Bhagya – Page 72 – Samacheer Kalvi

Samacheer Kalvi 9th English Speech Writing

Tamilnadu Samacheer Kalvi 9th English Speech Writing (on a given

topic)

Exercises
1. How do you prepare a speech for the morning assembly, stressing on the importance of ‘Sound mind in a sound body?
Answer:

Sound Mind In A Sound Body

Respected teachers and students,
Warm greetings to all of you.
I am here to deliver a speech on “Sound mind in a sound body”.

‘A sound mind in a sound body’ is the adage. It is true and worthy. The mind when free from pain or physical stress thinks clearly and remains active. It has the ability to march towards its goal as there are no stress psychologically and physically due to fear and illness. Active exercises, fresh air, activities like cycling, swimming and walking refresh the mind and keep it cheerful and relaxed. To overcome our stress we should play our favorite game, be it chess, cricket or any other game and remain calm and relaxed. Hence sports are an essential part of life.

Thank you for listening to my speech.

2. Prepare a speech in about 80-100 words of the morning assembly on highlighting the importance of Saving Water.
Answer:

Saving Water

Water is used by us for various purposes. Water is used for drinking. The human body depends on water for survival. Water is used for cooking food too. The water that we need for consuming must be clean, potable water. If humans use dirty or contaminated water for these purposes they can fall sick and die from the diseases. Therefore, for good health we need clean drinking water to be available for our use. Water is also needed for farming. We also use water to maintain personal hygiene. We also need water to wash our clothes, as also to keep our homes and surroundings clean.

3. Prepare a speech in about 80-100 words of the morning assembly on highlighting the importance conservation of Nature.
Answer:

Conservation Of Nature

Nature fulfills our basic requirement to live by providing us air, water, land, sunlight and plants. These resources are further used to manufacture various things that make life more convenient and comfortable for human beings. As a result, many of these resources are depleting at a fast pace and if it continues this way then the survival of human beings, as well as other living beings on Earth, would become very difficult. Conservation of nature means the preservation of forests, land, water bodies and conservation of resources such as minerals, fuels, natural gases, etc. to ensure that all these continue to be available in abundance.

4. Prepare a speech in about 80 – 100 words of the morning assembly on highlighting the importance of Saving Trees.
Answer:

Saving Trees

Respected teachers and students,
Warm greetings to all of you.
I am here to deliver a speech on “Saving Trees”.

Trees play a significant role in our daily life. Trees provide oxygen. They absorb carbon dioxide, thereby reducing the global warming. They provide food for us and also for forest animals. Trees offer habitation to birds, insects, lichen and fungi. Their trunks provide the hollow cover needed by species such as bats, wood-boring beetles, owls and woodpeckers. They benefit the environment by the way of preventing soil erosion and flooding. They reduce the speed of the wind. They act as natural air conditioners by cooling the roadsides by their shade. They are helpful for children to play in and discover their sense of adventure. They also help in protecting us from the harmful effect of ultra-violet rays. So it is our responsibility to save trees.

Thank you for listening to my speech.

5. Prepare a speech in about 80-100 words on Peacock.
Answer:

Peacock

Peacock is one of the most beautiful birds on the earth. It is particularly known for its colourful feathers that are a sight to behold. It looks best when it dances merrily in » the rain. Peacock is the national bird of India. It finds several references in the Indian mythology and history. It is known for its metallic blue and green colour and spectacular feather. These feathers are also considered auspicious and are used to bring in good luck and prosperity. Peacock has inspired many notable artists in the past and continues to do so.

6. Prepare a speech in about 80-100 words of the morning assembly on highlighting the Importance of Saving Pets.
Answer:

Saving Pets

Respected teachers and students,
Warm greetings to all of you.
I am, here to deliver a speech on “Saving Pets”.

Saving the life of a pet animal like a dog or a cat is our moral responsibility. There are so many ways in which we can save pets. A person can adopt his pet and encourage his friends and family to adopt pet. Most people don’t realize that they are unknowingly supporting the puppy mill industry when they purchase animals from pet stores or buy a puppy online. A person may talk with the owners of pet stores and ask them to feature homeless pets for adoption instead. We can add an advocacy message in the signature line of our email to help educate people about the issue.

Instead of buying pet animals, we can adopt them. We can use our social networking skills to promote adoptable animals. Most rescue groups need volunteers who can help with a variety of professional services. So as a volunteer, we can save a lot of lives. There are lots of ways to provide assistance without giving a donation. However, donations allow shelters and rescue groups to increase adoptions, promote spay/neuter, create lifesaving programs, educate the public about animal welfare issues, and much more. A person can save lives by fostering homeless animals.

Thank you for listening to my speech.

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Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.2

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.2

Question 1.
Fill in the blanks:

Question (i)
The solution of the equation ax + b = 0 is ………
Answer:
– \(\frac{b}{a}\)
Solution:
ax + b = 0
ax = – b
∴ x = – \(\frac{b}{a}\)

Question (ii)
If a and b are positive integers then the solution of the equation ax = b has to be always ………
Answer:
positive
Hint:
Since a & b are positive integers,
The solution to the equation ax = b is x = – \(\frac{b}{a}\) is also positive.

Question (iii)
One-sixth of a number when subtracted from the number itself gives 25. The number is ……..
Answer:
30
Hint:
Let the number be x.
As per question, when one sixth of number is subtracted from itself it gives 25
x – \(\frac{x}{6}\) = 25
∴ \(\frac{6x-x}{6}\) = 25
∴ \(\frac{5x}{6}\) = 25
∴ x = \(\frac{25×6}{3}\) = 5 x 6 = 30

Question (iv)
If the angles of a triangle are in the ratio 2 : 3 : 4 then the difference between the greatest and the smallest angle is
Answer:
40°
Hint:
Given angles are in the ratio 2 : 3 : 4
Let the angles be 2x, 3x & 4x
Since sum of the angles of a triangle is 180°,
We get
2x + 3x + 4x = 180
∴ 9x = 180
∴ X = \(\frac{180}{9}\) = 20°
∴ The angles are 2x = 2 x 20 = 40°
3x = 3 x 20 = 60°
4x = 4 x 20 = 80°
∴ Difference between greatest & smallest angle is
80° – 40° = 40°

Question (v)
In an equation a + b = 23. The value of a is 14 then the value of b is ……..
Answer:
b = 9
Hint:
Given equation is a + b = 23
a = 14
14 + b = 23
b = 23 – 14 = 9
b = 9

Question 2.
Say True or False

Question (i)
“Sum of a number and two times that number is 48” can be written as y + 2y = 48
Answer:
True
Hint:
Let the number be ‘y’
Sum of number & two times that number is 48
Can be written as y + 2y = 48 – True

Question (ii)
5(3x + 2) = 3(5x – 7) is a linear equation in one variable.
Answer:
True
Hint:
5 (3x + 2) = 3 (5x – 7) is a linear equation in one variable – ‘x’ – True

Question (iii)
x = 25 is the solution of one third of a number is less than 10 the original number.
Answer:
False
Hint:
One third of number is 10 less than original number.
Let number be ‘x’ Therefore let us frame the equation
\(\frac{x}{3}\) = x – 10
∴ x = 3x – 30
3x – x = 30
2x = 30
x = 15 is the solution

Question 3.
One number is seven times another. If their difference is 18, find the numbers.
Solution:
Let the numbers be x & y
Given that one number is 7 times the other & that the difference is 18.
Let x = 7y
also, x – y = 18 (given)
Substituting for x in the above
We get 7y – y = 18
∴ 6y = 18
y = \(\frac{18}{6}\) = 3
x = 7y = 7 x 3 = 21
The number are 3 & 21

Question 4.
The sum of three consecutive odd numbers is 75. Which is the largest among them?
Solution:
Given sum of three consecutive odd numbers is 75
Odd numbers are 1, 3, 5,1,9, 11, 13,……..
∴ The difference between 2 consecutive odd numbers is always 2. or in other words, if one odd number is x, the next odd number would be x + 2 and the next number would be x + 2 + 2 = x + 4
i.e x + 4
Since sum of 3 consecutive odd nos is 75
∴ x + x + 2 + x + 4 = 75
3x + 6 = 75 ⇒ 3x = 75 – 6
3x = 69
x = \(\frac{69}{3}\) = 23
The odd numbers are 23, 23 + 2, 23 + 4
i.e 23, 25, 27
∴ Largest number is 27.

Question 5.
The length of a rectangle is \(\frac{1}{3}\)rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Solution:

Let length & breadth of rectangle be l and ‘6’ respectively
Given that length is \(\frac{1}{3}\) of breadth,
∴ l = \(\frac{1}{3}\) x b ⇒ l = \(\frac{b}{3}\) ⇒ b = 3l …..(1)
Also given that perimeter is 64 m
Perimeter = 2 x (l + b)
2 x l + 2 x b = 64
Substituting for vahie of b from (1), we get
2l + 2(3l) = 64
∴ 2l + 6l = 64
8l = 64
∴ l = \(\frac{64}{8}\) = 8 m
b = 3l = 3 x 8 = 24 m
length l = 8 m & breadth h = 24 m

Question 6.
A total of 90 currency notes, consisting only of X5 and ?10 denominations, amount to ₹ 500. Find the number of notes in each denomination.
Solution:
Let the number of ₹ 5 notes be ‘x’
And number of ₹ 10 notes be ly ’
Total numbers of notes is x + y = 90 (given)
The total value of the notes is 500 rupees.
Value of one ₹ 5 rupee note is 5
Value of x ₹ 5 rupee notes is 5 × x = 5x
Value ofy ₹ 10 rupee notes is 10 × y = 10y
∴ The total value is 5x + 10y which is 500
we have 2 equations:
x + y = 90
5x + 1oy = 500
Multiplying both sides of (1) by 5, we get
5 × x + 5 x y = 90 x 5
5x + 5y = 450
Subtracting (3) from (2), we get

∴ y = \(\frac{50}{5}\) = 10
Substitute y = 10 in equation (1)
x + y = 90 ⇒ x + 10 = 90
⇒ x = 90 – 10 ⇒ x = 80
There are ₹ 5 denominations are 80 numbers and ₹ 10 denominations are 10 numbers

Question 7.
At present, Thenmozhi’s age is 5 years more than that of Murali’s age. Five years ago, the ratio of Thenmozhi’s age to Murali’s age was 3:2. Find their present ages.
Solution:
Let present ages ofThenmozhi & Murali be l’ & ‘m’
Given that at present
Thenmozhi’s age is 5 years more than Murali
∴ t = m + 5
5 years ago, Thenmozhi’s age would be t – 5
& Murali’s age would be m – 5
Ratio of their ages is given as 3 : 2
∴ \(\frac{t-5}{m-5}\) = \(\frac{3}{2}\) [∴ By cross multiplication]

2(t – 5) = 3(m – 5)
2 x t – 2 x 5 = 3 x m – 3 x 5 ⇒ 2t – 10 = 3m – 15
Substituting for t from (1)
2(m + 5) – 10 = 3m – 15
2m + 10 – 10 = 3m – 15
2m = 3m – 15
3m – 2m = 15
m = 15
t = m + 5 = 15 + 5 = 20
∴ Present ages of Thenmozhi & Murali are 20 & 15

Question 8.
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.
Solution:
Let the units/digit of a number be ‘u’ & tens digit of the number be ‘t’
Given that sum of it’s digits is 9
∴ t + u = 9 ….(1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as 10t + u
[Understand: Suppose a 2 digit number is 21,
it can be written as 2 x 10 + 1
∴ 32 = 3 x 10 + 2
45 = 4 x 10 + 5
tu = t x 10 + u = 10t + u]
Given that when 27 is subtracted, digits interchange
10t + u – 27 = 10u + t (number with interchanged digits)
∴ By transposition & bringing like variables together
10t – t + 10u =27
∴ 9t – 9u = 27
Dividing by ‘9’ throughout, we get
\(\frac{9t}{9}\) – \(\frac{9u}{9}\) = \(\frac{27}{9}\) ⇒ t – u = 3 ….(2)
Solving (1) & (2)

t = \(\frac{12}{2}\) = 6
∴ u = 3
t = 6 substitute in (1)
t + u = 9
⇒ 6 + u = 9
⇒ u = 9 – 6 = 3
Hence the number is 63.

Question 9.
The denominator of a fraction exceeds its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \(\frac{3}{2}\). Find the original fraction.
Solution:
Let the numerator & denominator be ‘n’ & ‘d’
Given that denominator exceeds numerator by 8
∴ d = n + 8
If numerator increased by 17 & denominator decreased by 1,
it becomes (n + 17) & (d – 1), fraction is \(\frac{3}{2}\).
i.e = \(\frac{n+17}{d-1}\) = \(\frac{3}{2}\) by cross multiplying, we get
\(\frac{n+17}{d-1}\) = \(\frac{3}{2}\)
2(n + 17) = 3(d – 1)
2n + 2 x 17 = 3d – 3

∴ 34 + 3 = 3d – 2n
∴ 3d – 2n = 37
Substituting eqn. (1) in (2), we get,
3 x (n + 8) – 2n = 37
3n + 3 x 8 – 2n = 37

∴ n = 37 – 24 = 13
d = n + 8 = 13 + 8 = 21
The fraction is \(\frac{n}{d}\) = \(\frac{13}{21}\)

Question 10.
If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Solution:
Let the distance to be covered by train be ‘d’
Using the formula, time take (t) = \(\frac{Distance}{Speed}\)
Case 1:
If speed = 60 km/h
The time taken is 15 minutes more than usual (t + \(\frac{15}{60}\))
Let usual time taken be ‘t’ hrs.
Caution:
Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in km.
Given that in case 1, it takes 15 min. more
15 m = \(\frac{15}{60}\) hr = \(\frac{1}{4}\) hr.
∴ Substituting in formula
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{60}\) = t + \(\frac{1}{4}\)
Since usually it takes ‘t’ hr, but when running at 60 km/h, it takes 15 min (\(\frac{1}{4}\)) extra.
Multiplying by 60 on both sides
d = 60 x t + 60 x \(\frac{1}{4}\) = 60t + 15 …..(1)
Case 2:
Speed is given as 85 km/h
Time taken is only 4 min (\(\frac{4}{60}\) hr) more than usual time
time taken = (t + \(\frac{1}{15}\)) hr. (\(\frac{4}{60}\) = \(\frac{1}{15}\))
Using the formula,
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{85}\) = t + \(\frac{1}{15}\)
Multiplying by 85 on both sides
\(\frac{d}{85}\) x 85 = 85 x t + 85 x \(\frac{1}{15}\)
∴ d = 85 t + \(\frac{17}{3}\)
From (1) & (2), we will solve for ‘t’
Equating & eliminating ‘d we get

By transposing, we get
15 – \(\frac{17}{3}\) = 85t – 60t
\(\frac{45-17}{3}\) = 25t
∴ 25t = \(\frac{28}{3}\)
∴ t = \(\frac{28}{3×25}\) = \(\frac{28}{75}\) hr (\(\frac{28}{75}\) x 60 = 22.4 min)
Substituting this value of ‘t’ in eqn. (1), we get
d = 60t+ 15 = 60 x \(\frac{28}{75}\) + 15 = \(\frac{1680}{75}\) + 15 = 22.4 + 15
= 37.4 km

Objective Type Questions

Question 11.
Sum of a number and its half is 30 then the number is ……..
(a) 15
(b) 20
(c) 25
(d) 40
Answer:
(b) 20
Hint:
Let number be V
half of number is \(\frac{x}{2}\)
Sum of number and it’s half is given by
x + \(\frac{x}{2}\) = 30 [Multiplying by 2 on both sides]
2x + x = 30 x 2
3x = 60
x = \(\frac{60}{3}\) = 20

Question 12.
The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is ………
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer:
(a) 62°

Hint:
As per property of ∆, exterior angle is equals to sum of interior opposite angles Let the other interior angle to be found be ‘x’
∴ x + 58 = 120°
∴ x = 120 – 58 = 62°

Question 13.
What sum of money will earn ₹ 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer:
(c) 10000
Hint:
Let sum of money be ‘P’
Time period (n) is given as 1 yr.
Rate of simple interest (r) is given as 5% p.a
∴ As per formula for simple interest.
S.I = \(\frac{Pxrxn}{100}\) = \(\frac{px5x1}{100}\)
P x 5 x 1 = 500 x 100
∴ P = \(\frac{500×100}{5}\) = 100 x 100 = 10,000

Question 14.
The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is ………
(a) 6
(b) 2
(c) 4
(d) 8
Answer:
(c) 4
Hint:
Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM & HCF is given as 24.
∴ Product of the 2 nos. is 24
Given one number is 6.
Let other number be ‘x’
∴ 6 × x = 24
∴ x = \(\frac{24}{6}\) = 4

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Samacheer Kalvi 9th English Letter Writing

Tamilnadu Samacheer Kalvi 9th English Letter Writing (to a friend)

Various parts of a formal letter

the heading
the inside address
the salutation
the body
the closing
the signature
the address on the envelope

The contents of a formal letter

Use a courteous, positive and professional tone. Maintain a respectful, constructive tone.
Use formal language. Avoid contractions or abbreviations.
State your purpose clearly and briefly. State the purpose of your writing in the first or second sentence of the letter.
Include all necessary information. Provide all the information your reader needs to understand and respond appropriately to your letter.

Format for Business and Official Letters:
Business : to dealers for buying and selling goods or comments on goods purchased.
Official : to Officials, to Principals, Teachers.

(a) Station (Place): y city
(b) Date: e.g. 20th Jan. 2019
(c) From Address
(d) To Address
(e) Salutation:
Sir, Dear Sir, Respected Sir, Madam (Principal, Teacher)
(f) Sub:
(i) Complaint on
(ii) Requisition for.
(iii) Orders placed on
(iv) Reminder on
(g) Ref:

Complaint letter
Bill No. _______ dated _______ (Reminder letter)
Previous letter dated
(h) Commencing lines :
(a) This is to inform you that
(b) I wish to bring to your kind notice that …………………………
(i) Content of the letter :
(j) Request:
(a) I shall be very grateful if the necessary measures are taken by the authorities concerned at the earliest
(b) I shall be obliged if early measures are taken in this regard by the authorities concerned
(k) Thank you,
(l) Subscription:
Yours truly,
Yours faithfully,
(m) Signature:
x x

Format For Personal Letter:

(a) From Address :
Y City

(b) Date : Eg. 20th June, 2019
(c) Salutation:
My Dearest __, Dearest __, Dear __,

(d) Commencing Lines:

Hope this letter finds you in the best of health
Hope you had received my previous letter
Thank you for your loving letter
Happy to meet you through this letter
Thank you for the you had sent me

(e) Concluding Lines:

Looking forward to your reply
Awaiting your reply / coming eagerly
Convey my love/regards to

(f) Subscription:
For Parents, Uncles,
Aunts – Yours lovingly,
For Friends – Yours lovingly, sincerely,
For Brother, Sisters, Cousins,
nephews, nieces – Yours affectionately,

(g) Signature

Exercises
1. Kumar wants to invite his friend Siva for his sister’s Marriage a week earlier, to assist him in the arrangments. He could not complete the content of his letter in about 100 words.
Answer:
Chennai,
23rd January 2019.

My Dear Siva,
Hope this letter finds you in good health. It’s been a few months since we’ve met. I have a good news. My sister, Shanthi’s marriage has been fixed on 30th of this month. We would like to have your presence two days ahead. So please be part of the family. Looking forward to be with you. We need your ideas for the reception hall and its decoration. All of us are expecting your arrival soon.

Your loving friend,
Kumar

Address on the envelope :
To
Mr. S. Siva,
20, Anna Salai,
Salem.

2. Kavya wanted to write a letter to her uncle to ask him if she could spend her summer holidays with him. She had started her letter but had not completed it. Complete the content of her letter in about 100 words.
Answer:
No. 20,4th Street,
Gandhi Nagar,
Chennai – 600 020.
13 th December, 2018.

Dear Uncle,
Hope this letter finds you in the best of health. Uncle, with my exams over and with the long wait for the results, I thought I could spend my summer holidays with you, aunt and brothers Ramu and Raju in the fields and plantation in the village.

It would be very refreshing after my exam and I am looking forward to put my hands to work. Shall I be there on Sunday? My father will drop me at the village.

Waiting for your reply or phone call at the earliest.

Your loving
niece,

Address on the envelope :

To
Mr. Vasanth,
20, Everon Heights, Ootacamund,
Nilgiris District.

3. Write a letter to your friend telling about how your house was burgled when your family was away on a holiday. (*^)
Answer:
36, Gandhi Road
Chennai – 45
04 August 2018

Dear Ramesh,
How are you? I feel sad to inform you that my house was burgled last week when I was on a holiday. Burglars might have known from the accumulated newspaper pile that I had gone away. When I came back last Sunday, I found the back-door lock broken. I could have forgotten to bolt the back-door from inside and they should have entered through it. My room was ransacked. They took my laptop and other valuables. I must have deposited the jewellery in a bank locker to avoid this loss. I should have informed my neighbours about my week-long trip. Well, I have registered an FIR with the police. They are investigating the case. They have assured that I would get my jewels back. The burglars will be caught very soon. Convey my regards to all at home.

Yours lovingly,
Murali

Address on the envelope :
Mr. Ramesh,
50, Bharathiyar Street,
Namakkal.

4. Suba wanted to write a letter to her friend congratulating her for her victory in Para Olympic Games. She had the format ready but Is yet to complete the letter. Complete the content of her letter in about 100 words.
Answer :
9/11, Kinely Street,
R City
2nd February, 2019.

Dear Rajalakshmi,
I am fine. How are you? I am very delighted to know that you have got victory in Para Olympic Games. Please accept my heartiest congratulations for this great achievement. Actually, your selfless dedication and devotion to the cause pf Sports and Games have earned you this deserving recognition. This is just the beginning and many such victories and medals should follow.

Your loving friend,
sudha

Address on the envelope :
To
Miss Rajalakshmi,
27, Woodland Street,
Ooty.

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Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.4

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.4

Miscellaneous Practice Problems

Question 1.
6² × 6^m = 6⁵, find the value of ‘m’
Solution:
6² × 6^m = 6⁵
6^2+m = 6⁵ [Since a^m × aⁿ= a^m+n]
Equating the powers, we get
2 + m = 5
m = 5 – 2 = 3

Question 2.
Find the unit digit of 124¹²⁸ × 126¹²⁴
Solution:
In 124¹²⁸, the unit digit of base 124 is 4 and the power is 128 (even power).
Therefore, unit digit of 124¹²⁸ is 4.
Also in 126¹²⁴, the unit digit of base 126 is 6 and the. power is 124 (even power).
Therefore, unit digit of 126¹²⁴ is 6.
Product of the unit digits = 6 × 6 = 36
∴ Unit digit of the 124¹²⁸ × 126¹²⁴ is 6.

Question 3.
Find the unit digit of the numeric expression: 16²³ + 71⁴⁸ + 59⁶¹
Solution:
In 16²³, the unit digit of base 16 is 6 and the power is 23 (odd power).
Therefore, unit digit of 16²³ is 6.
In 71⁴⁸, the unit digit of base 71 is 1 and the power is 48 (even power).
Therefore, unit digit of 71⁴⁸ is 1.
Also in 59⁶¹, the unit digit of base 59 is 9 and the power is 61 (odd power).
Therefore, unit digit of 59⁶¹ is 9.
Sum of the unit digits = 6 + 1 + 9 = 16
∴ Unit digit of the given expression is 6.

Question 4.
Find the value of

Solution:

Question 5.
Identify the degree of the expression, 2a³be + 3a³b + 3a³c – 2a²b²c²
Solution:
The terms of the given expression are 2a³bc, 3a³b + 3a³c – 2a²b²c²
Degree of each of the terms: 5,4,4,6.
Terms with the highest degree: – 2a²b²c²
Therefore degree of the expression is 6.

Question 6.
If p = -2, q = 1 and r = 3, find the value of 3p²q²r.
Solution:
Given p = -2; q = 1; r = 3
∴ 3p²q²r = 3 × (-2)² × (1)² × (3)
= 3 × (-2 × 1)² × (3) [Since a^m × b^m = (a × b)^m]
= 3 × (-2)² × (3)
= 3 × (-1)² × 2² × 3
= 3¹⁺¹ × 1 × 4 [Since a^m × aⁿ = a^m+n]
= 3² × 4 = 9 × 4
∴ 3p²q²r = 36

Challenge Problems

Question 7.
LEADERS is a WhatsApp group with 256 members. Every one of its member is an admin for their own WhatsApp group with 256 distinct members. When a message is posted in LEADERS and everybody forwards the same to their own group, then how many members in total will receive that message?
Solution:
Members of the groups LEADERS = 256
Members is individual groups of the members of LEADERS = 256
Total members who receive the message

= 256 × 256 = 2⁸ × 2⁸
2⁸⁺⁸ = 2¹⁶
= 65536
Totally 65536 members receive the message.

Question 8.
Find x such that 3^x+2 = 3^x + 216.
Solution:
Given 3^x+2 = 3^x + 216 ; 3^x+2 = 3^x+ 216
Dividing throught by 3^x, we get

Equating the powers of same base

Question 9.
If X = 5x² + 7x + 8 and Y = 4x² – 7x + 3, then find the degree of X + Y.
Solution:
Given x = 5x² + 7x + 8
X + Y = 5x² + 7x + 8 + (4x² – 7x + 3)
= (5x² + 4x²) + (7x – 7x) + (8 + 3)
= x² (5 + 4) + x(7 – 7) + (8 + 3) = 9x²+ 11
Degree of the expression is 2.

Question 10.
Find the degree of (2a² + 3ab – b²) – (3a² -ab- 3b²)
Solution:
(2a² + 3ab – b²) – (3a² – ab – 3b²)
= (2a² + 3ab – b²) + (- 3a² + ab + 3b²)
= 2a² + 3ab – b² – 3a² + ab + 3b²
= 2a² – 3a² + 3ab + ab + 3b² – b²
= 2a² – 3a² + ab (3 + 1) + b²(3 – 1)
= – a² + 4 ab + 2b²
Hence degree of the expression is 2.

Question 11.
Find the value of w, given that x = 4, y = 4, z = – 2 and w = x² – y² + z² – xyz.
Solution:
Given x = 3; y = 4 and z = -2.
w = x² – y² + z² – xyz
w = 3² – 4² + (-2)² – (3)(3)(-2)
w = 9 – 16 + 4 + 24
w = 37 – 16
w = 21

Question 12.
Simplify and find the degree of 6x² + 1 – [8x – {3x² – 7 – (4x² – 2x + 5x + 9)}]
Solution:
6x² + 1 – [8x – (3x²– 7 – (4x²– 2x + 5x + 9)}]
= 6x² + 1 – [8x – {3x² – 7 – 4x²– 2x + 5x + 9}]
= 6x² + 1 – [8x – 3x² + 7 + 4x² – 2x + 5x + 9}]
= 6x² – 1 – [8x + 3x² – 7 – 4x² + 2x – 5x – 9]
= 6x² + 3x² – 4x² – 8x + 2x – 5x – 1 – 7 – 9]
= x²(6 + 3 – 4) + x(8 + 2 – 5) – 15
= 5x² – 11x – 15
Degree of the expression is 2.

Question 13.
The two adjacent sides of a rectangle are 2x² – 5xy + 3z² and 4xy – x² – z². Find the perimeter and the degree of the expression.
Solution:
Let the two adjacent sides of the rectangle as
l = 2x² – 5xy + 3z² and b = 4xy – x²y + 3z²

Perimeter of the rectangle
= 2(l + b) = 2(2x² – 5xy + 3z² + 4xy – x² – z²)
= 4x² – 10xy + 6z² + 8xy – 2x² – 2z²
= 4x² – 2x² – 10xy + 8xy + 6z² – 2z²
= x²(4 – 2) + xy (-10 + 8) + z² (6 – 2z²)
Perimeter = 2x² – 2xy + 4z²
Degree of the expression is 2.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.4 Read More »

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1

Question 1.
Fill in the blanks

Question (i)
The value of x in the equation x +5 12 ¡s ……….
Answer:
7
Hint:
Given,
x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7

Question (ii)
The value ofy in the equation y – 9 = (-5) + 7 is ……….
Answer:
11
Hint:
Given,
y – 9 = (- 5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)

Question (iii)
The value of m in the equation 8m = 56 is ………
Answer:
7
Hint:
Given,
8m = 56
Divided by 8 on both sides
\(\frac{8xm}{8}\) = \(\frac{56}{8}\)
∴ m = 7

Question (iv)
The value ofp in the equation \(\frac{2p}{3}\) = 10 is ……….
Answer:
1
Hint:
Given,
\(\frac{2p}{3}\) = 10
Multiplying by 3 on both sides,

∴ p = 15

Question (v)
The linear equation in one variable has ……… Solution.
Answer:
one.

Question 2.
Say True or False.

Question (i)
The shifting of a number from one side of an equation to other is called transposition.
Answer:
True

Question (ii)
Linear equation in one variable has only one variable with power 2.
Answer:
False
[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.
Match the following :

(A) (i), (ii), (iv), (iii), (v)
(B) (iii), (iv), (i), (ii), (v)
(C) (iii), (i), (iv), (v), (ii)
(D) (iii), (i), (v), (iv), (ii)
Answer:
(C) (iii),(i), (iv), (v), (ii)
Hint:
a. \(\frac{x}{2}\) = 10,
multiplying by 2 on both sides, we get
\(\frac{x}{2}\) x 2 = 10 x 2 ⇒ x = 20

b. 20 = 6x – 4
by transposition ⇒ 20 + 4 = 6x
6x = 24
dividing by 6 on both sides,
\(\frac{6x}{6}\) = \(\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x
By transposing the variable ‘x’, we get
2x – 5 + x = 3
by transposing – 5 to other side,
2x + x = 3 + 5
∴ 3x = 8

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20
by transposing – 4 to other side,
7x – 8x = 20 + 4
– x = 24
∴ x = – 24
\(\frac{4}{11}\) – x = \(\frac{-7}{11}\)
Transposing \(\frac{4}{11}\) to other side,
– x = \(\frac{-7}{11}\)\(\frac{-4}{11}\) = \(\frac{-7-4}{11}\) = \(\frac{-11}{11}\) = – 1
∴ – x = – 1 ⇒ x = 1

Question 4.
Find x:

Question (i)
\(\frac{2x}{3}\) – 4 = \(\frac{10}{3}\)
Solution:
Transposing -4 to other side, it becomes +4
∴ \(\frac{2x}{3}\) = \(\frac{10}{3}\) + 4
Taking LCM & adding,
\(\frac{2x}{3}\) = \(\frac{10}{3}\) + \(\frac{4}{1}\) = \(\frac{10+12}{3}\) = \(\frac{22}{3}\)
\(\frac{2x}{3}\) = \(\frac{22}{3}\)
Multiplying by 3 on both sides

⇒ 2x = 22
dividing by 2 on both sides,
We get \(\frac{2x}{2}\) = \(\frac{22}{2}\)
∴ x = 11

Question (ii)
y + \(\frac{1}{6}\) – 3y = \(\frac{2}{3}\)
Solution:
Transposing \(\frac{1}{6}\) to the other side,
y – 3y = \(\frac{2}{3}\) – \(\frac{1}{6}\)
Taking LCM,
– 2y = \(\frac{2}{3}\) – \(\frac{1}{6}\) = \(\frac{2×2-1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ – 2y = \(\frac{1}{2}\) ⇒ 2y = – \(\frac{1}{2}\)
dividing by 2 or both sides.

Question (iii)
\(\frac{1}{3}\) – \(\frac{x}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\)
Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)
∴ \(\frac{1}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\) + \(\frac{x}{3}\)
Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)
\(\frac{1}{3}\) + \(\frac{5}{4}\) = \(\frac{7x}{12}\) + \(\frac{x}{3}\)
Multiply by 12 throughout
[we look at the denominators 3,4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4
-11 = 7x + 4x
11x = – 11
x = -1

Question 5.
Find x:

Question (i)
-3(4x + 9) = 21
Solution:

Expanding the bracket,
-3 × 4x + (-3) × 9 = 21
-12x + (-27) = 21
-12x – 27 = 21
Transposing – 27 to other side, it becomes +27
-12x = 21 + 27 = 48
12x = 48 ⇒ 12x = -48
Dividing by 12 on both sides

⇒ x = – 4

Question (ii)
20 – 2 ( 5 – p) = 8
Solution:

Expanding the bracket,
20 – 2 x 5 – 2 x (-p) = 8
20 – 10 + 2 + p = 8 (-2 x -P = 2p)
10 + 2p = 8 transporting 10 to other side
2P = 8 – 10 = -2
∴ 2p = -2
∴ p = -1

Question (iii)
(7x – 5) – 4(2 + 5x) = 10(2 – x)
Solution:

Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & -13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13,
Simplifying,
-3x = 33
∴ 3x = -33
x = \(\frac{-33}{3}\) = -11
x = -11

Question 6.
Find x and m:

Question (i)
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\) = -1
Solution:
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\)
Taking LCM on LHS, [LCM of 4 & 5 is 20]

∴ 11x + 2 = -20
∴ 11x = – 20 – 2 = – 22
x = \(\frac{-22}{11}\) = -2
x = -2

Question (ii)
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Solution:
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Cross multiplying, we get

∴ (m + 9) x 3 = 5 x (3m + 15)
m x 3 + 9 x 3 = 5 x 3m + 5 x 15

Transporting 3m & 75, we get
27 – 75 = 15m – 3m
-48 = 12m

⇒ m = -4

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1 Read More »

Samacheer Kalvi 10th English Report Writing

Tamilnadu Samacheer Kalvi 10th English Report Writing

Report writing is an attempt to gather information about an event, incident or accident from the persons concerned, the parties involved, the victims and authorities. The third-person point of view ensures objectivity in the report.

Guidelines
The heading is essential. The report may be in one or two paragraphs.
(i) Be objective.
(ii) Organise the details properly.
(iii) Present the material systematically
How should a report be written?
Answer:
A report should:

be in the form of a narrative
be in the past tense
include all relevant details
focus on one particular event only
mention the date and time of occurrence
mention the venue
mention the facts
mention the cause, result, etc.

Report For A School Magazine

Format:
(a) Heading
(b) Writer’s name and class.

Heading/Title Of The Happening
[by Dhanwanth/Aditi]……………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………….

Language : Should be semi-formal. Try to be simple but attractive and appealing. Avoid displaying your linguistic ability.
Content :

factual information about school activity/event
date, time, venue of the activity/event
sequence of event/programme
information about participants/chief guests/judges
kind of organisation, people responsible for programme/ arrangements
results, if describing a contest

Report For A Newspaper
Usual subjects : Political news, sports news, crimes, accidents, natural disasters, etc.
Format :
(a) Headline
(b) ‘Byline’, i.e., by a correspondent/reporter or an agency as its source and
(c) Date – line – date and place of occurrence

Headline
[by Sudhir/Sudha, TOI Correspondent/Staff reporter]

Chennai, 9 March

……………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………….

Language and Style :

quite formal
passive voice is preferred
journalistic jargon (vocabulary/expressions), e.g., according to ministry/party spokesman or according to government sources
use of words like ‘alleged’ or ‘suspected’ before ‘murderer’, ‘smuggler’, ‘thief’, etc.

Content : It is most important. The main information is given in the first two or three sentences. Other essential information like date, place, occasion, etc., follow. Presentation should be symmetrical.

Solved Questions
(a) The Principal of your school retired after rendering 35 years of meritorious service. A grand function was held to give him a hearty send-off. Write a report in 100 – 150 words for publication in your school magazine describing the celebrations. You are Sudha/ Sandeep of X A.
Answer:

A GRAND FAREWELL
(By Sudha, X A)

Our esteemed Principal Ms. R.R. Leela retired on 30th September this year after rendering 35 years of meritorious service. On the day of her retirement, a grand function was held to give her a hearty send-off. Students and teachers organised a cultural show. Songs and poems dealing with her qualities were recited. Our Vice Principal delivered a thought-provoking speech. He praised the outgoing principal for her contribution to education and sports. Students garlanded the principal and escorted her to the car. Some of them became emotional.
They began to weep. We bade her farewell with tears in our eyes.

(b) ’tour school has recently opened a computer wing. The facility of computer training and access to Internet connections has been made available to students. As Rakesh/ Rashmi of X B, write a report on the new computer wing in 100 – 150 words for your school magazine.
Answer:

NEW COMPUTER WING
[by Rashmi, XB]

The new computer wing of our school on 25th April constructed at a cost of two crore rupees was inaugurated by the Director of Education, yesterday. This wing with five halls on the first floor has twenty computers with access to internet connections. The ground floor has facility of computer training for the beginners. It has nine rooms with five computers in each room. Our school authorities have now fulfilled a long-felt need of the students. Computer training facilities will arouse and sustain the interest of the students in IT industry and new technology.

(c) You are Sneha/Gopal of Vivekananda School. Recently you had the honour of having participated as the contingent leader of your school team in the Republic Day parade in Delhi, in which your school was adjudged the best participating team.

Write a report in about 100 – 150 words about the memorable event for publication in your school magazine.

Republic Day Honours
[by Gopal, X B]

This year’s Republic Day will ever be cherished as a red-letter day by Vivekananda school.

On this historic occasion, our school team had the good fortune to participate in the Republic Day parade at Vijay Chowk in New Delhi. More than fifty teams representing different parts of India displayed their varied, colorful and romantic items of songs, dances and aerobics. Their feats made people glued to their feet. The enthusiasm and cheerfulness of the participants was beyond description. The celebrations were also a test of the performance of the participating teams. When the name of our school was announced as the best participating team, I felt overjoyed. For a moment, disbelief overpowered me. Then I as contingent leader went forward to receive the shield. I dedicated this award to our Principal and the Physical Instructor who had provided us such excellent training.

(d) The Golden Jubilee celebration of your school has just ended. It was a year-long programme featuring an inter-school debating contest, a play festival, a fun fair, an exhibition, special programme for ex-students and parents and concluding with a grand cultural programme. As Maya/Manoj of X A, write a report in 100 – 150 words for publication in your school magazine.

Golden Jubilee Celebrations
[by Manoj, X 5]

Our school completed fifty years of its existence on ‘Krishna Jayanthi’ – the 24th August, this year. To commemorate the event, a year-long programme of celebrations were planned.

Meetings of staff, and students’ council were held. An organising committee comprising the principal, five members of staff, office-bearers of the student council was constituted. A calendar of curricular activities was chalked out. It was decided to hold an exhibition of art and craft and science exhibition at school level. Fancy dress and dance competitions were held for the primary section. A Golden Jubilee issue of the school magazine was also brought out. Inter-school debate contest and a play festival were held in September and October. During November, special programmes were staged by ex-students. On the concluding function, a dance – drama ‘Eklavya’ was presented. It was acclaimed by the audience.

(c) You are Shekhar/Sarittia of Govt. Model Sr. Sec. School. The annual sports of your school were held on 12th February, 20XX. Write a report for your school magazine giving all details of the function.

The Annual Sports Day
[by Saritha, X C]

12th February, 20XX, was a big occasion for our school. The Annual Sports of our school were held on that day at our school playground. A colourful shamiana was erected for guests and teachers. The stands were also tastefully decorated. The athletes gathered in front of the stage at 8.30 a.m. There was a march – past. The Principal took the oath and declared the sports – meet open. At 9.30 the track events began with 200 metre race for boys and 100 metre race for girls. Field events like long jump, high jump, javelin throw and discus throw were held in between the races. The programme was beautifully planned. In the afternoon the cycle race, sack race and three-legged – race provided amusement. The musical chair race for
guests provided a lot of fun. Kamal Nath of XIIA was declared the best athlete.

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Samacheer Kalvi 10th English Speech Writing

Tamilnadu Samacheer Kalvi 10th English Speech Writing

1. In order to promote reading habits in the students, your school has organised a Library Week. You have to speak in the morning assembly and inform the students about the week-long programme.

Library Week
Respected Principal, Vice – Principal, teachers and friends!

It gives me great pleasure to speak to you about the Library Week that our school has organised from 2nd to 9th September this year. This week is slightly different from the others. Here we have to shift the focus to mental attainment. The aim of celebrating this week is to inculcate reading habits among students. Reading, you know, makes a man perfect. So new arrivals in the reference section such as the encyclopedia and dictionaries as well as general books will be displayed. National Book Trust and Oxford University Press are having an exhibition. We have invited Dr Kailash Vajpayee to inaugurate the exhibition. He will autograph the books and interact with students. Other authors like Khushwant Singh and Ruskin Bond will also pay us a visit. During the week, a quiz competition will be organised. The Reading section will have more new magazines of teenagers’ interest.

I appeal to all of you to spend as much free time in the library as you can.

Thank you.

2. Prepare a speech on the topic “Education Gives One Power”.

Education Gives Power
Respected Principal, teachers and friends!

Education provides us knowledge. It trains our mind and sharpens our skills and abilities. Education refines our tastes and temperaments and builds our thought process. Vocational courses help young boys to earn and learn together. They provide means of earning livelihood and open the route to employment. Professional courses, as is evident from the name itself, equip us for adopting various professions. Some of these highly skilled professionals seek placements or jobs in esteemed companies and business concerns. Thus education is important for our survival. Decent living is impossible without good income or high salary. Education improves the quality of our life and frees us of superstition, foolish, meaningless mind – blocks and rituals. If women are educated the whole family benefits as the food is hygienically prepared, children are healthy, well – mannered and disciplined. Education gives us power over our environment. We can control the situation and shape our destiny. Education spreads awareness among people and gives them freedom from social ills. It makes people independent by providing them means to learn their living. They become responsible citizens and realise their rights and duties. In short, education gives one power.

Thank you.

3. You are a member of the Environment Club of your school. After visiting many places you have realized that it is the need of the hour to protect environment. You decide to create awareness among the students. Write a speech in 150 – 200 words on ‘Environmental protection’ to be delivered in the morning assembly.

Environmental Protection
Respected Principal, teachers and friends!

Global warming has accelerated the rise of temperature on earth. The sea level is also rising and glaciers are melting away. Natural calamities are taking a toll of life on earth. Floods and scanty rainfall result in a crunch of food products, drinking water and disturb normal living conditions. The drought in Rajasthan has led to deaths and famine. Man is himself to blame for the deterioration of ecosystem. Depleting forests, industrial pollution, toxic – wastes, vehicular pollution, cutting of trees in cities, and lack of green cover are some of the contributory factors. The entire process of environmental pollution is becoming a vicious cycle. The urgent need of the hour is to protect environment. School children have begun to create awareness by campaigning against polythene bags and recycling waste material. Let us join hands to protect our forests, grow more trees, check toxic pollutants and change our lifestyle.

Thank you.

4. Suresh has been asked to deliver a speech on ‘The Brain Drain Problem’. Use your own ideas, and write the speech in about 150 – 200 words.

Respected Principal, teachers and dear friends!

Good morning to one and all present here. I Suresh of Std. X am thankful to the teachers for giving me this opportunity.
The problem of brain – drain has assumed serious proportion in the last thirty years or so. The nation spends its hard-earned meager resources on the education and training of its doctors, engineers, scientists, etc. But these highly talented and trained men and women of genius migrate to developed countries. They desert the ranks for the lure of money, better facilities and living conditions. Some of them get opportunities for the fulfilment of their ambitions and development of personality as there is full scope for it. They enjoy unlimited freedom for experiments and research and fear no resources crunch.

The parent countries become poorer by the depletion of resources as a result of migration of trained and talented persons. The migrants too feel maladjusted in the new country where they are considered second-grade citizens. Living and working in an alien culture among foreigners, they find themselves cut off from their own social modes and customs. They do suffer emotional vacuum as the memories of friends and relatives in their country haunt their minds. India too has been facing this problem. We must take steps to ensure them better facilities, improved living conditions, freedom for experimentation and research. In addition * to this, their talents must be utilised for proper work and their work must be given due recognition. These measures, if adopted seriously, can check the problem effectively.

Thank you.

5. As the secretary of School Red Cross Committee deliver a speech in the morning assembly to encourage schoolmates to join first – aid classes to be run by Indian Red – Cross Society. Use your own ideas to write the speech in about 150 – 200 words.

Respected Principal, teachers and friends!
First Aid is of immense help in case of an emergency or accident when the victim is to be given immediate help before some qualified doctor arrives or the victim is shifted to some nearby clinic or hospital. The importance of first aid can be judged from the fact that 1 many precious lives are saved because they had been administered proper first – aid before the patients received attention from a professional medical practitioner. The knowledge of first aid proves handy in a crisis. A person receiving bum injury, a victim of accident bleeding profusely, a drowned casualty or someone suffering from poisoning need urgent and immediate attention. Administration of immediate first aid provides the necessary relief till the case is attended by a competent doctor. Thus first aid can sometimes give a second: life to a person. Hence it is always beneficial to get acquainted with the basic fundamentals of first aid.

It is our moral duty to help the people in distress. But the aid rendered by an untrained person may sometimes prove harmful. Hence it is essential to have proper knowledge of first aid. Mere reading of books will not be sufficient. You must know the practical application too. Careful and correct use of first – aid can be beneficial as it can save precious human lives from the jaws of death. Hence I the Secretory of School Red Cross Committee exhort my fellow students to get the requisite knowledge by joining first aid classes.

Thank you.

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Samacheer Kalvi 10th English Article Writing

Tamilnadu Samacheer Kalvi 10th English Article Writing

Write an article in about 150 – 200 words on the following:

Honest Means of Livelihood
Every human being has to earn his bread and butter. Means vary from person to person, but one has to face many obstacles and odd situations in life. These means can be fair or foul, honest or dishonest. Unfortunately, the modem man hankers after money and has become commercial-minded. People are not afraid of the Almighty. They wish to accumulate riches by hook or by crook. They have no respect for human and moral values. The social norms and time – tested principles bemoan somewhere in a comer. The mortals of this computer age focus only on money. They want to become rich overnight. It is sure that no one can make easy money without resorting to corruption. One should always remember those who are honest get respect in society and feel satisfied. They don’t have to feel guilty. Those who are corrupt, hide behind the veils when caught. A person should always be honest and sincere. The factory workers, farmers, teachers and poor artisans live an honest life and are appreciated everywhere.

Advantages of Co – education
Co – education is the need of the hour. Many convent schools and public institutions are coeducational. Co – education has many advantages. It enhances the performance of the children. They feel that if they fail in any subject, it will be their insult. It is essential for a healthy competition. It also gives maximum exposure to teenagers. It is a bare fact that villagers don’t know how to talk to the womenfolk. However if they sit in a room and study together, they come to know about the various ways of interacting with girls. It increases the status of women in society. You can run separate schools for boys and girls. Will it be possible for a nation to make separate departments in government and private offices for men and women? Girls should be given ample opportunities to know the habits of men. It is beneficial for their married life also. Co – educational schools are indispensable for the harmonious growth of a child. There is no need to keep boys and girls apart. It will be an injustice to them.

The Joy Of Playing On The Street
Now – a – days, computers have become a craze for the present generation. Of late, we have become almost addicted to this fascinating machine of new age. As a result, we tend to overlook outdoor games. Recently, I had been to my cousin’s house in a small town. Wi-fi network connectivity was not available. Hence, most of the children were out on the road, playing hide – n – seek, seven stones, current or flying kite. That was the day when I went out and played Hide and Seek. Till then, I never realized how much we were missing in life. I enjoyed the game thoroughly finding new places to hide every time a person closed his/her eyes to count. There were times when we would go out even on sunny afternoons to play and we would not like to return home till late in the evening. I have understood that Outdoor games is very good and keeps one healthy and active, agile and alert. There is no scope of loneliness and tension in this joyful group activity. After a tiring day of sitting in school and concentrating on studies, an hour or two spent in playing takes off fatigue and refreshes us all over again. We can cultivate positive values like sportsman spirit, teamwork and cooperation. I have understood that playing outdoors is more joyful and enriching in many ways than sitting all alone and playing video games on computers and spoiling our eyes.

The Importance of Newspaper Reading
Newspaper reading is indeed a good habit that can provide a great sense of educational and information about politics, economy, entertainment, sports, business, industry, trade and commerce. This habit will improve your language skills and vocabulary. Many people have habits of reading daily newspapers that their day seems incomplete without taking hold of early morning newspapers. Here are some of the benefits that you can get by reading daily newspapers. They carry the news of the world and you feel you have visited many places reading news about various places. Reading a newspaper is a good habit and it is already a part of the modem life. This habit will broaden your viewpoint and will augment your information. It makes you well informed and enables you to take part in every conversation pertaining to the world’s contemporary events. Reading newspapers will improve your knowledge in general and it will be easy for you to relate to other people who often talk about current events and politics. Through newspapers, you will have an apparent thought and perceptive of what is happening in your country and the world around.

Disaster Management
Our country is prone to disasters like floods, drought, cyclones, or earthquakes. We do not have any clear – cut policy of disaster management nor any force to tackle the situation. Adhoc measures are adopted to cope with every disaster. We wait and watch for others to join the fray. There are heated arguments over jurisdiction – centre or state liability, official assessment and surveys before any help is rushed out to the affected area. The slow response results in the loss of precious human life and valuable property. We must have clear – cut, well defined guidelines for disaster management. A well – trained task – force having special equipment and trained personnel should be constituted. Its controlling officer should have the authority to •fake decisions and ensure their speedy implementation. Better transport and communication facilities will ensure better results. Bureaucratic set – up should not be allowed to interfere with the work of the disaster – management group.

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Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Additional Questions

Additional Questions and Answers

Question 1.
Fill in the blanks

Question (a)
Percent means ……….
Answer:
Per hundred or out of hundred.

Question (b)
Percent is useful in ………
Answer:
Comparing quantities easily

Question (c)
The formula to find the increased quantity ………
Answer:
I = (1 + \(\frac{x}{100}\))

Question (d)
The formula to find the decreased quantity ………
Answer:
D = (1 + \(\frac{x}{100}\))

Question (e)
Gain or profit % ……..
Answer:
(\(\frac{Profit}{C.P}\) x 100)%

Question (f)
Loss % = ………..
Answer:
(\(\frac{Loss}{C.P}\) x 100)%

Question (g)
S.P = ………. (if gain % is given)
Answer:

Question (h)
C.P = ……… (if gain % is given)
Answer:

Question (f)
S.P = ………. (if loss % is given)
Answer:

Question (h)
C.P = ………. (if loss % is given)
Answer:

Question (k)
Selling price = Marked price – …………
Answer:
Discount

Question (l)
Cost price = Cost price + ……….
Answer:
Over head expenses

Question 2.
If y% of ₹ 1000 is 600, find the value ofy.
Solution:
y% of 1000 = 600
\(\frac{y}{100}\) x 1000 = 600
y = \(\frac{600}{10}\)
y = 60

Question 3.
A number when decreased by 10% becomes 900. Then find the number.
Solution:
Let the number be ‘x’
Given x – \(\frac{10}{100}\)x = 900
\(\frac{100x-10x}{100}\) = 900
\(\frac{90x}{100}\) = 900
x = \(\frac{900×100}{90}\) = 1000

Question 4.
If the population in a city has increased from 5,00,000 to 7,00,000 in a year, find the percentage increase in population.
Solution:
Increase in population = 7,00,000 – 5,00,000 = 2,00,000
Percentage increase in population = \(\frac{2,00,000}{5,00,000}\) x 100 = 40%

Question 5.
If the selling price of a refrigerator is equal to \(\frac{10}{8}\) of its cost price, then find the gain/ profit percent.
Solution:
Let the C.P of the refrigerator be x
S.P = \(\frac{10}{8}\)x
Profit = S.P – C.P = \(\frac{10}{8}\)x – x = \(\frac{10x-8x}{8}\) = \(\frac{2x}{8}\)

Question 6.
Karnan bought a dishwasher for ₹ 32,300 and paid ₹ 2700 for its transportation. Then he sold it for X 38,500. Find his gain or loss percent.
Solution:
Total C.P of the dishwasher = C.P + Overhead expenses.
= ₹ 32300 + ₹ 2700 = ₹ 35000
S.P = ₹ 38500
Therefore, we find S.P > C.P

Question 7.
The value of a car 2 years ago was ₹ 1,40,000. It depreciates at the rate of 4% p.a. Find its present value.
Solution:
Depreciated value = P(1 + \(\frac{r}{100}\))ⁿ = 1,40,000(1 – \(\frac{4}{100}\))²
= 1,40,000(\(\frac{96}{100}\)) x (\(\frac{96}{100}\)) = ₹ 1,29,024

Question 8.
Find the difference in C.I and S.I for P = ₹ 10,000, r = 4% p.a, n = 2 years.
Solution:
C.I – S.I = P(\(\frac{r}{100}\))² = 10,000(\(\frac{4}{100}\))²
= 10,000 x \(\frac{4}{100}\) x \(\frac{4}{100}\) = ₹ 16

Question 9.
Find the C.I for the given Principal = ₹ 8,000, r = 5% p.a, n = 2 years
Amount A = P(1 + \(\frac{r}{100}\))ⁿ = 8000(1 + \(\frac{5}{100}\))²
= 8000 x \(\frac{105}{100}\) x \(\frac{105}{100}\)
= 8000 x \(\frac{21}{20}\) x \(\frac{21}{20}\)
A = ₹ 8820
Cl = A – P = 8820 – 8000 = 820

Question 10.
Find the S.I for the principal P = ₹ 16,000, r = 5% p.a, n = 3 years
Solution:
P = ₹ 16,000, n= 3 years, r = 5%
S.I = \(\frac{Pnr}{100}\) = \(\frac{16000x3x5}{100}\) = ₹ 2400

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Samacheer Kalvi 9th Science Solutions Chapter 12 Periodic Classification of Elements

Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 12 Periodic Classification of Elements

Samacheer Kalvi 9th Science Periodic Classification of Elements Textbook Exercises

I. Choose the correct answer.

Question 1.
If Dobereiner is related with ‘law of triads’, then Newlands is related with ……………………
(a) Modern periodic law
(b) Hund’s rule
(c) Law of octaves
(d) Pauli’s Exclusion principle
Answer:
(c) Law of octaves

Question 2.
Modem periodic law states that the physical and chemical properties of elements are the periodic functions of their ……………
(a) atomic numbers
(b) atomic masses
(c) similarities
(d) anomalies
Answer:
(a) atomic numbers

Question 3.
Elements in the modem periodic table are arranged in …………. groups and ………. periods.
(a) 7, 18
(b) 18, 7
(c) 17, 8
(d) 8, 17
Answer:
(b) 18, 7

II. Fill in the blanks.

In Dobereiner’s triads, the atomic weight of the middle element is the ……………….. of the atomic masses of 1st and 3rd elements.
Noble gases belong to ……………….. group of the periodic table.
The basis of the classifications proposed by Dobereiner, Newlands, and Mendeleev was…………………..
An example of liquid metal is …………………..

Answer:

average
18th
atomic masses
Mercury

III. Match the following.

Column – I	Column – II
1. Triads	(a) Newlands
2. Alkali metal	(b) Calcium
3. Law of octaves	(c) Henry Moseley
4. Alkali earth metal	(d) Sodium
5. Modem Periodic Law	(e) Dobereiner

Answer:

(e) Dobereiner
(d) Sodium
(a) Newlands
(b) Calcium
(c) Henry Moseley

IV. State whether True or False.

Newlands’ periodic table is based on atomic masses of elements and modem periodic table is based on the atomic number of elements – True
Metals can gain electrons – False.
Correct Statement: Metals tend to lose electrons to attain Noble Gas electron configuration.
Alloys bear the characteristics of both metals and nonmetals – False
Correct Statement: An alloy is a mixture of metals or a mixture of a metal and another element.
Lanthanides and actinides are kept at the bottom of the periodic table because they resemble each other but they do not resemble any other group elements – True
Group 17 elements are named as Halogens – True

V. Assertion and Reason

Statement: Elements in a group generally possess similar properties but elements along a period have different properties.
Reason: The difference in electronic configuration makes the element differ in their chemical properties along a period.
(a) Statement is true and the reason explains the statement.
(b) Statement is false but the reason is correct.
Answer:
(a) Statement is true and reason explains the statement.

VI. Answer the following.

Question 1.
State modern periodic law.
Answer:
“The Chemical and Physical properties of elements are periodic functions of their atomic number”.

Question 2.
What are groups and periods in the modern periodic table?
Answer:

The horizontal rows are called periods. There are seven periods in the periodic table.
Vertical columns in the periodic table starting from top to bottom are called groups. There are 18 groups in the periodic table.

Question 3.
What are the limitations of Mendeleev’s periodic table?
Limitations of Mendeleev’s periodic table:
Answer:

Elements with large difference in properties were included in the same group. Eg: Hard metals like copper (Cu) and silver (Ag) were included along with soft metals like sodium (Na) and potassium (K).
No proper position could be given to the element hydrogen. Non-metallic hydrogen was placed along with metals like lithium (Li), sodium (Na) and potassium (K).
The increasing order of atomic mass was not strictly followed throughout. Eg. Co & Ni, Te & I.
No place for isotopes in the periodic table.

Question 4.
State any five features of modern periodic table.
Answer:

All the elements are arranged in the increasing order of their atomic number
The horizontal rows are called periods. There are seven periods in the periodic table.
The elements are placed in periods based on the number of shells in their atoms
Vertical columns in the periodic table starting from top to bottom are called groups. There are 18 groups in the periodic table
Based on the physical and chemical properties of elements, they are grouped into various families.

Activity

Question 1.
Find the pair of elements having similar properties by applying Newlands law of Octaves (Example: Mg & Ca):
Set I: F, Mg, C, O, B
Set II: Al, Si, S, Cl, Ca
Answer:

Set I:	F	Mg	C	0	B
Set II:	Cl	Ca	Si	s	Al

Samacheer Kalvi 9th Science Periodic Classification of Elements Additional Questions

I. Short answers questions.

Question 1.
Why did the classification of elements evolve?
Answer:
Scientists found it difficult to organize all that was known about the elements. Therefore they started looking for some pattern in their properties, and the concepts of classification of elements were proposed by various scientists. This led to the evolution of the classification of elements from the early to modem period.

Question 2.
What is a triads group?
Answer:
Johann Wolfgang Dobereiner arranged the elements into groups containing three elements each based on their relative, atomic masses. He called these groups as ‘triads’ also known as Dobereiner’s Triads.

Question 3.
Explain law of octaves.
Answer:
John Newlands arranged 56 known elements in the increasing order of their atomic mass. He observed that every eighth element had properties similar to those of the first element like the eighth note in an octave of music is similar to the first and this arrangement was known as “law’ of octaves”.

Question 4.
What is a periodic table?
Answer:
Dmitri Mendeleev proposed the law of periodicity which states that “the physical and chemical properties of elements are the periodic functions of their atomic masses”. He arranged 56 elements known at that time according to his law of periodicity. This was best known as the short form of the periodic table.

Question 5.
What is IUPAC? Where is it located? Give examples of few elements named by IUPAC.
Answer:
IUPAC stands for-The International Union of Pure and Applied Chemistry. It is an international organization that represents Chemists from various countries. IUPAC is registered in Zurich, Switzerland and has its secretariat in the United States.

Question 6.
How are metals classified in the periodic table?
Answer:
Metals occupy a larger area in the periodic table and are categorized as follows –

Alkali metals eg. – Sodium and Potassium
Alkaline earth metals eg. – Calcium and Magnesium
Transition metals eg. – Iron and Nickel
Other metals eg. – Aluminium and Tin

Question 7.
What are alkali metals?
Answer:
The elements of group 1 (except hydrogen) are metals. They react with water to form solutions that are highly alkaline or basic. Hence they are called alkali metals.

Question 8.
Though they are found rare, they have many uses. Justify this statement.
Answer:
Helium, Neon, Argon, Krypton, Xenon, and Radon are called the Noble or rare gases and are placed in group 18 in the periodic table. They are monoatomic and do not react with other substances easily. Though these gases are chemically inert, they have stable electronic structures which are difficult to change. Their uses are as follows:

Helium has a very low density and used for filling weather ballon.
Neon gas is used in discharge lamps.
Argon is filled in electrical bulbs to prevent evaporation of the filament.
Radon is a radioactive gas.

II. Long Answers Questions.

Question 1.
Write any five advantages of the modern periodic law.
Answer:

The table is based on a more fundamental property i.e., atomic number.
It correlates the position of the element with its electronic configuration more clearly.
The completion of each period is more logical. In a period, as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached.
Each group is an independent group and the idea of subgroups has been discarded.
One position for all isotopes of an element is justified since the isotopes have the same atomic number.

Question 2.
Explain the limitations of Mendelev’s periodic table and why is the modern periodic table considered an extension of Mendeleev’s periodic table.
Limitations of Mendelev’s periodic table –
(a) Elements with large difference in properties were included in the same group. Eg. Elard metals like copper (Cu) and silver (Ag) were included along with soft metals like sodium and potassium
(b) No proper position could be given to the element hydrogen. Non-metallic Hydrogen was placed along with metals like lithium, sodium and potassium.
(c) The increasing order of atomic mass was not strictly followed throughout. Eg. – Co & Ni, te & I.
(d) No place for isotopes in the periodic table.

The modem periodic table is an extension of the original Mendeleev’s periodic table and is known as the long form of periodic table. Mendeleev’s initial table had 63 elements, modem table has 109 elements. It is also known important to note how the modem periodic table is arranged.

Although the columns which reflect a natural order have been retained, the row’s of the modem table show elements in the order of Mendeleev’s columns. In other words the elements of what we now call ‘a period’ were listed vertically by Mendeleev. ‘Groups are now shown vertically in contrast to their horizontal format in Mendeleev’s table.

Question 3.
What are the advantages of alloys?

Alloys do not get corroded or get corroded to very less extent.
They are harder and stronger than pure metals (example: gold is mixed with copper and it is harder than pure gold).
They have less conductance than pure metals (example: copper is good conductor of heat and electricity whereas brass and bronze are not good conductors).
Some alloys have lower melting point than pure metals (example: solder is an alloy of lead and tin which has lower melting point than each of the metals).
When metal is alloyed with mercury, it is called amalgam.

Question 4.
Explain the uniqueness of Hydrogen.
Answer:
Hydrogen is the lightest, smallest and first element of the periodic table. Its electronic configuration (Is1) is the simplest of all the elements. It occupies a unique position in the periodic table. It behaves like alkali metals as well as halogens in its properties. In the periodic table, it is placed at the top of the alkali metals.

Hydrogen can lose its only electron to form a hydrogen ion (H+) like alkali metals.
It can also gain one electron to form the hydride ion (H-) like halogens.
Alkali metals are solids while hydrogen is a gas.

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