Category: Class 9

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

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.

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 5^th value and 6^th 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

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

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 = \(\frac{1}{75}\)

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?

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

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

Students can download Maths Chapter 7 Mensuration Ex 7.3 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 7 Mensuration Ex 7.3

Question 1.
Find the volume of a cuboid whose dimensions are
(i) length = 12 cm, breadth = 8 cm, height = 6 cm
(ii) length = 60 m, breadth = 25 m, height = 1.5 m
Solution:
(i) Here l = 12 cm, b = 8 cm, h = 6 cm
Volume of a cuboid = l × b × h
= (12 × 8 × 6) cm³
= 576 cm³

(ii) Here l = 60 m, b = 25 m. h = 1.5 m
Volume of a cuboid = l × b × h
= 60 × 25 × 1.5 m³
= 2250 m³

Question 2.
The dimensions of a match box are 6 cm × 3.5 cm × 2.5 cm. Find the volume of a packet containing 12 such match boxes.
Solution:
Length of a match box (l) = 6 cm
Breadth of a match box (b) = 3.5 cm
Height of a match box (h) = 2.5 cm
Volume of one match box = l × b × h cu. units
= 6 × 3.5 × 2.5 cm³
= 52.5 cm³
Volume of 12 match box = 12 × 52.5 cm³
= 630 cm³

Question 3.
The length, breadth and height of a chocolate box are in the ratio 5 : 4 : 3. If its volume is 7500 cm³, then find its dimensions.
Solution:
Let the length of a chocolate be 5x, the breadth of a chocolate be 4x, and the height of a chocolate be 3x.
Volume of a chocolate = 7500 cm³
l × b × h = 7500
5x × 4x × 3x = 7500
5 × 4 × 3 × x³ = 7500
x³ = \(\frac{7500}{5×4×3}\)
x³ = 125 ⇒ x³ = 5³
x = 5
∴ Length of a chocolate = 5 × 5 = 25 cm
Breath of a chocolate = 4 × 5 = 20 cm
Height of a chocolate = 3 × 5 = 15 cm

Question 4.
The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres.
Solution:
Length of a pond (l) = 20.5 m
Breadth of a pond (b) = 16 m
Depth of a pond (h) = 8 m
Volume of the pond = l × b × h cu.units
= 20.5 × 16 × 8 m³
= 2624 m³ (1 cu. m = 1000 lit)
= (2624 × 1000) litres
= 2624000 lit

Question 5.
The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be required to build a wall of 20 m length, 48 cm breadth and 6 m height?
Solution:
Length of a brick (l) = 24 cm
Breadth of a brick (b) = 12 cm
Depth of a brick (h) = 8 cm
Volume of a brick = lbh cu.units
Volume of one brick = 24 × 12 × 8 cm³
Length of a wall (l) = 20 m = 2000 cm
Breadth of a wall (b) = 48 cm
Height of a wall (h) = 6 m = 600 cm
Volume of a wall = l × b × h cu. units
= 2000 × 48 × 600 cm³
Number of bricks

= 500 × 50 ( ÷ by 4)
= 25000 bricks
∴ Number of bricks = 25000

Question 6.
The volume of a container is 1440 m³. The length and breadth of the container are 15 m and 8 m respectively. Find its height.
Solution:
Let the height of the container be “h”
Length of the container (l) = 15 m
Breadth of the container (b) = 8 m
Volume of the container = 1440 m³
l × b × h = 1440
15 × 8 × h = 1440
h = \(\frac{1440}{15×8}\)
= 12 m
∴ Height of the container = 12 m

Question 7.
Find the volume of a cube each of whose side is
(i) 5 cm
(ii) 3.5 m
(iii) 21 cm
Solution:
(i) Side of a cube (a) = 5 cm
Volume of a cube = a³ cu. units
= 5 × 5 × 5 cm³
= 125 cm³

(ii) Side of a cube (a) = 3.5 m a³ cu. units
Volume of a cube = 3.5 × 3.5 × 3.5 m³
= 42.875 m³

(iii) Side of a cube (a) = 21 cm
Volume of a cube = a³ cu. units
= 21 × 21 × 21 cm³
= 9261 cm³

Question 8.
A cubical milk tank can hold 125000 litres of milk. Find the length of its side in metres.
Solution:
Volume of the cubical tank = 125000 liters
= \(\frac{125}{1000}\) m³ (1 cu.m = 1000 lit)
= 125 m³
a³ = 125 ⇒ a³ = 5³
a = 5
Side of a cube = 5 m

Question 9.
A metallic cube with side 15 cm is melted and formed into a cuboid. If the length and height of the cuboid is 25 cm and 9 cm respectively then find the breadth of the cuboid.
Solution:
Side of a cube (a) = 15 cm
Length of a cuboid (l) = 25 cm
Height of a cuboid (h) = 9 cm
Volume of the cuboid = Volume of the cube
l × b × h = a³
25 × b × 9 = 15 × 15 × 15
b = \(\frac{15 × 15 × 15}{25 × 9}\)
= 15 cm
Breadth of the cuboid = 15 cm

Students can download Maths Chapter 7 Mensuration Ex 7.2 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 7 Mensuration Ex 7.2

Question 1.
Find the Total Surface Area and the Lateral Surface Area of a cuboid whose dimensions are length = 20 cm, breadth = 15 cm, height = 8 cm.
Solution:
Here l = 20 cm, b = 15 cm, h = 8 cm
L.S.A. of the cuboid = 2(1 + b)h sq.m
= 2(20 + 15) × 8
= 2 × 35 × 8
= 560 sq.m
Total surface area of the cuboid = 2(lb + bh + lh) sq.units
= 2(20 × 15 + 15 × 8 + 8 × 20) sq. cm
= 2(300 + 120 + 160) sq. cm
= 2 × 580 sq. cm
= 1160 sq. cm

Question 2.
The dimensions of a cuboidal box are 6 m x 400 cm x 1.5 m. Find the cost of painting its entire outer surface at the rate of Rs 22 per m².
Solution:
Length of the cuboid box (l) = 6 m
Breadth of the cuboid box (b) = 400 cm = 4m
Height of the cuboid box (h) = 1.5 m
T.S.A. of the cuboid = 2(lb + bh + lh) sq.units
= 2(6 × 4 + 4 × 1.5 + 1.5 × 6) sq.units
= 2(24 + 6 + 9)
= 2 × 39 sq.m
= 78 sq.m
Cost of painting for one sq.m = Rs 22
Total cost of painting = Rs 78 × 22
= Rs 1716

Question 3.
The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of Rs 8.50 per m².
Solution:
Length of the hall (l) = 10 m
Breath of the hall (b) = 9 m
Height of the hall (h) = 8 m
Area to be white wash = L.S.A. + Ceiling of the building
= 2(l + b)h + lb sq.units
= 2(10 + 9)8 + 10 × 9 sq.m
= 2 × 19 × 8 + 10 × 9 sq. m
= (304 + 90) sq.m
= 394 sq.m
Cost of white washing one sq.m = Rs 8.50
Cost of white washing for 394 sq.m = Rs 394 × 8.50
= Rs 3349
Total cost of white washing = Rs 3349

Question 4.
Find the TSA and LSA of the cube whose side is
(i) 8 m
(ii) 21 cm
(iii) 7.5 cm
Solution:
(i) 8m
Side of a cube (a) = 8m
T.S.A. of the cube = 6a² sq.units
= 6 × 8 × 8 sq. m
= 384 sq.m
L.S.A. of the cube = 4a² sq.units
= 4 × 8 × 8 sq.m
= 256 sq.m

(ii) 21 cm
Solution:
Side of a cube (a) = 21 cm
T.S.A. of the cube = 6a² sq. units
= 6 × 21 × 21 cm²
= 2646 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 21 × 21 sq.cm
= 4 × 441 cm²
= 1764 cm²

(iii) 7.5 cm
Solution:
Side of a cube (a) = 7.5 cm
T.S.A. of the cube = 6a² sq.units
= 6 × 7.5 × 7.5 cm²
= 337.5 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 7.5 × 7.5 sq.cm
= 225 cm²

Question 5.
If the total surface area of a cube is 2400 cm² then, find its lateral surface area.
Solution:
T.S.A. of the cube = 2400 cm²
6a² = 2400
a² = \(\frac{2400}{6}\)
= 400 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 400 cm²
= 1600 cm²
(OR)
T.S.A. of the cube = 2400 cm²
6a² = 2400
a² = \(\frac{2400}{6}\)
= 400
a = \(\sqrt{400}\)
= 20 cm
Side of a cube (a) = 20 cm
L.S.A. of the cube = 4a² sq.units
= 4 × 20 × 20 cm²
= 1600 cm²

Question 6.
A cubical container of side 6.5 m is to be painted on the entire outer surface. Find the area to be painted and the total cost of painting it at the rate of Rs 24 per m².
Solution:
Side of a cube (a) = 6.5 m
Total surface area of the cube = 6a² sq.units
= 6 × 6.5 × 6.5 sq.m
= 253.50 sq.m
Cost of painting for 1 sq.m = Rs 24
Cost of painting for 253.5 sq.m = 253.5 × 24
= Rs 6084
∴ Area to be painted = 253.50 m²
Total cost of painting = Rs 6084

Question 7.
Three identical cubes of side 4 cm are joined end to end. Find the total surface area and lateral surface area of the new resulting cuboid.
Solution:
Joint the three identical cubes we get a new cuboid
Length of the cuboid (l) = (4 + 4 + 4) cm
l = 12 cm
Breadth of the cuboid (b) = 4 cm
Height of the cuboid (h) = 4 cm
Total surface area of the new cuboid = 2(lb + bh + lh) sq.units
= 2(12 × 4 + 4 × 4 + 4 × 12)
= 2(48 + 16 + 48) cm
= 2(112) cm²
= 224 cm²
Lateral surface area of the new cuboid = 2(l + b)h sq.units
= 2(12 + 4)4 cm²
= 2 × 16 × 4 cm²
= 128 cm²
∴ T.S.A of the new cuboid = 224 cm²
L.S.A of the new cuboid = 128 cm²