Samacheer Kalvi 9th Maths Guide Chapter 9 Probability Additional Questions

Students can download Maths Chapter 9 Probability 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 9 Probability Additional Questions

I. Choose the Correct Answer

Question 1.
Which of the following cannot be taken as probability of an event?
(a) 0
(b) 0.35
(c) $$\frac{7}{20}$$
(d) –$$\frac{7}{20}$$
Solution:
(d) –$$\frac{7}{20}$$

Question 2.
A letter is chosen at random from the word “MATHEMATICS” the probability of getting a vowel is ……..
(a) $$\frac{2}{11}$$
(b) $$\frac{3}{11}$$
(c) $$\frac{4}{11}$$
(d) $$\frac{5}{11}$$
Solution:
(c) $$\frac{4}{11}$$

Question 3.
If P(A) =$$\frac{1}{3}$$ then P(A)’ is ………
(a) $$\frac{1}{3}$$
(b) $$\frac{2}{3}$$
(c) $$\frac{3}{2}$$
(d) 1
Solution:
(b) $$\frac{2}{3}$$

Question 4.
An integer is chosen from the first twenty natural number, the probability that it is a prime number is ……..
(a) $$\frac{1}{5}$$
(b) $$\frac{2}{5}$$
(c) $$\frac{3}{5}$$
(d) $$\frac{4}{5}$$
Solution:
(b) $$\frac{2}{5}$$

Question 5.
From a well shuffled pack of 52 cards one card is drawn at random. The probability of getting not a king is ………
(a) $$\frac{12}{13}$$
(b) $$\frac{1}{13}$$
(c) $$\frac{4}{13}$$
(d) $$\frac{2}{13}$$
Solution:
(a) $$\frac{12}{13}$$

II. Answer the Following Questions

Question 6.
1500 families with 2 children were selected randomly and the following data were recorded.

Compute the probability of a family chosen at random having one girl.
Solution:
Total number of families = 1500
∴ n(S) = 1500
Let E be the event of getting one girl
n(E) = 814
P(E) = $$\frac{n(E)}{n(S)}$$
= $$\frac{814}{1500}$$
= $$\frac{407}{750}$$

Question 7.
The record of weather station shows that out of the part 250 consecutive days its whether forecast were correct 175 time. What is the probability that it was not correct on a given data?
Solution:
n(S) = 250
Let E be the event of getting whether forecast were correct
n(E) = 175
P(E) = $$\frac{n(E)}{n(S)}$$
= $$\frac{175}{250}$$
= $$\frac{7}{10}$$
Forecast was not correct on a given day = 1 – P(E)
= 1 – $$\frac{7}{10}$$
= $$\frac{10-7}{10}$$
= $$\frac{3}{10}$$

Question 8.
If A coin is tossed 200 times and is found that a tail comes up for 120 times. Find the probability of getting a tail.
Solution:
Number of trials = 200
n(S) = 200
Let E be the event of getting a tail
n(E) = 120
p(E) = $$\frac{n(E)}{n(S)}$$
= $$\frac{120}{200}$$
= $$\frac{12}{20}$$
= $$\frac{3}{5}$$

Question 9.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of drawing a red ball, then find the number of blue balls in the bag.
Solution:
Let the number of blue balls be “x”
Total number of balls = 5 + x
∴ n(S) = 5 + x
Let B be the event of drawing a blue ball and R be the event of drawing a red ball
Given P(B) = 3P(R)

∴ x = 15
Number of blue balls = 15

Question 10.
Find the probability that a non leap year selected at random will have 53 fridays.
Solution:
No. of days in a non leap year = 365 days
This year contain 52 weeks and one day
Sample space = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
Let A be the event of getting a friday
n(A) = 1
P(A) = $$\frac{n(A)}{n(S)}$$
= $$\frac{1}{7}$$

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

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

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

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

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

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

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 $$\frac{209}{400}$$

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 = $$\frac{15}{8}$$

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

Question 5.
1500 families were surveyed and following data was recorded about their maids at homes

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 = $$\frac{1}{6}$$

(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 9.1

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) = $$\frac{n(E)}{n(S)}$$
= $$\frac{1}{7}$$

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) = $$\frac{n(E)}{n(S)}$$
= $$\frac{3}{6}$$
= $$\frac{1}{2}$$

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) = $$\frac{n(E_{3})}{n(S)}$$
= $$\frac{16}{24}$$
= $$\frac{2}{3}$$

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) = $$\frac{n(E)}{n(S)}$$
= $$\frac{1}{4}$$

Question 6.
Two dice are rolled, find the probability that the sum is

(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

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) = $$\frac{n(E_{1})}{n(S)}$$
= $$\frac{25}{7000}$$
= $$\frac{1}{280}$$

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?

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) = $$\frac{n(E)}{n(S)}$$
= $$\frac{6}{8}$$
= $$\frac{3}{4}$$

Question 10.
Frame two problems in calculating probability, based on the spinner shown here.

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

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

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 …….

(a) 3
(b) 4
(c) 6
(d) 14
Solution:
(b) 4

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

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

II. Answer the Following Questions

Question 1.
Find the Arithmetic mean for the following data.

Solution:

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.

Solution:
Assumed mean = 35

Arithmetic mean ($$\bar { X }$$) = A + $$\frac{Σfd}{Σf}$$ × c
= 35 + $$\frac{(-20)}{100}$$ × 10
= 35 – 2
= 33
∴ Arithmetic mean = 33

Question 3.
Find the median for the following data.

Solution:
The given class intervals is inclusive type we convert it into exclusive type.

$$\frac{N}{2}$$ = $$\frac{70}{2}$$
= 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

Question 4.
Calculate the mode of the following data.

Solution:

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

∴ Arithmetic mean = 25 + 2.727
= 25 + 2.73
= 27.73
mode = 27.73

Question 5.
Find the mean, median and mode of marks obtained by 20 students in an examination. The marks are given below.

Solution:

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

= 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

= 33.33
Mean = 28, median = 30, mode = 33.33

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

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

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

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

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

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:

= 46

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 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:

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)

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

∴ 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

Question 4.
The demand of track suit of different sizes as obtained by a survey is given below:

Which size is demanded more?
Solution:
The highest frequency is 37
The corresponding value is the mode
∴ Mode = 40
Size 40 is demanded more.

Question 5.
Find the mode of the following data:

Solution:

The highest frequency is 46
20 – 30 is the modal class
Here l = 20, f = 46, f1 = 38, f2 = 34 and c = 10
Mode

= 20 + 4
= 24
∴ Mode = 24

Question 6.
Find the mode of the following distribution

Solution:
In the given table the class intervals are in inclusive form; convert them into exclusive form.

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

= 58.5
∴ Mode = 58.5

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

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

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

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

Question 5.
The following are the marks scored by the students in the Summative Assessment exam.

Calculate the median.
Solution:

$$\frac{N}{2}$$ = $$\frac{50}{2}$$
= 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

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 7 Mensuration Additional Questions

Students can download Maths Chapter 7 Mensuration 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 7 Mensuration Additional Questions

I. Choose the Correct Answer

Question 1.
If the sides of a triangles are 5 cm, 6 cm and 7 cm then the area is ……..
(a) 18 cm²
(b) 6 √2 cm²
(c) 6 √6 cm²
(d) 6 √3 cm²
Solution:
(c) 6 √6 cm²

Question 2.
The perimeter of an equilateral triangle is 60 cm then the area is ………
(a) 60 √3 cm²
(b) 20 √3 cm²
(c) 50 √3 cm²
(d) 100 √3 cm²
Solution:
(d) 100 √3 cm²

Question 3.
The total surface area of the cuboid with dimension 20 cm × 30 cm × 15 cm is ………
(a) 2700 cm²
(b) 1500 cm²
(c) 2500 cm²
(d) 3000 cm²
Solution:
(a) 2700 cm²

Question 4.
The number of bricks each measuring 70 cm × 80 cm × 40 cm that will be required to build a wall whose dimensions are 7 m × 8 m × 4 m is ……..
(a) 4000
(b) 3000
(c) 2000
(d) 1000
Solution:
(d) 1000

Question 5.
The volume of a cube is 4913 m² then the length of its side is ……..
(a) 13 m
(b) 17 m
(c) 34 m
(d) 27 m
Solution:
(b) 17 m

II. Answer the Following Questions

Question 6.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
The non parallel sides are 13 m and 14 m. Draw BE || AD. Such that BE = 13 m
∴ ABED is a parallelogram

To find Area of a ΔBCE
a = 13 m, b = 15 m and c = 14 m
s = $$\frac{a+b+c}{2}$$
= $$\frac{13+15+14}{2}$$
= $$\frac{42}{2}$$
= 21 m
s – a = 21 – 13 = 8 m
s – b = 21 – 15 = 6 m
s – c = 21 – 14 = 7 m
Area of a ΔBCE

= 2² × 3 × 7
= 84 m²
Let the height of the triangle BF be x
Area of the ΔBEC = 84 m²
= $$\frac{1}{2}$$ × b × h = 84
= $$\frac{1}{2}$$ × 15 × h = 84
x = $$\frac{84×2}{15}$$
= $$\frac{56}{5}$$ m
= 11.2 m
Area of parallelogram ABED = base × height sq. units
= 10 × 11.2 m²
= 112 m²
∴ Area of the field = Area of ΔBCE + Area of parallelogram ABED
= 84 m² + 112 m²
= 196 m²
(OR)
Area of the field = Area of the trapezium ABCD
= $$\frac{1}{2}$$ h (a + b)
= $$\frac{1}{2}$$ × 11.2 (25 + 10)
= $$\frac{1}{2}$$ × 11.2 (35)
= 196 m²

Question 7.
Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm and ⌊B = 90°.
Solution:

In ΔABC, ⌊B = 90°
∴ ABC is a right angle triangle
Area of the right ΔABC = $$\frac{1}{2}$$ × AB × BC sq.units
= $$\frac{1}{2}$$ × 8 × 6 cm²
= 24 cm²
In ΔACD a = 10 cm, b = 8 cm and c = 10 cm
s = $$\frac{a+b+c}{2}$$
= $$\frac{10+8+10}{2}$$
= $$\frac{28}{2}$$
= 14 cm
s – a = 14 – 10 = 4 cm
s – b = 14 – 8 = 6 cm
s – c = 14 – 10 = 4 cm
Area of ΔACD = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{14×4×6×4}$$
= $$\sqrt{2×7×4×2×3×4}$$
= 4 × 2 $$\sqrt{21}$$ cm²
= 8$$\sqrt{21}$$ cm²
= 8 × 4.58
= 36.64 cm²
Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 24 cm² + 36.64 cm²
= 60.64 cm²
Area of the quadrilateral = 60.64 cm²

Question 8.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area for white washing = Lateral surface area of four walls + Area of the ceiling
= 2(l + b) × h + (l × b)
= 2(5 + 4) × 3 + (5 × 4) m²
= (2 × 9 × 3 + 20) m²
= (54 + 20) m²
= 74 m²
Cost of white washing for one m² = Rs 7.50
Cost of white washing for 74 m² = Rs 74 × 7.50
= Rs 555
The required cost = Rs 555

Question 9.
How many hollow blocks of size 30 cm × 15 cm × 20 cm are needed to construct a wall 60 m in length 0.3 m in breadth and 2 m in height.
Solution:
Length of a wall = 60 m = 6000 cm
Breadth of a wall = 0.3 m = 30 cm
Height of a wall = 2 m = 200 cm
Volume of the wall = l × b × h sq. unit
= 6000 × 30 × 200 cm³
For hollow block
l = 30 cm, b = 15 cm, h = 20 cm
Volume of one hollow block = l × b × h
= 30 × 15 × 20 cm²
Number of hollow blocks required

= 4000
∴ Number of bricks = 4000

Question 10.
Find the number of cubes of side 3 cm that can be cut from a cuboid of dimensions 10 cm × 9 cm × 6 cm.
Solution:
Side of a cube = 3 cm
Volume of a cube = a³ cm
= 3 × 3 × 3 cm³
Length of the cuboid (l) = 10 cm
Breadth of the cuboid (b) = 9 cm
Height of the cuboid (h) = 6 cm
Volume of the cuboid = l × b × h cu. unit
= 10 × 9 × 6 cm
Number of cubes

∴ Number of cubes = 20

Samacheer Kalvi 9th Maths Guide Chapter 7 Mensuration Ex 7.4

Students can download Maths Chapter 7 Mensuration Ex 7.4 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 7 Mensuration Ex 7.4

Question 1.
The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is ……..
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
Solution:
(c) 30 cm
Hint:
l = 15 cm, b = 20 cm, h = 25 cm
Semi-perimeter = $$\frac{a+b+c}{2}$$
= $$\frac{15+20+25}{2}$$
= 30 cm

Question 2.
If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is ………
(a) 3 cm²
(b) 6 cm²
(c) 9 cm²
(d) 12 cm²
Solution:
(b) 6 cm²
Hint:
a- 3 cm, b = 4 cm, c = 5 cm
s = $$\frac{a+b+c}{2}$$
= $$\frac{3+4+5}{2}$$
= 6 cm
Area of the triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{6×3×2×1}$$
= $$\sqrt{36}$$
= 6 cm²

Question 3.
The perimeter of an equilateral triangle is 30 cm. The area is ……..
(a) 10 √3 cm²
(b) 12 √3 cm²
(c) 15 √3 cm²
(d) 25 √3 cm²
Solution:
(d) 25 √3 cm²
Hint:
Perimeter of an equilateral triangle = 30 cm
3a = 30 cm
a = $$\frac{30}{3}$$
= 10 cm
Area of an equilateral triangle = $$\frac{√3}{4}$$ a² sq.units
= $$\frac{√3}{4}$$ × 10 × 10
= 25 √3 cm²

Question 4.
The lateral surface area of a cube of side 12 cm is ……..
(a) 144 cm²
(b) 196 cm²
(c) 576 cm²
(d) 664 cm²
Solution:
(c) 576 cm²
Hint:
Side of a cube (a) = 12 cm
L.S.A. of a cube = 4a² sq.units
= 4 × 12 × 12 cm²
= 576 cm²

Question 5.
If the lateral surface area of a cube is 600 cm², then the total surface area is ………
(a) 150 cm²
(b) 400 cm²
(c) 900 cm²
(d) 1350 cm²
Solution:
(c) 900 cm²
Hint:
L.S.A. of a cube = 600 cm²
4a² = 600
a² = $$\frac{600}{4}$$
= 150
Total surface area of a cube = 6a² sq.units
= 6 × 150 cm²
= 900 cm²

Question 6.
The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is ………
(a) 280 cm²
(b) 300 cm²
(c) 360 cm²
(d) 600 cm²
Solution:
(a) 280 cm²
Hint:
T.S.A. of a cuboid = 2(lb + bh + lh) sq.units
= 2(10 × 6 + 6 × 5 + 10 × 5) cm²
= 2(60 + 30 + 50) cm²
= 2 × 140 cm²
= 280 cm²

Question 7.
If the ratio of the sides of two cubes are 2 : 3, then ratio of their surface areas will be ………
(a) 4 : 6
(b) 4 : 9
(c) 6 : 9
(d) 16 : 36
Solution:
(b) 4 : 9
Hint:
Ratio of the surface area of cubes = 4a12 : 4a22
= a12 : a22
= 4² : 9²
= 4 : 9

Question 8.
The volume of a cuboid is 660 cm and the area of the base is 33 cm². Its height is ………
(a) 10 cm
(b) 12 cm
(c) 20 cm
(d) 22 cm
Solution:
(c) 20 cm
Hint:
Volume of a cuboid = 660 cm³
l × b × h = 660
33 × h = 660 (Area of the base = l × b)
h = $$\frac{660}{33}$$
= 20 cm

Question 9.
The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is ………
(a) 75 litres
(b) 750 litres
(c) 7500 litres
(d) 75000 litres
Solution:
(d) 75000 litres
Hint:
The capacity of a tank = l × b × h cu.units
= (10 × 5 × 1.5) m³
= 75 m³
= 75 × 1000 litres [1m³ = 1000 lit]
= 75000 litres

Question 10.
The number of bricks each measuring 50 cm × 30 cm × 20 cm that will be required to build a wall whose dimensions are 5 m x 3 m x 2 m is ………
Solution:
(a) 1000
(b) 2000
(c) 3000
(d) 5000
Solution:
(a) 1000
Hint:
Volume of one brick = 50 × 30 × 20 cm³
Volume of the wall = l × b × h
[l = 5m = 500 cm]
[b = 3m = 300 cm]
[h = 2m = 200 cm]
= 500 × 300 × 200 cm³
No. of bricks

= 10 × 10 × 10
= 1000 bricks