Laxmi

Students can Download Maths Chapter 2 Algebra Ex 2.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.3

Question 1.
Fill in the blanks:

Question (i)
X – axis and Y – axis intersect at ……..
Answer:
Origin (0, 0)

Question (ii)
The coordinates of the point in third quadrant are always ……….
Answer:
negatives

Question (iii)
(0, -5) point lies on ………. axis.
Answer:
y – axis.

Question (iv)
The x coordinate is always ……… on the y – axis.
Answer:
Zero

Question (v)
Coordinates are the same for a line parallel to Y – axis.
Answer:
X.

Question 2.
Say True or False:

Question (i)
(-10, 20) lies in the second quadrant.
Answer:
True
Hint:
(-10, 20)
x = – 10,
y = 20
∴ (- 10, 20) lies in second quadrant – True

Question (ii)
(-9, 0) lies on the x – axis.
Answer:
True
Hint:
(-9, 0) on x – axis, Y – coordinate is always zero.
∴ (-9, 0) lies on x – axis – True

Question (iii)
The coordinates of the origin are (1, 1).
Answer:
False
Hint:
Coordinate of origin is (0, 0), not (1, 1). Hence – False

Question 3.
Find the quadrants without plotting the points on a graph sheet.
(3, -4), (5, 7), (2, 0), (- 3, – 5), (4, – 3), (- 7, 2), (- 8, 0), (0,10), (- 9, 50).
Solution:

• If x & y coordinate are positive – I quad
• If x is positive, y is negative – IV quad
• If x is negative, y is positive – II quad
• If both are negative, then – III quad

Question 4.
Plot the following points in a graph sheet.
A(5, 2), B(- 7, – 3), C(- 2, 4), D(- 1, – 1), E(0, – 5), F(2, 0), G(7, – 4), H(- 4, 0), 1(2,3), J(8, – 4) K (0, 7).
Solution:

Question 5.
Use the grid graph to determine the coordinates where each figure is located.

a) Star ……..
b) Bird ……..
c) Red Circle ……..
d) Diamond ……..
e) Triangle ……..
f) Ant ……..
g) Mango ……..
h) Housefly ……..
i) Medal ……..
j) Spider ……..
Solution:
a) Star (3, 2)
b) bird (-2, 0)
c) Red Circle (-2, -2)
d) Diamond (-2, 2)
e) Triangle (-1, -1)
f) Ant (3, -1)
g) Mango (0, 2)
h) Housefly (2, 0)
i) Medal (-3, 3)
j) Spider (0, -2)

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 2 Chapter 3 Algebra Intext Questions

Exercise 3.1

Try These (Text book Page No. 44)

Question 1.
Observe and complete the following table. First one is done for you.

Solution:

Try These (Text book Page No. 46)

Question 1.
Simplify and write the following in exponential form.
1. 2³ × 2⁵
2. p² × P⁴
3. x⁶ × x⁴
4. 3¹ × 3⁵ × 3⁴
5. (-1)² × (-1)³ × (-1)⁵
Solution:
1. 2³ × 2⁵ = 2³⁺⁵ = 2⁸ [since a^m × aⁿ = a^m+n]

2. p² × p⁴ = p²⁺⁴ = p⁶ [since a^m × aⁿ = a^m+n]

3. x⁶ × x⁴ = x⁶ + 4 = x¹⁰ [since a^m × aⁿ = a^m+n]

4. 3¹ × 3⁵ × 3⁴ = 3¹⁺⁵ × 3⁴ [since a^m × aⁿ = a^m+n]
= 3⁶ × 3⁴ [since a^m × aⁿ = a^m+n]
= 3¹⁰

5. (-1)² × (-1)³ × (-1)⁵
= (-1)²⁺³ × (-1)⁵ [Since a^m × aⁿ = a^m+n]
= (-1)⁵ × (-1)⁵
= (-1)⁵⁺⁵ [Since a^m × aⁿ = a^m+n]
= (-1)¹⁰

Try These (Text book Page No. 48)

Question 1.
Simply the following.
1. 23⁵ ÷ 23²
2. 11⁶ ÷ 11³
3. (-5)³ ÷ (-5)²
4. 7³ ÷ 7³
5. 15⁴ ÷ 15
Solution:

Try These (Text book Page No. 48)

Question 1.
Simplify and write the following in exponent form.
1. (3²)³
2. [(-5)³]²
3. (20⁶)²
4. (10³)⁵
Solution:
1. (3²)³ = 3^2×3 = 3⁶ [since (a^m)ⁿ = a^m×n]
2. [(-5)]² = (-5)^3×2 = (-5)⁶ [since (a^m)ⁿ = a^m×n]
3. (20⁶)² = 20^6×2 = 20¹² [since (a^m)ⁿ = a^m×n]
4. (10³)⁵ = 10^3×5 = 10¹⁵ [since (a^m)ⁿ = a^m×n]

Question 2.
Express the following exponent numbers using a^m × b^m = (a × b)^m.
(i) 5² × 3²
(ii) x³ × y³
(iii) 7⁴ × 8⁴
Solution:
(i) 5² × 3² = (5 × 3)² = 15² [since a^m × b^m = (a × b)^m]
(ii) x³ × y³ = (x × y)³ = (x y)³
(iii) 7⁴ × 8⁴ = (7 × 8)⁴ = 56⁴

Question 3.
Simplify the following exponent numbers by using (\(\frac { a }{ b } \))^m = \(\frac{a^{m}}{b^{m}}\)
(i) 5³ ÷ 2³
(ii) (-2)⁴ ÷ 3⁴
(iii) 8⁶ ÷ 5⁶
(iv) 6³ ÷ (-7)³
Solution:
(i) 5³ ÷ 2³ = (\(\frac { 5 }{ 2 } \))³ – [Since \(\frac{a^{m}}{b^{m}}\) = (\(\frac { a }{ b } \))^m]
(ii) (-2)⁴ ÷ 3⁴ = (\(\frac { -2 }{ 3 } \))⁴
(iii) 8⁶ ÷ 5⁶ = (\(\frac { 8 }{ 6 } \))⁶
(iv) 6³ ÷ (-7)³ = (\(\frac { 6 }{ -7 } \))³

Exercise 3.2

Try These (Text book Page No. 54)

Question 1.
Find the unit digit of the following exponential numbers:
(i) 106²¹
(ii) 25⁸
(iii) 31¹⁸
(iv) 20¹⁰
Solution:
(i) 106²¹ Unit digit of base 106 is 6 and the power is 21 and is positive.
Thus the unit digit of 106²¹ is 6.

(ii) 25⁸ Unit digit of base 25 is 5 and the power is 8 and is positive.
Thus the unit digit of 25⁸ is 5.

(iii) 31¹⁸ Unit digit of base 31 is 1 and the power 18 and is positive.
Thus the unit digit of 31¹⁸ is 1.

(iv) 20¹⁰ Unit digit of base 20 is 0 and the power 10 and is positive.
Thus the unit digit of 20¹⁰ is 0.

Try These (Text book Page No. 55)

Question 1.
Find the unit digit of the following exponential numbers:
(i) 64¹¹
(ii) 29¹⁸
(iii) 79¹⁹
(iv) 104³²
Solution:
(i) 64¹¹ Unit digit of base 64 is 4 and the power is 11 (odd power).
∴ Unit digit of 64¹¹ is 4.

(ii) 29¹⁸ Unit digit of base 29 is 9 and the power is 18 (even power).
Therefore, unit digit of 29¹⁸ is 1.

(iii) 79¹⁹ Unit digit of base 79 is 9 and the power is 19 (odd power).
Therefore, unit digit of 7919 is 9.

(iv) 104³² Unit digit of base 104 is 4 and the power is 32 (even power).
Therefore, unit digit of 104³² is 6.

Exercise 3.3

Try These (Text book Page No. 35)

Question 1.
Complete the following table:

Solution:

Question 2.
Identify the like terms from the following:
(i) 2x²y, 2xy²,3xy²,14x²y, 7yx
(ii) 3x³y², y³x, y³x², – y³x, 3y³x
(iii) 11pq, -pq, 11pqr, -11pq,pq
Solution:
(i) 2x²y, 2xy², 3xy², 14x²y, 7yx
(a) 2x²y and 14x²y are like terms.
(b) 2xy² and 3xy² are like terms.

(ii) 3x³y², y³x, y³x², – y³x, 3y³x
(a) y³x, – y³x and 3y³x are like terms.

(iii) 11 pq, -pq, 11pqr , -11 pq, pq
(a) 11 pq, -pq, -pq and pq are like terms.

You can Download Samacheer Kalvi 9th English Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th English Article Writing

An article is a piece of written work, the purpose of which is to propagate news, inform about results of research, analyze academic information, express opinions or persuade through suggestions.

Components of an article :
Title – it should be catchy enough to draw the reader’s attention
Byline – the name of the writer of the article Introduction – outlines the topic
Body – builds upon the topic; consists of two to four paragraphs
Conclusion – summarises and offers final opinions

Tips for writing articles :

• Be brief and concise.
• Cover the essential elements of who, what, when, where, how and why?
• Focus on the theme or topic of the article. Do not deviate from it.
• Use simple language.

Exercises
1. Write an article for your school magazine. Describing your experience about the Clean India Campaign.
Answer:

Clean India Campaign

– by Anitha

The ‘Clean India Campaign’ or ‘Swachh Bharat Abhiyan’ launched by our Prime Minister, Mr. Narendra Modi was implemented to fulfill the vision of a clean India. But it has not been as effective as it should have been opinion polls conducted in various cities of India suggested that nearly two-thirds of the residents surveyed felt there was no impact of the campaign on their city or surroundings. The fact remains that the government alone cannot bring about a change unless we the people take initiatives to keep our surroundings clean. It is all about changing our attitude which cannot be changed by imposing fines of punishments.

There is a significant percentage of our population who are totally unaware of and neglect basic cleanliness hygiene. These are people who don’t even think once before littering on the road or throwing garbage anywhere and everywhere. Their lack of awareness has led to the ineffectiveness of such cleanliness initiatives. It is important to sensitize people on the importance of hygiene and cleanliness through awareness programs in electronic, print and social media. Enlightened people need to work with not only individuals but also organizations to make those people aware of the importance of cleanliness.

2. Write an article for your school magazine. Describing your experience about River Pollution.
Answer:

River Pollution

– by Rama

It is indeed unfortunate that rivers, which are our largest source of water are victims of pollution. Countless tanneries, chemical plants, textile mills, slaughter houses, hospitals contribute to pollution of rivers by dumping untreated waste into it. People who do not have the proper living conditions in India, resort to carrying out all their daily activities like bathing, washing and defecation, on the bank of rivers, and thus pollute them. Discharge of sewage and other domestic waste into the rivers is yet another environmentally hazardous practice, that is polluting the rivers. Religious practices also demand that floral offerings be cast-off in rivers and the ashes, after cremating a person should also be poured into rivers.
Unfortunately one concludes despite laws dealing with the task of prevention and control of river pollution there is no end to the misery we humans inflict on our rivers.

3. Write an article for your school magazine. Describing your experience about Serve Mankind.
Answer:

Serve Mankind

– by Bala

Men represent society in a Microform. God made man to serve those who need the service. It may be in any field or in any form. But the best form is “Voluntary Community Work” which means that the work should be done not for an individual but for all. For example, I see that the village is full of garbage. Let us together dig pits far away from the village and bury these garbage there, which will turn to manure in a couple of years’ time.

At the same time, the removal of the garbage will give a better look to the village. The second community work for our village is to make drains. Let us control the dirty water and direct it to the pits where the garbage has been buried. But this is not the work done by one or two men. Let the people realise that the best service to God is to serve His creation.

4. Write an article for your school magazine. Describing your experience about school picnic.
Answer:

SCHOOL PICNIC TO MAMALLAPURAM

– by Srinivasan

School picnics are the most awaited ones. They are with great fun. The memories of school picnics are the one to be cherished for a lifetime. This year’s picnic was destined to be at Mamallapuram which is situated around 60 km from Chennai. On the day of the picnic, we reported to the school assembly hall at 5 a.m. Since it was the last week of December, the weather was very cold. The school bus departed at sharp 5.10 a.m. in the morning. Almost one and half hour travel to the heritage town felt like a matter of few minutes. We reported to the guest house at around 6.30 a.m., we were all provided welcome drinks. Then we left for the Shore Temple.

Sunrise at Shore Temple is an unforgettable sight. Five Rathas, Cave Temple, Tiger Cave, Elephant and other creatures carved in granite, Arjuna’s Penance, Lord Krishna’s Butter Ball (a huge stone almost spherical in shape) are the important sculptures to be seen there. It was the most exciting part of the picnic. Then we had lunch, followed by some fun activities. The time had passed so soon that we couldn’t even realize the day was over. This was indeed a day I will remember for a long time.

You can Download Samacheer Kalvi 9th English Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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.

Students can Download Maths Chapter 2 Algebra Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 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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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.

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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

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Question 1.
Fill in the blanks.

(i) The degree of the term a³b²c⁴d² is _______
(ii) Degree of the constant term is _______
(iii) The coefficient of leading term of the expression 3z²y + 2x – 3 is _______
Answers:
(i) 11
(ii) 0
(iii) 3

Question 2.
Say True or False.

(i) The degree of m² n and mn² are equal.
(ii) 7a²b and -7ab² are like terms.
(iii) The degree of the expression -4x² yz is -4
(iv) Any integer can be the degree of the expression.
Answers:
(i) True
(ii) False
(iii) False
(iv) True

Question 3.
Find the degree of the following terms.
(i) 5x²
(ii) -7 ab
(iii) 12pq² r²
(iv) -125
(v) 3z
Solution:
(i) 5x²
In 5x², the exponent is 2. Thus the degree of the expression is 2.

(ii) -7ab
In -7ab, the sum of powers of a and b is 2. (That is 1 + 1 = 2).
Thus the degree of the expression is 2.

(iii) 12pq² r²
In 12pq² r², the sum of powers of p, q and r is 5. (That is 1 +2 + 2 = 5).
Thus the degree of the expression is 5.

(iv) -125
Here – 125 is the constant term. Degree of constant term is 0.
∴ Degree of -125 is 0.

(v) 3z
The exponent is 3z is 1.
Thus the degree of the expression is 1.

Question 4.
Find the degree of the following expressions.
(i) x³ – 1
(ii) 3x² + 2x + 1
(iii) 3t⁴ – 5st² + 7s²t²
(iv) 5 – 9y + 15y2 – 6y3
(v) u⁵ + u⁴v + u³v² + u₂v³ + uv⁴
Solution:
(i) x³ – 1
The terms of the given expression are x³, -1
Degree of each of the terms: 3,0
Terms with highest degree: x³.
Therefore, degree of the expression is 3.

(ii) 3x² + 2x + 1
The terms of the given expression are 3x², 2x, 1
Degree of each of the terms: 2, 1, 0
Terms with highest degree: 3x²
Therefore, degree of the expression is 2.

(iii) 3t⁴ – 5st² + 7s²t²
The terms of the given expression are 3t⁴, – 5st², 7s³t²
Degree of each of the terms: 4, 3, 5
Terms with highest degree: 7s²t²
Therefore, degree of the expression is 5.

(iv) 5 – 9y + 15y² – 6y³
The terms of the given expression are 5, – 9y , 15y², – 6y³
Degree of each of the terms: 0, 1, 2, 3
Terms with highest degree: – 6y³
Therefore, degree of the expression is 3.

(v) u⁵ + u⁴v + u³v² + u₂v³ + uv⁴
The terms of the given expression are u⁵, u⁴v , u³v², u²v³, uv⁴
Degree of each of the terms: 5, 5, 5, 5, 5
Terms with highest degree: u⁵, u⁴v , u³v², u²v³, uv⁴
Therefore, degree of the expression is 5.

Question 5.
Identify the like terms : 12x³y²z, – y³x²z, 4z³y²x, 6x³z²y, -5y³x²z
Solution:
-y³ x²z and -5y³x²z are like terms.

Question 6.
Add and find the degree of the following expressions.
(i) (9x + 3y) and (10x – 9y)
(ii) (k² – 25k + 46) and (23 – 2k² + 21 k)
(iii) (3m²n + 4pq²) and (5nm² – 2q²p)
Solution:
(i) (9x + 3y) and (10x – 9y)
This can be written as (9x + 3y) + (10x – 9y)
Grouping the like terms, we get
(9x + 10x) + (3y – 9y) = x(9 + 10) + y(3 – 0) = 19x + y(-6) = 19x – 6y
Thus degree of the expression is 1.

(ii) (k² – 25k + 46) and (23 – 2k² + 21k)
This can be written as (k² – 25k + 46) + (23 – 2k² + 21k)
Grouping the like terms, we get
(k² – 2k²) + (-25 k + 21 k) + (46 + 23)
= k² (1 – 2) + k(-25 + 21) + 69 = – 1k² – 4k + 69
Thus degree of the expression is 2.

(iii) (3m²n + 4pq²) and (5nm² – 2q²p)
This can be written as (3m²n + 4pq²) + (5nm² – 2q²p)
Grouping the like terms, we get
(3m²n + 5m²n) + (4pq² – 2pq²)
= m²n(3 + 5) + pq²(4 – 2) = 8m²n + 2pq²
Thus degree of the expression is 3.

Question 7.
Simplify and find the degree of the following expressions.
(i) 10x² – 3xy + 9y² – (3x² – 6xy – 3y²)
(ii) 9a⁴ – 6a³ – 6a⁴ – 3a² + 7a³ + 5a²
(iii) 4x² – 3x – [8x – (5x² – 8)]
Solution:
(i) 10x² – 3xy + 9y² – (3x² – 6xy – 3y²)
= 10x² – 3xy + 9y² + (-3x² + 6xy + 3y²)
= 10x² – 3xy + 9y² – 3x² + 6xy + 3y²
= (10x² – 3x²) + (- 3xy + 6xy) + (9y² + 3y²)
= x²(10 – 3) + xy(-3 + 6) + y²(9 + 3)
= x²(7) + xy(3) + y²(12)
Hence, the degree of the expression is 2.

(ii) 9a⁴ – 6a³ – 6a⁴ – 3a² + 7a³ + 5a²
= (9a⁴ – 6a⁴) + (- 6a³ + 7a³) + (- 3a² + 5a²)
= a⁴(9-6) + a³ (- 6 + 7) + a²(-3 + 5)
= 3a⁴ + a³ + 2a²
Hence, the degree of the expression is 4.

(iii) 4x² – 3x – [8x – (5x² – 8)]
= 4x² – 3x – [8x + -5x² + 8)]
= 4x² – 3x – [8x – 5x² – 8]
= 4x² – 3x – 8x + 5x² – 8
(4x² + 5x²) + (- 3x – 8x) – 8
= x²(4+ 5) + x(-3-8) – 8
= x²(9) + x(- 11) – 8
= 9x² – 11x – 8
Hence, the degree of the expression is 2.

Objective Type Question

Question 8.
3p² – 5pq + 2q² + 6pq – q² +pq is a
(i) Monomial
(ii) Binomial
(iii) Trinomial
(iv) Quadrinomial
Answer:
(iii) Trinomial

Question 9.
The degree of 6x⁷ – 7x³ + 4 is
(i) 7
(ii) 3
(iii) 6
(iv) 4
Answer:
(i) 7

Question 10.
If p(x) and q(x) are two expressions of degree 3, then the degree of p(x) + q(x) is
(i) 6
(ii) 0
(iii) 3
(iv) Undefined
Answer:
(iii) 3

Students can Download Maths Chapter 1 Life Mathematics Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

Students can Download Maths Chapter 3 Algebra Ex 3.1 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 2 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks.

1. The exponential form 14⁹ should be read as ______
2. The expanded form of p³ q² is ______
3. When base is 12 and exponent is 17, its e×ponential form is _____
4. The value of (14 × 21)⁰ is _____

Answers:

1. 14 Power 9
2. p × p × p × q × q
3. 12¹⁷
4. 1

Question 2.
Say True or False.

1. 2³ × 3² = 65
2. 2⁹ × 3² = (2 × 3)^9×2
3. 3⁴ × 3⁷= 3¹¹
4. 2⁰ × 1000⁰
5. 2³ < 3²

Answers:

1. False
2. False
3. True
4. True
5. True

Question 3.
Find the value of the following.

1. 2⁶
2. 11²
3. 5⁴
4. 9³

Solution:

1. 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64
2. 11² = 11 × 11 = 121
3. 5⁴ = 5 × 5 × 5 × 5 = 625
4. 9³ = 9 × 9 × 9 = 729

Question 4.
Express the following in e×ponential form.

1. 6 × 6 × 6 × 6
2. t × t
3. 5 × 5 × 7 × 7 × 7
4. 2 × 2 × a × a

Solution:

1. 6 × 6 × 6 × 6 = 6^1+1+1+1 = 64 [Since a^m × aⁿ = a^m+n]
2. t × t = t¹⁺¹ = t²
3. 5 × 5 × 7 × 7 × 7 = 5¹⁺¹ × 7¹⁺¹⁺¹ = 5² × 7³
4. 2 × 2 × a × a = 2¹⁺¹ × a¹⁺¹ = 2² × a² = (2a)²

Question 5.
E×press each of the following numbers using e×ponential form,
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
(i) 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹ [Using product rule]

(ii) 343

343 = 7 × 7 × 7 = 7¹⁺¹⁺¹
= 7³ [Using product rule]

(iii) 729

729 = 3 × 3 × 3 × 3 × 3 × 3
= 3⁶ [Using product rule]

(iv) 3125

3125 = 5 × 5 × 5 × 5 × 5
= 5⁵ [Using product rule]

Question 6.
Identify the greater number in each of the following.
(i) 6³ or 3⁶
(ii) 5³ or 3⁵
(iii) 2⁸ or 8²
Solution:
(i) 6³ or 3⁶
6³ = 6 × 6 × 6 = 36 × 6 = 216
3⁶ = 3 × 3 × 3 × 3 × 3 × 3 = 729
729 > 216 gives 3⁶ > 6³
∴ 36 is greater.

(ii) 5³ or 3⁵
5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243
243 > 125 gives 3⁵ > 5³
∴ 3⁵ is greater.

(iii) 2⁸ or 8²
2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8² = 8 × 8 = 64
256 > 64 gives 2⁸ > 8²
∴ 2⁸ is greater.

Question 7.
Simplify the following
(i) 7² × 3⁴
(ii) 3² × 2⁴
(iii) 5² × 10⁴
Solution:
(i) 7² × 3⁴ = (7 × 7) × (3 × 3 × 3 × 3)
= 49 × 81 = 3969

(ii) 3² × 2⁴ = (3 × 3) × (2 × 2 × 2 × 2)
= 9 × 16 = 144

(iii) 5² × 10⁴ = (5 × 5) × (10 × 10 × 10 × 10)
= 25 × 10000 = 2,50,000

Question 8.
Find the value of the following.
(i) (-4)²
(ii) (-3) × (-2)³
(iii) (-2)³ × (-10)³
Solution:
(i) (-4)² = (-1)² × (4)² [since a^m × b^m = (a × b)^m]
= 1 × 16 = 16 [since (-1)ⁿ = 1 if n is even]

(ii) (-3) × (-2)³ = (-1) × (-3) × (-1)³ × (-2)³
= (-1)⁴ × 24 [Grouping the terms of same base]
= 24

(iii) (-2)³ × (-10)³ = (-1)³ × (-2)³ × (-1)³ × (-10)³
= (-1)³⁺³ × 2³ × 10³ [Grouping the terms of same base]
= (-1)⁶ × (2 × 10)³
[∵ a^m × b^m = (a × b)^m]
= 1 × 20³ [since (-1)ⁿ = 1 if n is even]
= 8000

Question 9.
Simplify using laws of exponents.
(i) 3⁵ × 3⁸
(ii) a⁴ × a¹⁰
(iii) 7x × 7²
(iv) 2⁵ ÷ 2³
(v) 18⁸ ÷ 18⁴
(vi) (6⁴)³
(vii) (x^m)⁰
(viii) 9⁵ × 3⁵
(ix) 3^y × 12^y
(x) 25⁶ × 5⁶
Solution:

Question 10.
If a = 3 and b = 2, then find the value of the following.
(i) a^b + b^a
(ii) a^a – b^b
(iii) (a + b)^b
(iv) (a – b)^a
Solution:
(i) a^b + b^a
a = 3 and b = 2
we get 3² + 2³ = (3 × 3) + (2 × 2 × 2) = 9 + 8 = 17

(ii) (a^a – b^b)
Substituting a = 3 and b = 2
we get 3² – 2² = (3 × 3 × 3) – (2 × 2) = 27 – 4 = 23

(iii) (a + b)^b
Substituting a = 3 and b = 2
we get (3 + 2)² = 5² = 5 × 5 = 25

(iv) (a – b)^a
Substituting a = 3 and b = 2
we get (3 – 2)³ = 1³ = 1 × 1 × 1 = 1

Question 11.
Simplify and express each of the following in exponential form:
(i) 4⁵ × 4² × 4⁴
(ii) (3² × 3³)⁷
(iii) (5² × 5⁸) ÷ 5^s
(iv) 2⁰ × 3⁰ × 4⁰
(v) \(\frac{5^{5} \times a^{8} \times b^{3}}{4^{3} \times a^{5} \times b^{2}}\)
Solution:

Objective Type Questions

Question 12.
a × a × a × a × a equal to
(i) a⁵
(ii) 5 a
(iii) 5a
(iv) a + 5
Answer:
(i) a⁵

Question 13.
The exponential form of 72 is
(i) 72
(ii) 27
(iii) 2² × 3³
(iv) 2³ × 3²
Answer:
(iv) 2³ × 3²

Question 14.
The value of x in the equation a¹³ = x³ × a¹⁰ is
(i) a
(ii) 13
(iii) 3
(iv) 10
Answer:
(i) a

Question 15.
How many zeros are there in 100¹⁰ ?
(i) 2
(ii) 3
(iii) 100
(iv) 20
Answer:
(iv) 20

Question 16.
2⁴⁰ + 2⁴⁰ is equal to
(i) 4⁴⁰
(ii) 2⁸⁰
(iii) 2⁴¹
(iv) 4⁸⁰
Answer:
(iii) 2⁴¹