + 25π mm2
= 70π mm2
\(=70 \times \frac{22}{7}\)
= 220 mm2

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1m and 4m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m2. (note that the base of the tent will not be covered with canvas.)
Solution:
(i) Radius of base of cylinder, r= 2 mm.
Height of base of cylinder, h= 2.1 m.
∴ Curved Surface area of cylinder = 2πrh
= 2π × 2 × (2.1)
\(=4 \times \frac{22}{7} \times 2.1\)
= 26.4 m2.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 7
(ii) Radius of conical tent,
r = 2 m.
length, l = 2.8 m.
∴ Curved Surface area of cylinder = πrl
\(=\frac{22}{7} \times 2 \times 2.8\)
= 17.6 m2.
∴ Total area of the tent = 26.4 + 17.6
= 44m2
Cost of 1 aq.m. of canvas is Rs. 500.
Cost of 44 sq. m. canvas … ??
500 × 44
= Rs. 22,000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Height of the cylinder, h = 2.4 m.
Diameter of the cylinder, d = 1.4 m.
∴ Radius, r = 0.7 m.
(i) Outer Curved surface area of cylinder = 2πrh
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 8
\(=2 \times \frac{22}{7} \times(0.7) \times 2.4\)
= 44. × 0.24
= 10.56 cm2.
(ii) Area of base of = πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 8.1
= 1.54 cm2.
(iii) Inner surface area of cylinder, = πrl
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 8.2
= 5.5 cm2
∴ Total surface area of newly formed cube = 10.56 + 1.54 + 5.5 = 17.6 cm2.
∴ Nearest value is 18 sq.cm.

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm. and its base is of radius 3.5 cm. find the total surface area of the article.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 9
Solution:
Radius of base of cylinder, r= 3.5 cm
Height, h = 10 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 9.1
Total area of article, = Curved Surface area of Cylinder + 2 × Area of Hemisphere
= 2πrh + 2 × 2πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 9.2
= 374 sq.cm.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1

   

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 1
Which method did you use for finding the mean, and why ?
Solution:
Average by Direct Method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 1.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 1.2

Question 2.
Consider the following distribution of daily wages of 30 workers of a factory.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 2
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Step deviation method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 tabe 2.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 2.2
∴ Average = 545.20.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 3
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 3.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 3.2

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 4
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 4.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 4.2
∴ Mean heartbeats per minute = 75.9

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 5
Find the mean number of mangeos kept in a packing box. Which method of finding the mean did you choose ?
Solution:
Step Deviation Method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 5.1
∴ Mean Number of mangoes in the boxes = 57.19

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 6
Solution:
By Step deviation method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 6.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 6.2
∴ Mean Daily expenditure for food = Rs. 211.

Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 7
Find the mean concentration of SO2 in the air.
Solution:
By step Deviation method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 7.1
∴ Mean Concentration of SO2 = 0.099 ppm.

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 8
Solution:
Mean Deviation method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 8.1
Here, a = 17
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 8.2
∴ Mean number of days a student was absent = 12.48 Days.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 9
Solution:
Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 9.1
By Step Deviation Method
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 9.2
∴ Mean Literacy rate = 69.43%

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1, drop a comment below and we will get back to you at the earliest.

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