Bhagya

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Samacheer Kalvi 11th Commerce Solutions Chapter 32 Direct Taxes

Samacheer Kalvi 11th Commerce Direct Taxes Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
Income Tax is ……………
(a) a business tax
(b) a direct tax
(c) an indirect tax
(d) none of these
Answer:
(b) a direct tax

Question 2.
Period of assessment year is ……………
(a) 1^st April to 31^st March
(b) 1^st March to 28^st Feb
(c) 1^st July to 30^st June
(d) 1^st Jan to 31^st Dec
Answer:
(a) 1^st April to 31^st March

Question 3.
The year in which income is earned is known as ……………
(a) Assessment Year
(b) Previous Year
(c) Light Year
(d) Calendar Year
Answer:
(b) Previous Year

Question 4.
The aggregate income under five heads is termed as ……………
(a) Gross Total Income
(b) Total Income
(c) Salary Income
(d) Business Income
Answer:
(b) Total Income

Question 5.
Agricultural income earned in India is ……………
(a) Fully Taxable
(b) Fully Exempted
(c) Not Considered for Income
(d) None of the above
Answer:
(b) Fully Exempted

II. Very Short Answer Questions

Question 1.
What is Income tax?
Answer:
Income tax is a direct tax under which tax is calculated on the income, gains or profits earned by a person such as individuals and other artificial entities.

Question 2.
What is meant by the previous year?
Answer:
The year in which income is earned is called the previous year. It is also normally consisting of a period of 12 months commencing on 1^st April every year and ending on 31^st March of the following year. It is also called a financial year immediately following the assessment year.

Question 3.
Define the term person?
Answer:
According to the Income Tax 1961, the term Person includes, “ individual, Hindu Undivided Family, Fìrrn, Company, local authority, Association of person or Body of Individual or any other artificial juridical persons.

Question 4.
Define the term assessee?
Answer:
Assessee means a person by whom any tax or any other sum of money is payable under this Act. It includes every person in respect of whom any proceeding has been taken for the assessment of his income or assessment of fringe benefits.

Question 5.
What is an assessment year?
Answer:
The term has been defined under section 2(9). The year in which tax is paid iš called the assessment year. It normally consisting of a period óf 12 months còrnmencing on 1st April every year and ending on 31 st March of the following year.

III. Short Answer Questions

Question 1.
What is Gross Total Income?
Answer:
Income from the five heads, namely – Salaries, House Property, Profits and Gains of Business or Profession, Capital Gains, and Other Sources – is computed separately according to the provisions given in the Act. Income computed under these heads shall be aggregated after adjusting past and present losses and the total so arrived at is known as ‘Gross Total Income’.

Question 2.
List out the five heads of income.
Answer:

1. Income from Salaries
2. Income from House Property
3. Income from Business or Profession
4. Income from Capital Gain
5. Income from Other Sources

Question 3.
Write a note on Agricultural Income.
Answer:
Any rent or revenue derived from land which is situated in India and is used for agriculture purposes. Agricultural income is fully exempted from tax u/s 10(1) and as such does not form part of total income.

Question 4.
What do you mean by Total Income?
Answer:
According to Section 2 (45) of the Income Tax Act 1961, the income which is earned out of the Gross Total Income, after allowing the deductions under section 80 is known as total income.

Question 5.
Write short notes on:

1. Direct Tax
2. Indirect Tax

Answer:
1. Direct Tax:
If a tax levied on the income or wealth of a person and is paid by that person (or his office) directly to the Government, , it is called a direct tax, for example, Income – Tax, Wealth Tax, Capital Gains Tax, Securities Transaction Tax, Fringe Benefits Tax (from 2005), Banking Cash Transaction Tax (for Rs,50,000 and above – from 2005), etc. In India, all direct taxes are levied and administered by the Central Board of Direct Taxes.

2. Indirect Tax:
If tax is levied on the goods or services of a person (seller). It is collected from, the buyers and is paid by the seller to the Government. It is called indirect tax example GST.

IV. Long Answer Questions

Question 1.
Elucidate any five features of Income Tax.
Answer:
Meaning: Income tax is a direct tax under which tax is calculated on the income, gains or profits earned by a person such as individuals and other artificial entitled.
Levied as Per the Constitution
Income tax is levied in India by virtue of entry No. 82 of list 1 (Union List) of Seventh Schedule to Article 246 of the Constitution of India.

Levied by Central Government
Income tax is charged by the Central Government on all incomes other than agricultural income. However, the power to charge income tax on agricultural income has been vested with the State Government as per entry 46 of List II, i.e., State List.

Direct Tax
Income tax is a direct tax. It is because the liability to deposit and ultimate burden are on the same person. The person earning income is liable to pay income tax out of his own pocket and cannot pass on the burden of tax to another person.

Annual Tax
Income tax is an annual tax because it is the income of a particular year which is chargeable to tax.

Tax on Person
It is a tax on income earned by a person. The term ‘person’ has been defined under the Income-tax Act. It includes individual, Hindu Undivided Family, Firm, Company, local authority, Association of person or Body of Individual or any other artificial juridical persons.

Question 2.
Define Tax. Explain the term direct tax and indirect tax with an example.
Answer:
Tax is a compulsory contribution to state revenue by the Government. It is levied on the income or profits from business of individuals and institutions. It may be added to the price of goods, services or transactions. Tax is the basic source of revenue to the Government. This revenue is utilised for the expenses of civil administration, internal and external security, building infrastructure, etc.

There are two types of taxes – direct taxes and indirect, taxes.

1. Direct Tax:
If a tax levied on the income or wealth of a person and is paid by that person (or his office) directly to the Government, it is called direct tax, e.g., Income – Tax, Wealth Tax, Capital Gains Tax, Securities Transaction Tax, Fringe Benefits Tax (from 2005), Banking Cash Transaction Tax (for Rs.50,000 and above – from 2005), etc. In India all direct taxes are levied and administered by Central Board of Direct Taxes.

2. Indirect Tax:
If tax is levied on the goods or services of a person (seller). It is collected from the buyers and is paid by seller to the Government. It is called indirect tax. example GST.

Question 3.
List out any ten kinds of incomes chargeable under the head income tax.
Answer:
According to section 2(24) of the Income Tax Act, Income includes the following:

1. Profits and gains of business or profession.
2. Dividend
3. Export incentives, like duty drawback, cash compensatory support, sale of licenses, etc.,
4. Interest, salary, bonus, commission or remuneration earned by a partner of a firm from such firm.
5. Capital gains chargeable u/s 45. viii. Profits and gains from the business of banking carried on by a co-operative society with its members.
6. Winnings from lotteries, crossword puzzles, races including horse races, card games, and other games of any sort or from gambling or betting of any form or nature whatsoever.
7. Deemed income u/s 41 or 59.
8. Amount received under key man. insurance policy including bonus thereon.
9. Benefit or perquisite received from a company, by a director or a person holding substantial interest or a relative of the director or such person.
10. Gift as defined u/s 56 (2) (vi) and others.

Question 4.
Discuss the various kinds of assessments.
Answer:
Assessee means a person by whom any tax or any other sum of money is payable under this Act. It includes every person in respect of whom any proceeding has been taken for the assessment of his income or assessment of fringe benefits. The term ‘person’ includes the following:

1. an individual
2. a Hindu Undivided Family (HUE)
3. a company
4. a firm
5. an Association Of Persons or a Body Of Individual, whether incorporated or not
6. a local authority, and
7. every artificial juridical person example an idol or deity.

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Tamilnadu Samacheer Kalvi 11th English Speaking Group Discussion

(i) Intelligent Quotient (IQ) Versus Emotional Quotient (EQ)

RAJ:
According to my opinion, those who have a good IQ level will always have the ability to reason out things and find an amicable solution to any problem that he/she faces in life. Even in today’s world, when a young aspirant is made to face an interview with a board of panelist, one of the key methods to finalise the situation is a problem solving abilities a candidate possesses. Nowadays, your mental appearance is more important than your physical appearance. Though eloquence in communication was given weight age, the skill and ability to convince others is much more important than flamboyant language. Presence of mind, common sense and practicality is more important than bookish knowledge. Therefore people with more IQ is needed than people with better EQ.

Chandran:
According to my understanding, I strongly contradict the views of Raj. It is the character that brings you huge success and I think good character, good personalities and good leaders are mostly driven by emotions.

Emotional quotient and intelligence quotient are undoubtedly the most important aspects of a person’s life and can make a huge difference between an average life and a successful life. To be successful, one may want to have a great IQ but, that success will not last long if your EQ is not appropriate. But on the contrary, if you have a good EQ and a little less IQ, then the chances of you becoming successful to carry on a successful career increases.

LEELA:
Everyone is equal. All should be recognised. One needs to be balanced with all qualities just as a balanced diet is needed for healthy living. It is illogical for learned people to say that IQ is much more needed than EQ or vice versa. I feel everyone is unique and is important. There is always something that one can contribute to. It needs to be balanced. Team play is vital at such a situation. A person with better IQ/EQ has his own workspace and methodology he contributes to the work differently as a person with different IQ/EQ. One who has a high IQ level can work smartly to do even the toughest tasks easily and hence procure popularity and wealth. Likewise, one who has better EQ can develop the same popularity and also leadership qualities. He/She can withstand adverse condition and overcome them.

Let me detail you with examples. To manage a team, a person of high EQ is needed whereas someone who needs to plan tasks and perform it in the best possible way, a person with high IQ should be there.

A person with higher IQ may work on any area and be extremely successful but may lack good social circle. This will surely lead to depression or emotional setbacks.

A person with higher EQ may win a large social circle of friends and supporters who are a big pillar of support at difficult times but due to their lack of technical abilities that needs better IQ, will turn out to be failure.

Hence, I want to reiterate that a person with an average IQ level and an equal amount of EQ level will be the best to survive in the present society. He can work anywhere happily with a fair wealth and fair emotional stability. He would be the most satisfied person.

Therefore a balanced IQ & EQ is the most desirable to be successful.

RAJ:
But I strongly feel that IQ is more important than EQ. Emotions are the biggest barrier to decision-making abilities. If we think about what others say and starts taking into consideration of others opinion, we get messed up and hence lack in making the right decision. Aren’t we living in a world where everything changes in an instant. So, we need someone who is confident of themselves and not get carried away by emotions. Hence I would prefer IQ over EQ,

Chandran:
I strongly opine that you consider the weight age of EQ more than IQ because if you . are perfect in EQ then you can manage with less IQ. For instance, if you have the qualities of EQ such as good personality skills, a better way of communication, and good attitude you can face any competitive situation. On the contrary, although you have strong IQ but poor EQ, then no one will take you seriously or you will be less focused leading to a few people joining you. So overall conclusion is that you don’t need to have a balanced or same level of IQ or EQ. If you are the best in EQ, then you can easily survive in this competitive era with less IQ.

LEELA:
There is no point in Mr. Raj arguing for IQ and Mr.Chandran taking sides with those who have better EQ, IQ and EQ have their own importance in our life and both should be in balanced proportion. I feel IQ makes us ambitious but EQ put limits on these ambitions. High IQ and Low EQ are the traits that can be observed in terrorists. So today, our topic is IQ vs EQ. I stay strong in both needed for a successful person. EQ is important for our social . life. AS we live in a society, we need good EQ levels for our personal and social behaviour and character to be well accepted in the society. Whereas, better IQ is surely the need for enhancing our knowledge to take the right decisions in this society and be a successful person. A balance between both is the need of the hour for both act as the two wheels in our journey of a successful life cycle.

(ii) To Live or Not

Kiruba:
We live in a world where most of us have a pet at home. When they fall sick, we rush them to the hospital. When our heart melts to see an animal or a bird suffer because of sickness, how will our conscience allow us to kill a fellow human being just because he is incurably ill? Moreover, he has done no harm to the society and his illness is not because of his fault.Therefore, we must provide him proper treatment and allow him to live as long as nature and God has willed his life span to be in this world.

Suganthi:
Exactly. Well said. God has gifted us life. So, he alone has the right to take it back. No human being has the right to interfere in His decisions or will for us in this world. Every man/woman has the right to live as long as God intends him/her to lead a life in this world.

Therefore, the illogical reasoning that a man suffering from an incurable disease, should be put to death is inexcusable beyond reason and a thought to be condemned.

Babu:
This world is governed by Darwin’s survival of the fittest principle. Darwin drew parallels between his own economic theories and his biological ones: “This survival of the fittest”, which he had sought to express in mechanical terms, is that which he has also called ‘natural selection’, or the preservation of favoured races in the struggle for life. An incurably diseased person is weak and has no value whatsoever to the society. More so, he has no means to live. Therefore, it would be in the fitness of things to kill him even against his wish. Therefore I bring to your kind notice that it is the natural selection to kill a person who cannot live befitting the needs of the society.

Kiruba:
Now a days, new vistas of progress have been opened in medical sciences and alternative ‘ medicine like Acupuncture, Acupressure, Reiki Pranik healing, Touch therapy, Herbal therapy, Diet therapy, etc. hold a ray of hope for the so called incurably diseased persons. So, why snatch life from them? When the kith and kin are ready to try all the means for survival, why should one interfere and speak things agaainst their survival. Miracles do happen and ‘ prayers are surely answered. Nothing wrong in giving a chance for someone fighting and struggling to live. We have no right to speak about putting another person to death.

Babu:
An incurably diseased person is the cause of constant worry to his family. His demands are unending and not withstanding the best possible attention, care and treatment given to him, he always remains dissatisfied and disgruntled. This adversely affects the peace of mind and comfort of the family members and all those who are trying to help them. Many a times, the physical worry and mental agony faced by the caretakers fall a prey to many other difficult situations and even fall sick being in an unhealthy environment for weeks and months. Therefore, the best way out of such a situation is to put an end to his life.

Suganthi:
It is not always the case that incurably diseased persons spread contagious diseases as some might argue. Even in those rare cases where it may be true, these persons are not real health hazards because it is medically established now that all incurable diseases are not contagious. However, as a precautionary measure, we should open separate hospitals or isolation wards for persons suffering from incurable contagious diseases and thus quarantine them. Isn’t that a better solution than thinking of putting an end to their life? To me, it’s a murder under legal permits.

Babu:
But these days we are burdened with the responsibility of reducing our rapidly increasing population. The many diseased persons constitute a good part of it. Even otherwise their contribution to society is a burden. It would be justified and reasonable not to allow them to drag on their agonizing life. Killing an incurably diseased person will put an end to research work in medical science. As I have already stated, there are many hospitals and many doctors who start experimenting with diseased persons. Even otherwise, suffering people have been the subject of research work for those who are into higher education. Thereby, its appropriate to free them from pain and suffering!

Kiruba:
Such an act only points out to the degenerated values in this society. It is almost compared to that of an individual throwing an appliance when it is old and doesn’t yield good results. Isn’t there a difference between living and non-living? How can a living human being be treated like a material object? What has happened to the values that students were taught or imbibed in their life? Such a thinking is only an assertion of degenerated moral values.

Conclusion:
India is a land of values. We are a secular society. We believe in respecting each other. Let us not be carried away by degenerated moral values and learn to respect one another and help them to live in peace without a feeling of guilt.

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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

Identifying the Degree and Leading Coefficient Calculator of Polynomials.

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

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 1 Number System Intext Questions

Exercise 1.1
Try These (Text book Page No. 1)

Question 1.
Represent the fraction \(\frac { 1 }{ 4 } \) in decimal form
Solution:
\(\frac { 1 }{ 4 } \) = \(\frac{1 \times 25}{4 \times 25}\) = \(\frac { 25 }{ 100 } \) = 0.25

Question 2.
What is the place value of 5 in 63.257.
Solution:
Place value of 5 in 63.257 is 5 hundredths (Hundreth place)

Question 3.
Identify the digit in the tenth place of 75.036.
Solution:
0

Question 4.
Express the decimal number 3.75 as a fraction.
Solution:
3.75 = \(\frac { 375 }{ 100 } \) = \(\frac { 15 }{ 4 } \)

Question 5.
Write the decimal number for the fraction 5 \(\frac { 1 }{ 5 } \)
Solution:
5 \(\frac { 1 }{ 5 } \) = \(\frac { 26 }{ 5 } \) = \(\frac{26 \times 2}{5 \times 2}\) = \(\frac { 52 }{ 10 } \) = 5.2

Question 6.
Identify the biggest number : 0.567 and 0.576.
Solution:
Comparing the digits of 0.567 and 0.576 from left to right, we have the tenths place same comparing the hundredths place 7 > 6.
⇒ 0.576 > 0.567

Question 7.
Compare 3.30 and 3.03 and identify the smaller number.
Solution:
The whole number is equal in both the numbers.
Now comparing the tenths place we have 3 > 0
⇒ 3.03 < 3.30 Smaller number is 3.03

Question 8.
Put the appropriate sign (<, >, =). 2.57 [ ] 2.570
Solution:
2.57 [=] 2.570

Question 9.
Arrange the following decimal numbers in ascending order.
5.14, 5.41, 1.54, 1.45, 4.15, 4.51.
Solution:
Comparing the numbers from left to right. Ascending order : 1.45, 1.54, 4.15, 4.51, 5.14, 5.41

Exercise 1.2
Try These (Text book Page No. 6)

Question 1.
Find the following using grid models:
(i) 0.83 + 0.04
(ii) 0.35 – 0.09
Solution:
(i) 0.83 + 0.04
0.83 = \(\frac { 83 }{ 100 } \) and 0.04 = \(\frac { 4 }{ 100 } \)

Shading the regions
0.83 and 0.04
The sum is the total shaded region.
S = 0.83 + 0.04 = 0.87

(ii) 0.35 – 0.09
0.35 = \(\frac { 35 }{ 100 } \) and 0.09 = \(\frac { 9 }{ 100 } \)

Shading the regions 0.35 by shading 35 boxes out of 100. Striking off 9 boxes out of 35 shaded boxes to subtract 0.09 from 0.35.
The left over shaded boxes represent the required value.
∴ 0.35 – 0.09 = 0.26

Try These (Text book Page No. 7)

Question 1.
Using the area models solve the following
(i) 1.2 + 3.5
(ii) 3.5 – 2.3
Solution:
(i) 1.2 + 3.5

Here 1.2 is represented in blue colour and 3.5 is represented in Green colour. Sum of 1.2 and 3.5 is 4.7.

(ii) 3.5 – 2.3

Representing 3.5 using 3 squares and 5 rectangular strips. Crossing out 2 squares from 3 squares and 3 rectangular strips from 5 to get the difference. So 3.5 – 2.3 = 1.2.

5/9 as a decimal is 0.56

Try These (Text book Page No. 9)

Question 1.
Complete the magic square in such a way that rows, columns and diagonals give the same sum 1.5.

Solution:

Exercise 1.3
Think (Text book Page No. 13)

Question 1.
How are the products 2.1 × 3.2 and 21 × 32 alike? How are they different.
Solution:
2.1 × 3.2 = 6.72 and 21 × 32 = 672.
In both the cases the digits ambers are the same. But the place value differs.

Try These (Text book Page No. 13)

Question 1.
Shade the grid to multiply 0.3 × 0.6.

Solution:

3 rows of Yellow represent 0.3, 6 columns of Red colour represent 0.6 Double shaded 18 squares of orange colour represent.
∴ 0.3 × 0.6 = 0.18

Question 2.
Use the area model to multiply

Solution:

Here each row contains 1 whole and 2 tenths. Each column contains 2 wholes and 5 tenths. The entire area model represents 2 wholes 9 tenths and 10 hundredths ( = 1 tenths). So 1.2 × 2.5 = 3.

Try These (Text book Page No. 14)

Question 1.
Complete the following table

Solution:

Try These (Text book Page No. 15)

Question 1.
Find:

1. 9.13 × 10
2. 9.13 × 100
3. 9.13 × 1000

Solution:

1. 9.13 × 10 = 91.3
2. 9.13 × 100 = 913
3. 9.13 × 1000 = 9130

Try These (Text book Page No. 16)

Question 1.
Complete the following table

Solution:

Exercise 1.4
Try These (Text book Page No. 19)

Question.

Solution:

Try These (Text book Page No. 19)

Question 1.
Divide the following
(i) 17.237 ÷ 10
(ii) 17.237 ÷ 100
(iii) 17.237 ÷ 1000
Solution:
(i) 17.237 ÷ 10
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 17237 }{ 1000 } \)
= 1.7237

(ii) 17.237 ÷ 100
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 17237 }{ 100000 } \)
= 0.17237

(iii) 17.237 ÷ 1000
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 17237 }{ 1000000 } \)
= 0.017237

Try These (Text book Page No. 21)

Question 1.
Find the value of the following:
(i) 46.2 ÷ 3 = ?
(ii) 71.6 ÷ 4 = ?
(iii) 23.24 ÷ 2 = ?
(iv) 127.35 ÷ 9 = ?
(v) 47.201 ÷ 7 = ?
Solution:
(i) 46.2 ÷ 3
= \(\frac { 462 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 1 }{ 10 } \) × \(\frac { 462 }{ 3 } \)
= \(\frac { 1 }{ 10 } \) × 15.4
= \(\frac { 154 }{ 10 } \)
= 15.4

(ii) 71.6 ÷ 4
= \(\frac { 716 }{ 10 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 1 }{ 10 } \) × \(\frac { 716 }{ 4 } \)
= \(\frac { 1 }{ 10 } \) × 179
= 17.9

(iii) 23.24 ÷ 2
= \(\frac { 2324 }{ 100 } \) × \(\frac { 1 }{ 2 } \)
= \(\frac { 2324 }{ 2 } \) × \(\frac { 1 }{ 100 } \)
= 1162 × \(\frac { 1 }{ 100 } \)
= \(\frac { 1162 }{ 100 } \)
= 11.62

(iv) 127.35 ÷ 9
= \(\frac { 12735 }{ 100 } \) × \(\frac { 1 }{ 9 } \)
= \(\frac { 12735 }{ 9 } \) × \(\frac { 1 }{ 100 } \)
= 1415 × \(\frac { 1 }{ 100 } \)
= \(\frac { 1415 }{ 100 } \)
= 14.15

(v) 47.201 ÷ 7
= \(\frac { 47201 }{ 1000 } \) × \(\frac { 1 }{ 7 } \)
= \(\frac { 47201 }{ 7 } \) × \(\frac { 1 }{ 1000 } \)
= 6743 × \(\frac { 1 }{ 1000 } \)
= \(\frac { 6743 }{ 1000 } \)
= 6.743

Try These (Text book Page No. 22)

Question 1.
Divide the following
(i) \(\frac { 9.25 }{ 0.25 } \)
(ii) \(\frac { 8.6 }{ 4.3 } \)
(iii) \(\frac { 44.1 }{ 0.21 } \)
(iv) \(\frac { 9.6 }{ 1.2 } \)
Solution:
(i) \(\frac { 9.25 }{ 0.25 } \)

(ii) \(\frac { 8.6 }{ 4.3 } \)

(iii) \(\frac { 44.1 }{ 0.21 } \)

(iv) \(\frac { 9.6 }{ 1.2 } \)

Think (Text book Page No. 22)

Question 1.
The price of a tablet strip containing 30 tablets is 22.63 Then how will you find the price of each tablet?
Solution:
Price of 30 tablets = ₹ 22.63 = ₹ \(\frac { 2263 }{ 100 } \)
∴ Price of 1 tablet

= \(\frac { 2263 }{ 100 } \) × \(\frac { 1 }{ 30 } \)
= \(\frac { 2263 }{ 30 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 2263 }{ 3 } \) × \(\frac { 1 }{ 1000 } \)
= 754.33 × \(\frac { 1 }{ 1000 } \)
= \(\frac { 754.33 }{ 1000 } \)
= 0.75433
Price of each tablet is ₹ 0.7543

Students can Download Maths Chapter 1 Number System Ex 1.1 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 3 Chapter 1 Number System Ex 1.1

Question 1.
Round each of the following decimals to the nearest whole number.
(i) 8.71
(ii) 26.01
(iii) 69.48
(iv) 103.72
(v) 49.84
(vi) 101.35
(vii) 39.814
(viii) 1.23
Solution.
(i) 8.71
Underlining the digit to be rounded 8.71. Since the digit next to the underlined digit, 7 which is greater than 5, adding 1 to the underlined digit.
Hence the nearest whole number 8.71 rounds to is 9.

(ii) 26.01
Underlining the digit to be rounded 26.01. Since the digit next to the underlined digit, 0 which is less than 5, the underlined digit 6 remains the same.
∴ The nearest whole number 26.01 rounds to is 26.

(iii) 69.48
Underlining the digit to be rounded 69.48. Since the digit next to the underlined digit, 4 which is less than 5, the underlined digit 9 remains the same.
∴ The whole number is 69.48 rounds to is 69.

(iv) 103.72
Underlining the digit to be rounded 103.72 since the digit next to the underlined digit, 7 which is greater than 5, we add 1 to the under lined digit.
Hence the nearest whole number 103.72 rounds to is 104.

(v) 49.84
Underlining the digit to be rounded 49.84. Since the digit next to the underlined digit 8 which is greater than 5, we add 1 to the underlined digit.
Hence the nearest whole number 49.84 rounds to 50.

(vi) 101.35
Underlining the digit to be rounded 101.35. Since the digit next to the underlined digit 3 is less than 5, the underlined digit 1 remains the same.
Hence the nearest whole number 101.35 rounds to is 101.

(vii) 39.814
Underlining the digit to be rounded 39.814. Since the digit next to the underlined digit 8 is greater than 5, we add 1 to the underlined digit.
Hence the nearest whole number 39.814 rounds to is 40.

(viii) 1.23
Underlining the digit to be rounded 1.23. Since the digit next to the underlined digit 2, is less than 5, the underlined digit 1 remains the same.
Hence the nearest whole number 1.23 rounds to is 1.

1/8 as a decimal is 0.125.

Question 2.
Round each decimal number to the given place value.
(i) 5.992; tenths place
(ii) 21.805; hundredth place
(iii) 35.0014; thousandth place
Solution:
(i) 992; tenths place
Underlining the digit to be rounded 5.992. Since the digit next to the underlined digit is 9 greater than 5, we add 1 to the underlined digit.
Hence the rounded number is 6.0.

(ii) 21.805; hundredth place
Underlining the digit to be rounded 21.805 since the digit next to the underlined digit is 5, we add 1 to the underlined digit.
Hence the rounded number is 21.81.

(iii) 35.0014; thousandth place
Underlining the digit to be rounded 35.0014. Since the digit next to the underlined digit is 4 less than 5 the underlined digit remains the same.
Hence the rounded number is 35.001.

One Decimal is equal to 435.56 Square Feet.

Question 3.
Round the following decimal numbers upto 1 places of decimal.
(i) 123.37
(ii) 19.99
(iii) 910.546
Solution:
(i) 123.37
Rounding 123.37 upto one places of decimal means round to the nearest tenths place. Underling the digit in the tenths place of 123.37 gives 123.37. Since the digit next to the tenth place value is 7 which is greater than 5, we add 1 to the underlined digit to get 123.4. Hence the rounded value of 123.37 upto one places of decimal is 123.4.

(ii) 19.99
Rounding 19.99 upto one places of decimal means round to the nearest tenth place. Underling the digit in the tenths place of 19.99 gives 19.99. Since the digit next to the tenth place value is 9 which is greater than 5, we add 1 to the underlined digit to get 20.
Hence the rounded value of 19.99 upto one places of decimal is 20.0.

(iii) 910.546
Rounding 910.546 upto one places of decimal means round to the nearest tenths place underlining the digit in the tenths place of 910.546 gives 910.546. Since the digit next to the tenth place value is 4, which is less than 5 the underlined digit remains the same. Hence the rounded value of 910.546 upto one places of decimal is 910.5.

Rounding to the nearest hundredth is 838.27.

Question 4.
Round the following decimal numbers upto 2 places of decimal.
(i) 87.755
(ii) 301.513
(iii) 79.997
Solution:
(i) 87.755
Rounding 87.755 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 87.755 gives 87.755. Since the digit next to the hundredth place value is 5, we add 1 to the underlined digit.

Hence the rounded value of 87.755 upto two places of decimal is 87.76.

(ii) 301.513
Rounding 301.51 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 301.513 gives 301.513. Since the digit next to the underlined digit 3 is less than 5, the underlined digit remains the same.

∴ The rounded value of 301.513 upto 2 places of decimal is 301.51.

(iii) 79.997
Rounding 79.997 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 79.997 gives 79.997. Since the digit next to the underlined digit 7 is greater than 5, we add 1 to the underlined number.

Hence the rounded value of 79.997 upto 2 places of decimal is 80.00.

3/5 as a decimal is 0.6

Question 5.
Round the following decimal numbers upto 3 place of decimal
(a) 24.4003
(b) 1251.2345
(c) 61.00203
Solution:
(a) 24.4003
Rounding 24.4003 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 24.4003 gives 24.4003. In 24.4003 the digit next to the thousandths value is 3 which is less than 5.

∴ The underlined digit remains the same. So the rounded value of24.4003 upto 3 places of decimal is 24.400.

(b) 1251.2345
Rounding 1251.2345 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 1251.2345 gives 1251.2345, the digit next to the thousandths place value is 5 and so we add 1 to the underlined digit. So the rounded value of 1251.2345 upto 3 places of decimal is 1251.235.

(c) 61.00203
Rounding 61.00203 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandth place of 61.00203 gives 61.00203. In 61.00203, the digit next to the thousandths place value is 0, which is less than 5.

Hence the underlined digit remains the same. So the rounded value of 61.00203 upto 3 places of decimal is 61.002.

When we write a decimal number with three places, we are representing the thousandths place.

Students can Download Commerce Chapter 11 Employee Selection Process Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Commerce 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 12th Commerce Solutions Chapter 11 Employee Selection Process

Samacheer Kalvi 12th Commerce Employee Selection Process Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
The recruitment and Selection Process aimed at right kind of people.
(a) At right people
(b) At right time
(c) To do right things
(d) All of the above
Answer:
(d) All of the above

Convert CGPA into Percentage out of 4 … You must divide the percentage by 25 to get the percentage

Question 2.
The poor quality of selection will mean extra cost on _________ and supervision.
(a) Training
(b) Recruitment
(c) work quality
(d) None of these
Answer:
(a) Training

Question 3.
_________ refers to the process of identifying and attracting job seekers so as to build a pool of qualified job applicants.
(a) Selection
(b) Training
(c) Recruitment
(d) Induction
Answer:
(c) Recruitment

Question 4.
Selection is usually considered as a _________ process.
(a) Positive
(b) Negative
(c) Natural
(d) None of these
Answer:
(b) Negative

Question 5.
Which of the following test is used to measure the various characteristics of the candidate?
(a) physical Test
(b) Psychological Test
(c) attitude Test
(d) Proficiency tests
Answer:
(b) Psychological Test

Question 6.
Wfifich of the following orders is followed in a typical selection process? .
(a) application form test and or interview, reference check and physical examination
(b) Application form test and or interview, reference check, and physical examination
(c) Reference check, application form, test and interview and physical examination
(d) physical examination test and on interview application term and reference check.
Answer:
(b) Application form test and or interview, reference check, and physical examination

Question 7.
The purpose of an application blank is to gather information about the
(a) Company
(b) Candidate
(c) Questionnaire or Interview Schedule
(d) Competitors
Answer:
(b) Candidate

Question 8.
Identify the test that acts as an instrument to discover the inherent ability of a candidate.
(a) Aptitude Test
(b) Attitude Test
(c) Proficiency Test
(d) Physical Test
Answer:
(a) Aptitude Test

Question 9.
The process of eliminating unsuitable candidate is called _________
(a) Selection
(b) Recruitment
(c) Interview
(d) Induction
Answer:
(a) Selection

Question 10.
Scrutiny of application process is the _________
(a) Last step in Selection process
(b) First step in Selection process
(c) Third step in Selection Process Selection process
(d) None of the above
Answer:
(b) First step in Selection process

Question 11.
Scrutiny of application process is the
(a) Locating candidates
(b) Determinining the suitable of the candidates
(c) preparing employees for training
(d) None of the above
Answer:
(b) Determinining the suitable of the candidates

Question 12.
The process of placing the right man on the right job is called _________
(a) Training
(b) Placement
(c) Promotion
(d) Transfer
Answer:
(b) Placement

Question 13.
Probation/Trial period signifies _________
(a) one year to two years
(b) One year to three years
(c) Two years to four years
(d) None of the above
Answer:
(a) one year to two years

Question 14.
Job first man next is one of the principles of _________
(a) Test
(b) Interview
(c) Training
(d) placement
Answer:
(d) placement

II. Very Short Answer Questions

Question 1.
What is selection?
Answer:
Selection is the process of choosing the most suitable person for the vacant position in the organization.

Question 2.
What is an interview?
Answer:
According to Scott and others “an interview is a purpose full exchange of ideas, the answering of questions and communication between two or more persons.”

Question 3.
What is intelligence test?
Answer:
Intelligence tests are one of the psychological tests, that is designed to measure a variety of mental ability, individual capacity of a candidate.

Question 4.
What do you mean by test?
Answer:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualification to fit into various positions in the organization.

Question 5.
What do you understand about bio data?
Answer:
Most of the Public sector undertakings Recruitment boards supply applications forms for selection of jobs. In this form, the candidate fill the details about family background, educational qualifications, experience, co-curricular activities.

Question 6.
What do you mean by placement?
Answer:
Placement is a process of assigning a specific job to each and every candidate selected. It includes initial assignment of new employees and promotion, transfer or demotion of present employees.

III. Short Answer Questions

Question 1.
What is stress interview?
This type of interview is conducted to test the temperament and emotional balance of the candidate interviewed. Interviewer deliberately creates stressful situation and tries to assess the suitability of the candidate by observing his reaction and response to the stressful situations.

Question 2.
What is structured interview?
Answer:
Under this method, a series of questions to be asked by the interviewer are pre-prepared by the interviewer and only these questions are asked in the interview.

Question 3.
Name the types of selection test?
Answer:
Selection tests are of two types: Ability Tests and Personality Tests. Ability tests can further be divided into: aptitude test, achievement test, intelligence test, and judgement test. Personality tests can further be divided into: interest test, personality inventory test, projective test or thematic appreciation test, and attitude test.

Question 4.
What do you mean by achievement test?
Answer:
This test measures a candidate’s capacity to achieve in a particular field. In other words this test measures a candidate’s level of skill in certain areas, accomplishment and knowledge in a particular subject. It is also called proficiency test.

Question 5.
Why do you think the medical examinations of a candidate is necessary?
Answer:
The last technique used in selection process is medical examination. This is the most important step in the selection because a person of poor health cannot work competently and any investment on him may go waste, if he/she is unable to discharge duties efficiently on medical grounds.

Question 6.
What is aptitude test?
Answer:
Aptitude test is a test to measure suitability of the candidates for the post/role. It actually measures whether the candidate possess a set of skills required to perform a given job. It helps in predicting the ability and future performance of the candidate.

Question 7.
How is panel interview conducted?
Answer:
Where a group of people interview the candidate, it is called panel interview. Usually panel comprises chair person, subject expert, psychological experts, representatives of minorities/ underprivileged groups, nominees of higher bodies and so on. All panel members ask different types of questions on general areas of specialization of the candidate.

Question 8.
List out the various selection interviews.
Answer:
Interview represents a face to face interaction between the interviewer and interviewee

1. Preliminary Interview
2. Structured Interview
3. Unstructured Interview
4. In-depth Interview
5. Panel Interview
6. Stress Interview
7. Telephone Interview
8. Online Interview
9. Group interview
10. Video Conference Interview

Question 9.
List out the significance of placement.
Answer:
The significance of the placement is as follows:

1. It improves employee morale.
2. It helps in reducing employee turnover.
3. It helps in reducing conflict rates or accidents.
4. It avoids misfit between the candidates and the job.
5. It helps the candidate to work as per the predetermined objectives of the organization.

IV. Long Answer Questions

Question 1.
Briefly explain the various types of tests.
Answer:
Several tests are conducted in the selection process to ensure whether the candidate possesses the necessary qualifications.

A. Ability Test: Ability test may be divided into:

• Aptitude Test: Aptitude test is a test to measure suitability of the candidates for the post.
• Achievement Test: This test measures a candidate’s capacity to achieve in a particular field.
• Intelligence Test: Intelligence test is designed to measure a variety of mental ability, individual capacity of a candidate.
• Judgment Test: This test is conducted to test the presence of mind and reasoning capacity of the candidates.

B. Personality Test: It refers to the test conducted to find out the non-intellectual traits of a candidate. It can be further divided into:

• Interest Test: Interest test measures a candidate’s extent of interest in a particular area.
• Projective Test: This test measures the candidate’s values, personality of the candidate.
• Attitude Test: measures candidate’s tendencies towards the people, situation, action and related things.

Question 2.
Explain the important methods of interview.
Interview means a face to face interaction between the interviewer and interviewee. Interview may be of various types:-

1. Preliminary Interview: It is conducted to know the general suitability of the candidates who have applied for the job.
2. Structured Interview: In this method, a series of questions is to be asked by the interviewer. The questions may be pre-prepared.
3. In depth Interview: This interview is conducted to test the level of knowledge of the interviewee in a particular field.
4. Panel Interview: Where a group of people interview the candidate. The panel usually comprises the chair person, subject expert, psychological experts and so on.
5. Stress Interview: This type of interview is conducted to test the temperament and emotional balance of the candidate.
6. Online Interview: Due to tremendous growth in information and communication technology, interviews are conducted by means of internet via Skype, Google duo, Whatsapp.

Question 3.
Explain the principles of placement.
Answer:
The following are the principles of placement:

1. Job First, Man Next: Man should be placed on the job according to the requirements of the job.
2. Job Offer: The job should be offered to the man based on his qualification..
3. Terms and conditions: The employee should be informed about the terms and conditiqns of the organisation.
4. Aware about the Penalties: The employee should also be made aware of the penalties if he / she commits a mistake.
5. Loyalty and Co-operation: When placing a person in a new job, an effort should be made to develop a sense of loyalty and co-operation in him.

Samacheer Kalvi 12th Commerce Employee Selection Process Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
The types of Aptitude Test are
(i) Numerical Reasoning Test
(ii) Attitude Test
(iii) Vocabulary Test
(iv) Interest Test
(a) (i) and (ii)
(b) (i) and (iii)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(b) (i) and (iii)

Question 2.
Which one of the following is not correctly matched?

(a)	In-depth interview	Skype
(b)	Online interview	Group of people interview the candidate
(c)	Video conferencing interview	Level of knowledge
(d)	Telephone interview	Face to face interview

Answer:
(d) Telephone interview – Face to face interview

II. Very Short Answer Questions

Question 1.
What is meant by personality test?
Answer:
Personality test refers to the test conducted to find out the non-intellectual traits of a candidate namely temperament, emotional response, capability and stability.

Question 2.
Write a note on attitude test.
Answer:
Attitude test measures candidate’s tendencies towards the people, situation, action and related things. For example: morale study, values study, etc.

III. Short Answer Questions

Question 1.
What do you mean by online interview?
Answer:
Due to tremendous growth in information and communication technology, these days interviews are conducted by means of internet via Skype. We chat, Google duo, Viber, Whatsapp or Video chat applications. This enables the interviewers to conduct interview with the candidates living in faraway places.

Students can Download Maths Chapter 1 Number System Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.2

Question 1.
Fill in the following place value table.

Answer:

1/2 as a decimal is 0.5.

Question 2.
Write the decimal numbers from the following place value table.

Answer:

Question 3.
Write the following decimal numbers in the place value table.
(i) 25.178
(ii) 0.025
(iii) 428.001
(iv) 173.178
(v) 19.54
Solution:
(i) 25.178

(ii) 0.025

(iii) 428.001

(iv) 173.178

(v) 19.54

Question 4.
Write each of the following as decimal numbers.
(i) 20 + 1 + \(\frac { 2 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 7 }{ 1000 } \)
(ii) 3 + \(\frac { 8 }{ 10 } \) + \(\frac { 4 }{ 100 } \) + \(\frac { 5 }{ 1000 } \)
(iii) 6 + \(\frac { 0 }{ 10 } \) + \(\frac { 0 }{ 100 } \) + \(\frac { 9 }{ 1000 } \)
(iv) 900 + 50 + 6 + \(\frac { 3 }{ 100 } \)
(v) \(\frac { 6 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 1 }{ 1000 } \)
Solution:
(i) 20 + 1 + \(\frac { 2 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 7 }{ 1000 } \) = 21 + 2 × \(\frac { 1 }{ 10 } \) + 3 × \(\frac { 1 }{ 100 } \) + 7 × \(\frac { 1 }{ 1000 } \) = 21.237

(ii) 3 + \(\frac { 8 }{ 10 } \) + \(\frac { 4 }{ 100 } \) + \(\frac { 5 }{ 1000 } \) = 3 + 8 × \(\frac { 1 }{ 10 } \) + 4 × \(\frac { 1 }{ 100 } \) + 5 × \(\frac { 1 }{ 1000 } \) = 3.845

(iii) 6 + \(\frac { 0 }{ 10 } \) + \(\frac { 0 }{ 100 } \) + \(\frac { 9 }{ 1000 } \) = 6 + 0 × \(\frac { 1 }{ 10 } \) + 0 × \(\frac { 1 }{ 100 } \) + 9 × \(\frac { 1 }{ 1000 } \) = 6.009

(iv) 900 + 50 + 6 + \(\frac { 3 }{ 100 } \) = 956 + 0 × \(\frac { 1 }{ 10 } \) + 3 × \(\frac { 1 }{ 100 } \) = 956.03

(v) \(\frac { 6 }{ 10 } \) + \(\frac { 3 }{ 100 } \) + \(\frac { 1 }{ 1000 } \) = 6 × \(\frac { 1 }{ 10 } \) + 3 × \(\frac { 1 }{ 100 } \) = 0.631

Question 5.
Convert the following fractions into decimal numbers.
(i) \(\frac { 3 }{ 10 } \)
(ii) 3 \(\frac { 1 }{ 2 } \)
(iii) 3 \(\frac { 3 }{ 5 } \)
(iv) \(\frac { 3 }{ 2 } \)
(v) \(\frac { 4 }{ 5 } \)
(vi) \(\frac { 99 }{ 100 } \)
(vii) 3 \(\frac { 19 }{ 25 } \)
Solution:
(i) \(\frac { 3 }{ 10 } \) = 0.3
(ii) 3 \(\frac { 1 }{ 2 } \) = \(\frac { 7 }{ 2 } \) = \(\frac{7 \times 5}{2 \times 5}\) = \(\frac { 35 }{ 10 } \) = 3.5
(iii) 3 \(\frac { 3 }{ 5 } \) = \(\frac { 18 }{ 5 } \) = \(\frac{18 \times 2}{5 \times 2}\) = \(\frac { 36 }{ 10 } \) = 3.6
(iv) \(\frac { 3 }{ 2 } \) = \(\frac{3 \times 5}{2 \times 5}\) = \(\frac { 15 }{ 10 } \) = 1.5
(v) \(\frac { 4 }{ 5 } \) = \(\frac{4 \times 2}{5 \times 2}\) = \(\frac { 8 }{ 10 } \) = 0.8
(vi) \(\frac { 99 }{ 100 } \) = 0.99
(vii) 3 \(\frac { 19 }{ 25 } \) = \(\frac { 94 }{ 25 } \) = \(\frac{94 \times 4}{25 \times 4}\) = \(\frac { 376 }{ 100 } \) = 3.76

Question 6.
Write the following decimals as fractions.
(i) 2.5
(ii) 6.4
(iii) 0.75
Solution:
(i) 2.5 = 2 + \(\frac { 5 }{ 10 } \) = \(\frac { 25 }{ 10 } \)
(ii) 6.4 = 6 + \(\frac { 4 }{ 10 } \) = \(\frac { 64 }{ 10 } \)
(iii) 0.75 = 0 + \(\frac { 7 }{ 10 } \) + \(\frac { 5 }{ 100 } \) = \(\frac { 70+5 }{ 100 } \) = \(\frac { 75 }{ 100 } \)

Question 7.
Express the following decimals as fractions in lowest form.
(i) 2.34
(ii) 0.18
(iii) 3.56
Solution:
(i) 2.34 = 2 + \(\frac { 34 }{ 100 } \) = 2 + \(\frac{34 \div 2}{100 \div 2}\) = 2 + \(\frac { 17 }{ 50 } \) = 2\(\frac { 17 }{ 50 } \) = \(\frac { 117 }{ 50 } \)
(ii) 0.18 = 0 + \(\frac { 18 }{ 100 } \) = \(\frac{18 \div 2}{100 \div 2}\) = \(\frac { 9 }{ 50 } \)
(iii) 3.56 = 3 + \(\frac { 56 }{ 100 } \) = 3 + \(\frac{56 \div 4}{100 \div 4}\) = 3 + \(\frac { 14 }{ 25 } \) = 3 \(\frac { 14 }{ 25 } \) = \(\frac { 89 }{ 25 } \)

Objective Questions

Question 8.
3 + \(\frac { 4 }{ 100 } \) + \(\frac { 9 }{ 1000 } \) = ?
(i) 30.49
(ii) 3049 9
(iii) 3.0049
(iv) 3.049
Answer:
(iv) 3.049
Hint: = 3 × 1 + \(\frac { 0 }{ 10 } \) + \(\frac { 4 }{ 100 } \) + \(\frac { 9 }{ 1000 } \) = 3.049

Question 9.
\(\frac { 3 }{ 5 } \) = _______
(i) 0.06
(ii) 0.006
(iii) 6
(iv) 0.6
Answer:
(iv) 0.6
Hint: \(\frac { 3 }{ 5 } \) = \(\frac{3 \times 2}{5 \times 2}\) = \(\frac { 6 }{ 10 } \) = 0.06

Question 10.
The simplest form of 0.35 is
(i) \(\frac { 35 }{ 1000 } \)
(ii) \(\frac { 35 }{ 10 } \)
(iii) \(\frac { 7 }{ 20 } \)
(iv) \(\frac { 7 }{ 100 } \)
Answer:
(iii) \(\frac { 7 }{ 20 } \)
Hint: 0.35 = \(\frac { 35 }{ 100 } \) = \(\frac{35 \div 5}{100 \div 5}\) = \(\frac { 7 }{ 20 } \)

Students can Download Maths Chapter 1 Life Mathematics Ex 1.1 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 1 Chapter 1 Life Mathematics Ex 1.1

Question 1.
Fill in the blanks:

Question (i)
If 30% of x is 150, then x is ………….
Answer:
500
Hint:
Given 30% of x is 150

Question (ii)
2 minutes is ………… % to an hour.
Answer:
3\(\frac{1}{3}\)%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60 min
x% = \(\frac{2}{60}\) x 100 = \(\frac{200}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)
x = 3\(\frac{1}{3}\)

Question (iii)
If x % of x = 25, then x = …………
Answer:
50
Hint:
Given that x % of x is 25
∴ \(\frac{x}{100}\) = 25
x² = 25 x 100 = 2500
∴ x = \(\sqrt{2500}\) = 50

Question (iv)
In a school of 1400 students, there are 420 girls. The percentage of boys in the school is ………….
Answer:
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
Number of boys in school = 1400 – 420 = 980

% of boys = 70%

Question (v)
0.5252 is ………….. %.
Answer:
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 x 100 = 52.52%

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 2.
Rewrite each underlined portion using percentage language.

Question (i)
One half of the cake is distributed to the children.
Answer:
Answer:
50% of the cake is distributed to the children
Hint:
One half is nothing but \(\frac{1}{2}\) as percentage, we need to multiply by 100
\(\frac{1}{2}\) x 100 = 50%

Question (ii)
Aparna scored 7.5 points out of 10 in a competition.
Answer:
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is \(\frac{7.5}{10}\) = 0.75
For percentage, we need to multiply by 100
∴ We get 0.75 x 100 = 75%

Question (iii)
The statue was made of pure silver.
Answer:
96% students participated in sports
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = \(\frac{100}{100}\)
∴ to express as percentage, \(\frac{100}{100}\) x 100% = 100%

Question (iv)
48 out of 50 students participated in sports
Answer:
96% students participated in sports
Hint:
48 out of 50 students in fraction form is \(\frac{48}{50}\).
As a percentage, we need to multiply by 100

Question (v)
Only 2 persons out of 3 will be selected in the interview
Answer:
Only 66\(\frac{2}{3}\)% persons will be selected in the interview.
Hint:
2 out of 3 in fraction form is \(\frac{2}{3}\) to express as percentage, we need to multiply by 100
\(\frac{2}{3}\) x 100 = \(\frac{200}{3}\) = 66\(\frac{2}{3}\)%

Online percentage off calculator tool makes the calculation faster, and it displays the percentage off of the product in a fraction of seconds.

Question 3.
48 is 32% of what number?
Solution:
Let the number required to be found be ‘x’
Given that 32% of x is 48

Question 4.
A bank pays ₹ 240 as interest for 2 years for a sum of ? 3000 deposited as savings. Find the rate of interest given by the bank.
Solution:
The formula for simple interest is given by
Interest (I) = \(\frac{PxRxT}{100}\) 100
Where P is principal amount; R is rate of interest in percentage; t is time period Substituting given values in formula, we get
P = 3000, t = 2 yrs, I = Rs 240

Question 5.
A Welfare Association has a sports club where 30% of the members play cricket, 28% play volleyball, 22% play badminton and the rest play indoor games. If 30 member play indoor games.

1. How many members are there in the sports club?
2. How many play cricket, volleyball and badminton?

Solution:
Given data
Members who play Cricket = 30% …..(1)
Members who play Vollyball = 28% …..(2)
Members who play Batminton = 22% ……(3)
Members who play outdoor games = (1) + (2) + (3) = 30
Members who play indoor games = 100 – 80 = 20%
Also given 30 members play indoor games
Let total no of members be ‘x’
∴ \(\frac{30}{x}\) = 20
∴ x = \(\frac{30×100}{20}\) = 15

1. Total number of members is 150

2. Members who play Cricket = 30% of total = \(\frac{30}{100}\) x 150

Question 6.
What is 25% of 30% of 400?
Solution:
Required to find 25% of 30% of 400

Question 7.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of the number.
Solution:
Given that 75% of number less 60% of number is 82.5. Let the number be ‘x’
∴ \(\frac{75}{100}\) × x = 82.5
∴ 0.75 – 0.6.60x = 82.5
∴ 0.15x = 82.5
∴ x = \(\frac{82.5}{0.15}\) = \(\frac{8250}{15}\) = 550
Required to find 20% of number ie 20% of x.

Question 8.
If a car is sold for ₹ 2,00,000 from its original price of? 3,00,000 find the percentage for decrease in the value of the car.

Solution:
Original price of car = ₹ 3,00,000
actual selling price of car = ₹ 2,00,000
Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000

Question 9.
A number when increased by 18% gives 236. Find the number.
Solution:
Let the number be x. Given that when it is increased by 18%, we get 236.
x + \(\frac{18}{100}\) = 236
\(\frac{100x+18x}{100}\) = 236
\(\frac{118}{100}\)x = 236
The number = x = \(\frac{236×100}{118}\) = 200

Question 10.
A number when decreased by 20% gives 80. Find the number.
Solution:
Let the number be x. Given that when it is increased by 20% we get 80.

Question 11.
A number is increased by 25% and then decreased by 20%. Find the change in that number.
Solution:
Method 1:
Let the number be x. First it is increased by 25%

Method 2:
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we get

We get back 100 ⇒ No change

Question 12.
If the numerator of a fraction is increased by 25% and the denominator is increased by 10%, it becomes \(\frac{2}{5}\). Find the original fraction.
Solution:
Let the fraction be \(\frac{N}{D}\) where N is numerator & D is denominator
It is given that numerator is include by 25%

Question 13.
A fruit vendor bought some mangoes of which 10% were rotten. He sold 33 \(\frac{1}{3}\)% of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.

Solution:
Let the number of mangoes bought by fruit seller initially be x.
Given that 10% of mangoes were rotten
∴ Number of rotten mangoes = \(\frac{10}{100}\) × x
Number of good mangoes = x – no. of rotten mangoes
= x – \(\frac{10}{100}\)x = \(\frac{100x-10x}{100}\) = \(\frac{90}{100}\)x …(1)
Number of mangoes sold = 33\(\frac{1}{3}\)% of good mangoes = \(\frac{100}{3}\)%
∴ Mangoes sold = \(\frac{100}{3}\) x \(\frac{90}{100}\)x × \(\frac{1}{100}\) = \(\frac{30}{100}\)x
Number of mangoes remaining = No. of good mangoes – No. of mangoes sold
From (1) and (2)

Question 14.
A student gets 31% marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.
Solution:
Let the maximum marks in the exam be ‘x’. Pass percentage is given as 35%
∴ Pass mark = \(\frac{35}{100}\) × x = \(\frac{35}{100}\)x
Student gets 31 % marks = \(\frac{31}{100}\) × x = \(\frac{31}{100}\)x
But student fails by 12 marks → meaning his mark is 12 less than pass mark.
∴ \(\frac{31}{100}\)x = \(\frac{35}{100}\)x – 12
∴ \(\frac{35}{100}\)x – \(\frac{31}{100}\)x = 12
∴ \(\frac{35x-31x}{100}\) = 12 ⇒ \(\frac{4x}{100}\) = 12
∴ \(\frac{12×100}{4}\) = 300

Question 15.
The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.
Solution:
Let number of boys be ‘6’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ \(\frac{b}{g}\) = \(\frac{5}{3}\) …..(A)
Failure in boys = 16% = \(\frac{16}{100}\) x b = \(\frac{16b}{100}\)
Failure in girls = 8% = \(\frac{8}{100}\) x g = \(\frac{8g}{100}\)
Pass in boys = 100 – 16% = 84% \(\frac{84}{100}\)b …..(1)
Pass in girls = 100 – 8% = 92% = \(\frac{92}{100}\)g
From A, we have \(\frac{b}{g}\) = \(\frac{5}{3}\), = adding 1 on both sides, we get
\(\frac{b}{g}\) + 1 = \(\frac{5}{3}\) + 1
\(\frac{b+g}{g}\) = \(\frac{5+3}{3}\) = \(\frac{8}{3}\)
∴ g = \(\frac{3}{8}\)(b + g)
Similarly b = \(\frac{5}{8}\)(b + g)
Total pass = Pass in girls + Pass in boys
= (1) + (2) = \(\frac{84}{100}\)b + \(\frac{92}{100}\)g

Objective Type Questions

Question 16.
12% of 250 litres is the same as ………. of 150 liters
(a) 10%
(b) 15%
(c) 20%
(d) 30%
Answer:
(c) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) x 250 = 30 lit
Percentage : \(\frac{30}{150}\) x 100 = 20%

Question 17.
If three candidates A, B and C is 3 in a school election got 153, 245 and 102 votes respectively, the percentage of votes for the winner is ……..
(a) 48%
(b) 49%
(c) 50%
(d) 45%
Answer:
(b) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes]
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500

Question 18.
15% of 25% of 10000 = ………..
(a) 375
(b) 400
(c) 425
(d) 475
Answer:
(a) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is

Next 15% of the above is \(\frac{15}{100}\) x 2500 = 375

Question 19.
When 60 is subtracted from 60% of a number to give 60, the number is –
(a) 60
(b) 100
(c) 150
(d) 200
Answer:
(d) 200
Hint:
Let the number be ‘x’.
60% of the number is \(\frac{60}{100}\) × x = \(\frac{60x}{100}\)
Given that when 60 is subtracted from 60%, we get 60
i.e \(\frac{60}{100}\)x – 60 = 60
∴ \(\frac{60}{100}\)x = 60 + 60 = 120
∴ x = \(\frac{120×100}{60}\) = 200

Question 20.
If 48% of 48 = 64% of x, then x =
(a) 64
(b) 56
(c) 42
(d) 36
Answer:
(d) 36
Hint:
Given that 48% of 48 48 = 64% of x

Students can Download Maths Chapter 1 Number System Ex 1.5 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 1 Number System Ex 1.5

5/8 as a decimal is 0.625.

Question 1.
Write the following decimal numbers in the place value table.
(i) 247.36
(ii) 132.105
Solution:
(i) 247.36

(ii) 132.105

11/16 as a decimal is 0.6875.

Question 2.
Write each of the following as decimal number.
(i) 300 + 5 + \(\frac { 7 }{ 10 } \) + \(\frac { 9 }{ 100 } \) + \(\frac { 2 }{ 100 } \)
(ii) 1000 + 400 + 30 + 2 + \(\frac { 6 }{ 10 } \) + \(\frac { 7 }{ 100 } \)
Solution:
(i) 300 + 5 + \(\frac { 7 }{ 10 } \) + \(\frac { 9 }{ 100 } \) + \(\frac { 2 }{ 100 } \) = 305.792
(ii) 1000 + 400 + 30 + 2 + \(\frac { 6 }{ 10 } \) + \(\frac { 7 }{ 100 } \) = 1432.67

Question 3.
Which is greater?
(i) 0.888 (or) 0.28
(ii) 23.914 (or) 23.915
Solution:
(i) 0.888 (or) 0.28
The whole number parts is equal for both the numbers.
Comparing the digits in the tenths place we get, 8 > 2.
0.888 > 0.28 ∴ 0.888 is greater.

(ii) 23.914 or 23.915
The whole number part is equal in both the numbers.
Also the tenth place and hundredths place are also equal.
∴ Comparing the thousandths place, we get 5 > 4.
23.915 > 23.914 ∴ 23.915 is greater.

Question 4.
In a 25 m swimming competition, the time taken by 5 swimmers A, B, C, D and E are 15.7 seconds, 15.68 seconds, 15.6 seconds, 15.74 seconds and 15.67 seconds respectively. Identify the winner.
Solution:
The winner is one who took less time for swimming 25 m.
Comparing the time taken by A, B, C, D, E the whole number part is equal for all participants.
Comparing digit in tenths place we get 6 < 7.
∴ Comparing 15.68, 15.6, 15.67, that is comparing the digits in hundredths place we get 15.60 < 15.67 < 15.68
One who took 15.6 seconds is the winner. ∴ C is the winner.

Question 5.
Convert the following decimal numbers into fractions
(i) 23.4
(ii) 46.301
Solution:
(i) 23.4 = \(\frac { 234 }{ 10 } \) = \(\frac{234 \div 2}{10 \div 2}\) = \(\frac { 117 }{ 5 } \)
(ii) 46.301 = \(\frac { 46301 }{ 1000 } \)

Question 6.
Express the following in kilometres using decimals,
(i) 256 m
(ii) 4567 m
Solution:
1 m = \(\frac { 1 }{ 1000 } \) km = 0.001 Km
(i) 256 m = \(\frac { 256 }{ 1000 } \) km = 0.256 km
(ii) 4567 m = \(\frac { 4567 }{ 1000 } \) km = 4.567 km

Question 7.
There are 26 boys and 24 girls in a class. Express the fractions of boys and girls as decimal numbers.
Solution:
Boys = 26; Girls = 24; Total = 50
Fraction of boys = \(\frac { 26 }{ 50 } \) = \(\frac{26 \times 2}{50 \times 2}\) = \(\frac { 52 }{ 100 } \) = 0.52
Fraction of girls = \(\frac { 24 }{ 50 } \) = \(\frac{24 \times 2}{50 \times 2}\) = \(\frac { 48 }{ 100 } \) = 0.48

Challenge Problems

Question 8.
Write the following amount using decimals.
(i) 809 rupees 99 paise
(ii) 147 rupees 70 paise
Solution:
100 paise = 1 rupee; 1 paise = \(\frac { 1 }{ 100 } \) rupee

(i) 809 rupees 99 paise = 809 rupees + \(\frac { 99 }{ 100 } \) rupees
= 809 + 0.99 rupees = ₹ 809.99

(ii) 147 rupees 70 paise = 147 rupees + \(\frac { 70 }{ 100 } \) rupees
= 147 rupees + 0.70 rupees = ₹ 147.70

Question 9.
Express the following in metres using decimals.
(i) 1328 cm
(ii) 419 cm
Solution:
100 cm = 1 m; 1 cm = \(\frac { 1 }{ 100 } \) m
(i) 1328 cm = \(\frac { 1328 }{ 100 } \) m = 13.28 m
(ii) 419 cm = \(\frac { 419 }{ 100 } \) m = 4.19 m

Question 10.
Express the following using decimal notation.
(i) 8 m 30 cm in metres
(ii) 24 km 200 m in kilometres
Solution:
(i) 8 m 30 cm in metres
8 m + \(\frac { 30 }{ 100 } \) m = 8 m + 0.30 m = 8.30 m

(ii) 24 km 200 m in kilometres
24 km + \(\frac { 200 }{ 1000 } \) km = 24 km + 0.200 km = 24.200 km

Question 11.
Write the following fractions as decimal numbers.
(i) \(\frac { 23 }{ 10000 } \)
(ii) \(\frac { 421 }{ 100 } \)
(iii) \(\frac { 37 }{ 10 } \)
Solution:
(i) \(\frac { 23 }{ 10000 } \) = 0.0023
(ii) \(\frac { 421 }{ 100 } \) = 4.21
(iii) \(\frac { 37 }{ 10 } \) = 3.7

Question 12.
Convert the following decimals into fractions and reduce them to the lowest form,
(i) 2.125
(ii) 0.0005
Solution:
(i) 2.125 = \(\frac { 2125 }{ 1000 } \) = \(\frac{2125 \div 25}{1000 \div 25}\) = \(\frac { 85 }{ 40 } \) = \(\frac{85 \div 5}{40 \div 5}\) = \(\frac { 17 }{ 8 } \)

(ii) 0.0005 = \(\frac { 5 }{ 1000 } \) = \(\frac{5 \div 5}{10000 \div 5}\) = \(\frac { 1 }{ 2000 } \)

Question 13.
Represent the decimal numbers 0.07 and 0.7 on a number line.
Solution:

0.07 lies between 0.0 and 0.1
The unit space between 0 and 0.1 is divided into 10 equal parts and 7th part is taken. Also 0.7 lies between 0 and 1.
The unit space between 0 and 1 is divided into 10 equal parts, and the 7th part is taken.

Question 14.
Write the following decimal numbers in words.
(i) 4.9
(ii) 220.0
(iii) 0.7
(iv) 86.3
Solution:
(i) 4.9 = Four and nine tenths
(ii) 220.0 = Two hundred and twenty
(iii) 0.7 = Seven tenths
(iv) 86.3 = Eighty six and three tenths.

Question 15.
Between which two whole numbers the given numbers lie?
(i) 0.2
(ii) 3.4
(iii) 3.9
(iv) 2.7
(v) 1.7
(vi) 1.3
Solution:
(i) 0.2 lies between 0 and 1.
(ii) 3.4 lies between 3 and 4.
(iii) 3.9 lies between 3 and 4.
(iv) 2.7 lies between 2 and 3.
(v) 1.7 lies between 1 and 2.
(vi) 1.3 lies between 1 and 2.

Question 16.
By how much is \(\frac { 9 }{ 10 } \) km less than 1 km. Express the same in decimal form.
Solution:
Given measures are 1 km and \(\frac { 9 }{ 10 } \) km. i.e., 1 km and 0.9 km.
Difference = 1.0 – 0.9 = 0.1 km.

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.5

To convert CGPA to percentage, all you need to do is multiply your CGPA by 9.5.

Miscellaneous Practice Problems

Question 1.
When Mathi was buying her flat she had to put down a deposit of \(\frac { 1 }{ 10 } \) of the value of the flat. What percentage was this?
Solution:
Percentage of \(\frac { 1 }{ 10 } \) = \(\frac { 1 }{ 10 } \) × 100 % = 10 %
Mathi has to put down a deposit of 10 % of the value of the flat.

Question 2.
Yazhini scored 15 out of 25 in a test. Express the marks scored by her in percentage.
Solution:
Yazhini’s score = 15 out of 25 = \(\frac { 15 }{ 25 } \)
Score in percentage = \(\frac { 15 }{ 25 } \) × 100% = 60%

Question 3.
Out of total 120 teachers of a school 70 were male. Express the number of male teachers as percentage.
Solution:
Total teachers of the school = 120
Number of male teachers = 70
∴ Percentage of male teacher = \(\frac { 70 }{ 120 } \) × 100 % = \(\frac { 700 }{ 12 } \) %
Score in percentage = 58.33%
Percentage of male teachers = 58.33%

The percentage difference calculator is here to help you compare two numbers.

Question 4.
A cricket team won 70 matches during a year and lost 28 matches and no results for two matches. Find the percentage of matches they won.
Solution:
Number of Matches won = 70
Number of Matches lost = 28
“No result” Matches = 2
Total Matches = 70 + 28 + 2 = 100
Percentage of Matches won = \(\frac { 70 }{ 100 } \) × 100 % = 70 %
The won 70% of the matches

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 5.
There are 500 students in a rural school. If 370 of them can swim, what percentage of them can swim and what percentage cannot?
Solution:
Total number of students = 500
Number of students who can swim = 370
Percentage of students who can swim = \(\frac { 370 }{ 500 } \) × 100 % = 74 %
Number of students who cannot swim = 500 – 370 = 130
Percentage of students who cannot swim = \(\frac { 130 }{ 500 } \) × 100 % = 26 %
i.e. 74% can swim and 26% cannot swim

Question 6.
The ratio of Saral’s income to her savings is 4 : 1. What is the percentage of money saved by her?
Solution:
Total parts of money = 4 + 1 = 5
Part of money saved = 1
∴ Percentage of money saved = \(\frac { 1 }{ 5 } \) × 100% = 20%
∴ 20% of money is saved by Saral

Question 7.
A salesman is on a commission rate of 5%. How much commission does he make on sales worth ₹ 1,500?
Solution:
Total amount on sale = ₹ 1,500
Commission rate = 5 %
Commission received = 5 % of ₹ 1,500 = \(\frac { 5 }{ 100 } \) × 1500 = ₹ 75
∴ Commission received = ₹ 75

Question 8.
In the year 2015 ticket to the world cup cricket match was ₹ 1,500. This year the price has been increased by 18%. What is the price of a ticket this year?
Solution.
Price of a ticket in 2015 = ₹ 1500
Increased price this year = 18% of price in 2015
= 18 % of ₹ 1500 = \(\frac { 18 }{ 100 } \) × 1500
= ₹ 270
Price of ticket this year = last year price + increased price
= ₹ 1500 + ₹ 270 = ₹ 1770
Price of ticket this year = ₹ 1770

Question 9.
2 is what percentage of 50?
Solution:
Let the required percentage be x
x% of 50 = 2
\(\frac { x }{ 100 } \) × 50 = 2
x = \(\frac{2 \times 100}{50}\) = 4 %
∴ 4 % of 50 is 2

Question 10.
What percentage of 8 is 64?
Solution:
Let the required percentage be x
So x % of 8 = 64
\(\frac { x }{ 100 } \) × 8 = 64
x = \(\frac{64 \times 100}{8}\) = 800
∴ 800 % of 8 is 64

Question 11.
Stephen invested ₹ 10,000 in a savings bank account that earned 2% simple interest. Find the interest earned if the amount was kept in the bank for 4 years.
Solution:
Principal (P) = ₹ 10,000
Rate of interest (r) = 2%
Time (n) = 4 years
∴ Simple Interest I = \(\frac { pnr }{ 100 } \)
= \(\frac{10000 \times 4 \times 2}{100}\)
= ₹ 800
Stephen will earn ₹ 800

Question 12.
Riya bought ₹ 15,000 from a bank to buy a car at 10% simple interest. If she paid ₹ 9,000 as interest while clearing the loan, find the time for which the loan was given.
Solution:
Here Principal (P) = ₹ 15,000
Rate of interest (r) = 10 %
Simple Interest (I) = ₹ 9000
I = \(\frac { pnr }{ 100 } \)
9000 = \(\frac{15000 \times n \times 10}{100}\)
n = \(\frac{9000 \times 100}{15000 \times 10}\)
n = 6 years
∴ The loan was given for 6 years

Question 13.
In how much time will the simple interest on ₹ 3,000 at the rate of 8% per annum be the same as simple interest on ?4,000 at 12% per annum for 4 years?
Solution:
Let the required number of years be x
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P₁ = ₹ 3000
Rate of interest (r) = 8 %
Time (n₁) = n₁ years
Simple Interest I₁ = \(\frac{3000 \times 8 \times n_{1}}{100}\) = 240 n₁
Principal (P₂) = ₹ 4000
Rate of interest (r) = 12 %
Time n₂ = 4 years
Simple Interest I₂ = \(\frac{4000 \times 12 \times 4}{100}\)
I₂ = 1920
If I₁ = I₂
240 n₁ = 1920
n₁ = \(\frac { 1920 }{ 240 } \) = 8
∴ The required time = 8 years

Challenge Problems

Question 14.
A man travelled 80 km by car and 320 km by train to reach his destination. Find what percent of total journey did he travel by car and what per cent by train?
Solution:
Distance travelled by car = 80 km.
Distance travelled by train = 320 km
Total distance = 80 + 320 km = 400 km
Percentage of distance travelled by car = \(\frac { 80 }{ 400 } \) × 100 % = 20 %
Percentage of distance travelled by train = \(\frac { 320 }{ 800 } \) × 100 % = 40 %

Question 15.
Lalitha took a math test and got 35 correct and 10 incorrect answers. What was the percentage of correct answers?
Solution:
Number of correct answers = 35
Number of incorrect answers = 10
Total number of answers = 35 + 10 = 45
Percentage of correct answers = \(\frac { 35 }{ 45 } \) × 100 %
= 77.777 % = 77.78 %

Question 17.
The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. Find the percentage of literate women to the total population?
Solution:
Population of the village = 8000 people
literate people = 80 % of population
= 80 % of 8000 = \(\frac { 80 }{ 100 } \) × 8000
literate people = 6400
Percentage of women = 40 %
Number of women = 40 % of literate people
= \(\frac { 40 }{ 100 } \) × 6400 = 2560
∴ literate women : Total population
= 8000 : 2560
= 25 : 8

Question 18.
A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer correctly?
Solution:
Total number of problems in the test = 20
Students score = 80 %
Number of problem answered = \(\frac { 80 }{ 100 } \) × 20 = 16

Question 19.
A metal bar weighs 8.5 kg. 85% of the bar is silver. How many kilograms of silver are in the bar?
Solution:
Total weight of the metal = 8.5 kg
Percentage of silver in the metal = 85%
Weight of silver in the metal = 85% of total weight
= \(\frac { 85 }{ 100 } \) × 8.5 kg
= 7.225 kg
7.225 kg of silver are in the bar.

Question 20.
Concession card holders pay ₹ 120 for a train ticket. Full fare is ₹ 230. What is the percentage of discount for concession card holders?
Solution:
Train ticket fare = ₹ 230
Ticket fare on concession = ₹ 120
Discount = Ticket fare – concession fare = 230 – 120 = ₹ 110

Percentage of discount = 47.83%

Question 21.
A tank can hold 200 litres of water. At present, it is only 40% full. How many litres of water to fill in the tank, so that it is 75 % full?
Solution:
Capacity of the water tank = 200 litres
Percentage of water in the tank = 40%
Percentage of water to fill = Upto 75%
Difference in percentage = 75 % – 40 % = 35 %
∴ Volume of water to be filled = Percentage of difference × total capacity
= \(\frac { 35 }{ 100 } \) × 200 = 70 l
70 l of water to be filled

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 22.
Which is greater 16 \(\frac { 2 }{ 3 } \) or \(\frac { 2 }{ 5 } \) or 0.17 ?
Solution:
16 \(\frac { 2 }{ 3 } \) = \(\frac { 50 }{ 30 } \)
= \(\frac { 50 }{ 30 } \) × 100 % = 1666.67 %
⇒ \(\frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \) × 100 = 40 %
0.17 = \(\frac { 17 }{ 100 } \) = 17 %
∴ 1666.67 is greater
∴ 16 \(\frac { 2 }{ 3 } \) is greater

Question 23.
The value of a machine depreciates at 10% per year. If the present value is ₹ 1,62,000, what is the worth of the machine after two years.
Solution:
Present value of the machine = ₹ 1,67,000
Rate of depreciation = 10 % Per annum
Time (n) = 2 years
For 1 year depreciation amount = \(\frac{1,62,000 \times 1 \times 10}{100}\) = ₹ 16,200
Worth of the machine after one year = Worth of Machine – Depreciation
= 1,67,000 – 16,200 = 1,45,800
Depreciation of the machine for 2nd year = 145800 × 1 × \(\frac { 10 }{ 100 } \) = 14580
Worth of the machine after 2 years = 1,45,800 – 14,580 = 1,31,220
∴ Worth of the machine after 2 years = ₹ 1,31,220

Question 24.
In simple interest, a sum of money amounts to ₹ 6,200 in 2 years and ₹ 6,800 in 3 years. Find the principal and rate of interest.
Solution:
Let the principal P = ₹ 100
If A = 6200
⇒ Principal + Interest for 2 years = 6200
A = ₹ 7400
⇒ Principal + Interest for 3 years = 7400
∴ Difference gives the Interest for 1 year
∴ Interest for 1 year = 7400 – 6200
I = 1200
\(\frac { pnr }{ 100 } \) = 1200 ⇒ \(\frac{P \times 1 \times r}{100}\) = 1200
If the Principal = 10,000 then
\(\frac{10,000 \times 1 \times r}{100}\) = 1200 ⇒ r = 12 %
Rate of interest = 12 % Per month

Question 25.
A sum of ₹ 46,900 was lent out at simple interest and at the end of 2 years, the total amount was ₹ 53,466.Find the rate of interest per year.
Solution:
Here principal P = ₹ 46900
Time n = 2 years
Amount A = ₹ 53466
Let r n be the rate of interest per year p
Intrest I = \(\frac { pnr }{ 100 } \)
A = P + I
53466 = 46900 + \(\frac{46900 \times 2 \times r}{100}\)
53466 – 46900 = \(\frac{46900 \times 2 \times r}{100}\)
6566 = 469 × 2 × r
r = \(\frac{6566}{2 \times 469}\) % = 7 %
Rate of interest = 7 % Per Year

Question 26.
Arun lent ₹ 5,000 to Balaji for 2 years and ₹ 3,000 to Charles for 4 years on simple interest at the same rate of interest and received ₹ 2,200 in all from both of them as interest. Find the rate of interest per year.
Solution:
Principal lent to Balaji P₁ = ₹ 5000
Time n₁ = 2 years
Let r be the rate of interest per year
Simple interest got from Balaji = \(\frac { pnr }{ 100 } \) ⇒ I₁ = \(\frac{5000 \times 25 \times r}{100}\)
Again principal let to Charles P₂ = ₹ 3000
Time (n₂) = 4 years
Simple interest got from Charles (I₂) = \(\frac{3000 \times 4 \times r}{100}\)
Altogether Arun got ₹ 2200 as interest.
∴ I₁ + I₂ = 2200
\(\frac{5000 \times 2 \times r}{100}+\frac{3000 \times 4 \times r}{100}\) = 2200
100r + 120r = 2200
220r = 2200 = \(\frac { 2200 }{ 220 } \)
r = 10 %
Rate of interest per year = 10 %

Question 27.
If a principal is getting doubled after 4 years, then calculate the rate of interest. (Hint: Let P = ₹ 100)
Solution:
Let the principal P = ₹ 100
Given it is doubled after 4 years
i.e. Time n = 4 years
After 4 years A = ₹ 200
∴ A = P + I
A – P = I
200 – 100 = I
After 4 years interest I = 100
I = \(\frac { pnr }{ 100 } \) ⇒ 100 = \(\frac{100 \times 4 \times r}{100}\)
4r = 100 ⇒ r = 25 %
Rate of interest r = 25 %