Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.3

Students can download Maths Chapter 9 Probability Ex 9.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.3

I. Multiple choice questions.

Question 1.
A number between 0 and 1 that is used to measure uncertainty is called ……….
(a) Random variable
(b) Trial
(c) Simple event
(d) Probability
Solution:
(d) Probability

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.3

Question 2.
Probability lies between ………
(a) -1 and +1
(b) 0 and 1
(c) 0 and n
(d) 0 and ∞
Solution:
(b) 0 and 1

Question 3.
The probability based on the concept of relative frequency theory is called ………
(a) Empirical probability
(b) Classical probability
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Solution:
(a) Empirical probability

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.3

Question 4.
The probability of an event cannot be ……….
(a) Equal to zero
(b) Greater than zero
(c) Equal to one
(d) Less than zero
Solution:
(d) Less than zero

Question 5.
The probability of all possible outcomes of a random experiment is always equal to ……..
(a) One
(b) Zero
(c) Infinity
(d) Less than one
Solution:
(a) One

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.3

Question 6.
If A is any event in S and its complement is A’ then P(A’) is equal to ………
(a) 1
(b) 0
(c) 1 – A
(d) 1 – P(A)
Solution:
(d) 1 – P(A)

Question 7.
Which of the following cannot be taken as probability of an event?
(a) 0
(b) 0.5
(c) 1
(d) -1
Solution:
(d) -1

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.3

Question 8.
A particular result of an experiment is called ………
(a) Trial
(b) Simple event
(c) Compound event
(d) Outcome
Solution:
(d) Outcome

Question 9.
A collection of one or more outcomes of an experiment is called ……….
(a) Event
(b) Outcome
(c) Sample point
(d) None of the above
Solution:
(a) Event

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.3

Question 10.
The six faces of the dice are called equally likely if the dice is ………
(a) Small
(b) Fair
(c) Six-faced
(d) Round
Solution:
(b) Fair

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

Students can download Maths Chapter 9 Probability Ex 9.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.2

Question 1.
A company manufactures 10000 Laptops in 6 months. In that 25 of them are found to be defective. When you choose one Laptop from the manufactured, what is the probability that selected Laptop is a good one?
Solution:
Total number of laptops = 10000
∴ n(S) = 10000
Number of good laptops = 10000 – 25 = 9975
Let E be the event of getting good laptops
n(E) = 9975
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{9975}{10000}\)
= \(\frac{399}{400}\) (or)
= 0.9975

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

Question 2.
In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.
Solution:
Here n(S) = 400
Number of persons having voter ID = 191
Number of persons does not have their voter ID
= 400 – 191
= 209
n(E) = 209
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{209}{400}\)
∴ The required probability is \(\frac{209}{400}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

Question 3.
The probability of guessing the correct answer to a certain question is \(\frac{x}{3}\). If the probability of not guessing the correct answer is \(\frac{x}{5}\), then find the value of x.
Solution:
Probability of guessing the correct answer = \(\frac{x}{3}\)
P(E) = \(\frac{x}{3}\)
Probability of not guessing the correct answer = \(\frac{x}{5}\)
P(E)’ = \(\frac{x}{5}\)
But P(E) + P(E)’ = 1
\(\frac{x}{3}\) + \(\frac{x}{5}\) = 1
\(\frac{5x+3x}{15}\) = 1 ⇒ \(\frac{8x}{15}\) = 1
8x = 15
x = \(\frac{15}{8}\)
∴ The value of x = \(\frac{15}{8}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

Question 4.
If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?
Solution:
Probability of a player winning a tennis match = 0.72
P(E) = 0.72
Probability of a player loosing the match be P(E)’
But P(E) + P(E)’ = 1
0.72 + P(E)’ = 1
P(E)’ = 1 – 0.72
= 0.28
Probability of the player loosing the match = 0.28

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

Question 5.
1500 families were surveyed and following data was recorded about their maids at homes
Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2 1
A family selected at random. Find the probability that the family selected has
(i) Both types of maids
(ii) Part time maids
(iii) No maids
Solution:
Total number of families surveyed = 1500
n(S) = 1500
Number of families used maids = 860 + 370 + 250
= 1480
Number of families not using any maids = 1500 – 1480
= 20

(i) Let E1 be the event of getting families use both types of maids
n(E1) = 250
p(E1) = \(\frac{n(E_{1})}{n(S)}\)
= \(\frac{250}{1500}\)
= \(\frac{25}{150}\)
= \(\frac{1}{6}\)
Probability of getting both types of maids = \(\frac{1}{6}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

(ii) Let E2 be the event of getting families use part time maids
n(E2) = 860
p(E2) = \(\frac{n(E_{2})}{n(S)}\)
= \(\frac{860}{1500}\)
= \(\frac{43}{75}\)
Probability of getting part time maids = \(\frac{43}{75}\)

(iii) Let E3 be the event of getting family use no maids
n(E3) = 20
p(E3) = \(\frac{n(E_{3})}{n(S)}\)
= \(\frac{20}{1500}\)
= \(\frac{2}{150}\)
= \(\frac{1}{75}\)
Probability of getting no maids = \(\frac{1}{75}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.2

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Students can download Maths Chapter 9 Probability Ex 9.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.1

Question 1.
You are walking along a street. If you just choose a stranger crossing you, what is the probability that his next birthday will fall on a Sunday?
Solution:
Sample space = {M, T, W, Th, F, S, Sun}
n(S) = 7
Let E be the event of getting birthday on Sunday
n(E) = 1
p(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{1}{7}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 2.
What is the probability of drawing a King or a Queen or a Jack from a deck of cards?
Solution:
The outcomes n(S) = 52
Let E be the event of getting a king or a queen or a jack
= 4 + 4 + 4
n(E) = 12
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{12}{52}\)
= \(\frac{3}{13}\)

Question 3.
What is the probability of throwing an even number with a single standard dice of six faces?
Solution:
Sample space (S) = {1, 2, 3, 4, 5, 6}
n(S) = 6
Let E be the event of getting an even number
E = {2, 4, 6}
n(E) = 3
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{3}{6}\)
= \(\frac{1}{2}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 4.
There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining are Green then, what is the probability of picking out (i) a Blue ball,
(ii) a Red ball and
(ii) a Green ball?
Solution:
Sample space n(S) = 24
Number of green ball = 24 – (3 + 5)
= 24 – 8
= 16

(i) Let E1 be the event of getting a blue ball
n(E1) = 5
p(E1) = \(\frac{n(E_{1})}{n(S)}\)
= \(\frac{5}{24}\)

(ii) Let E2 be the event of getting a red ball
n(E2) = 3
p(E2) = \(\frac{n(E_{2})}{n(S)}\)
= \(\frac{3}{24}\)
= \(\frac{1}{8}\)

(iii) Let E3 be the event of getting a green ball
n(E3) = 16
p(E3) = \(\frac{n(E_{3})}{n(S)}\)
= \(\frac{16}{24}\)
= \(\frac{2}{3}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 5.
When two coins are tossed, what is the probability that two heads are obtained?
Solution:
Sample space (S) = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Let E be the event of getting two heads
n(E) = 1
p(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{1}{4}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 6.
Two dice are rolled, find the probability that the sum is
Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1 1
(i) equal to 1
(ii) equal to 4
(iii) less than 13
Solution:
Sample space (S) = {(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)}
n(S) = 36

(i) Let E1 be the event of getting the sum is equal to 1
n(E1) = 0
p(E1) = \(\frac{n(E_{1})}{n(S)}\)
= \(\frac{0}{36}\)
= 0

(ii) Let E2 be the event of getting the sum is equal to 4
E2 = {(1,3), (2, 2) (3, 1)}
n(E2) = 3
p(E2) = \(\frac{n(E_{2})}{n(S)}\)
= \(\frac{3}{36}\)
= \(\frac{1}{12}\)

(iii) Let E3 be the event of getting the sum is less than 13
n(E3) = 36
p(E3) = \(\frac{n(E_{3})}{n(S)}\)
= \(\frac{36}{36}\)
= 1

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 7.
A manufacturer tested 7000 LED lights at random and found that 25 of them were defective. If a LED light is selected at random, what is the probability that the selected LED light is a defective one.
Solution:
Sample space n(S) = 7000
Let E1 be the event of getting selected LED light is a defective one
n(E1) = 25
p(E1) = \(\frac{n(E_{1})}{n(S)}\)
= \(\frac{25}{7000}\)
= \(\frac{1}{280}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 8.
In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40 attempts tried by a team. Find the probability that the opponent team can convert the attempt into a goal.
Solution:
Sample space n(S) = 40
Opponent team trying to attempt the goal = 40 – 32 = 8
Let E be the event of getting opponent team convert the attempt into a goal
n(E) = 8
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{8}{40}\)
= \(\frac{1}{5}\)

Question 9.
What is the probability that the spinner will not land on a multiple of 3?
Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1 2
Solution:
Sample space (S) = {1, 2, 3, 4, 5, 6, 7, 8}
n(S) = 8
Land in a multiple of 3 = {3, 6}
Land not a multiple of 3 = 8 – 2
= 6
Let E be the event of getting not a multiple of 6
n(E) = 6
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{6}{8}\)
= \(\frac{3}{4}\)

Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1

Question 10.
Frame two problems in calculating probability, based on the spinner shown here.
Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Ex 9.1 3
Solution:
(i) What is the probability that the spinner will get an odd number?
(ii) What is the probability that the spinner will not land on a multiple of 2?

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Students can download Maths Chapter 8 Statistics Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions

I. Choose the Best Answer

Question 1.
The Arithmetic mean of all the factors of 10 is ………
(a) 4.5
(b) 5.5
(c) 10
(d) 55
Solution:
(a) 4.5

Question 2.
The mean of five numbers is 27, if one number is excluded, then mean is 25. Then the excluded number is ……..
(a) 0
(b) 15
(c) 25
(d) 35
Solution:
(d) 35

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 3.
The mean of 8 numbers is 15. If each number is multiplied by 2, then the new mean will be ……..
(a) 7.5
(b) 30
(c) 10
(d) 25
Solution:
(b) 30

Question 4.
The median of 11, 8, 4, 9, 7, 5, 2, 4, 10 is ……..
(a) 1
(b) 8
(c) 4
(d) 11
Solution:
(a) 1

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 5.
Median is …….
(a) the most frequent value
(b) the least frequent value
(c) middle most value
(d) mean of first and last values
Solution:
(c) middle most value

Question 6.
The mode of the distribution is …….
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 1
(a) 3
(b) 4
(c) 6
(d) 14
Solution:
(b) 4

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 7.
Mode is …….
(a) the middle value
(b) extreme value
(c) minimum value
(d) the most repeated value
Solution:
(d) the most repeated value

Question 8.
The mode of the data 72, 33, 44, 72, 81, 72, 15 is ……..
(a) 12
(b) 33
(c) 81
(d) 15
Solution:
(a) 12

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 9.
The Arithmetic mean of 10 number is -7. If 5 is added to every number, then the new arithmetic mean is ……..
(a) 17
(b) 12
(c) -2
(d) -7
Solution:
(c) -2

Question 10.
The Arithmetic mean of integers from -5 to 5 is ……..
(a) 25
(b) 10
(c) 3
(d) 0
Solution:
(d) 0

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

II. Answer the Following Questions

Question 1.
Find the Arithmetic mean for the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 2
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 3
Arithmetic mean (\(\bar { X }\)) = \(\frac{Σfx}{Σf}\)
= \(\frac{5540}{96}\)
= 57.7
∴ Arithmetic mean = 57.7

Question 2.
Calculate the Arithmetic mean of the following data using step deviation method.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 4
Solution:
Assumed mean = 35
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 5
Arithmetic mean (\(\bar { X }\)) = A + \(\frac{Σfd}{Σf}\) × c
= 35 + \(\frac{(-20)}{100}\) × 10
= 35 – 2
= 33
∴ Arithmetic mean = 33

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 3.
Find the median for the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 6
Solution:
The given class intervals is inclusive type we convert it into exclusive type.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 7
\(\frac{N}{2}\) = \(\frac{70}{2}\)
= 35
Median class is 25.5 – 30.5
Here l = 25.5, f = 26, m = 30, c = 5
Median = l + \(\frac{\frac{N}{2}-m}{f}\) × c
= 25.5 + \(\frac{35-30}{26}\) × 5
= 25.5 + \(\frac{5×5}{26}\)
= 25.5 + 0.96
= 26.46
∴ Median = 26.46

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 4.
Calculate the mode of the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 8
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 9
The highest frequency is 30. Corresponding class interval is the modal class.
Here l = 25, f = 30, f1 = 18, f2 = 20 and c = 5
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 10
∴ Arithmetic mean = 25 + 2.727
= 25 + 2.73
= 27.73
mode = 27.73

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 5.
Find the mean, median and mode of marks obtained by 20 students in an examination. The marks are given below.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 11
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 12
Arithmetic mean:
Here Σfx = 560, Σf = 20
\(\bar { X }\) = \(\frac{Σfx}{Σf}\)
= \(\frac{560}{20}\)

Median:
Median class is 20 – 30
Here l = 20, \(\frac{N}{2}\) = 10, m = 5, c = 10, f = 5
Median
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 13
= 20 + 10
= 30

Mode:
The highest frequency is 8
Modal class is 30 – 40
Here, l = 30, f = 8, f1 = 5, f2 = 2, c = 10
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 14
= 33.33
Mean = 28, median = 30, mode = 33.33

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Students can download Maths Chapter 8 Statistics Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.4

Question 1.
Let m be the mid point and b be the upper limit of a class in a continuous frequency distribution. The lower limit of the class is …….
(a) 2m – b
(b) 2m + b
(c) m – b
(d) m – 2b
Solution:
(a) 2m – b

Question 2.
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is ……..
(a) 101
(b) 100
(c) 99
(d) 98
Solution:
(c) 99
Hint:
Total of 8 numbers = 81 × 7 = 567
Total of 7 numbers = 78 × 6 = 468
The number is = 567 – 468
= 99

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 3.
A particular observation which occurs maximum number of times in a given data is called its ………
(a) frequency
(b) range
(c) mode
(d) median
Solution:
(c) mode

Question 4.
For which set of numbers do the mean, median and mode all have the same values?
(a) 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 2, 1, 5
Solution:
(b) 1, 3, 3, 3, 5

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 5.
The algebraic sum of the deviations of a set of n values from their mean is ……..
(a) 0
(b) n – 1
(c) n
(d) n + 1
Solution:
(a) 0

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 6.
The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is ……..
(a) 24
(b) 36
(c) 26
(d) 34
Solution:
(d) 34
Hint:
Mean = 28
a + b + c + d + e = 28 × 5 = 140
= 140 …… (1)
But \(\frac{a+c+e}{3}\) = 24
a + c + e = 72
a + b + c + d + e = 140
b + d + 72 = 140
b + d = 140 – 72
= 68
Mean = \(\frac{68}{2}\)
= 34

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 7.
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is …….
(a) 9
(b) 11
(c) 13
(d) 15
Solution:
(a) 9
Hint:
Mean = \(\frac{x+x+2+x+4+x+6+x+8}{5}\)
11 = \(\frac{5x+20}{5}\)
5x + 20 = 55
5x = 55 – 20
= 35
x = \(\frac{35}{5}\)
= 7
Mean of first 3 observation is = \(\frac{7+9+11}{3}\)
= 9

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 8.
The mean of 5, 9, x, 17 and 21 is 13, then find the value of x.
(a) 9
(b) 13
(c) 17
(d) 21
Solution:
(b) 13
Hint:
Mean = \(\frac{5+9+x+17+21}{5}\)
13 = \(\frac{52+x}{5}\)
65 = 52 + x
x = 65 – 52
= 13

Question 9.
The mean of the square of first 11 natural numbers is
(a) 26
(b) 46
(c) 48
(d) 52
Solution:
(b) 46
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4 1
= 46

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 10.
The mean of a set of numbers is \(\bar { X }\). If each number is multiplied by z, the mean is ……..
(a) \(\bar { X }\) + z
(b) \(\bar { X }\) – z
(c) z\(\bar { X }\)
(d) \(\bar { X }\)
Solution:
(c) z\(\bar { X }\)
Hint:
If each observation is multiplied by k, k ≠ 0 then the arithmetic mean is also multiplied by the same quantity.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Students can download Maths Chapter 4 Geometry Ex 4.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.4

Question 1.
The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?
Solution:
Let the radius AB be r. In the right ∆ ABO,
OB2 = OA2 + AB2
252 = 242 + r2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 1
252 – 242 = r2
(25 + 24) (25 – 24) = r2
r = \(\sqrt { 49 }\) =7
Radius of the circle = 7 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 2.
∆ LMN is a right angled triangle with ∠L = 90°. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle.
Solution:
LN = 6; ML = 8. In the right ∆ LMN,
MN2 = LN2 + LM2
= 62 + 82 = 36 + 64 = 100
MN = \(\sqrt { 100 }\) = 10
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 2
OA= OB = OC = r
AN = CN (Tangent of the circle)
LN – AL= CN
LN – r = CN
8 – r = CN ……(1)
MC = MB (tangent of the circle)
MC = ML – LB
MC = 6 – r …….(2)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 3
Add (1) and (2)
MC + CN = (6 – r) + (8 – r)
MN = 14 – 2r
10 = 14 – 2r
2r = 14 – 10 = 4
r = \(\frac { 4 }{ 2 } \) = 2 cm
radius of the circle = 2 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 3.
A circle is inscribed in ∆ ABC having sides 8 cm, 10 cm and 12 cm as shown in figure, Find AD, BE and CF.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 4
Solution:
AD = AF = x (tangent of the circle)
BD = BE = y (tangent of the circle)
CE = CF = z (tangent of the circle)
AB = AD + DB
x + y = 12 ……(1)
BC = BE + EC
y + z= 8 …….(2)
AC = AF + FC
x + z = 10 ……(3)
Add (1) (2) and (3)
2x + 2y + 2z = 12 + 8 + 10
x + y + z = \(\frac { 30 }{ 2 } \) = 15 …….(4)
By x + y = 12 in (4)
z = 3
y + z = 8 in (4)
x = 7
x + z = 10 in (4)
y = 5
AD = 7 cm; BE = 5 cm and CF = 3 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 4.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120° . Find ∠OPQ.
Solution:
∠POQ = 180° – 120° = 60°
In ∆OPQ, we know
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 5
∠POQ + ∠OQP + ∠OPQ = 180°
(Sum of the angles of a ∆ is 180°)
60° + 90° + ∠OPQ = 180°
∠OPQ = 180° – 150° = 30°
∠OPQ = 30°

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 5.
A tangent ST to a circle touches it at B. AB is a chord such that ∠ABT = 65°. Find ∠AOB, where “O” is the centre of the circle.
Solution:
Given ∠ABT = 65°
∠OBT = 90°(TB is the tangent of the circle)
∠ABO = 90° – 65° = 25°
∠ABO + ∠BOA + ∠OAB = 180°
25° + x + 25° = 180° (Sum of the angles of a ∆)
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 6
(OA and OB are the radius of the circle.
∴ ∠ABO = ∠BAO = 25°
x + 50 = 180°
x = 180° – 50° = 130°
∴ ∠BOA = 130°

Question 6.
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 7
Solution:
In the right ∆ OTP,
PT2 = OT2 – OP2
= 132 – 52
= 169 – 25
= 144
PT = \(\sqrt { 144 }\) = 12 cm
Since lengths of tangent drawn from a point to circle are equal.
∴ AP = AE = x
AT = PT – AP
= (12 – x) cm
Since AB is the tangent to the circle E.
∴ OE ⊥ AB
∠OEA = 90°
∠AET = 90°
In ∆AET, AT2 = AE2 + ET2
In the right triangle AET,
AT2 = AE2 + ET2
(12 – x)2 = x2 + (13 – 5)2
144 – 24x + x2 = x2 + 64
24x = 80 ⇒ x = \(\frac { 80 }{ 24 } \) = \(\frac { 20 }{ 6 } \) = \(\frac { 10 }{ 3 } \)
BE = \(\frac { 10 }{ 3 } \) cm
AB = AE + BE
= \(\frac { 10 }{ 3 } \) + \(\frac { 10 }{ 3 } \) = \(\frac { 20 }{ 3 } \)
Lenght of AB = \(\frac { 20 }{ 3 } \) cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 7.
In two concentric circles, a chord of length 16 cm of larger circle becomes a tangent to the smaller circle whose radius is 6 cm. Find the radius of the larger circle.
Solution:
Here AP = PB = 8 cm
In ∆OPA,
OA2 = OP2 + AP2
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 8
= 62 + 82
= 36 + 64
= 100
OA = \(\sqrt { 100 }\) = 10 cm
Radius of the larger circle = 10 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 8.
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’ P are tangents to the two circles. Find the length of the common chord PQ.
Solution:
In ∆ OO’P
(O’O)2 = OP2 + O’P2
= 32 + 42
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 9
= 9 + 16
(OO’)2 = 25
∴ OO’ = 5cm
Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.
OR ⊥ PQ and PR = RQ
Let OR be x, then O’R = 5 – x again Let PR = RQ = y cm
In ∆ ORP, OP2 = OR2 + PR2
9 = x2 + y2 …(1)
In ∆ O’RP, O’P2 = O’R2 + PR2
16 = (5 – x)2 + y2
16 = 25 + x2 – 10x + y2
16 = x2 + y2 + 25 – 10x
16 = 9 + 25 – 10x (from 1)
16 = 34 – 10x
10x = 34 – 16 = 18
x = \(\frac { 18 }{ 10 } \) = 1.8 cm
Substitute the value of x = 1.8 in (1)
9 = (1.8)2 + y2
y2 = 9 – 3.24
y2 = 5.76
y = \(\sqrt { 5.76 }\) = 2.4 cm
Hence PQ = 2 (2.4) = 4.8 cm
Length of the common chord PQ = 4.8 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 9.
Show that the angle bisectors of a triangle are concurrent.
Solution:
Given: ABC is a triangle. AD, BE and CF are the angle bisector of ∠A, ∠B, and ∠C.
To Prove: Bisector AD, BE and CF intersect
Proof: The angle bisectors AD and BE meet at O. Assume CF does not pass through O. By angle bisector theorem.
AD is the angle bisector of ∠A
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \) …..(1)
BE is the angle bisector of ∠B
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 10
\(\frac { CE }{ EA } \) = \(\frac { BC }{ AB } \) …….(2)
CF is the angle bisector ∠C
\(\frac { AF }{ FB } \) = \(\frac { AC }{ BC } \) …….(3)
Multiply (1) (2) and (3)
\(\frac { BD }{ DC } \) × \(\frac { CE }{ EA } \) × \(\frac { AF }{ FB } \) = \(\frac { AB }{ AC } \) × \(\frac { BC }{ AB } \) × \(\frac { AC }{ BC } \)
So by Ceva’s theorem.
The bisector AD, BE and CF are concurrent.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 10.
In ∆ABC , with ∠B = 90° , BC = 6 cm and AB = 8 cm, D is a point on AC such that AD = 2 cm and E is the midpoint of AB. Join D to E and extend it to meet at F. Find BF.
Solution:
Consider ∆ABC. Then D, E and F are respective points on the sides AC, AB and BC.
By construction D, E, F are collinear.
By Menelaus’ theorem \(\frac { AE }{ EB } \) × \(\frac { BF }{ FC } \) × \(\frac { CD }{ DA } \) = 1 ……(1)
AD = 2 cm; AE = EB = 4 cm; BC = 6 cm; FC = FB + BC = x + 6
In ∆ABC, By Pythagoras theorem.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 11
AC2= AB2 + BC2
AC2 = 82 + 62 = 64 + 36 = 100
AC = \(\sqrt { 100 }\) = 10
CD = AC – AD
= 10 – 2 = 8 cm
Substituting the values in (1) we get
\(\frac { 4 }{ 4 } \) × \(\frac { x }{ x+6 } \) × \(\frac { 8 }{ 2 } \) = 1
\(\frac { x }{ x+6 } \) × 4 = 1
4x = x + 6
3x = 6 ⇒ x = \(\frac { 6 }{ 3 } \) = 2
The value of BF = 2 cm

Question 11.
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 12
Solution:
Given that AE = 3 cm, EC = 4 cm, CD = 10 cm, DB = 3 cm, AF = 5 cm.
Let FB be x
Using Ceva’s theorem we have
\(\frac { AE }{ EC } \) × \(\frac { CD }{ DB } \) × \(\frac { BF }{ AF } \) = 1
\(\frac { 3 }{ 4 } \) × \(\frac { 10 }{ 3 } \) × \(\frac { x }{ 5 } \) = 1
\(\frac { 2x }{ 4 } \) = 1
2x = 4 ⇒ x = \(\frac { 4 }{ 2 } \) = 2
The value of BF = 2

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 12.
Draw a tangent at any point R on the circle of radius 3.4 cm and centre at P ?
Answer:
Given Radius = 3.4 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 13
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 14
Steps of construction:

  1. Draw a circle with centre “O” of radius 3.4 cm.
  2. Take a point P on the circle Join OP.
  3. Draw a perpendicular line TT’ to OP which passes through P.
  4. TT’ is the required tangent.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 13.
Draw a circle of radius 4.5 cm. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem.
Answer:
Radius of the circle = 4.5 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 15
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 16
Steps of construction:

  1. With O as centre, draw a circle of radius 4.5 cm.
  2. Take a point L on the circle. Through L draw any chord LM.
  3. Take a point M distinct from L and N on the circle, so that L, M, N are in anti-clockwise direction. Join LN and NM.
  4. Through “L” draw tangent TT’such that ∠TLM = ∠MNL
  5. TT’ is the required tangent.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 14.
Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 5 cm; Distance = 10 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 17
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 18
Steps of construction:

  1. With O as centre, draw a circle of radius 5 cm.
  2. Draw a line OP =10 cm.
  3. Draw a perpendicular bisector of OP, which cuts OP at M.
  4. With M as centre and MO as radius draw a circle which cuts previous circle at A and B.
  5. Join AP and BP. AP and BP are the required tangents.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Verification: In the right ∆ OAP
PA2 = OP2 – OA2
= 102 – 52 = \(\sqrt { 100-25 }\) = \(\sqrt { 75 }\) = 8.7 cm
Lenght of the tangent is = 8.7 cm

Question 15.
Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point.
Answer:
Radius = 4 cm; Distance = 11 cm
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 19
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 20
Steps of construction:

  1. With O as centre, draw a circle of radius 4 cm.
  2. Draw a line OP = 11 cm.
  3. Draw a perpendicular bisector of OP, which cuts OP at M.
  4. With M as centre and MO as radius, draw a circle which cuts previous circle A and B.
  5. Join AP and BP. AP and BP are the required tangents.

This the length of the tangents PA = PB = 10.2 cm
Verification: In the right angle triangle OAP
PA2 = OP2 – OA2
= 112 – 42 = 121 – 16 = 105
PA = \(\sqrt { 105 }\) = 10.2 cm
Length of the tangents = 10.2 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 16.
Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 3cm; Distance = 5cm.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 21
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 22
Steps of construction:

  1. With O as centre, draw a circle of radius 3 cm.
  2. Draw a line OP = 5 cm.
  3. Draw a perpendicular bisector of OP, which cuts OP at M.
  4. With M as centre and MO as radius draw a circle which cuts previous circles at A and B.
  5. Join AP and BP, AP and BP are the required tangents.

The length of the tangent PA = PB = 4 cm
Verification: In the right angle triangle OAP
PA2 = OP2 – OA2
= 52 – 32
= 25 – 9
= 16 PA
= \(\sqrt { 16 }\) = 4 cm
Length of the tangent = 4 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Question 17.
Draw a tangent to the circle from the point P having radius 3.6 cm, and centre at O. Point P is at a distance 7.2 cm from the centre.
Answer:
Radius = 3.6; Distance = 7.2 cm.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 23
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4 24
Steps of construction:

  1. With O as centre, draw a circle of radius 3.6 cm.
  2. Draw a line OP = 7.2 cm.
  3. Draw a perpendicular bisector of OP which cuts OP at M.
  4. With M as centre and MO as radius draw a circle which cuts the previous circle at A and B.
  5. Join AP and BP, AP and BP are the required tangents.

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.4

Length of the tangents PA = PB = 6.26 cm
Verification: In the right triangle ∆OAP
PA2 = OP2 – OA2
= 7.22 – 3.62 =(7.2 + 3.6) (7.2 – 3.6)
PA2 = 10.8 × 3.6 = \(\sqrt { 38.88 }\)
PA = 6.2 cm
Length of the tangent = 6.2 cm

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Students can download Maths Chapter 8 Statistics Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.3

Question 1.
The monthly salary of 10 employees in a factory are given below:
Rs 5000, Rs 7000, Rs 5000, Rs 7000, Rs 8000, Rs 7000, Rs 7000, Rs 8000, Rs 7000, Rs 5000
Find the mean, median and mode.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 1
Mean = 6600
Median:
Arrange in ascending order we get.
5000, 5000, 5000, 7000, 7000, 7000, 7000, 7000, 8000, 8000
The number of values = 10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\)
= Average of 5th value and 6th value
= \(\frac{7000+7000}{2}\)
∴ Median = 7000
Mode: 7000 repeated 5 times
∴ Mode = 7000

Question 2.
Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution:
3.1 occuring two times
3.3 occuring two times
∴ 3.1 and 3.3 are the mode (bimodal)

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 3.
For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14, find the value of x. Also find the mode of the data.
Solution:
Arithmetic mean
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 2
∴ 2x + 64 = 14 × 7
2x = 98 – 64
2x = 34
x = \(\frac{34}{2}\)
= 17
The given numbers are 11, 15, 17, 18, 19, 15 and 3
15 occuring two times
∴ Mode = 15
The value of x = 17 and mode = 15

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 4.
The demand of track suit of different sizes as obtained by a survey is given below:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 3
Which size is demanded more?
Solution:
The highest frequency is 37
The corresponding value is the mode
∴ Mode = 40
Size 40 is demanded more.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 5.
Find the mode of the following data:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 4
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 5
The highest frequency is 46
20 – 30 is the modal class
Here l = 20, f = 46, f1 = 38, f2 = 34 and c = 10
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 6
= 20 + 4
= 24
∴ Mode = 24

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 6.
Find the mode of the following distribution
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 7
Solution:
In the given table the class intervals are in inclusive form; convert them into exclusive form.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 8
The highest frequency is 14
Modal class is 54.5 – 64.5
Here l = 54.5, f = 14, f1 = 10, f2 = 8 and c = 10
mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 9
= 58.5
∴ Mode = 58.5

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Students can download Maths Chapter 8 Statistics Ex 8.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.2

Question 1.
Find the median of the given values: 47, 53, 62, 71, 83, 21, 43, 47, 41
Solution:
Arrange the values in ascending order we get
21, 41, 43, 47, 47, 53, 62, 71, 83
The number of values = 9 which is odd
Median = (\(\frac{9+1}{2})^{th}\) variable
= 5th variable
∴ Median = 47

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 2.
Find the median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51
Solution:
Arrange the values in ascending order we get
31, 35, 36, 37, 44, 44, 51, 60, 86, 86
The number of values = 10 which is even
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\) value
= Average of 5th and 6th value
= \(\frac{44+44}{2}\)
= \(\frac{88}{2}\)
= 44
∴ Median = 44

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 3.
The median of observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the values of x.
Solution:
The given observation is 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 (is ascending order)
The number of values =10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\) value
= Average of 5th and 6th value
24 = \(\frac{x+2.+x+4}{2}\)
24 = \(\frac{2x+6}{2}\)
2x + 6 = 48
2x = 48 – 6
2x = 42
x = \(\frac{42}{2}\)
= 21
The value of x = 21

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 4.
A researcher studying the behavior of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching its food.
Solution:
Arrange the value in ascending order we get
27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63
The number of values = 13 which is odd
Median = (\(\frac{13+1}{2})^{th}\) value
= (\(\frac{14}{2})^{th}\)
= 7th value
7th value is = 32
∴ Median = 32

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 5.
The following are the marks scored by the students in the Summative Assessment exam.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2 1
Calculate the median.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2 2
\(\frac{N}{2}\) = \(\frac{50}{2}\)
= 25
Here l = 30, f = 10; m = 24 and c = 10
Median = l + \(\frac{(\frac{N}{2}-m)×c}{f}\)
= 30 + \(\frac{(25-24)10}{10}\)
= 30 + 1
= 31
∴ Median = 31

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 6.
The mean of five positive integers is twice their median. If four of the integers are 3, 4, 6, 9 and median is 6, then find the fifth integer.
Solution:
Let the 5th positive integer be x
\(\bar { x }\) = \(\frac{3+4+6+9+x}{5}\)
= \(\frac{22+x}{5}\)
Median = 6
Mean = 2 × median
\(\frac{22+x}{5}\) = 2 × 6
22 + x = 60
x = 60 – 22
= 38
The fifth integer is 38.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Students can download Maths Chapter 5 Coordinate Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
On which quadrant does the point (- 4, 3) lie?
(a) I
(b) II
(c) III
(d) IV
Solution:
(b) II

Question 2.
The point whose abscissa is 5 and lies on the x-axis is …….
(a) (-5, 0)
(b) (5, 5)
(c) (0, 5)
(d) (5, 0)
Solution:
(d) (5, 0)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 3.
A point which lies in the III quadrant is ……..
(a) (5, 4)
(b) (5, -4)
(c) (-5, -4)
(d) (-5, 4)
Solution:
(c) (-5, -4)

Question 4.
A point on the y-axis is ……..
(a) (1, 1)
(b) (6, 0)
(c) (0, 6)
(d) (-1, -1)
Solution:
(c) (0, 6)

Question 5.
The distance between the points (4, -1) and the origin is ……..
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(d) \(\sqrt{17}\)

Question 6.
The distance between the points (-1, 2) and (3, 2) is ……..
(a) \(\sqrt{14}\)
(b) \(\sqrt{15}\)
(c) 4
(d) 0
Solution:
(c) 4

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 7.
The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is …….
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)

Question 8.
The point (0, -3) lies on
(a) + ve x-axis
(b) + ve y-axis
(c) – ve x-axis
(d) – ve y-axis
Solution:
(d) – ve y-axis

Question 9.
The point which is on y-axis with ordinate -5 is ……..
(a) (0, -5)
(b) (-5, 0)
(c) (5, 0)
(d) (0, 5)
Solution:
(a) (0, -5)

Question 10.
The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is …….
(a) 2
(b) 4
(c) √2
(d) 8
Solution:
(a) 2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 11.
The distance between the points (-2, 2) and (3, 2) is ……..
(a) 10 units
(b) 5 units
(c) 5√3 units
(d) 20 units
Solution:
(b) 5 units

Question 12.
The midpoint of the line joining the points (1, -1) and (-5, 3) is ……..
(a) (2, 1)
(b) (2, -1)
(c) (-2, -1)
(d) (-2, 1)
Solution:
(d) (-2, 1)

Question 13.
If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ……..
(a) (-2, 2)
(b) (2, -2)
(c) (-2, -2)
(d) (2, 2)
Solution:
(b) (2, -2)

Question 14.
The ratio in which the X-axis divides the line segment joining the points (6, 4) and (1, -7) is ……..
(a) 1 : 2
(b) 2 : 3
(c) 4 : 7
(d) 7 : 4
Solution:
(c) 4 : 7

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 15.
The centroid of a triangle (3, -5), (-7, 4) and (10, -2) is …….
(a) (2, -1)
(b) (2, 1)
(c) (-2, 1)
(d) (1, -2)
Solution:
(a) (2, -1)

II. Answer the Following Questions.

Question 1.
Show that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
Solution:
Let A (1, 1), B (5, 4) and G (-2, 5)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 1
AB = 5, AC = 5
∴ ABC is an isosceles triangle …….. (1)
BC² = AB² + AC²
50 = 25 + 25 ⇒ 50 = 50
∴ ∠A = 90° ……… (2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Question 2.
Show that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 2
= \(\sqrt{16}\)
= 4
AB = BC = CD = DA = 4. All the four sides are equal.
∴ ABCD is a Rhombus ……..(1)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 3
Diagonal AC = Diagonal BD = \(\sqrt{32}\) ……..(2)
From (1) and (2) we get ABCD is a square.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 3.
Show that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 4
AB + BC = AC ⇒ \(\sqrt{13}\) + 3\(\sqrt{13}\) = 4\(\sqrt{13}\)
∴ The points A, B, C are collinear.

Question 4.
Find the type of triangle formed by (-1, -1), (1, 1) and (-√, √3)
Solution:
Let the point A (-1, -1), B (1, 1) and C (-√3, √3)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 5
AB = BC = AC = √8
∴ ABC is an equilateral triangle.

Question 5.
Find x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 6
But PQ = QR
\(\sqrt{(x-1)^{2}+25}\) = \(\sqrt{41}\)
Squaring on both sides
(x – 1)² + 25 = 41
(x – 1)² = 41 – 25 = 16
x – 1 = \(\sqrt{16}\) = ± 4
x – 1 = 4 (or) x – 1 = – 4
x = 5 (or) x = -4 + 1 = -3
The value of x = 5 or – 3

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 6.
Find the coordinate of the point of trisection of the line segment joining (4, -1) and
Solution:
Let A (4, -1) and B (-2, -3) are the given points
Let P (a, b) and Q (c, d) be the points of trisection of AB.
∴ AP = PQ = QB
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 7
The required coordinate P is (2, –\(\frac{5}{3}\)) and Q is (0, –\(\frac{7}{3}\))

Question 7.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given points are A(-3, 10), B(6, -8) and P(-1, 6)
divides AB internally in the ratio m : n
By section formula.
A line divides internally in the ratio m : n the point P =
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 8
∴ \(\frac{6m-3n}{m+n}\) = -1
6m – 3n = -m – n
6m + m = 3n – n
7m = 2n ⇒ \(\frac{m}{n}\) = \(\frac{2}{7}\)
∴ m : n = 2 : 7
and
\(\frac{-8m+10n}{m+n}\) = 6
-8m + 10n = 6m + 6n
-8m – 6m = 6n – 10n
14m = 4n
∴ \(\frac{m}{n}\) = \(\frac{14}{4}\) = \(\frac{2}{7}\)
Hence P divides AB internally in the ratio 2 : 7

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 8.
If (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find “x” and “y”.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 9
Since ABCD is a parallelogram the diagonal bisect each other
Mid point of AC = Mid point of BD
(\(\frac{1+x}{2}\), 4) = (\(\frac{7}{2}\), \(\frac{y+5}{2}\))
\(\frac{1+x}{2}\) = \(\frac{7}{2}\)
1 + x = 7
x = 7 – 1
= 6
and
\(\frac{y+5}{2}\) = 4
y + 5 = 8
y = 8 – 5
= 3
∴ The value of x = 6 and y = 3

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Students can download Maths Chapter 3 Algebra Ex 3.19 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.19

Question 1.
A system of three linear equations in three variables is inconsistent if their planes
(1) intersect only at a point
(2) intersect in a line
(3) coincides with each other
(4) do not intersect
Answer:
(4) do not intersect

Question 2.
The solution of the system x + y -3z = -6, -7y + 7z = 7,3z = 9 is
(1) x = 1, y = 2, z = 3
(2) x = -1, y = 2, z = 3
(3) x = -1, y = -2, z = 3
(4) x = 1, y = 2, z = 3
Solution:
(4) x = 1, y = 2, z = 3
Hint:
x + y – 3z = -6
-7y + 7z = 7
3z = 9
z = 3
-7y + 21 = 7
-7y = -14
y = 2
x + 2 -3 × 3 = -6
x + 2 – 9 = -6 .
x = -6 + 7 = 1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 3.
If (x – 6) is the HCF of x2 – 2x – 24 and x2 – kx – 6 then the value of k is ……………..
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5
Hint.
HCF = x – 6
p(x) = x2 – 2x – 24
= (x – 6) (x + 4)
g(x) = x2 – kx – 6
∴ x – 6 is the common factor.
g (6) = 62 – k(6) – 6
= 36 – 6k – 6
= 30 – 6k
g(6) = 0
30 – 6k = 0
30 = 6k ⇒ k = \(\frac { 30 }{ 6 } \) = 5
The value of k = 5

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 4.
\(\frac { 3y-3 }{ y } \) + \(\frac { 7y-7 }{ 3y2 } \) is …………
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 1
Answer:
(1) \(\frac{9 y}{7}\)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 2

Question 5.
y2 + \(\frac{1}{y^{2}}\) is not equal to ………….
(1) \(\frac{y^{4}+1}{y^{2}}\)
(2) (y + \(\frac { 1 }{ y } \))2
(3) (y – \(\frac { 1 }{ y } \))2
(4) (y + \(\frac { 1 }{ y } \))2 – 2
Answer:
(2) (y + \(\frac { 1 }{ y } \))2
Hint.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 4

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 6.
\(\frac{x}{x^{2}-25}-\frac{8}{x^{2}+6 x+5}\) gives …………..
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 5
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 6
Hint.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 7

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 7.
The square root of \(\frac{256 x^{8} y^{4} z^{10}}{25 x^{6} y^{6} z^{6}}\) is equal to ………….
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 8
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 9
Hint.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 10

Question 8.
Which of the following should be added to make x4 + 64 a perfect square
(1) 4x2
(2) 16x2
(3) 8x2
(4) -8x2
Solution:
(2) 16x2
Hint:
x4 + 64 = (x2)2 + 82 + 2 × 8x2
= (x2 + 8)2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 9.
The solution of (2x – 1)2 = 9 is equal to ………..
(1) -1
(2) 2
(3) -1, 2
(4) None of these
Answer:
(3) -1, 2
Hint.
(2x – 1)2 = 9 ⇒ (2x – 1) = \(\sqrt { 9 }\)
2x – 1 = ± 3 ⇒ 2x – 1 = 3 or 2x – 1 = 3
2x – 1 = 3
2x = 4 ⇒ x = \(\frac { 4 }{ 2 } \) = 2
2x – 1 = -3 ⇒ 2x = -3 + 1
x = -1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 10.
The values of a and b if 4x4 – 24x3 + 76x2 + ax + b is a perfect square are ………..
(1) 100,120
(2) 10,12
(3) -120,100
(4) 12,10
Answer:
(3) -120,100
Hint.
Since it is a perfect square
a + 120 = 0
a = -120
b – 100 = 0
b = 100
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 11

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 11.
If the roots of the equation q2x2 + p2x + r2 = 0 are the squares of the roots of the equation qx2 + px + r = 0 , then q,p,r are in ……………..
(1) A.P.
(2) G.P.
(3) Both A.P and G.P
(4) none of these
Answer:
(2) G.P.
Hint.
q2x2 + p2x + r2 = 0
Let the roots be α2 + β2
α2 + β2 = \(\frac{-p^{2}}{q^{2}}\)
α2β2 = \(\frac{r^{2}}{q^{2}}\)
qx2 + px + r = 0
Let the root be α and β
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 12
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 13

Question 12.
Graph of a linear polynomial is a
(1) straight line
(2) circle
(3) parabola
(4) hyperbola
Solution:
(1) straight line

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 13.
The number of points of intersection of the quadratic polynomial x2 + 4x + 4 with the X axis is …………….
(1) 0
(2) 1
(3) 0 or 1
(4) 2
Answer:
(2) 1
Hint:
(x + 2)2 = 0 ⇒ (x + 2) (x + 2) = 0
x + 2 = 0 or x + 2 = 0 ⇒ x = -2 or x = -2
Number of points of intersection is 1 (both the values are same)

Question 14.
For the given matrix
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 14
the order of the matrix AT is ……………
(1) 2 × 3
(2) 3 × 2
(3) 3 × 4
(4) 4 × 3
Answer:
(3) 3 × 4

Question 15.
If A is a 2 × 3 matrix and B is a 3 × 4 matrix, how many columns does AB have ………..
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4
Hint.
The order of AB is 2 × 4
∴ Number of columns is 4.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 16.
If the number of columns and rows are not equal in a matrix then it is said to be a
(1) diagonal matrix
(2) rectangular matrix
(3) square matrix
(4) identity matrix
Solution:
(2) rectangular matrix

Question 17.
Transpose of a column matrix is …………..
(1) unit matrix
(2) diagonal matrix
(3) column matrix
(4) row matrix
Answer:
(4) row matrix

Question 18.
Find the matrix X if
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 15
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 16
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 17
Hint.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 18

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

Question 19.
Which of the following can be calculated from the given matrices?
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 19
(i) A2
(ii) B2
(iii) AB
(iv) BA
(1) (i) and (ii) only
(2) (ii) and (iii) only
(3) (ii) and (iv) only
(4) all of these
Answer:
(3) (ii) and (iv) only
Hint: (i) A2 is possible to find
(ii) B2 is also possible to find
(iii) not possible: order of A= (3 × 2) order of B is (3 × 3). AB is not possible number of column of matrix A ≠ number of rows of the matrix B.
(iv) Possible number column of the matrix is equal to the number of the matrix A.

Question 20.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 20
Which of the following statements are correct?
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19 21
(1) (i) and (ii) only
(2) (ii) and (iii) only
(3) (iii) and (iv)
(4) all of these
Answer:
(1) (i) and (ii) only
Hint.
(i) AB + C : order of A = 2 × 3
order of B = 3 × 2
order of AB = 2 × 2
order of C = 2 × 2
It is possible to add AB + C

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.19

(ii) BC :
order of B = 3 × 2
order of C = 2 × 2
It is possible to find BC

(iii) BA is not possible
order of B = 3 × 2
order of A = 3 × 2
BA does not exist.
BA + C is not a correct statement

(iv) ABC is not possible
order of A = 2 × 3 order of
B = 3 × 2
order of AB = 2 × 2
order of C = 3 × 2
It is not possible to multiply AB and C.
∴ The statement ABC is not correct.