# Class 7

## Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

Students can Download Maths Chapter 2 Measurements Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

Question 1.
Find the area of rhombus PQRS shown in the following figures.

Solution:
(i) Given the diagonals d1 = 16 cm ; d2 = 8 cm
Area of the rhombus = $$\frac{1}{2}$$(d1 × d2) sq. units
= $$\frac{1}{2}$$ × 16 × 8 cm2 = 64 cm2
Area of the rhombus = 64 cm2

(ii) Given base b = 15 cm ; Height h = 11 cm
Area of the rhombus = (base × height) sq. units
= 15 × 11 cm2 = 165 cm2
Area of the rhombus = 165 cm2

Question 2.
Find the area of a rhombus whose base is 14 cm and height is 9 cm.
Solution:

Given base b = 14 cm ; Height h = 9 cm
Area of the rhombus = b × h sq. units
= 14 × 9 cm2 = 126 cm2

Question 3.
Find the missing value.

Solution:
(i) Given diagonal d1 = 19 cm ; d2 = 16 cm
Area of the rhombus = $$\frac{1}{2}$$(d1 × d2) sq. units = $$\frac{1}{2}$$ × 19 × 16
= 152 cm2

(ii) Given diagonal d1 = 26 m ; Area of the rhombus = 468 sq. m
= $$\frac{1}{2}$$(d1 × d2) = 468 ; (26 × d2) = 468 × 2
d2 = $$\frac{468 \times 2}{26}$$ = d2 = 36 m

(iii) Given diagonal d2 = 12 mm; Area of the rhombus = 180 sq. m
$$\frac{1}{2}$$(d1 × d2) = 180
$$\frac{1}{2}$$(d1 × 12) = 180
d1 × 12 = 180 × 2
d1 = $$\frac{180 \times 2}{12}$$
d1 = 30 mm
Diagonal d1 = 30 mm
Tabulating the results we have

Question 4.
The area of a rhombus is 100 sq. cm and length of one of its diagonals is 8 cm. Find the length of the other diagonal.
Solution:
Given the length of one diagonal d1 = 8 cm ; Area of the rhombus = 100 sq. cm
$$\frac{1}{2}$$(d1 × d2) = 100
$$\frac{1}{2}$$ × 8 × d2 = 100
8 × d2 = 100 × 2
d2 = $$\frac{100 \times 2}{8}$$ = 25 cm
Length of the other diagonal d2 = 25 cm

Question 5.
A sweet is in the shape of rhombus whose diagonals are given as 4 cm and 5 cm. The surface of the sweet should be covered by an aluminum foil. Find the cost of aluminum foil used for 400 such sweets at the rate of ₹ 7 per 100 sq. cm.
Solution:
Diagonals d1 = 4 cm and d2 = 5 cm
Area of one rhombus shaped sweet = $$\frac{1}{2}$$(d1 × d2) sq. units = $$\frac{1}{2}$$ × 4× 5 cm2 = 10 cm2
Aluminum foil used to cover 1 sweet = 10 cm2
∴ Aluminum foil used to cover 400 sweets = 400 × 10 = 4000 cm2
Cost of aluminum foil for 100 cm2 = ₹ 7
∴ Cost of aluminum foil for 4000 cm2 = $$\frac{4000}{100}$$ × 7 = ₹ 280
∴ Cost of aluminum foil used = ₹ 280.

Objective Type Questions

Question 6.
The area of the rhombus with side 4 cm and height 3 cm is
(i) 7 sq. cm
(ii) 24 sq. cm
(iii) 12 sq. cm
(iv) 10 sq. cm
Solution:
(iii) 12 sq. cm
Hint:
Area = Base × Height = 4 × 3 = 12 cm2

Question 7.
The area of the rhombus when both diagonals measuring 8 cm is
(i) 64 sq. cm
(ii) 32 sq. cm
(iii) 30 sq. cm
(iv) 16 sq. cm
Solution:
(ii) 32 sq. cm
Hint:
Area = $$\frac{1}{2}$$(d1 × d2) = $$\frac{1}{2}$$ × 8 × 8 = 32

Question 8.
The area of the rhombus is 128 sq. cm. and the length of one diagonal is 32 cm. The length of the other diagonal is
(i) 12 cm
(ii) 8 cm
(iii) 4 cm
(iv) 20 cm
Solution:
(ii) 8 cm
Hint:
$$\frac{1}{2}$$ × d1 × d2 = 128 ⇒ d2 = $$\frac{128 \times 2}{32}$$ = 8cm

Question 9.
The height of the rhombus whose area 96 sq. m and side 24 m is
(i) 8 m
(ii) 10 m
(iii) 2 m
(iv) 4 m
Solution:
(iv) 4 m
Hint:
Area = Base × height = 96 ⇒ height = $$\frac{96}{24}$$ = 4

Question 10.
The angle between the diagonals of a rhombus is
(i) 120°
(ii) 180°
(iii) 90°
(iv) 100°
Solution:
(iii) 90°
Hint:
Angles of a rhombus bisect at right angles.

## Samacheer Kalvi 7th English Solutions Term 3 Poem Chapter 1 Sea Fever

Students can Download English Poem 1 Sea Fever Questions and Answers, Summary, Notes Pdf, Activity, Samacheer Kalvi 7th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th English Solutions Term 3 Poem Chapter 1 Sea Fever

Sea Fever Poem Line By Line Explanation By John Masefield, Sea Fever Exercise Question Answer.

Read And Understand

B. Choose the best answer

Question 1.
The title of the poem ‘Sea Fever means _________
(a) flu fever
(b) the poet’s deep wish to be at sea
(c) the poet s fear of the sea
Answer:
(b) the poet’s deep wish to be at sea

Question 2.
The poet asks for _________
(a) a fishing net
(b) a big boat
(c) a tall ship
Answer:
(c) a tall ship

Sea Fever Questions Answers  Question 3.
The poet wants to lead a life at sea like _________
(a) the gulls and whales
(b) the penguins and sharks
(c) the pelicans and dolphins
Answer:
(a) the gulls and whales

C. Read the lines and answer the questions.

Question 1.
I must go down to the sea again, to the lonely sea and the sky
Where does the poet want to go?
Answer:
The poet wants to go to the sea again.

Question 2.
And the wheel’s kick and the winds song and the white sail’s shaking
What according to the poet are the pleasures of sailing?
Answer:
Will Watching from the shore, the wind’s song, the ship’s steering wheel and the shaking of the sail in the breeze are the pleasures of sailing.

Question 3.
And all I ask is a windy day with the white clouds flying
Why does the poet ask for a windy day?
Answer:
The poet asks for a windy day, as it would blow away the thick white clouds from the sky and take the sail forward throughout the day.

Question 4.
And all I ask is a merry yarn from a laughing fellow rover.
What kind of human company does the poet want?
Answer:
The poet wants the company of a fellow sailor or wanderer like him.

Question 5.
And quiet sleep and a sweet dream when the long trick’s over What does the poet want to do after his voyage is over?
Answer:
The poet wants to sleep soundly with pleasant dreams at the end of his long shift on watch.

D. Poem Appreciation

Question 1.
Fill in the blanks with correct rhyming words from the poem.
Answer:
sky – by, knife – life. rover – over.

Question 2.
Quote the line that has been repeated in the poem.
Answer:
“I must go down to the seas again”.

Question 3.
Write the poetic device which is used in the line below.
And a grey mist on the sea’s face _________
Answer:
Imagery and personification.

Question 4.
What poetic device is used for the comparison in the below line?
Where the wind’s like a whetted life _________
Answer:
Simile

E. The poem has many visual and sound images. Complete the table with examples from the poem. The first one has been done for you.
Answer:

 Visual images Sound images Wheels kick Wind s song white sail’s shaking wild call grey dawn clear call running tide windy day white clouds flying blown spume sea – gulls crying

### Sea Fever Additional Questions

I. Poem Comprehension:

Question 1.
And all I ask is a tall ship and a star to steer her by;
What does the word ‘her’ refer to?
Answer:
‘Her’ refers to the tall ship.

Question 2.
To the gulls way and the whales way where the wind’s like a whetted knife;
What is like a whetted knife?
Answer:
The wind is strong like a whetted knife.

II. Poetic Devices:

Question 1.
I must go down to the seas again, to the lonely sea and the sky,
Pick out the alliterated words.
Answer:
sea –sky are the alliterated words.

Question 2.
And the wheel’s kick and the wind’s song and the white sail’s shaking,
What poetic device is used here?
Answer:
Personification : The wind, the wheel and the white sail are given the human qualities.

Question 3.
And quiet sleep and a sweet dream when the long trick’s over.
Name the literary device used in this line.
Answer:
Metaphor : “when the long trick’s over’. This is comparing the end of a long voyage to the end of a long life.

III. Sea Fever Short Questions and Answers.

Question 1.
Why does the poet ask for a star?
Answer:
The poet asks for a star to guide the tall ship.

Question 2.
According to the poet, how is the call of the running tide?
Answer:
The call of the running tide is wild and clear.

Question 3.
What does the word ‘vagrant’ mean?
Answer:
‘Vagrant’ means wandering.

Question 4.
What does he want from a fellow-wanderer?
Answer:
He wants to share a happy tale and laughter with a fellow-wanderer.

Question 5.
What does the word ‘trick’ in sailing terms refer to?
Answer:
It refers to ‘a watch at sea’.

IV. Paragraph Question with Answer.

Question 1.
What message does the poet convey through the poem?
Answer:
John Masefield’s poem ‘Sea Fever’ is a work of art that brings beauty to the English language through its use of rhythm, imagery and many complex figures of speech. The imagery in ‘Sea Fever’ suggests an adventurous ocean that appeals to all five senses. Along with an adventurous ocean, ‘Sea Fever’ also sets a mood of freedom through the imagery of travelling gypsies. ‘Sea Fever’ not only depicts a strong longing for the sea through its theme, but also through the use of complex figures of speech.

Warm Up

Rescue the sinking words!

Answer:

1. Voyage
2. Vaves
3. Island
4. Navigate
5. Sailor
6. Explore
7. Harbour
8. Shore

### Sea Fever Summary

In the poem ‘Sea Fever’, the poet, John Masefield expresses his desire for sailing once again in the quiet sea, under the quiet sky. He expresses his desire for the need of a well-built ship to sail and a star in the dark sky to act as a guide. Soon the adventures would begin and as each day dawns, he wakes up to see the early grey mist rise from the sea. He wishes that there would be strong winds that would blow away the thick white clouds in the sky and take sail forward throughout the day. Watching from the shore, fills his heart with the adventure and spirit. The poet finally wishes to be a wandering gypsy.

He loves to share the tales and laughter with a fellow sailor. He wanted to sleep soundly with pleasant dreams at the end of his long shift on watch.

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 6 Information Processing Additional Questions

Students can Download Maths Chapter 6 Information Processing Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 6 Information Processing Additional Questions

Exercise 6.1

Question 1.
Draw a flowchart to calculate simple interest for your bank deposit.
Solution:
Flow Chart:

Algorithm:
(i) Read principal
(ii) Read years
(iii) Read rate of interest per year
(iv) Calculate the interest with formula Interest = Principal × Years × Rate/100
(v) Print Interest

Question 2.
Draw a flowchart to convert the given Fahrenheit temperature to Celsius. Formula is C = $$\frac { 5 }{ 9 }$$ × (F – 32)
Solution:
Algorithm:
(i) Read temperature in Fahrenheit
(ii) Calculate temperature in Celsius using formula C = $$\frac { 5 }{ 9 }$$ × (F – 32)
(iii) Print C

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Additional Questions

Students can Download Maths Chapter 5 Statistics Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Additional Questions

Additional Questions and Answers

Exercise 5.1

Question 1.
The class scores on an exam are 97, 85, 92, 78 and 90. What is the average score?
Solution:

= $$\frac { 442 }{ 5 }$$ = 88.4

Question 2.
Mean of 20 observations is 27. Find the sum of 20 observations.
Solution:

Sum of observations = 27 × 20
Sum of observations = 540

Question 3.
Mean of same observations is 15. Their sum is 405. Find the number of values.
Solution:

Number of values = $$\frac { 405 }{ 15 }$$ = 27

Exercise 5.2

Question 1.
Find the mode of the following data. 1,2,5, 7, 3, 4, 2, 5, 7, 6, 2, 3.
Solution:
Arranging the data in ascending order: 1, 2, 2, 2, 3, 3,4, 5, 5, 6, 7, 7
Here 2 occurs maximum number of times.
∴ Mode is 2.

Exercise 5.3

Question 1.
Find the median of the data 1, 3, 7, 16, 0, 19, 7, 4, 3.
Solution:
Arranging the given data in ascending order: 0, 1, 3, 3, 4, 7, 7, 16, 19
Number of terms n = 9, which is odd.
∴ Median = ($$\frac { n+1 }{ 2 }$$)th term
= ($$\frac { 9+1 }{ 2 }$$)th term = ($$\frac { 10 }{ 2 }$$)th term
= 5th term
Hence Median = 4

Question 2.
Find the median of first 10 even number.
Solution:
First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Here number of data n = 10, which is even

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Intext Questions

Students can Download Maths Chapter 5 Statistics Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Intext Questions

Exercise 5.1
Try These (Text book Page No. 96)

Question 1.
Collect the height of students of your class. Organise the data in ascending order.
Solution:
Height of 15 students in our class.
130cm, 150 cm, 155 cm, 142 cm, 138 cm, 145 cm, 148 cm, 147 cm, 148cm, 143 cm, 141cm, 152 cm, 147 cm, 139 cm, 155 cm.

Ascending order:
130cm, 138cm, 139cm, 141cm, 142cm, 143cm, 145cm, 147cm, 147cm, 148cm, 148 cm, 150cm, 152 cm, 155cm, 155cm.

Try These (Text book Page No. 97)

Find the Arithmetic Mean or average of the following data.

Question 1.
The study time spent by Kathir in a week is 3 hrs, 4 hrs, 5 hrs, 3 hrs, 4 hrs, 3:45 hrs; 4:15 hrs.
Solution:

mean = 3 : 52 hrs

Question 2.
The marks scored by Muhil in five subjects are 75, 91, 48, 63, 51.
Solution:

Arithmetic Mean = 65.6

Question 3.
Money spent on vegetables for five days is ₹ 120, ₹ 80, ₹ 75, ₹ 95 and ₹ 86.
Solution:

Arithmetic Mean = 91.2

Think (Text book Page No. 99)

Check the properties of arithmetic mean for the example given below:

Question 1.
If the mean is increased by 2, then what happens to the individual observations.
Solution:
Given number are 3, 6, 9, 12, 15

If mean is increased by 2 then,

Sum of observations = 5 × 11 = 55
Difference in sum = 55 – 45 = 10
∴ Each number is increased by 2 if the mean is increased by 2.

Question 2.
If first two items are increased by 3 and last two items are reduced by 3, then what will be the new mean?
Solution:
If the first two items is increased by 3, then the numbers will be 3 + 3, 6 + 3 ⇒ 6, 9.
If last two numbers are decreased by 3, then the numbers will be 12 – 3, 15 – 3 ⇒ 9,12.

= $$\frac { 45 }{ 5 }$$ = 9
There is no change in the mean.

Exercise 5.2
Try These (Text book Page No. 101)

Question 1.
Find the mode of the following data. 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2
Solution:
Arranging the numbers in ascending order we get 0, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6
Since 2 and 3 occurs the maximum of 3 times. So mode of this data is 2 and 3.

Question 2.
Find the mode of the following data set. 3, 12, 15, 3, 4, 12, 11, 3, 12, 9, 19.
Solution:
Arranging the given data in ascending order : 3, 3, 3, 4, 9,11, 12, 12,12, 15, 19.
The data 3 and 12 occurs the maximum of 3 times.
So mode of this data is 3 and 12.

Question 3.
Find the mode of even numbers within 20.
Solution:
Even numbers within 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18.
There is no mode for this data.

Think (Text book Page No. 102)

Question 1.
A toy factory making variety of toys for kids, wants to know the most popular toy liked by all the kids. Which average will be the most appropriate for it?
Solution:
Mode.

Question 2.
Is there a mode exists between the odd numbers from 20 to 40? Discuss.
Solution:
Odd number between 20 to 40 are 21, 23, 25, 27, 29, 31, 33, 35, 37, 39.
As all numbers occurs only once there is no mode for this data.

Think (Text book Page No. 103)

Question 3.
Which average will be most aprropriate for the companies producing the following goods? why?
(i) Diaries and notebooks
(ii) School bags
(iii) Jeans and T-shirts.
Solution:
for all the above data mode will be more appropriate.

Exercise 5.3
Try These (Text book Page No. 106)

Question 1.
Find the median of 3, 8, 7, 8,4, 5, 6.
Solution:
Arranging in ascending order: 3, 4, 5, 6, 7, 8, 8.
Here n = 7, which is odd.

Hence the median is 6.

Question 2.
Find the median: 11, 14, 10, 9, 14, 11, 12, 6, 7, 7.
Solution:
Arranging in ascending order: 6, 7, 7, 9, 10, 11, 11, 12, 14, 14
Here n = 10, which is even.

Think (Text book Page No. 108)

Complete the table given below and observe it to answer the following questions.

Solution:

Question 1.
Which are all the series having common mean and median?
Solution:
Solution:
A, B and C.

Question 2.
Why median is same for all the 4 series?
Solution:
Since the middle value is 100.

Question 3.
How mean is unchanged in the series A, B and C ?
Solution:
The difference between the given numbers are equal.

Question 4.
What change is to be made in the data, so that mean and median of ‘D’ series is equal to other series?
Solution:
If 99 becomes 0 or 200 becomes 101 then mean becomes 100.

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Intext Questions

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Intext Questions

Exercise 3.1
Think (Text book Page No. 50)

Question 1.
Is it the only way to decompose the numbers representing length and breadth? Discuss.
Solution:
No, for example 15 can be decompose into 1 × 15, 3 × 5, 5 × 3, 15 × 1

Try These (Text book Page No. 52)

Question 1.
Observe the following figures and try to find its area, geometrically. Also verify the same by multiplication of monomial.

Solution:
Area of each box = xy
Totally 12 boxes
∴ Total area = 12 × xy = 12xy
Also multiplying the length 4x and breadth 3y
We have area of the rectangle = 4x × 3y = 12xy

(ii) Area of each small box = x2
Total number of boxes = 3
∴ Total area = 3x2
Also length of the rectangle = 3x
breadth of the rectangle = x
Area of the rectangle = length × breadth
= 3x × x
= 3x2

(iii) Area of each small box is ay, by, cy
∴ Total area = ay + by + cy = y (a + b + c)
Area of the rectangle = length × breadth
= (a + b + c) y

(iv) Area of each small square = x2
There are 4 small squares
∴ Total area of the given square = 4x2
Also side of the big square = 2x
∴ Area of the square = (2x)2 = 4x2

(v) Area of each small rectangle = xy
There are 9 such rectangles
∴ Total area = 9xy
Area of big rectangle = lenght × breath
= 3x × 3y = 9xy

Question 2.
Let the length and breadth of a tile be x and y respectively. Using such tiles construct as many rectangles as you can and find out the length and breadth of the rectangles so formed such that its area is
(i) 12 xy
(ii) 8xy
(iii) 9xy
Solution:

Try These (Text book Page No. 58)

Question 1.
Consider a square shaped paddy field with side of 48 m. A pathway with uniform breadth is surrounded the square field and the length of the outer side is 52 m. Can you find the area of the pathway by using identities?
Solution:
Let a = 52
b = 4

(a – b)2 = a2 – 2ab + b2 = 522 – 2 (52) (4) + 42
= 2704 – 416 + 16 = 2304

Think (Text book Page No. 60)

Question 1.
Can we factorize the following expressions using any basic identities? Justify your answer.
(i) x2 + 5x + 4
(ii) x2 – 5x + 4
Solution:
(i) x2 + 5x + 4 = x2 + (1 + 4)x + (1 × 4)
Which is of the form x2 + (a + b) x + ab
= (x + a) (x + b)
x2 + (1 + 4)x + (1 × 4) = (x + 1) (x + 4)
∴ x2 + 5x + 4 = (x + 1) (x + 4)

(ii) x2 – 5x + 4 = x2 + ((-1) + (- 4))x + (-1) (- 4)
Which is of the form x2 + (a + b) x + ab
= (x + a) (x + b)
x2 + ((-1) + 4))x + ((-1)(-4)) = (x + (-4)) = (x – 1) (x – 4)
x2 – 5x + 4 = (x – 1) (x – 4))

Exercise 3.2
Try These (Text book Page No. 63)

Question 1.
Construct inequations for the following statements:
1. Ramesh’s salary is more than ₹ 25,000 per month.
2. This lift can carry maximum of 5 persons.
3. The exhibition will be there in town for at least 100 days.
Solution:
1. x > 25,000, where x is Ramesh’s Salary per month.
2. y < 5, where y is the maximum number of persons the left can carry.
3. z > 100, where z is the number of days when the exhibition is there.

Think (Text book Page No. 65)

Question 1.
Hameed saw a stranger in the street. He told his parent, “The stranger’s age is between 40 to 45 years, and his height is between 160 to 170 cm”
Convert the above verbal statement into algebraic inequations by using x and y as variables of age and height.
Solution:
Let x be the age and y be the height then
40 < x < 45 and 160 < y < 170

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.3

Students can Download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice problems

Question 1.
The bishop, in given picture of chess board, can move diagonally along dark squares. Describe the translations of the bishop after two moves as shown in the figure.

Solution:
For first: 2 →, 2↓; For second move: 5 ←, 5↓

Question 2.
Write a possible translation for each of chess piece for a single move.

Solution:
Pawn – 1 ↑ or 2↑
Rook – 1 to 8 ↑
Knight – 2 →,1 ↑or 2 ←,1 ↑ or 1 →,2 ↑or 1 ←,2↑
Bishop – 1 →,1 ↑ or 2 →,2↑or 3 →,3↑ or 4 →,4↑ or 5 → 5 ↑1 ←,1↑ or 2 ←,2↑or 3 ←,3↑or 4 ←,4↑ or 5 ← 5↑
Queen – 1 to 8 ,1 →, 1 ↑ or 2 →,2↑ or 3 → ,3↑or 4 → ,4↑or 5 →,5↑or 1 ←,↑1 or 2 ←,2 ↑or
3 ←,3 ↑ or 4 ←,4 ↑ or 5 ← 5↑
King – 1 → or ← or ↑

Question 3.
Referring the graphic given, answer the following questions. Each bar of the category is made up of boy-girl-boy unit, (i) Which categories show a boy- girl-boy unit that is translation within the bar? (ii) Which categories show a boy-girl-boy unit that is reflected within the bar?

Solution:
(i) Essay Writing category shows translation
(ii) Essay Writing and Mono Acting categories shows reflection

Question 4.
Given figure is a floor design in which the length of the small red equilateral triangle is 30 cm. All the triangles and hexagons are regular. Describe the translations in cm, represented by the (i) yellow line (ii) black line (iii) blue line.

Solution:
(i) 120cm →, 210cm ↓
(ii) 270cm ← ,330cm ↑
(iii) 150 cm →

Question 5.
Describe the transformation involved in the following pair of figures (letters). Write translation, reflection or rotation.

Solution:
(i) rotation
(ii) reflection
(iii) translation
(iv) reflection
(v) rotation
(vi) reflection
(vii) rotation
(viii) translation

Challenge problems

Question 1.
In chess, a knight can move only in an L-shaped pattern:

• two vertical squares, then one horizontal square;
• two horizontal squares, then one vertical square;
• one vertical square, then two horizontal squares; or
• one horizontal square, then two vertical squares.

Write a series of translations to move the knight from g8 to g5 (at most two moves)
Solution:.
2 ←,1↓ and then 1 ←, 2↓ (or) 2 ←, 1↓and then 1 ← ,2↓

Question 2.
The pink shape is congruent to blue shape. Describe a sequence of transformations in which the blue shape is the image of pink shape.

Solution:
(i) Translation 3 ←, 5↑ and 90° counter clockwise rotation about the green point and translates 5 ←, 2↓,
(ii) Translation 2 ← 90° counter clockwise rotation about the green point and translates 2 ←, 2↓.

Question 3.

Solution:

Question 4.
Draw concentric circles given that radius of inner circle is 4.5 cm and width of circular ring is 2.5 cm.
Solution:
Give radius of inner circle = 4.5 cm
Width of circular ring is 2.5 cm
Radius of outer circle = 4.5 + 2.5 = 7 cm

Step 1 : Drawn a rough diagram and marked the given measurements
Step 2 : Taken any point O and marked it as the center.
Step 3 : With O as center and drawn a circle of radius OA = 4.5 cm.
Step 4 : With O as center drawn a circle of radius OB = 4.5 + 2.5 = 7 cm. Thus the concentric circles C1 and C2 are drawn.
Width of the circular ring = OB – OA
= 7 – 4.5 = 2.5 cm

Question 5.
Draw concentric circles given that radius of outer circle is 5.3 cm and width of circular ring is 1.8 cm.
Solution:
Give radius of outer circle = 5.3 cm
Width of circular ring = 1.8 cm
Radius of the inner circle = 5.3 – 1.8
= 3.5 cm

Step 1: Drawn a rough diagram and marked the given measurements
Step 2 : Taken any point O and marked it as the center.
Step 3 : With O as center drawn a circle of radius OA =3.5 cm.
Step 4 : With O as center, drawn a circle of radius OB = 5.3 cm. Thus concentric circles
C1 and C2 are drawn.
Width of the circular ring = OB – OA
= 5.3 – 3.5 = 1.8 cm

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.4

Students can Download Maths Chapter 5 Statistics Ex 5.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice problems

Question 1.
Arithmetic mean of 15 observations was calculated as 85. In doing so an observation was wrongly taken as 73 for 28. What would be correct mean?
Solution:

85 × 15 = sum of 15 observations
1275 = sum of 15 observations
Wrong observation = 73
Correct observation = 28

Correct mean = 82

Question 2.
The median of 25,16,15,10, 8, 30.
Solution:
Arranging is ascending order : 8, 10, 15, 16, 25, 30
Here n = 6, even

∴ Median = 15.5

Question 3.
Find the mode of 2, 5, 5, 1, 3, 2, 2, 1, 3, 5, 3.
Solution:
Arranging the data in ascending order: 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5
Here 2, 3 and 5 occurs 3 times each.
Which is the maximum number of times.
∴ Mode is 2, 3 and 5.

Question 4.
The marks scored by the students in social test out of 20 marks are as follows.
12, 10, 8, 18, 14, 16. Find the mean and the median?
Solution:
Arranging the given data in ascending order: 8, 10, 12, 14, 16, 18.

There are n = 6 observations, which is even

Question 5.
The number of goals scored by a football team is given below.
Find the mode and median for the data of 2 ,3, 2, 4, 6, 1, 3, 2, 4, 1, 6.
Solution:
Arranging the given data in ascending order: 1, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6
Clearly 2 occurs at the maximum of 3 times and so mode = 2
Here number of data of data n = 11, odd.
∴ Median = ($$\frac { n+1 }{ 2 }$$)th term
= ($$\frac { 11+1 }{ 2 }$$)th
term = ($$\frac { 12 }{ 2 }$$)th term
= 6th term
Median = 3

Question 6.
Find the mean and mode of 6, 11, 13, 12, 4, 2.
Answer:
Arranging is ascending order : 2, 4, 6, 11, 12, 13

Mean = $$\frac { 48 }{ 6 }$$ = 8
All observation occurs only once and so there is no mode for this date.

Challenge Problems

Question 1.
The average marks of six students is 8. One more student mark is added and the mean is still 8. Find the student mark that has been added.
Solution:

Sum of observation = 6 × 8 = 48
If one more mark is added then number of observations = 6 + 1 = 7
Let the number be x
Still average = 8
∴ 8 = $$\frac { 48+x }{ 7 }$$
48 + x = 7 × 8
48 + x = 56
48 + x = 56 – 48
x = 8
∴ The number that is added = 8

Question 2.
Calculate the mean, mode and median for the following data: 22, 15, 10, 10, 24, 21.
Solution:
Arranging in ascending order: 10, 10, 15, 21, 22, 24

Here n = 6, even

∴ Median = 18
Clearly the data 10 occurs maximum number of times and so 10 is the mode.
∴ Mode = 10

Question 3.
Find the median of the given data: 14, -3, 0, -2, -8, 13, -1, 7.
Solution:
Arranging the data is ascending order: -8, -3, -2, -1, 0, 7, 13, 14
Here number of data n = 8, even
∴ Median

∴ Median = – 0.5

Question 4.
Find the mean of first 10 prime numbers and first 10 composite numbers.
Solution:
First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Mean = 12.9
Mean of first 10 prime numbers = 12.9
First 10 composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Mean of first 10 composite numbers = 11.2

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.2

Students can Download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 4 Geometry Ex 4.2

Question 1.
Draw circles for the following measurements of radius (r)/ diameters(d).
(i) r = 4 cm
(ii) d = 12 cm
(iii) r = 3.5 cm
(iv) r = 6.5 cm.
(v) d = 6 cm
Solution:
(i) r = 4 cm

Step 1 : Market a point ‘O’ on the paper.
Step 2 : Extended the compass distance equal to radius 4 cm.
Step 3 : At center ‘O’, helded the compass firmly and placed the pointed end of the compass.
Step 4 : Slowly rotated the compass around to get the circle.

(ii) d = 12 cm
given d= 12 cm
∴ radius r = $$\frac { d }{ 2 }$$ = $$\frac { 12 }{ 2 }$$ = 6 cm

Step 1: Marked a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 6 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

(iii) r = 3.5 cm

Step 1: Market a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 3.5 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

(iv) r = 6.5 cm

Step 1: Market a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 6.5 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

(v) d = 6 cm
∴ radius r = $$\frac { d }{ 2 }$$ = $$\frac { 6 }{ 2 }$$ = 3 cm

Step 1: Market a point ‘O’ on the paper.
Step 2: Extended the compass distance equal to radius 3 cm.
Step 3: At center ‘O’, held the compass firmly and placed the pointed end of the compass.
Step 4: Slowly rotated the compass around to get the circle.

Question 2.
Draw concentric circles for the following measurements of radii / diameters. Find out the width of each circular ring.
(i) r = 3 cm and r = 5 cm.
(ii) r = 3.5 cm and r = 6.5 cm.
(iii) d = 6.4 cm and d = 11.6 cm.
(iv) r = 5 cm and r = 7.5 cm.
(vi) r = 7.1 cm and d = 12 cm.
Solution:
(i) r = 3 cm and r = 5 cm.

Width of the circular ring = OB – OA = 5 – 3 = 2 cm
Step 1: Drawn a rough diagram and market the given measurements
Step2: Taken any point O and marked it as center.
Step 3: With O as center drawn a circle of radius OA = 3 cm.
Step 4: With O as center drawn a circle of radius OB = 5 cm. Thus concentric circles C1 and C2 are drawn.
Width of the circular ring = OB – OA = 5 – 3 = 2 cm

(ii) r = 3.5 cm and r = 6.5 cm

Step 1: Drawn a rough diagram and market the given measurements
Step 2: Taken any point O and marked it as the center.
Step 3: With O as center drawn a circle of radius OA = 3.5 cm.
Step 4: With O as center drawn a circle of radius OB = 6.5 cm. Thus the concentric circles C1 and C2 are drawn.
Width of the circular ring = OB – OA = 6.5 – 3.5 = 3 cm

(iii) d = 6.4 cm and d = 11.6 cm
r = $$\frac { d }{ 2 }$$
r = $$\frac { 6.4 }{ 2 }$$ = 3.2 cm; r = $$\frac { 11.6 }{ 2 }$$ = 5.8 cm

Step 1: Drawn a rough diagram and market the given measurements
Step 2: Taken any point O and marked it as the center.
Step 3: With O as center drawn a circle of radius OA =3.2 cm.
Step 4: With O as center and drawn a circle of radius OB = 5.8 cm. Thus the concentric circles C1 and C2 are drawn.

Width of the circular ring = OB – OA = 5.8 – 3.2 = 2.6 cm

(iv) r = 5 cm and r = 7.5 cm

Step 1 : Drawn a rough diagram and market the given measurements
Step 2 : Taken any point O and marked it as the center.
Step 3 : With O as center drawn a circle of radius OA = 5 cm.
Step 4 : With O as center drawn a circle of radius OB = 7.5 cm. Thus the concentric circles C1 and C2 are drawn.
Width of the circular ring = OB – OA = 7.5 – 5 = 2.5 cm

(v) d = 6.2 cm and r = 6.2 cm.
r = $$\frac { d }{ 2 }$$
∴ r = $$\frac { 6.2 }{ 2 }$$ = 3.1 cm and r = 6.2 cm

Step 1: Drawn a rough diagram and market the given measurements
Step 2: Taken any point O and marked it as the center.
Step 3: With O as center, drawn a circle of radius OA = 3.1 cm.
Step 4: With O as center, drawn a circle of radius OB = 6.2 cm.
Thus the concentric circles C1 and C2 are drawn.
Width of the circular ring = OB – OA = 6.2 – 3.1 = 3.1 cm

(vi) r = 7.1 cm and d = 12 cm.
r = $$\frac { d }{ 2 }$$
∴ r = $$\frac { 12 }{ 2 }$$ = 6 cm and r = 7.1 cm

Step 1: Drawn a rough diagram and marked the given measurements
Step 2 : Taken any point O and marked it as center.
Step 3 : With O as center, drawn a circle of radius OA = 6 cm.
Step 4 : with O as center, drawn a circle of radius OB = 7.1 cm. Thus concentric circles
C1 and C2 are drawn.
Width of the circular ring = OB – OA = 7.1 – 6 = 1.1 cm

## Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.3

Students can Download Maths Chapter 5 Statistics Ex 5.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 5 Statistics Ex 5.3

Question 1.
Fill in the blanks.

(i) The median of the data 12, 14, 23, 25, 34, 11, 42, 45, 32, 22, 44 is ________
(ii) The median of first ten even natural numbers is ________
Answers:
(i) 25
(ii) 11

Question 2.
Find the median of the given data: 35, 25, 34, 36, 45, 18, 28.
Solution:
Arranging the given data in ascending order 18, 25, 28, 34, 35, 36, 45.
Here the number of observations n = 7, which is odd.

Hence Median = 34

Question 3.
The weekly sale of motor bikes in a showroom for the past 14 weeks given below.
10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7.
Find the median of the data.
Solution:
Arranging the given data in ascending order 3, 4, 4, 5, 6, 6, 7, 7, 8, 10, 10, 12, 13, 16.
Here number of data n = 14, which is even

∴ Median = 7

Question 4.
Find the median of the 10 observations 36, 33, 45, 28, 39, 45, 54, 23, 56, 25.
If another observation 35 is added to the above data, what would be the new median?
Solution:
Arranging the given 10 observations in ascending order 23, 25, 28, 33, 36, 39, 45, 45, 54, 56.
Here number of data n = 10, which is even

∴ Median = 37.5
If 35 is added to the above data then it will be the 5th term then number of data n = 11, odd
∴ Median = ($$\frac { n+1 }{ 2 }$$)th term = ($$\frac { 11+1 }{ 2 }$$)th term
= ($$\frac { 12 }{ 2 }$$)th term = 6th term
New median = 36

Objective Type Questions

Question 1.
If the median of a, 2a, 4a, 6a, 9a is 8, then find the value of a is
(i) 8
(ii) 6
(iii) 2
(iv) 10
Answer:
(iii) 2
Hint:

Median = 4a = 8
a = 2

Question 2.
The median of the data 24, 29, 34, 38, 35 and 30, is ________
(i) 29
(ii) 30
(iii) 34
(iv) 32
Answer:
(iv) 32
Hint:

Median = $$\frac { 30+34 }{ 2 }$$
= $$\frac { 64 }{ 2 }$$
= 32

Question 3.
The median first 6 odd natural numbers is
(i) 6
(ii) 7
(iii) 8
(iv) 14
Answer:
(i) 6
Hint:

Median = $$\frac { 5+7 }{ 2 }$$
= $$\frac { 12 }{ 2 }$$
= 6