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Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using Factor Theorem:

Question 1.
Show that \(\left| \begin{matrix} x & a & a \\ a & x & a \\ a & a & x \end{matrix} \right| \) = (x – a)² (x + 2a)
Answer:

By putting x = a , we have three rows of |A| are identical. Therefore (x – a)² is a factor of |A|
Put x = – 2a in |A|

∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)² (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.

a³ [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k . 4a³
a³ [o + 2 + 2 ] = 4 ka³
4a³ = 4 ka³
k = 1

 

Free handy Remainder Theorem Calculator tool displays the remainder of a difficult polynomial expression in no time.

Question 2.
Show that \(\left| \begin{matrix} b\quad +\quad c & a\quad -\quad c & a\quad -\quad b \\ b\quad -\quad c & c\quad +\quad a & b\quad -\quad a \\ c\quad -\quad b & c\quad -\quad a & a\quad +\quad b \end{matrix} \right| \) = 8 abc
Answer:


since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|

since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|

since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.

Question 3.
Solve that \(\left| \begin{matrix} x\quad +\quad a & b & c \\ a & x\quad +\quad b & c \\ a & b & x\quad +\quad c \end{matrix} \right| \) = 0
Answer:

Put x = 0

x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)

∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0 , – (a + b + c)

Question 4.
Show that \(\left| \begin{matrix} b\quad +\quad c & a & { a }^{ 2 } \\ c\quad +\quad a & b & { b }^{ 2 } \\ a\quad +\quad b & c & { c }^{ 2 } \end{matrix} \right| \) = (a + b + c) (a – b) (b – c) (c – a)
Answer:

Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a , b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c² is 4.
Therefore, the other factor is k (a + b + c).

5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1

Question 5.
Solve \(\left| \begin{matrix} 4\quad -\quad x & 4\quad +\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad -\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad +\quad x & 4\quad -\quad x \end{matrix} \right| \) = 0
Answer:

Put x = 0

∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.

Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.


∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0 , 0 , – 12

Question 6.
Show that \(\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \) = (x – y) (y – z) (z – x)
Answer:

|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z² is 3. Therefore, the other factor is the constant factor k.

Put x = 0, y = 1, z = -1 we get

Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Having trouble working out with the Binomial theorem? You’ve come to the right place, our Binomial Expansion Calculator is here to save the day for you.

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:

(iii) (5 + x²)^2/3
Answer:

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:

Expanding binomials calculator is a free online tool that displays the expansion of the given binomial term.

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
Answer:

(ii) e^-2x
Answer:

(iii) e^x/2
Answer:

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )

(ii) log (1 – 2x)

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)

Question 7.

Answer:

Question 8.

Answer:

Question 9.
Find the coefficient of x⁴ in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:

Question 10.
Find the value of
Answer:

Students can download 5th Maths Term 2 Chapter 2 Numbers Additional Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 2 Chapter 2 Numbers Additional Questions

What is the factor of 50 between 11 and 20? The factors of 50 are: 1, 2, 5, 10, 25, 50 None of them is between 11 and 20.

Question 1.
Find the common factors for:
(i) 6 and 9
(ii) 15 and 20
Answer:
(i) 6 and 9
Factors of 6 is 1, 2, 3, 6
Factors of 9 is 1, 3, 9
Common factors of 6 and 9 is 1, 3

(ii) 15 and 20
Factors of 15 is 1, 3, 5, 15
Factors of 20 is 1, 2, 4, 5, 10, 20
Common factors of 15 and 20 are 1, 5.

Question 2.
Draw a picture of factor tree.
(i) 45
(ii) 100
Answer:
(i) 45

(ii) 100

Question 3.
Write down the first 5 multiple of the given numbers.
(i) 15
(ii) 25
Answer:
(i) 15
15 × 1 = 15
15 × 2 = 30
15 × 3 = 45
15 × 4 = 60
5 × 5 = 75
5, 30, 45, 60, 75

(ii) 25
25 × 1 = 25
25 × 2 = 50
25 × 3 = 75
25 × 4 = 100
25 × 5 = 125
25, 50, 75, 100, 125

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 3 Books of Prime Entry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 3 Books of Prime Entry

11th Accountancy Guide Books of Prime Entry Text Book Back Questions and Answers

prime factorization of 38 are numbers that, when multiplied in pairs give the product as 38.

I. Choose the correct answer.

Question 1.
Accounting equation signifies …………….
(a) Capital of a business is equal to assets
(b) Liabilities of a business are equal to assets
(c) Capital of a business is equal to liabilities
(d) Assets of a business are equal to the total of capital and liabilities
Answer:
(d) Assets of a business are equal to the total of capital and liabilities

Question 2.
‘Cash withdrawn by the proprietor from the business for his personal use’ causes …………….
(a) Decrease in assets and decrease in owner’s capital
(b) Increase in one asset and decrease in another asset
(c) Increase in one asset and increase in liabilities
(d) Increase in asset and decrease in capital
Answer:
(a) Decrease in assets and decrease in owner’s capital

Question 3.
A firm has assets of Rs. 1,00,000 and the external liabilities of Rs. 60,000. Its capital would be ……………….
(a) Rs. 1,60,000
(b) Rs. 60,000
(c) Rs. 1,00,000
(d) Rs. 40,000
Answer:
(d) Rs. 40,000

Question 4.
The incorrect accounting equation is ………………….
(a) Assets = Liabilities + Capital
(b) Assets = Capital + Liabilities
(c) Liabilities = Assets + Capital
(d) Capital = Assets – Liabilities
Answer:
(c) Liabilities = Assets + Capital

Question 5.
Accounting equation is formed based on the accounting principle of ………………
(a) Dual aspect
(b) Consistency
(c) Going concern
(d) Accrual
Answer:
(a) Dual aspect

Question 6.
Real account deals with ……………..
(a) Individual persons
(b) Expenses and losses
(c) Assets
(d) Incomes and gains
Answer:
(c) Assets

Question 7.
Which one of the following is representative personal account?
(a) Building A/c
(b) Outstanding salary A/c
(c) Mahesh A/c
(d) Balan & Co
Answer:
(b) Outstanding salary A/c

Question 8.
Prepaid rent is a ……………….
(a) Nominal A/c
(b) Personal A/c
(c) Real A/c
(d) Representative personal A/c
Answer:
(d) Representative personal A/c

Question 9.
Withdrawal of cash from business by the proprietor should be credited to
(a) Drawings A/c
(b) Cash A/c
(c) Capital A/c
(d) Purchases A/c
Answer:
(b) Cash A/c

Question 10.
In double entry system of book keeping, every business transaction affects
(a) Minimum of two accounts
(b) Same account on two different dates
(c) Two sides of the same account
(d) Minimum three accounts
Answer:
(a) Minimum of two accounts

II. Very Short Answer Type Questions

Question 1.
What are source documents?
Answer:
“Source documents are the authentic evidence of financial transactions. These documents show the nature of the transaction, the date, the amount, and the parties involved. Source documents include cash receipt, invoice, debit note, credit note, pay-in-slip, salary bills, wage bills, cheque record slips, etc.

Question 2.
What Is the accounting equation?
Answer:
The relationship of assets with that of liabilities to outsiders and to owners in the equation form is known as the accounting equation.
The accounting equation is a mathematical expression which shows that the total of assets is equal to the total of liabilities and capital.
This is based on the dual aspect concept of accounting.
This means that the total claims of outsiders and the proprietor against a business enterprise will always be equal to the total assets of the business enterprise.
Capital + Liabilities = Assets

Question 3.
Write any one transaction which

1. Decreases the assets and decreases the liabilities
2. Increases one asset and decreases another asset

Answer:

1. Paid creditors
2. Gash sales

Question 4.
What is meant by journalizing?
Answer:
The process of analyzing the business transactions under the heads of debit and credit and recording them in the journal is called journalizing.

Question 5.
What is the real account?
Answer:
All accounts relating to tangible and intangible properties and possessions are called real accounts.

Question 6.
How are personal accounts classified?
Answer:

Question 7.
State the accounting rule for a nominal account.
Answer:

Question 8.
Give the golden rules of double-entry accounting system.
Answer:

III. Short Answer Questions

Question 1.
Write a brief note on the accounting equation approach of recording transactions.
Answer:
The relationship of assets with that of liabilities to outsiders and to owners in the equation form is known as the accounting equation.
The accounting equation is a mathematical expression which shows that the total of assets is equal to the total of liabilities and capital.
This is based on the dual aspect concept of accounting.

This means that the total claims of outsiders and the proprietor against a business enterprise will always be equal to the total assets of the business enterprise. The equation is given under:
Capital + Liabilities = Assets
Capital can also be called owner’s equity and liabilities as outsider’s equity.

As the revenues and expenses will affect capital, the expanded equation may be given as under:
Assets = Liabilities + Capital + Revenues – Expenses
Accounts are classified into five categories: (i) Asset account, (ii) Liability account, (iii) Capital account, (iv) Revenue account and (v) Expense account as follows:

Question 2.
What is an Account? Classify the accounts with suitable examples.
Answer:
Every transaction has two aspects and each aspect affects a minimum of one account.
An account is the basic unit of identification in accounting.
A ledger account is a summary of relevant transactions at one place relating to a particular head.
The account is the systematic presentation of all material information regarding a particular person or item at one place, under one head.

Classification of Accounts:
Personal Account;
Accounting relating to persons is called a personal account. The personal account may be a natural, artificial, or representative personal account.
E.g: Malini account, BHEL account, Prepaid Rent account.

Impersonal Account:
All accounts which do not affect persons are called impersonal accounts. These are further classified into Real and Nominal Accounts.
E.g; Building account, Goodwill, Salaries account.

Question 3.
What are the three different types of personal accounts?
Answer:
Under the double-entry system of book-keeping, for the purpose of recording the various financial transactions, the accounts are classified as personal accounts and impersonal accounts.

1. Natural person’s account: Natural person means human beings. Example: Vinoth account, Malini account.
2. Artificial person’s account: Artificial person refers to persons other than human beings recognized by law as persons. They include business concerns, charitable institutions, etc. Example: BHEL account, Bank account.
3. Representative personal accounts: These are the accounts which represent persons natural or artificial or a group of persons. Example: Outstanding salaries account, Prepaid rent account. When expenses are outstanding, it is payable to a person. Hence, it represents a person.

Question 4.
What is the accounting treatment for insurance premium paid on the of the proprietor?
Answer:
The proprietor has to pay an agreed premium on a monthly or annual basis. The amount should be treated as prepayment which is considered a current asset.
The following entry should be made:
Insurance A/c Dr –
To Cash/Bank A/c –
(Being insurance premium paid.)

Question 5.
State the principles of the double-entry system of book-keeping.
Answer:
Following are the principles of the double-entry system:

1. In every business transaction, there are two aspects.
2. The two aspects involved are the benefit or value receiving aspect and the benefit or value giving aspect.
3. These two aspects involve a minimum of two accounts; at least one debit and at least one credit.
4. For every debit, there is a corresponding and equivalent credit.
5. If one account is debited the other account must be credited.

Question 6.
Briefly explain about steps in journalizing.
Answer:
The process of analyzing the business transactions under the heads of debit and credit and recording them in the journal is called journalizing. An entry made in the journal is called a journal entry. The following steps are followed in journalizing:

Analyze the transactions and identify the accounts (based on aspects) which are involved in the transaction.

Classify the above accounts under Personal account, Real account or Nominal account, Apply the rules of debit and credit for the above two accounts.

Find which account is to be debited and which account is to be credited by the application of rules of the double-entry system. Record the date of the transaction in the date column.

Enter the name of the account to be debited in the particulars column very close to the left hand side of the particulars column followed by the abbreviation ‘Dr/ at the end in the same line. Against this, the amount to be debited is entered in the debit amount column in the same line.

Write the name of the account to be credited in the second line starting with the word ‘To’ prefixed a few spaces away from the margin in the particulars column. Against this, the amount to be credited is entered in the credit amount column in the same line.

Write the narration within brackets in the next line in the particulars column.

Question 7.
What is the double-entry system? State its advantages.
Answer:
A double-entry system of book keeping is a scientific and complete system of recording the financial transactions of an organization. According to this system, every transaction has a two-fold effect. That is, there are two aspects involved, namely, receiving aspect and the giving aspect. It is denoted by debit (Dr.) and credit (Cr.). The basic principle of the double-entry system is that for every debit there must be an equivalent and corresponding credit. Debit denotes an increase in assets or expenses or a decrease in liabilities, income, or capital. Credit denotes an increase in liabilities, income, or capital or a decrease in assets or expenses.

IV. Exercises

Question 1.
Complete the accounting equation.
(a) Assets = Capital + Liabilities
1,00,000 = 80,000 + ?
Answer:
(a) Liabilities = Assets – Capital
Liabilities = 1,00,000 – 80,000
Liabilities = 20,000

(b) Assets = Capital + Liabilities
2,00,000 = ? + 40,000
Answer:
Capital = Assets – Liabilities
Capital = 2,00,000 – 40,000
Capital = 1,60,000

(c) Assets = Capital + Liabilities
? = 1,60,000 + 80,000
Answer:
Assets = Capital + Liabilities
Assets = 1,60,000 + 80,000
Assets = 2,40,000

Question 2.
For the following transactions, show the effect on the accounting equation.
(a) Raj Started business with cash – 40,000
(b) Opened bank account with a deposit of – 30,000
(c) Bought goods from Hari on credit for – 12,000
(d) Raj withdrew cash for personal use – 1,000
(e) Bought furniture by using a debit card for – 10,000
(f) Sold goods to Murugan and cash received – 6,000
(g) Money is withdrawn from bank for office use – 1,000
Answer:
(a) Raj started the business with cash Rs 40,000
Increasing in capital and increasing in Asset Effects:
Cash comes in → Increase in asset
Capital provided by the owner → Increase in capital of owner
Capital = Asset
Capital = Cash
(+)Rs. 40,000 = (+)Rs. 40,000

(b) Opened bank account with a deposit of Rs 30,000
Decrease in one asset and increase in another asset Effects:
Bank is the receiver → Increases in Assets
Cash goes out → Decrease in Asset
Capital = Asset
Capital = Bank + cash
Capital = (+) Rs. 30,000 + (-) Rs. 30,000

(c) Bought goods from Hari on credit for Rs. 12,000
Increasing Asset and Increase in Liabilities Effects:
Stock comes in → Increase in Asset
Creditors arises → Increase in Liabilities
Asset = Liabilities
Stock = Creditors
(+) Rs. 12,000 = (+) Rs. 12,000

(d) Raj withdrew cash for personal use Rs 1,000
Decrease in Asset and decrease in capital Effects:
Cash goes out → Decrease in cash
Drawings proprietor is the receiver → Decrease in capital
Asset = Capital
Cash = Capital
(-) Rs. 1,000 = (-) Rs. 1,000

(e) Bought furniture by using debit card for Rs 10,000
Decrease in one asset and Increase in another asset Effects:
Furniture comes in → Increase in Furniture
Bank is giver → Decrease in a Bank
Asset = Liabilities
Furniture + Bank = Liabilities
(+) Rs. 10,000 + (-) Rs. 10,000 = Liabilities

(f) Sold goods to Murugan and cash received Rs 6,000
Decrease in one asset and increase in another asset Effects:
Cash comes in → Increase in cash
<$> Stock goes out → Decrease in stock
Asset = Liabilities
Cash + Stock = Liabilities
(+) Rs. 6,000 + (-) Rs. 6,000 = Liabilities

(g) Money is withdrawn from bank for office use Rs 1,000
Decrease in one asset and increase in another asset Effects:
Cash comes in → Increase in Cash
The bank is a giver → Decrease in Bank
Asset = Liabilities
Cash + Bank = Liabilities
(+) Rs. 1,000 + (-) Rs. 1,000 = Liabilities

Accounting Equation:

Question 3.
Prepare accounting equation for the following transactions,
(a) Murugan commenced business with cash Rs 80,000
(b) Purchased goods for cash Rs 30,000
(c) Paid salaries by cash Rs 5,000
(d) Bought goods from Kumar for Rs 5,000 and deposited the money in CDM.
(e) Introduced additional capital of Rs 10,000
Answer:
Accounting Equation:

Question 4.
What will be the effect of the following on the accounting equation?
(a) Sunil started the business with Rs 1,40,000 cash and goods worth Rs 60,000
(b) Purchased furniture worth Rs 20,000 in cash
(c) Depredation on furniture Rs 800
(d) Deposited into bank Rs 40,000
(e) Paid electricity charges through net banking Rs 500
(f) Sold goods to Ravi costing Rs 10,000 for Rs 15,000
(g) Goods returned by Ravi Rs 5,000
Answer:
(a) Sunil started the business with Rs 1,40,000 cash and goods worth Rs 60,000
Transactions affecting more than two accounts Effects:
Cash comes in → Increase in asset
Stock comes in → an Increase in asset
Capital provide by the owner → Increase in capital
Capital = Asset
Capital = Cash + Stock
(+) 2,00,000 = (+) 1,40,000 + (+) 60,000

(b) Purchased furniture worth Rs 20,000 in cash
Decrease in one asset and Increase in Another asset Effects:
Furniture comes in → Increase in asset
Cash goes out → Decrease in asset
Liabilities = Assets
Liabilities = Cash + Furniture
Liabilities = (-) 20,000 + (-) 20,000

(c) Depreciation on furniture Rs 800
Decrease in asset and decrease in capital effects:
Furniture goes out → Decrease in Asset
Capital → Decrease in Capital
Capital = Asset
Capital = Furniture
(-) 800 = (-) 800

(d) Deposited into bank Rs 40,000
Increase in one Asset and decrease in another asset Effects:
Cash goes out → Decrease in asset
The bank is a receiver → Increase in asset
Liability = Asset
Liability = Cash + bank
= (-) 40,000 + (+) 40,000

(e) Paid electricity charges through net banking Rs 500
decrease in Asset and Decrease in capital Effects:
The bank is a given → Decrease in Asset
Capital (Expenses) → Decrease in capital
Capital = Asset
Capital = Bank
(-) 500 = (-) 500

(f) Sold goods to Ravi costing Rs 10,000 for Rs 15,000
Increase in asset and decreases in another asset and Increase in Capital Effects:
Creditors → Increases in Liabilities
Stock goes out → Decreases in asset
Capital (Incom(e) → Increase in capital
Liabilities + Capital = Asset
Creditors + Capital = Stock
(+) 15,000 + (+) 5,000 = (-) 5,000

(g) Goods returned by Ravi Rs 5,000
Increase in asset and decrease in Liabilities Effects:
Stock comes in → an Increase in asset
Reducing in creditors → Decreases in Liabilities
Liability = Asset
Creditors = Stock
(-) 5,000 = (+) 5,000
Accounting Equation:

Question 5.
Create an accounting equation on the basis of the following transactions.
(i) Rakesh started the business with a capital of Rs 1,50,000
(ii) Deposited money with the bank Rs 80,000
(iii) Purchased goods from Mahesh and paid through credit card Rs 25,000
(iv) Sold goods (costing Rs 10,000) to Mohan for Rs 14,000 who pays through debit card
(v) Commission received by cheque and deposited the same in the bank Rs 2,000
(vi) Paid office rent through ECS Rs 6,000
(vii) Sold goods to Raman for Rs 15,000 of which Rs 5,000 was received at once
Answer:
Accounting Equation:

Question 6.
Create an accounting equation on the basis of the following transactions.
(i) Started business with cash 80,000 and goods Rs 75,000
(ii) Sold goods to Shanmugam on credit for Rs 50,000
(iii) Received cash from Shanmugam in full settlement Rs 49,000
(iv) Salary outstanding Rs 3,000
(v) Goods costing Rs 1,000 given as charity
(vi) Insurance premium paid Rs 3,000
(vii) Out of insurance premium paid, prepaid is Rs 500
Answer:
Accounting Equation:

Question 7.
Create an accounting equation on the basis of the following transactions:
(i) Opening balance on 1st January, 2018 cash Rs 20,000; stock Rs 50,000 and bank Rs 80,000
(ii) Bought goods from Suresh Rs 10,000 on credit
(iii) Bank charges Rs 500
(iv) Paid Suresh Rs 9,700 through credit card in full settlement.
(v) Goods purchased on credit from Philip for Rs 15,000
(vi) Goods returned to Philip amounting to Rs 4,000
Answer:

Question 8.
Enter the following transactions in the journal of Manohar who is dealing in textiles. 2018 March
Manohar started the business with cash – 60,000
Purchased furniture for cash – 10,000
Bought goods for cash – 25,000
Bought goods from Kamaiesh on credit – 15,000
Sold goods for cash – 28,000
Sold goods to Hari on credit – 10,000
Paid Kamaiesh – 12,000
Paid rent – 500
Received from Hari – 8,000
Withdrew cash for personal use – 4,000
Answer:
journal of Mr. Manohar

Question 9.
Pass journal entries in the books of Sasi Kumar who is dealing in automobiles. 2017 Oct.
1. Commenced business with goods – 40,000
3. Cash introduced in the business – 60,000
4. Purchased goods from Arul on credit – 70,000
6. Returned goods to Arul – 10,000
10. Paid cash to Arul on account – 60,000
15. Sold goods to Chandar on credit – 30,000
18. Chandar returned goods worth – 6,000
20. Received cash from Chandar in full settlement – 23,000
25. Paid salaries through ECS – 2,000
30. Sahil took for personal use goods worth – 10,000
Answer:
journal of Mr. Sasi Kumar:

Question 10.
Pass Journal entries in the books of Hart who is deafer in sports items. 2017 Jan.
1. Commenced business with cash 50,000
2. Purchased goods from Subash on credit of 20,000
4. Sold goods to Ramu on credit of 15,000
8. Ramu paid the amount through cheque
10. Cheque received from Ramu is deposited with the bank
15. Sports items purchased from Gopal on credit 10,000
18. Paid rent for the proprietor’s residence 1,500
20. Paid Gopal in a full settlement after deducting a 5% discount
25. Paid Subash Rs 4,750 and discount received 250
28. Paid by cash: wages Rs 500; electricity charges Rs 3,000 and trade expenses 1,000
Answer:
journal of Hari:

Question 11.
Karthick opened a provisions store on 1st April 2017, Journalise the following transactions in this books:
2017 April?
1. Paid into the bank for opening a current account – 2,00,000
3. Goods purchased by cheque – 40,000
5. Investments made in securities – 40,000
6. Goods sold to Radha for Rs 20,000 and cheque received and deposited into bank
7. Amount withdrawn from bank for office use – 15,000
10. Purchased goods from Kamala and cash deposited in CDM – 10,000
12. Sold goods to Vanitha who paid through the debit card – 10,000
15. Interest on securities directly received by the bank – 1,000
20. Insurance paid by the bank as per standing instructions – 2,000
25. Sales made to Kunal who made payment through CDM – 6,000
Answer:
journal of Karthik:

Question 12.
Journalise the following transactions in the books of Ramesh who is dealing in computers:
2018, May 1. Ramesh started business with cash Rs 3,00,000, Goods Rs 80,000 and Furniture Rs 27,000.
2. Money deposited into bank Rs 2,00,000
3. Bought furniture from M/s Jayalakshmi Furniture for Rs 28,000 on credit.
4. Purchased goods from Asohan for Rs 5,000 by paying through debit card.
5. Purchased goods from Guna and paid through net banking for cash Rs 10,000
6. Purchased goods from Kannan and paid through credit card Rs 20,000
7. Purchased goods from Shyam on credit for Rs 50,000
8. Bill drawn by Shyam was accepted for Rs 50,000
9. Paid half the amount owed to M/s Jayalakshmi Furniture by cheque
10. Shyam’s bill was paid
Answer:
journal of Mr. Ramesh

Question 13.
Journalise the following transactions in the books of Sundar who is a bookseller.
Dec 2017.
1. Commenced business with cash – 2,00,000
2. Bought goods from X and Co. on credit – 80,000
4. Opened a bank account with – 50,000
5. Sold goods to Naresh who paid the amount through net banking – 5,000
6. Sold goods to Devi who paid through credit card – 7,000
7. Sold goods to Ashish on credit – 700
8. Money withdrawn from the bank through ATM for office use – 1,000
9. Purchased furniture and paid through the debit card – 2,000
10. Salaries paid by cash – 6,000
11. Furniture purchased from Y for Rs 25,000 and advance given – 5,000
Answer:
Journal of Mr. Sundar

Question 14.
Raja has a hotel. The following transactions took place in his business. Journalize them.
Jan. 1. Started business with cash – 3,00,000
2. Purchased goods from Rajiv on credit – 1,00,000
3. Cash deposited with the bank – 2,00,000
20. Borrowed loan from the bank – 1,00,000
22. Withdrew from the bank for personal us – 800
23. Amount paid to Rajiv in full settlement through NEFT – 99,000
25. Paid club bill of the proprietor by cheque – 200
26. Paid electricity bill of the proprietor’s house through the debit card – 2,000
31. Lunch provided at free of cost to a charity – 1,000
31. Bank levied charges for locker rent – 1,000
Answer:
journal of Mr. Raja

Question 15.
From the following transactions of Shyam, a stationery dealer, pass journal entries for the month of August 2017,
Aug 1. Commenced business with cash Rs 4,00,000, Goods Rs 5,00,000
2. Sold goods to A and money received through RTGS 12,50,000
3. Goods sold to Z on credit for Rs 20,000
5. Bill drawn on Z and accepted by him Rs 20,000
8. Bill received from Z is discounted with the bank for 119,000
10. Goods sold to M on credit Rs 12,000
12. Goods distributed as free samples for Rs 2,000
16. Goods taken for office use Rs 5,000
17. M became insolvent and only 0.80 per rupee is received in the final settlement
20. Bill of Z discounted with the bank is dishonored
Answer:
Journal of Mr. Shyam

Question 16.
Mary is a cement dealer having a business for more than 5 years. Pass journal entries in her books for the period of March 2018,
1. Cement bags bought on credit from Sibi – 20,000
2. Electricity charges paid through net banking – 500
3. Returned goods bought from Sibi – 5,000
4. Cement bags are taken for personal use – 1,000
5. Advertisement expenses paid – 2,000
6. Goods sold to Mano – 20,000
7. Goods returned by Mano – 5,000
8. Payment received from Mano through NEFT
Answer:
Journal of Mr. Mary

11th Accountancy Guide Conceptual Framework of Accounting Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
A debit note is also called…………….
(a) Credit note
(b) Debit memo
(c) Vouchers
(d) Cash memo
Answer:
(b) Debit memo

Question 2.
Journal is a book of ……….
(a) Original Entry
(b) Final Entry
(c) Assets
(d) None of the above
Answer:
(a) Original Entry

Question 3.
When cash or cheque is deposited in a bank, a form is to be filled by a customer is called…………….
(a) Pay – in – slip
(b) Voucher
(c) Cash memo
(d) Invoice
Answer:
(a) Pay – in – slip

Question 4.
Accounts of persons with whom the business deals is known as ………….
(a) Personal A/c
(b) Real A/c
(c) Nominal A/c
(d) Profit & Loss A/c
Answer:
(a) Personal A/c

Question 5.
Outsider’s equity is otherwise called as …………….
(a) Capital
(b) liabilities
(c) debtors
(d) Assets
Answer:
(b) liabilities

Question 6.
Which one of the following equations is correct?
(a) Owner’s Equity = Liability + Asset
(b) Owner’s Equity = Asset – Liability
(c) Liability = Owner’s Equity + Asset
(d) Asset = Owner’s Equity – Liability
Answer:
(b) Owner’s Equity = Asset – Liability

Question 7.
Goodwill is an example of accounts …………….
(a) Real
(b) Nominal
(c) Tangible real
(d) Intangible real
Answer:
(d) Intangible real

Question 8.
Atul purchased a car for Rs. 5,00,000, by making a down payment of Rs. 1,00,000 and signing an Rs. 4,00,000 bill payable due in 60 days. As a result of this transaction
(a) Total assets increase by Rs. 5,00,000
(b) Total liabilities increase by Rs.4,00,000
(c) Total assets increase by Rs. 4,00,000
(d) Total assets increased by Rs. 4,00,000 with a corresponding increase in liabilities by Rs. 4,00,000
Answer:
(d) Total assets increased by Rs. 4,00,000 with a corresponding increase in liabilities by Rs. 4,00,000

Question 9.
Journal means …………….
(a) daily
(b) monthly
(c) yearly
(d) weekly
Answer:
(a) daily

Question 10.
Amount paid to a supplier is ……….
(a) An event
(b) A Transaction
(c) Both (a) and (b)
(d) Neither (a) Nor (b)
Answer:
(c) Both (a) and (b)

Question 11.
Goods worth Rs. 2,000 were distributed as free samples in the market. The journal entry will be
(a) Drawings A/c Dr. 2,000
To Purchases A/c 2,000
(b) Sales A/c Dr. 2,000
To Cash A/c 2,000
(c) Advertisement A/c Dr. 2,000
To Purchases A/c 2,000
(d) No entry
Answer:
(c) Advertisement A/c Dr. 2,000
To Purchases A/c 2,000

Question 12.
If two or more transactions of the same nature are journalized together, it is known as;
(a) Compound journal entry
(b) Separate journal entry
(c) Posting
(d) None of the above
Answer:
(a) Compound journal entry

Question 13.
An amount of Rs.200 received from A credited to B would affect ………….
(a) Accounts of A and B
(b) A’s Account
(c) Cash Account
(d) B’s Account
Answer:
(a) Accounts of A and B

Question 14.
Contra entries are passed only when …………..
(a) Double Column cash book is prepared
(b) Three-Column Cashbook is prepared
(c) Single Cashbook is prepared
(d) None of the above
Answer:
(b) Three-Column Cashbook is prepared

Question 15.
Which of the following statements is correct?
(a) Accounting profit is the difference between cash receipts and cash paid in a period.
(b) Accounting profit is the total cash sales in the year is the expenses for the period.
(c) Accounting profit is the difference between revenue income and expenses for the period.
(d) Accounting profit is the difference between revenue income and cash payment for the period.
Answer:
(c) Accounting profit is the difference between revenue income and expenses for the period.

Question 16.
Direct Expenses are those which can be identified with the particular products. Which of the following items is a direct cost?
(a) The rent of the factory in which products are made
(b) The wages of the production supervisor
(c) A royalty paid to the holder of a patent after each item is produced
(d) The cost of the glue used to attach labels to the products.
Answer:
(c) A royalty paid to the holder of a patent after each item is produced

Question 17.
Rent payable to the landlord Rs. 5,00,000 is credited to
(a) Cash Account
(b) Landlord Account
(c) Outstanding Rent Account
(d) None of the above
Answer:
(c) Outstanding Rent Account

Question 18.
Bad debts entry is passed in …………
(a) Sales Book
(b) Cash Book
(c) journal Book
(d) None of the above
Answer:
(c) journal Book

Question 19.
Return of cash sales is recorded in …………..
(a) Sales Return book
(b) cash book
(c) Journal proper
(d) None of the above
Answer:
(b) cash book

Question 20.
John purchased goods ‘ Rs, 5900 for cash at 20% trade discount and 5% cash discount. Purchase account Is to fee debited by Rs ………….
(a) Rs, 3,800
(b) Rs. 5,000
(c) Rs. 3,750
(d) Rs. 4,000
Answer:
(d) Rs. 4,000

Question 21.
Out of the following Identify the wrong statement.
(a) Real & Personal accounts are transferred to Balance Sheet
(b) Nominal accounts are transferred to profit and loss account
(c) Cash account is not opened separately in the ledger
(d) Rent account is a personal account and outstand rent account is a Nominal Account
Answer:
(a) Real & Personal accounts are transferred to Balance Sheet

Question 22.
The financial position of a business concern is ascertained on the basis of ………….
(a) Records Prepared under book-keeping process
(b) Trial Balance
(c) Accounting reports
(d) None of the above
Answer:
(c) Accounting reports

Question 23.
Salary “payable to an employee Rs.50,000. Which account is to be credited?
(a) Cash Account
(b) Salaries Account
(c) Outstanding Salaries
(d) None
Answer:
(b) Salaries Account

Question 24.
Loss on sale of machinery Is credited to …………..
(a) Machinery Account
(b) Purchase Account
(c) Profit and Loss Account
(d) None of the above
Answer:
(a) Machinery Account

Question 25.
The Capital of a sole trader Is affected by ……………..
(a) Purchase of raw material
(b) Commission received
(c) Cash received from trade receivables
(d) Purchase of an asset for cash
Answer:
(b) Commission received

II. Very Short Answer Type Questions

Question 1.
From the following information below journalise its
i. Salary paid Rs. 5,000
ii. Rent paid to house owner Rs. 1,000
iii. Cash sales Rs. 10,000
Answer:
Journal Entries

Question 2.
Journalise the following transactions:
2017, Jan 11 Purchased goods for Rs. 1,500
13. Bought furniture for cash Rs. 2,000
19. Drew for private use Rs. 500
Answer:
Journal Entries

III. Short Answer Questions

Question 1.
Journalise the following transitions in the books of G.
i. Started business with cash Rs. 9,000
ii. Deposited into Canara Bank Rs, 3,000
iii. Paid Salary Rs, 5,000
iv. Received commission Rs. 200
v. Cash received from Rajan Rs. 400
Answer:
Journal Entries

IV. Additional Sums

Question 1.
Prepare accounting equation on the basis of the following.
1) Ramya started the business with cash Rs. 1,20,000
2) She Paid Rent Rs. 4,000
3) She Purchases Furniture Rs. 20,000
4) She Purchased goods on credit from Mr. Parthiban Rs. 60,000
5) She Sold goods (Cost Price Rs. 40,000) for Cash Rs. 50,000
Answer:
Accounting Equation

Question 2.
Show the Accounting equation the basis of the following transaction.
1. Pailavan Started business with cash Rs. 50,000
2. Purchased goods from Monisha Rs. 40,000
3. Sold goods to Aruna (Costing Rs. 36,000) for Rs. 50,000
4. wan withdraw from business Rs. 10,000
Answer:
Accounting Equation

Question 3.
Journalise the following transaction of Mrs. Padmini
2018 May 1. Received Cash from Siva – 75,000
8. Paid Cash to Sayed – 45,000
11. Bought Goods for cash – 27,000
12. Bought Furniture – 48,000
15. Sold Goods for cash – 70,000
18. Sold Furniture – 50,000
Answer:
Journal of Mrs. Padmini

Question 4.
Journalise the following transactions of Mr. Dharani.
2018 Jan
1. Mr. Dharani Started Business with Cash – 1,00,000
4. Sold Goods to Mohan – 70,000
10. Purchased goods from Bashyam – 50,000
20. Sold goods for cash from Natarajan – 70,000
25. Paid to Bashyam – 50,000
28. Received from Mohan – 70,000
Answer:
Journal of Mr. Dharani

Question 5.
Journalise the following transaction of Tmt. Rani
2018 Feb
1. Tmt Rani Started Business With cash – 300000
5. Opening a current account with Indian Overseas Bank – 50000
10. Bought goods from Sumathi – 90000
18. Paid to Sumathi – 90000
20. Sold Goods to Chitra – 126000
28. Chitra Settled her account
Answer:
Journal of Tmt. Rani

Question 6.
Journalise the following transactions of Mr.Jeevan.
2018 Mar 5. Sold goods to Arun on Credit – 17,500
10. Bought Goods for cash from Ravi – 22,500
12. Met Travelling Expenses – 2,500
18. Received Rs.80000 from Sivakumar as a loan
21. Paid wages to workers – 3,000
Answer:
Journal of Mr. Jeevan

Question 7.
Give Transaction with imaginary figure involving the following
(i) Increase in assets and capital
(ii) Increase and decrease in assets
(iii) Increase in assets and a liabilities
(iv) Decrease of an asset and owners capital
Answer:
(i) Mr. Raja Commenced Business with Rs. 2,00,000
(ii) Purchased Computer Rs. 15,000
(iii) Purchased goods from Suresh Rs. 80,000
(iv) Drawings Rs. 15,000

Question 8.
Journalise the following transaction of Mrs. Rama
2018 April 1. Mrs. Rama Commenced Business with cash – 30,000
2. Paid into Bank – 21,000
3. Purchases goods and paid through Debit Card – 15,000
7. Drew Cash from Bank for Personal Use – 3,000
15. Purchased goods from Siva – 15,000
20. Cash Sales – 30,000
23. Money Withdrawn from Bank thought ATM for office use – 1,000
25. Paid to Siva – 14,750
Discount Received – 250
31 Paid Rent – 500
Paid Salaries – 2,000
Answer:
Journal of Mrs. Rama

Question 9.
Record the transactions in the books of Shri. Ganesh & Co.
2012 Jan
1. Ganesh started a business with cash – 90,000
1. Paid into bank – 20,000
2. Purchased goods for cash – 50,000
4. Purchased furniture by issuing cheque – 24,000
5. Sold goods to Siva – 17,000
6. Purchased goods from Anand – 30,000
9. Returned goods to Anand – 3,000
10. Cash received from Siva – 16,000
11. Withdraw cash for personal use – 1,500
18. Purchase of stationary – 500
Answer:
Journal Entries in the Books of Shri. Ganesh & Co.

Question 10.
Record the transactions in the books of XYZ & Co:
2017
May 1. Furniture purchased for Rs. 10,000
2. Old Computer worth Rs. 13,700 sold to Sankar on credit
3. Rent paid Rs. 5,000
6. Depreciation on Machinery Rs. 6,400 paid by cheque
10. Paid Rs. 250 for the installation charges of Machinery
13. Cash taken for personal use Rs. 7,000
Answer:
Journal Entries in the books XYZ & Co.

Question 11.
Record the transaction in the books of Y.
2017, April 1. Started Business with cash Rs, 9,000
2. Purchased goods for cash Rs. 21,000
3. Sold goods for cash Rs. 8,000
4. Deposited into Canara Bank Rs. 3,000
5. Cash received from Rajan Rs. 4,000
7. Paid salary Rs. 3,000
8. Paid Rent Rs. 4,000
9. Received Commission Rs. 500
10. Withdrew from Canara Bank Rs. 2,700
Answer:
Journal Entries in the books of Y

Question 12.
Journalize the following transactions:
2016, Dec 1. Purchased goods for cash Rs. 14,000
8. Purchased goods from Anand Stores Rs. 600
12. Sold goods for Rs. 12,500
16. Sold goods to Sridhar Rs. 6,450
18. Bought Machinery for Cash Rs. 1,20,000
20. Goods returned to Anand Stores Rs. 300
22. Sridhar returned goods worth Rs. 450
25. Drew for personal use Rs. 500
25. Electric charges amounted Rs. 1,000
30. Got a loan from Joice Rs. 750
Answer:

Question 13.
Journalise the following transactions in the books of Z.
2016, july 1. Sold goods for Cash Rs. 2,800
2. Purchased goods for cash Rs, 5,900
5. Purchased goods from K Rs. 3,500
8. Sold goods to P Rs. 6,500
5. Received cash from P Rs. 6,200
9. Paid to K Rs. 3,200
9. Paid Telephone Charges Rs: 1,200
10. Paid Salary Rs. 2,600
10. Received Commission Rs. 3,000
10. Computer purchased from S for Rs. 20,000 and an advance of Rs. 8,000 is given as cash
Answer:

Question 14.
Varna is a women entrepreneur, dealing in textiles. From the following transactions, pass journal entries for the month of March 2017.
2017,
March, 1. Commenced business with cash Rs. 50,000 & with goods Rs. 45,000
2. Purchased 20 sarees from V & Co on credit Rs. 60,000
3. Cash deposited into bank Rs. 48,000
4. Paid to V & Co through net banking
5. Sold 10 sarees @ Rs. 3,000 each on credit to X & Co.
6. Paid Travelling expenses Rs. 450
8. Bank Charges levied Rs. 150
9. X becomes insolvent and only Rs. 2,000 per saree received by cash in final settlement.
10. Electricity charges amount to Rs. 1,500
10. Received Commission Rs. 2,000
Answer:

Question 15.
Journalise the following transactions:
2018, March 1. Salary paid Rs. 8,000
2. Rent paid to the house owner Rs. 7,000
3. Credit purchases from Mr. A Rs. 5,800
8. Cash sales Rs. 10,000
9. Deposited cash into bank Rs. 2,500
10. Purchased furniture Rs. 2,400
Answer:

Question 16.
Complete the accounting equation.
(a) Assets = Capital + Liabilities
Rs.2,00,000 = Rs. 1,60,000 + ?
Answer:
Rs. 40,000

(b) Assets = Capital + Liabilities
Rs. 1,00,000 = ? + Rs. 20,000
Answer:
Rs. 80,000

(c) Assets = Capital + Liabilities
? = Rs. 1,60,000 + Rs. 80,000
Answer:
Rs. 2,40,000

Question 17.
Journalise the following transactions in the books of S.
2017, March
1. M commenced business with Rs. 20,000
2. Remitted into bank account Rs. 18,000
3. Paid to K by cheque Rs. 5,000 and discount allowed Rs. 100
10. Cash sales Rs. 4,500
15. Issued a cheque to L for furniture Rs. 2,000
18. Withdrew from bank Rs. 500
25. Cash purchase Rs. 800
31. Paid salaries by cheque Rs. 6,000
Answer:

Question 18.
Write down the transactions for the following.

Answer:
1. Sale of building for Rs. 2,50,000
2. Paid Telephone Charges by Cheque Rs. 2,000
3. Withdrew Cash from Bank Rs. 5,000
4. Withdrew cash for personal use Rs. 3,000
5. Purchased goods for Rs. 6,400 and paid by cheque.

Students can download 5th Maths Term 1 Chapter 4 Measurements Ex 4 Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 1 Chapter 4 Measurements Ex 4

The ultimate Mile Calculator.

Question 1.
Fill in the blanks.

a. 7m 5cm = ______ cm
Answer:
7m 5cm = 7 × 100 + 5cm: 700 + 5 = 705 cm

b. 505 mm = ______ cm ______ mm
Answer:
505 mm = 505 + 100 = 50 cm 5 mm

c. 326m = ______ cm
Answer:
326 m = 326 × 100 = 32600 cm

d. 5km 30m = _____ m
Answer:
5 km 30 m : 5 × 1000 + 30 = 5000 + 30 = 5030 m

e. 650cm: ______ m ______ cm
Answer:
650 cm = 650 ÷ 100 = 6 m 50 cm

Question 2.
True or false.

a) 600 m is 6mm.
Answer:
False (∵6 cm)

b) 7000 m is 7 km.
Answer:
True

c) 400 cm is 4 km.
Answer:
False (∵4m)

d) 770 mm is 77 cm.
Answer:
True

e) 9000m is 90 mm.
Answer:
False (∵ 9000 × 1000 = 9000000 mm)

Question 3.
Find the sum of the following

a. 17 m 45 cm + 52 m 30 cm
Answer:

= 69 m 750 cm

b. 75 km 400 m + 37 km 300 m + 52 km 750 m
Answer:
400 + 300 + 750 = 1450 m
1450 m = 1 km 450 m

= 165 km 450 m

c. 4 cm 8 mm + 5 cm 9 mm
Answer:
8mm + 9mm = 17 mm = 1 cm 7 mm

= 1 cm 7mm

Question 4.
Subtract the following

a. 15 km 450 m – 13 km 200 m.
Answer:
Difference = 15 km 450 m – 13 km 200 m
= 2 km 250 m

b. 750 m 840 mm – 370m 480 mm.
Answer:
Difference = 750 m 840 mm – 370m 480 mm
= 380 m 360 mm

c. 5 km 400 m – 3 km 350 m
Answer:
Difference = 5 km 400 m – 3 km 350 m
= 2 km 50 mm

Question 5.
Multiply the following.

a. 350 m 45c m × 7
Answer:
45 × 7 = 315 m
= 3 m 15 m

350 m 45 cm × 7 = 2453 km 45 cm

b. 25 km 300 m × 6
Answer:
300 × 6 = 1800 m
= 1 km 800 m

25 km 300 m × 6 = 151 km 800 mm

c. 37 m 350 mm × 8
Answer:
350 × 8 = 2800 mm
= 2 m 800 mm

37 m 350 mm × 8 = 298 m 800 mm

Question 6.
Divide the following

a. 950 km 800 m ÷ 5
Answer:

950 km 800 ÷ 5 = 190 km 160 m

b. 49 m 770 mm ÷ 7
Answer:

49 m 770 mm ÷ 7 = 7 m 110 mm

c. 172 m 48 cm ÷ 4
Answer:

172 m 48 cm ÷ 4 = 43 m 12 cm

Life Related Problems

Question 7.
Answer the foliowing:

a. Saravanan had chosen to drive his vehicle from puducherry to Chennai for a distance of 165 Km. While starting his vehicle, the odometre showed 000157 Km, Find the reading of the odometre, when he reach Chennai?
Answer:
Distance form Pudicherry to Chennai = 165 km
Odometer showed = 000157 km
Reading of odometer after reached Chennai = 165 km + 000157 km
= 000322 km.

b. Karthik Raja decided to travel from A. He moves lKm in east to reach B. Then he goes 2 Km towards north and reaches C. Then he goes 1 Km towards west and reaches D. If he goes 2 Km towards South, Where would he reach? Draw apt Diagram and Justify. Also find out the total distance travelled by him.
Answer:

He would reach = A
Total Distance =1 + 2 + 1 +2 = 6 km

c. Sangeetha has just finished building a new house with garden area. She measured the garden area and found it to be 6m x 6m. Suppose she has to put a pole every lm, how many poles are required? Each pole is of height 1.5m from the ground, What should be the total length of the fencing material to fence the garden?
Answer:
Side of the Garden = 6m
No. of poles required = 6 + 6 +6+ 6= 24
The total length of the fencing material to fence the garden
= 24 × 1.5
= 36 m

d. One student needs lm 25 cm cloth to stich a shirt. What is the total length of clothes need to stitch a class of 22 students?
Answer:
Cloth is needed for one student = 1 m 25 cm
Total cloth is needed for 22 students
= 1 m 25 cm × 22
= 27 m 50 cm

e. The distance from village A to village B is 3 km 450 m. The distance from village B to village C is 5 km 350 m. What will be the total length of the road, if the road is laid from village A to village C?
Answer:
Distance from village A to Village B = 3 km 450 m
Distance from village B to village C = 5 km 350 m
The distance from A to C = 8 km 800 m

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements Ex 2.4

Question 1.
Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet distance from the wall to which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (π = 3.14).

Amswer:
Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.
Now the arc length BC gives the distance moved by the wheel.

Length of the arc = \(\frac{\theta}{360^{\circ}}\) × 2πr units
= \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 6 feets
= 3.14 × 3 feets
= 9.42 feets
∴ Distance moved by the wheel = 9.42 feets.

Question 2.
With his usual speed, if a person covers a circular track of radius 150 m in 9 minutes, find the distance that he covers in 3 minutes (π = 3.14).
Answer:
Radius of the circular track = 150m
Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units
Distance covered in 9 min = 2 × 3.14 × 150m
Distance covered in 1 min = \(\frac{2 \times 3.14 \times 150}{9} \mathrm{m}\)

Distance he covers in 3min = 314m

Question 3.
Find the area of the house drawing given in the figure.

Answer:
Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, b = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm
= (side × side) + (l × b) + (\(\frac { 1 }{ 2 }\) × b × h) + bh cm²
= (6 × 6) + (8 × 6) + (\(\frac { 1 }{ 2 }\) × 6 × 4) + (8 × 4) cm²
= 36 + 48 + 12 + 32 cm²
= 128 cm²
Required Area = 128 cm²

Question 4.
Draw the top, front and side view of the following solid shapes
(i)

Answer:
(a) Top view

(b) Front view

(c) Side view

(ii)

Answer:
(a) Top view

(b) Front view

(c) Side view

Challenging Problems

Expert Guide to Buying and Selling Feet on Feet Finder ·

Question 5.
Guna has fixed a single door of width 3 feet in his room where as Nathan has fixed a double door, each of width 1 \(\frac { 1 }{ 2 }\) feet in his room. From the closed position, if each of the single and double doors can open up to 120°, whose door takes a minimum area?
Answer:
Width of the door that Guna fixed = 3 feet.
When the door is open the radius of the sector = 3 feet

Angle covered = 120°
∴ Area required to open the door

= 3π feet²

(b) Width of the double doors that Nathan fixed = 1 \(\frac { 1 }{ 2 }\) feet.
Angle described to open = 120°
Area required to open = 2 × Area of the sector
z
= \(\frac { 1 }{ 2 }\) (3π) feet²
∴ The double door requires the minimum area.

Question 6.
In a rectangular field which measures 15 m x 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (π = 3.14).
Answer:
Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circle

Area of the rectangle = l × b units²
= 15 × 8m² = 120m²
Area of 4 quadrant circles = 4 × \(\frac { 1 }{ 4 }\) πr² units
Radius ofthe circle = 3m
Area of 4 quadrant circles = 4 × \(\frac { 1 }{ 4 }\) × 3.14 × 3 × 3 = 28.26 m²
Area of the circle at the middle = πr² units
= 3.14 × 3 × 3m² = 28.26m²
∴ Area where none of the cows can graze
= [120 – 28.26 – 28.26]m² = 120 – 56.52 m² = 63.48 m²

Question 7.
Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins. (π = 3.14) (√3 = 1.732)

Answer:
Given diameter of the coins = 6 cm
∴ Radius of the coins = \(\frac { 6 }{ 2 }\) = 3 cm
Area of the shaded region = Area of equilateral triangle – Area of 3 sectors of angle 60°

Area of the equilateral triangle = \(\frac{\sqrt{3}}{4}\) a² units² = \(\frac{\sqrt{3}}{4}\) × 6 × 6 cm²
= \(\frac{1.732}{4}\) × 6 × 6 cm² = 15.588 cm²
Area of 3 sectors = 3 × \(\frac{\theta}{360^{\circ}}\) × πr² sq.units
= 3 × \(\frac{60^{\circ}}{360^{\circ}}\) × 3.14 × 3 × 3 cm² = 1.458 cm²
∴ Area of the shaded region = 15.588 – 14.13 cm² = 1.458 cm²
Required area = 1.458 cm² (approximately)

Question 8.
Using Euler’s formula, find the unknowns.

Answer:
Euler’s formula is given by F + V – E = 2
(i) V = 6, E = 14
By Euler’s formula = F + 6 – 14 = 2
F = 2 + 14 – 6
F = 10

(ii) F = 8,E = 10
By Euler’s formula = 8 + V – 10 = 2
V = 2 – 8 + 10
V = 4

(iii) F = 20, V = 10
By Euler’s formula = 20 + 10 — E = 2
30 – E = 2
E = 30 – 2.
E = 28
Tabulating the required unknowns

Students can download 5th Maths Term 2 Chapter 2 Numbers Ex 2.2 Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 2 Chapter 2 Numbers Ex 2.2

Question 1.
Choose the best answer:

(i) The number divisible by 5 with no remainder
(a) 14
(b) 535
(c) 447
(d) 316
Answer:
(b) 535

(ii) Pick the number which is not a multiple of 6.
(a) 18
(b) 26
(c) 72
(d) 36
Hint:
(a) 18 = 6 × 3
(c) 72 = 8 × 9
(d) 36 = 6 × 6
Answer:
(b) 26

(iii) The common multiple of 4 and 8 among the given number is.
(a) 32
(b) 84
(c) 68
(d) 76
Hint:
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 ….
Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, …
Common multiples of 4, 8 is 8, 16, 24, 32, 40,…
Answer:
(a) 32

(iv) Factors of 6
(a) 1, 2, 3
(b) 1, 6
(c) 1,2,3,6
(d) 2,3
Hint: Multiple of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ……..
Answer:
(c) 1, 2, 3, 6

(v) Multiple of 9 is
(a) 79
(b) 87
(c) 29
(d) 72
Hint: Multiple of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, …….
Answer:
(d) 72

Question 2.
Fill in the blanks:

Question 1.
Factors of 7 ______
Answer:
1, 7

Question 2.
The only even prime number is ______
Answer:
1,2

Question 3.
L.C.M of 4, 12 ______
Answer:
12

Question 4.
L.C.M of 5, 15 ______
Answer:
15

Question 5.
The numbers which divides 35 without reminders are _____, ______, ______.
Answer:
1, 5, 7

Question 3.
Write down the factors of the given numbers.
(i) 25
(ii) 36
(iii) 14
(iv) 16
(v) 12
Answer:
(i) 25
Factors of 25 is 1, 5, 25

(ii) 36
Factors of 36 is 1, 2, 3, 4. 6, 9, 12, 18, 36

(iii) 14
Factors of 14 is 1, 2, 7, 14

(iv) 16
Factors of 16 is 1, 2, 4, 8, 16

(v) 12
Factors of 1, 2, 3. 4, 6, 12

Question 4.
Draw a picture of factor tree.

(i) 18
(ii) 33
(iii) 16
(iv) 50
Answer:
(i) 18

(ii) 33

(iii) 16

(iv) 50

Question 5.
Write down the first 5 multiples of the given numbers.
(i) 7
(ii) 9
(iii) 16
(iv) 11
(v) 21
Answer:
(i) 7
= 7 Multiple of 7 – 7, 14, 21, 28, 35

(ii) 9
= 9 Multiple of 9 – 9, 18, 27, 36, 45

(iii) 16
= 16 Multiple of 16 – 16, 32, 48, 64, 80

(iv) 11
= 11 Multiple of 11 is 11, 22, 33, 44, 55

(V) 21
= 21 Multiple of 21 is 21, 42, 63, 84, 105

Question 6.
Find first three common multiples of the given numbers.
(i) 24, 16
(ii) 12,9
(iii) 24,36
Answer:
(i) 24, 16
Multiples of 24 = 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264
Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192. 208, 224, 240, 256
Common multiples of 24 and 16 is 48, 96, 144, 192, 240

(ii) 12, 9
Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120,
Multiples of 9 – 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117
Common multiples of 12 and 9 is 36, 72, 108

(iii) 24, 36
Multiples of 24 = 24, 48, 72, 96, 120, 144, 168, 192, 216,
Multiples of 36 = 36, 72, 108, 144, 180, 216
Common multiples of 24 and 36 is 72, 144, 216

Question 7.
Find L. C.M of the given numbers.
(i) 12 and 28
(ii) 16 and 24
(iii) 8 and 14
Answer:
(i) 12 and 28
Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180
Multiples of 28 = 28, 56, 84, 112, 140, 168, 196, ……
Common multiples of 12 and 28 is 84, 168, ……..
LCM is 84

(ii) 16 and 24
Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 114, …….
Multiples of 24 = 24, 48, 72, 96, 120, 144, ……
Common multiples of 16 and 24 is 48, 96, 144, ……..
LCM is 48

(iii) 8 and 14
Multiples of 8 = 8, 16, 24, 32, 4O48, 56, 64, 72, 80, 88, 96, 104, 112, …
Multiples of 14 = 14, 28, 42, 56, 70,84 98 112 ……
Common multiples of 8 and 14 is 56, 112, …….

(iv) 30 and 20
Multiples of 30 = 30, 60, 90, 120, 150, 180, 210, 240, 270, …
Multiples of 20 = 20, 40, 60, 80, 100, 120, 140, 160, 180, …
Common multiples of 30 and 20 is 60, 120, 180
LCM of 30 and 20 is 60.

The least common multiple (LCM) of 12 and 18 is 36.

Question 8.
Ramya visits the gym five days once and Kavitha visits the gym six days once. In which day will they meet each other?
Answer:
Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, …
Multiples of 6 – 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, …
Common multiples of 5 and 6 is 30, 60
LCM of 5 and 6 is 30.

Question 9.
Arun and shahjahan goes for walking in a circular path of a park in the same direction. Arun takes 6 minutes to complete one round, while shahjahan takes 8 minutes to completed one round. In how many minutes will they meet each other?
Answer:
Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54,…
Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, …
Common multiples of 6 and 8 is 24, 48, …
LCM of 6 and 8 is 24.
They will meet each other after 24 minutes.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.2

Question 1.
Which of the following data can be represented in a histogram?
(i) The number of mountain climbers in the age group 20 to 60 in TamilNadu.
Answer:
Yes

(ii) Production of cycles in different years.
Answer:
No

(iii) The number of students in each class of a school.
Answer:
No

(iv) The number votes polled from 7 am to 6 pm in an election.
Answer:
Yes

(v) The wickets fallen from 1 over to 50th over in a one day cricket match.
Answer:
Yes

Question 2.
Fill in the blanks:
(i) The total area of the histogram is _________ to the total frequency of the given data.
Answer:
proportional

(ii) A graph that displays data that changes continuously over the periods of time is _________ .
Answer:
Histogram

(iii) Histogram is a graphical representation of_________ data.
Answer:
grouped

Question 3.
In a village, there are 570 people who have cell phones. An NGO survey their cell phone usage. Based on this survey a histogram is drawn. Answer the following questions.

(i) How many people use the cell phone for less than 3 hours?
Answer:
330 people (110 + 220)

(ii) How many of them use the cell phone for more than 5 hours?
Answer:
150 of them (100 + 50)

(iii) Are people using cell phone for less than 1 hour?
Answer:
No

Question 4.
Draw a histogram for the following data.

Answer:
The given (tata IS continuous frequency distribution taking class intervals on X axis and No. of students on Y-axis, the histogram is given below.

Question 5.
Construct a histogram from the following distribution of total marks of 40 students in a class.

Answer:
The given distribution is continuous taking marks on X axis and No. of students on Y-axis the histogram is constructed.

Question 6.
The distribution of heights (in cm ) of loo people is given below. Construct a histogram and the frequency polygon imposed on it.

Answer:
The given distribution is discontinuous.
Converting into continuous distribution we have
Lower boundary = lower limit – \(\frac { 1 }{ 2 }\) (gap between adjacent class interval)
= 125 – \(\frac { 1 }{ 2 }\) (1) = 124.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 }\) (gap between the adjacent class interval)
= 135 + \(\frac { 1 }{ 2 }\) = 135.5
∴ The new frequency table is

Question 7.
In a study of dental problem, the following data were obtained.

Represent the above data by a frequency polygon.
Answer:
Finding the midpoints of the class interval we get.

Ages	Mid point (x)	No. of patients
0 – 10	5	5
10 – 20	15	13
20 – 30	25	25
30 – 40	35	14
40 – 50	45	30
50 – 60	55	35
60 – 70	65	43
70 – 80	75	50

The points tobe plotted are A(5,5), B(15, 13), C(25, 25), D(35, 14), E(45, 30), G(65, 43.), H(75, 50) to obtain the frequency polygon ZABCDEFGHI.
Where I imagined class between 80 and 90.

Question 8.
The marks obtained by 50 students in Mathematics are given below (j) Make a frequency distribution table taking a class size of 10 marks (ii) Draw a histogram and a frequency polygon.

Answer:
Maximum marks obtained = 89
Minimum marks obtained = 08
Range = Maximum marks – Minimim marks
= 89 – 08
= 81
Taking the class size = 10, then

= \(\frac{81}{10}\) = 8.1 = 9

Now we have the continuous frequency table.

We will draw the histogram taking class interval in x-axis and frequency in y-axis as follows.

Objective Type Questions

Question 9.
Data is a collection of _________
(A) numbers
(B) words
(C) measurements
(D) all the three
Answer:
(D) all the three

Question 10.
The number of times an observation occurs in the given data is called _________
(A) tally marks
(B) data
(C) frequency
(D) none of these
Answer:
(C) frequency

This age difference calculator lets you quickly determine the age gap between two people.

Question 11.
The difference between the largest value and the smallest value of the given data is ________
(A) range
(B) frequency
(C) variable
(D) none of these
Answer:
(A) range

Question 12.
The data that can take values between a certain range is called_________
(A) ungrouped
(B) grouped
(C) frequency
(D) none of these
Answer:
(B) grouped

Question 13.
Inclusive series is a _________series.
(A) continuous
(B) discontinuous
(C) both
(D) none of these
Answer:
(B) discontinuous

Question 14.
In a class interval the upper limit of one class is the lower limit of the other class. This is _________ series.
(A) Inclusive
(B) exclusive
(C) ungrouped
(D) none of these
Answer:
(B) exclusive

Question 15.
The graphical representation of ungrouped data is ________
(A) histogram
(B) frequency polygon
(C) pie chart
(D) all the three
Answer:
(C) pie chart

Question 16.
Histogram is a graph of a ________ frequency distribution.
(A) continuous
(B) discontinuous
(C) discrete
(D) none of these
Answer:
(A) continuous

Question 17.
A _________ is a line graph for the graphical representation of the continuous frequency
distribution.
(A) frequency polygon
(B) histogram
(C) pie chart
(D) bar graph
Answer:
(A) frequency polygon

Question 18.
The graphical representation of grouped data is _________
(A) bar graph
(B) pictograph
(C) pie chart
(D) histogram
Answer:
(D) histogram

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.1

Question 1.
Fill in the blanks:
(i) If 30 % of x is 150, then x is _______ .
Answer:
500
Hint:
Given 30% of x is 150
i.e \(\frac{30}{100}\) × x = 150

∴ x = 500

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

(iii) If x% of x = 25, then x = _______ .
Answer:
50
Hint:
Given that x% of x is 25
∴ \(\frac{x}{100}\) × x = 25
∴ x² = 25 × 100 = 2500
∴ x = √2500 = 50

(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

= \(\frac{980}{14}\) = 70
% of boys = 70%

(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 × 100 = 52.52%

percent difference Calculator can be found by first finding the difference between the numbers, then finding the average of the numbers, and then dividing.

Question 2.
Rewrite each underlined part using percentage language.
(i) One half of the cake is distributed to the children.
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
∴ \(\frac { 1 }{ 2 }\) × 100 = 50%

(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 × 100 = 75%

(iii) The statue was made of pure silver.
Answer:
The statue was made of 100% pure silver
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, \(\frac{100}{100}\) × 100% = 100%

(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

(v) Only 2 persons out of 3 will be selected in the interview.
Answer:
Only 66\(\frac{2}{3}\)% 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
\(\frac{2}{3} \times 100=\frac{200}{3}=66 \frac{2}{3} \%\)

To convert CGPA to Percentage, we need to multiply CGPA by 9.5, which will give us the percentage.

Question 3.
48 is 32% of which number?
Answer:
Let the number required to be found be ‘x’
Given that 32% of x is 48
i.e., \(\frac{32}{100}\) × x = 48

∴ x = 150

Question 4.
What is 25% of 30% of 400?
Answer:
Required to find 25% of 30% of 400

The percentage decrease calculator determines the change from one amount to a lesser amount in terms of percent decrease.

Question 5.
If a car is sold for ₹ 2,00,000 from its original price of ₹ 3,00,000, then find the percentage of decrease in the value of the car.

Answer:
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 6.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of that number.
Answer:
Given that 75% of number less 60% of number is 82.5
Let the number be ‘x’
∴ \(\frac{75}{100}\) x x – \(\frac{60}{100}\) x x = 82.5
∴ 0.75 x – 0.60 x = 82.5
∴ 0.15 x = 82.5
∴ x = \(\frac{82.5}{0.15}=\frac{8250}{15}\) = 550
Required to find 20% of number ie 20% of x.

Question 7.
A number when increased by 18% gives 236. Find the number.
Answer:
Let the number be x. Given that when it is increased by 18%, we get 236.

Question 8.
A number when decreased by 20% gives 80. Find the number.
Answer:
Let the number be x. Given that when it is increased by 20% we get 80.

x = 100

It is important to learn to convert CGPA into percentage because both the systems are interconnected.

Question 9.
A number is increased by 25% and then decreased by 20%. Find the percentage change in that number.
Answer:
Method 1.
Let the number be x.
First it is increased by 25%
∴ It becomes x + \(\frac{25}{100}\) × x = \(\frac{125}{100}\)
Secondary it is decreased by 20%
\(\frac{125 x}{100}-\frac{20}{100} \times \frac{125}{100} x=\frac{125}{100} x \times \frac{80}{100}=x\)
Now we get back x, therefore there is no change.
Hence percentage change in that number is 0%

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

Now this 125 is decreased by 20%, we get
125 – \(\frac{25}{100}\) × 125 = 125 – 25 = 100
∴ We get back 100 ⇒ No change
Hence percentage change in that number is 0%

Question 10.
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.
Answer:
Let number of boys be ‘b’ 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% = \(\frac{16}{100}\) × b = \(\frac{16b}{100}\)
Failure in girls = 8% = \(\frac{8}{100}\) × g = \(\frac{8g}{100}\)
Pass in boys = 100 – 16% = 84% = \(\frac{84}{100} b\) …… (1)
Pass in girls = 100 – 8% = 92% = \(\frac{92}{100} g\) ……. (2)
From A, we have \(\frac{b}{g}=\frac{5}{3}\) , adding I 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) ……. (3)
Similarly b = \(\frac{5}{8}\) (b + g) ……. (4)
Total pass = pass in girls + pass in boys
= (1) + (2) = \(\frac{84}{100} b+\frac{92}{100} g\)

Objective Type Questions

Question 11.
12% of 250 litre is the same as ________ of 150 litre.
(A) 10%
(B) 15%
(C) 20%
(D) 30%
Answer:
(C) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) × 250 = 30 lit.
Percentage: \(\frac{30}{150}\) × 100 = 20%

Question 12.
If three candidates A, B and C in a school election got 153,245 and 102 votes respectively, then the percentage of votes got by 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

= \(\frac{245}{500}\) × 100 = 49%

Question 13.
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 \(\frac{15}{100}\) × 2500 = 375

Question 14.
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 = \(\frac{120 \times 100}{60}\) = 200

Question 15.
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 = 64% of x

x = 36

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 Numbers InText Questions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers InText Questions

Recap Exercise (Text Book Page No. 3)

Question 1.
The simplest form of \(\frac{125}{200}\) is
Answer:
\(\frac{125}{200}=\frac{125 \div 25}{200 \div 25}=\frac{5}{8}\)
= \(\frac{5}{8}\)

Question 2.
Which of the following is not an equivalent fraction of \(\frac{8}{12}\) ?
(A) \(\frac{2}{3}\)
(B) \(\frac{16}{24}\)
(C) \(\frac{32}{60}\)
(D) \(\frac{24}{36}\)
Answer:
(C) \(\frac{32}{60}\)
\(\frac{8}{12}=\frac{8+4}{12 \div 4}=\frac{2}{3}\)
\(\frac{8}{12}=\frac{8 \times 2}{12 \times 2}=\frac{16}{24}\)
\(\frac{8}{12}=\frac{8 \times 3}{12 \times 3}=\frac{24}{36}\)
But \(\frac{32}{60}=\frac{32 \div 5}{60 \div 5}=\frac{6.4}{12}\)
∴ \(\frac{32}{60}\) is not equivalent fraction of \(\frac{8}{12}\)

Question 3.
Which is bigger \(\frac{8}{9}\) or \(\frac{4}{5}\) ?
Answer:
LCM of 5 and 9 = 45

\(\frac{8}{9}\) is bigger than \(\frac{4}{5}\)

The LCM of 12 and 16 is 48.

Question 4.
Add the fractions : \(\frac{3}{5}+\frac{5}{8}+\frac{7}{10}\).
Answer:
LCM of 5, 8, 10 = 5 × 2 × 4
= 40

Question 5.
Simplify: \(\frac{1}{8}-\left(\frac{1}{6}-\frac{1}{4}\right)\)
Answer:

Question 6.
Multiply: \(2 \frac{3}{5}\) and \(1 \frac{4}{7}\).
Answer:
\(2 \frac{3}{5} \times 1 \frac{4}{7}=\frac{13}{5} \times \frac{11}{7}=\frac{143}{35}=4 \frac{3}{35}\)

Question 7.
Divide \(\frac{7}{36}\) by \(\frac{35}{81}\).
Answer:
\(\frac{7}{36}+\frac{35}{81}=\frac{7}{36} \times \frac{81}{35}=\frac{9}{20}\)

Question 8.

Answer:

Question 9.
In a city \(\frac{7}{20}\) of the population are women and \(\frac{1}{4}\) are children. Find the fraction of the population of men.
Answer:
Let the total population = 1
Population of men = Total population – Women – Children

∴ Population of men = \(\frac{2}{5}\)

Question 10.
Represent \(\left(\frac{1}{2}+\frac{1}{4}\right)\) by a diagram.
Answer:

Try These (Text Book Page No. 3)

Question 1.
Is the number -7 a rational number? Why?
Answer:
A rational number, Because – 7 = \(\frac{-14}{2}=\frac{p}{q}\)

Question 2.
Write any 6 rational numbers between 0 and 1.
Answer:
\(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}\)

Try These (Text Book Page No. 4)

That’s literally all there is to it! 1/8 as a decimal is 0.125.

Write the decimal forms of the following rational numbers:

Question 1.
\(\frac{4}{5}\)
Answer:
\(\frac{4}{5}=\frac{4 \times 20}{5 \times 20}=\frac{80}{100}=0.80\)

Question 2.
\(\frac{6}{25}\)
Answer:
\(\frac{6}{25}=\frac{6 \times 4}{25 \times 4}=\frac{24}{100}=0.24\)

5/9 as a decimal is 0.56

Question 3.
\(\frac{486}{1000}\)
Answer:
\(\frac{486}{1000}\) = 0.486

Question 4.
\(\frac{1}{9}\)
Answer:

\(\frac{1}{9}\) = 0.11….

Question 5.
\(3 \frac{1}{4}\)
Answer:

\(3 \frac{1}{4}\) = \(\frac{13}{4}\) = 3.25

That’s literally all there is to it! 5/8 as a decimal is 0.625.

Question 6.
\(-2 \frac{3}{5}\)
Answer:

\(-2 \frac{3}{5}\) = \(\frac{-13}{5}\) = -2.6

Try These (Text Book Page No. 6)

Question 1.
\(\frac{7}{3}=\frac{?}{9}=\frac{49}{?}=\frac{-21}{?}\)
Answer:

Question 2.
\(\frac{-2}{5}=\frac{?}{10}=\frac{6}{?}=\frac{-8}{?}\)
Answer:

Try These (Text Book Page No. 7)

Question 1.
Which of the following pairs represents equivalent rational numbers?
(i) \(\frac{-6}{4}, \frac{18}{-12}\)
(ii) \(\frac{-4}{-20}, \frac{1}{-5}\)
(iii) \(\frac{-12}{-17}, \frac{60}{85}\)
Answer:
(i) \(\frac{-6}{4}, \frac{18}{-12}\)
\(\frac{-6}{4}=\frac{-6 \times 3}{4 \times 3}=\frac{-18}{12}\)
∴ \(\frac{-6}{4}\) equivalent to \(\frac{-18}{12}\)

(ii) \(\frac{-4}{-20}, \frac{1}{-5}\)
\(\frac{-4}{-20}=\frac{-4 \div(-4)}{-20 \div(-4)}=\frac{1}{5} \neq-\frac{1}{5}\)
∴ \(\frac{-4}{-20}\) equivalent to \(\frac{1}{-5}\)

(iii) \(\frac{-12}{-17}, \frac{60}{85}\)
\(\frac{-12}{-17}=\frac{-12 x-5}{-17 x-5}=\frac{60}{85}\)
∴ \(\frac{-12}{-17}\) equivalent to \(\frac{60}{85}\)

Question 2.
Find the standard form of
(i) \(\frac{36}{-96}\)
(ii) \(\frac{-56}{-72}\)
(iii) \(\frac{27}{18}\)
Answer:
(i) \(\frac{36}{-96}\)
= \(\frac{-36 \div 12}{96 \div 12}=\frac{-3}{8}\)

(ii) \(\frac{-56}{-72}\)
= \(\frac{-56 \div(-8)}{-72 \div(-8)}=\frac{7}{9}\)

(iii) \(\frac{27}{18}\)
= \(1 \frac{9}{18}=1 \frac{1}{2}\)

Question 3.
Mark the following rational numbers on a number line.
(i) \(\frac{-2}{3}\)
Answer:
\(\frac{-2}{3}\) lies betveen 0 and -1.
T?ìe unit part between O and —lis divided into 3 equal parts and second part is taken.

(ii) \(\frac{-8}{-5}\)
Answer:
\(\frac{-8}{-5}\) = \(1 \frac{3}{5}\)
\(1 \frac{3}{5}\) lies between I and 2, The unit part between I and 2 is divided into 5 equal parts and the third part is taken.

(iii) \(\frac{5}{-4}\)
Answer:
\(\frac{5}{-4}\) = \(-\frac{5}{4}\) = \(-1 \frac{1}{4}\)
\(-1 \frac{1}{4}\) lies between -1 and -2. The unit part between -1 and -2 is divided into four equal parts and the first part is taken.

Think (Text Book Page No. 15)

Is zero a rational number? If so, what is its additive inverse
Answer:
Yes zero a rational number Additive inverse of zero is zero.

Think (Text Book Page No. 16)

What is the multiplicative inverse of 1 and -1?
Answer:
Multiplicative inverse of 1 is 1 and -1 is -1.

Try These (Text Book Page No. 16)

Divide
(i) \(\frac{-7}{3}\) by 5
(ii) 5 by \(\frac{-7}{3}\)
(iii) \(\frac{-7}{3}\) by \(\frac{35}{6}\)
Answer:
(i) \(\frac{-7}{3}\) by 5

(ii) 5 by \(\frac{-7}{3}\)

(iii) \(\frac{-7}{3}\) by \(\frac{35}{6}\)

Try These (Text Book Page No. 20)

The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Answer:
Let 0 and \(\frac{1}{2}\) te two rational numbers 0 – \(\frac{1}{2}\) = –\(\frac{1}{2}\) is a rational numter
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number \(\frac{5}{2}\) and 0.
\(\frac{5}{2}\) + 0 = \(\frac{5}{2 \times 0}=\frac{5}{0}\)
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.

Try These (Text Book Page No. 22)

(i) Is \(\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}\) ?
Answer:

LHS ≠ RHS
∴ \(\frac{3}{5} \div \frac{7}{8}\) ≠ \(\frac{7}{8}-\frac{3}{5}\)
∴ Subtraction of rational numbers is not commutative.

(ii) \(\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}\)? So, what do you conclude?
Answer:

LHS ≠ RHS
∴ \(\frac{3}{5} \div \frac{7}{8}\) ≠ \(\frac{7}{8} \div \frac{5}{3}\)
∴ Commutative property not hold good br division of rational numbers.

Try This (Text Book Page No. 22)

Check whether associative property holds for subtraction and division.
Answer:
Consider for rational numbers \(\frac{2}{3}, \frac{1}{2}\) and \(\frac{3}{4}\)

∴ Associative property not holds for subraction of rational numbers


∴ Associative property not holds for division of rational numbers

Try This (Text Book Page No. 25)

Question 1.
Observe that,
\(\frac{1}{1.2}+\frac{1}{2.3}=\frac{2}{3}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=\frac{3}{4}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}=\frac{4}{5}\)
Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.
Answer:

Think (Text Book Page No. 26)

Question 1.
Is the square of a prime number, prime?
Answer:
No, the square of a prime number ‘P’ has at Least 3 divisors 1, P and P². But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.

Question 2.
Will the sum of two perfect squares always be a perfect square? What about their difference and their product?
Answer:
The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. Bu the product of two perfect square is a perfect square.

Try These (Text Book Page No. 26)

Question 1.
Which among 256, 576, 960, 1025,4096 are perfect square numbers? (Hint: Try to extend the table of squares already seen).
Answer:
256 = 16²
576 = 24²
4096 = 64²
∴ 256, 576, and 4096 are perfect squares

Question 2.
One can judge just by look that each of the following numbers 82, 113, 1972, 2057, 8888, 24353 is not a perfect square. Explain why?
Answer:
Because the unit digit ola perfect square will be 0, 1,4, 5, 6, 9. But the given numbers have unit digits 2, 3, 7, 8. So they are not perfect squares.

Think (Text Book Page No. 27)

Consider the claim: “Between the squares of the consecutive numbers n and (n + 1), there are 2n non-square numbers’ Can it be true? FInd how many non-square numbers are there
(i) between 4 and 9 ?
(ii) between 49 and 64? and Verify the claim.
Answer:

Therefore we conclude that there are 2n non-square numbers between two consecutive square numbers.

Think (Text Book Page No. 30)

In this quick guide we’ll describe what the factors of 96 are, how you find them and list out the factor pairs of 96 for you to prove the calculation works.

In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?
Answer:
108 = 2 × 2 × 3 × 3 = 2² × 3² × 3
If we multiply the factors by 2, then we get
2² × 3² × 3 × 3 = 2² × 3² × 3² = (2 × 3 × 3)²
Which is perfect square.
∴ Again if we divide by 3 then we get 2² × 3² ⇒ (2 × 3)², a perfect square.
∴ We have to multiply or divide 108 by 3 to get a perfect square.

Try These (Text Book Page No. 32)

Find the square root by long division method.
Question 1.
400
Answer:

√400 = 20

Question 2.
1764
Answer:

√1764 = 42

Question 3.
9801
Answer:

√9801 = 66

Try These (Text Book Page No. 32)

without calculating the square root, guess the number of digits in the square root of the following numbers:
Question 1.
14400
Answer:
\(\sqrt{14400}\) = \(\sqrt{144 \times 100}\)
= \(\sqrt{144} \times \sqrt{100}\)
= 12 × 10
= 120

Question 2.
390625
Answer:
\(\sqrt{390625}\) = \(\sqrt{25 \times 25 \times 25 \times 25}\)
= \(\sqrt{25 \times 25} \times \sqrt{25 \times 25}\)
= 25 × 25
= 625

Question 3.
100000000
Answer:
\(\sqrt{100000000}\) = \(\sqrt{10000 \times 10000}\)
= \(\sqrt{10000} \times \sqrt{10000}\)
= 100 × 100
= 10,000

Try These (Text Book Page No. 33)

Find the square root of
Question 1.
5.4756
Answer:

Question 2.
19.36
Answer:

Question 3.
116.64
Answer:

Think (Text Book Page No. 33)

Try to fill in the blanks using √ab = √a × √b

Answer:

Try These (Text Book Page No. 34)

Using this method, find the square root of the numbers 1.2321 and 11.9025.
Answer:
(i) 1.2321
√1.2321 = \(\sqrt{\frac{12321}{10000}}\)
= \(\frac{111}{100}\) = 1.11

(ii) 11.9025
√11.9025 = \(\frac{\sqrt{119025}}{\sqrt{10000}}\)
= \(\frac{345}{100}\) = 3.45

Try These (Text Book Page No. 34)

Write the numbers in ascending order.
Question 1.
4, √14, 5
Answer:
Squaring all the numbers we get 4², (√14)², 5² ⇒ 16, 14, 25
∴ Ascending order: 14, 16, 25
Ascending order: √14, 4, 5

Question 2.
7, √65, 8
Answer:
Squaring 7, √65 and 8 we get 7², (√65)², 8² ⇒ 49, 65, 64
Ascending order : 49, 64, 65
Ascending order: 7, 8, √65

Try These (Text Book Page No. 37)

Find the ones digit in the cubes of each of the following numbers.
(i) 12
(ii) 27
(iii) 38
(iv) 53
(v) 71
(vi) 84
Answer:
(i) 12
12 ends with 2, so its cube ends with 8 i.e. ones digit in 12³ is 8.

(ii) 27
27 ends with 7, so its cube end with 3. i.e., ones digit in 27³ is 3.

(iii) 38
38 ends with 8, so its cube ends with 2 i.e. ones digit in 38³ is 2.

(iv) 53
53 ends with 3, so its cube ends with 7. i.e, ones digit in 53³ is 7.

(v) 71
71 ends with 1, so its cube ends with 1. i.e. ones digit in 71³ is 1

(vi) 84
84 ends with 4, so its cube ends with 4. i.e, ones digit in 84³ is 4.

Try These (Text Book Page No. 41)

Expand the following numbers using exponents:
(i) 8120
(ii) 20,305
(iii) 3652.01
(iv) 9426.521
Answer:
(i) 8120
8120 = (8 × 1000) + (1 × 100) + (2 x×10) + 0 × 1
= (8 × 10³) + (1 × 10²) + (2 × 10¹)

(ii) 20,305
20305 = (2 × 10000) + (0 × 1000) + (3 × 100) + (0 × 10) + (5 × 1)
= (2 × 10⁴) + (3 × 10²) + 5

(iii) 3652.01
3652.01 = 3000 + 600 + 50 + 2 + \(\frac{0}{10}+\frac{1}{100}\)
= (3 × 1000) + (6 × 100) + (5 × 10) + (2 × 1) + (1 × \(\frac{1}{100}\))
= (3 × 10³) + (6 × 10²) + (5 × 10¹) + 2 + (1 × 10^-2)

(iv) 9426.521
= (9 × 1000) + (4 × 100) + (2 × 10) + (6 × 1) + \(\left(\frac{5}{10}\right)+\left(\frac{2}{100}\right)+\left(\frac{1}{1000}\right)\)
= (9 × 10³) + (4 × 10²) + (2 × 10¹) + 6 + (5 × 10^-1) + (2 × 10^-2) + (1 × 10^-3)

Try These (Text Book Page No. 42)

Verify the following rules (as we did above). Here, a,b are non-zero integers and are any integers.
1. Product of same powers to power of product rule: a^m × b^m = (ab)^m
2. Quotient of same powers to power of quotient rule: \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)
3. Zero exponent rule: a⁰ = 1.
Answer:
let a = 2; b = 3; m = 2
1. a^m × b^m = 2² × 3²
= 4 × 9 = 36 = (2 × 3)²

2. \(\frac{a^{m}}{b^{m}}=\frac{2^{2}}{3^{2}}=\frac{4}{9}=\left(\frac{2}{3}\right)^{2}\)

3. a⁰ = 2⁰ = 1.

Try These (Text Book Page No. 44)

Question 1.
Write in standard form: Mass of planet Uranus is 8.68 × 10²⁵ kg.
Answer:
Mass of Planet Uranus = 86800000000000000000000000 kg
[23 zeros after 88]

Question 2.
Write in scientific notation:
(i) 0.000012005
Answer:
0.000012005 = 1.2005 × 10^-5

(ii) 43 12.345
Answer:
43 12.345 = 4.312345 × 10³

(iii) 0.10524
Answer:
0.10524 = 1.0524 × 10^-1

(iv) The distance between the Sun and the planet Saturn 1.4335 × 10¹² miles.