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Students can Download Accountancy Chapter 3 Accounts of Partnership Firms-Fundamentals Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Accountancy 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 Accountancy Solutions Chapter 3 Accounts of Partnership Firms-Fundamentals

Samacheer Kalvi 12th Accountancy Accounts of Partnership Firms-Fundamentals Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
In the absence of a partnership deed, profits of the firm will be shared by the partners in …………….
(a) Equal ratio
(b) Capital ratio
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Equal ratio

Question 2.
In the absence of an agreement among the partners, interest on capital is …………….
(a) Not allowed
(b) Allowed at bank rate
(c) Allowed @ 5% per annum
(d) Allowed @ 6% per annum
Answer:
(a) Not allowed

Question 3.
As per the Indian Partnership Act, 1932, the rate of interest allowed on loans advanced by partners is …………….
(a) 8% per annum
(b) 12% per annum
(c) 5% per annum
(d) 6% per annum
Answer:
(d) 6% per annum

Question 4.
Which of the following is shown in Profit and loss appropriation account?
(a) Office expenses
(b) Salary of staff
(c) Partners’ salary
(d) Interest on bank loan
Answer:
(c) Partners’ salary

Question 5.
When fixed capital method is adopted by a partnership firm, which of the following items will appear in capital account?
(a) Additional capital introduced
(b) Interest on capital
(c) Interest on drawings
(d) Share of profit
Answer:
(a) Additional capital introduced

Question 6.
When a partner withdraws regularly a fixed sum of money at the middle of every month, period for which interest is to be calculated on the drawings on an average is …………….
(a) 5.5 months
(b) 6 months
(c) 12 months
(d) 6.5 months
Answer:
(b) 6 months

Question 7.
Which of the following is the incorrect pair?
(a) Interest on drawings – Debited to capital account
(b) Interest on capital – Credited to capital account
(c) Interest on loan – Debited to capital account
(d) Share of profit – Credited to capital account
Answer:
(c) Interest on loan – Debited to capital account

Question 8.
In the absence of an agreement, partners are entitled to …………….
(a) Salary
(b) Commission
(c) Interest on loan
(d) Interest on capital
Answer:
(c) Interest on loan

Question 9.
Pick the odd one out …………….
(a) Partners share profits and losses equally
(b) Interest on partners’ capital is allowed at 7% per annum
(c) No salary or remuneration is allowed
(d) Interest on loan from partners is allowed at 6% per annum.
Answer:
(b) Interest on partners’ capital is allowed at 7% per annum

Question 10.
Profit after interest on drawings, interest on capital and remuneration is ₹ 10,500. Geetha, a partner, is entitled to receive commission @ 5% on profits after charging such commission. Find out commission. …………….
(a) ₹ 50
(b) ₹ 150
(c) ₹ 550
(d) ₹ 500
Answer:
(d) ₹ 500

II. Very short answer questions

Question 1.
Define partnership.
Answer:
According to section 4 of the Indian Partnership Act, 1932, the partnership is defined as, “the relation between persons who have agreed to share the profits of a business carried on by all or any of them acting for all.

Question 2.
What is a partnership deed?
Answer:
A partnership deed is a document in writing that contains the terms of the agreement among the partners. It is not compulsory for a partnership to have a partnership deed as per the Indian Partnership Act, 1932. But, it is desirable to have a partnership deed as it serves as evidence of the terms of the agreement among the partners.

Question 3.
What is meant by the fixed capital method?
Answer:
Under the fixed capital method, the capital of the partners is not altered and it remains generally fixed. Two accounts are maintained for each partner namely

• Capital account
• Current account.

The transactions relating to initial capital introduced, additional capital introduced, and capital permanently withdrawn are entered in the capital account and all other transactions are recorded in the current account.

Question 4.
What is the journal entry to be passed for providing interest on capital to a partner?
Answer:
(a) For providing interest on capital

(b) For closing interest on capital account

Question 5.
Why is the Profit and loss appropriation account prepared?
Answer:
The profit and loss appropriation account is an extension of the profit and loss account prepared for the purpose of adjusting the transactions relating to amounts due to and amounts due from partners. It is a nominal account in nature. It is credited with net profit, interest on drawings and it is debited with interest on capital, salary, and other remuneration to the partners. The balance being the profit or loss is transferred to the partners’ capital or current account in the profit-sharing ratio.

III. Short answer questions

Question 1.
State the features of a partnership.
Answer:

1. A partnership is an association of two or more persons. The maximum number of partners is limited to 50.
2. There should be an agreement among the persons to share the profit or loss of the business. The agreement may be oral or written or implied.
3. The agreement must be to carry on a business and to share the profits of the business.
4. The business may be carried on by all the partners or any of them acting for all.

Question 2.
State any six contents of a partnership deed.
Answer:
The contents of the partnership deed are:

1. The name of the firm and nature and place of business.
2. Date of commencement and duration of business.
3. Names and addresses of all partners.
4. Capital contributed by each partner.
5. Profit-sharing ratio.
6. Amount of drawings allowed to each partner.

Question 3.
State the differences between the fixed capital method and the fluctuating capital method.
Answer:

Question 4.
Write a brief note on the applications of the provisions of the Indian Partnership Act, 1932 in the absence of a partnership deed.
Answer:

1. Remuneration to partners: No salary or remuneration is allowed to any partner; op [ Section 13 (a)]
2. Profit-sharing ratio: Profits and losses are to be shared by the partners equally. [ Section 13 (b)]
3. Interest on capital: No interest is allowed on the capital. Where a partner is entitled to interest on capital contributed as per partnership deed, such interest on capital will be payable only out of profits. [ Section 13 (c)]
4. Interest on loans advanced by partners to the firm: Interest on the loan is to be allowed at the rate of 6 percent per annum. [ Section 13 (d)]
5. Interest on drawings: No interest is charged on the drawings of the partners.

Question 5.
Jayaraman is a partner who withdrew ₹ 10,000 regularly in the middle of every month. Interest is charged on the drawings at 6% per annum. Calculate interest on drawings for the year ended 31st December 2018.
Answer:
Jayaraman:
Interest on drawings: 10,000 × \(\frac { 12 }{ 24 }\) × \(\frac { 6 }{ 100 }\) × 12 = ₹ 3600

IV. Exercises

Question 1.
Akash, Bala, Chandru, and Daniel are partners in a firm. There is no partnership deed. How will you deal with the following?

1. Akash has contributed maximum capital. He demands interest on capital at 10% per annum.
2. Bala has withdrawn ₹ 3,000 per month. Other partners ask Bala to pay interest on drawings @ 8% per annum to the firm. But, Bala did not agree to it.
3. Akash demands the profit to be shared in the capital ratio. But, others do not agree.
4. Daniel demands a salary at the rate of ₹ 10,000 per month as he spends full time for the business.
5. The loan advanced by Chandru to the firm is ₹ 50,000. He demands interest on loan @ 12% per annum.

Answer:

1. No interest on capital is payable to any partner.
2. No interest is charged on the drawing made by the partner.
3. Profit should be distributed equally.
4. No remuneration is payable to any partner.
5. Interest on the loan is payable at 6% per annum.

Question 2.
From the following information, prepare capital accounts of partners Rooban and Deri, when their capitals are fixed.

Answer:
Capital Account

Current Account

Question 3.
Arun and Selvam are partners who maintain their capital accounts under fixed capital method. From the following particulars, prepare capital accounts of partners.

Answer:
Capital Account

Current Account

Question 4.
From the following information, prepare capital accounts of partners Padmini and Padma, when their capitals are fluctuating.

Answer:

Question 5.
Mannan and Ramesh share profits and losses in the ratio of 3:2 and their capital on 1^st April, 2018 was Mannan ₹ 1,50,000 and Ramesh ₹ 1,00,000 respectively and their current accounts show a credit balance of’ 25,000 and ₹ 20,000 respectively. Calculate interest on capital at 6% p.a. for the year ending 31^st March, 2019 and show the journal entries.
Answer:

Question 6.
Prakash and Supria were partners who share profits and losses in the ratio of 5:3. Balance in their capital account on 1^st April, 2018 was Prakash ₹ 3,00,000 and Supria ₹ 2,00,000. On 1^st July, 2018 Prakash introduced additional capital of ₹ 60,000. Supria introduced additional capital of ₹ 30,000 during the year. Calculate interest on capital at 6% p.a. for the year ending 31^st March, 2019 and show the journal entries.
Answer:

Journal Entries

Question 7.
The capital account of Begum and Fatima on 1^st January, 2018 showed a balance of ₹ 50,000 and ₹ 40,000 respectively. On 1^st October, 2018, Begum introduced an additional capital of? 10,000 and on 1^st May, 2018 Fatima introduced an additional capital of ₹ 9,000.
Answer:
Calculate interest on capital at 4% p.a. for the year ending 31^st December 2018.

Question 8.
From the following balance sheets of Subha and Sudha who share profits and losses in 2:3, calculate interest on capital at 5% p.a. for the year ending 31st December 2018.
Balance sheet as on 31^st December, 2018

Drawings of Subha and Sudha during the year were ₹ 8,000 and ₹ 10,000, respectively. Profit earned during the year was ₹ 30,000.
Answer:

Question 9.
From the following balance sheets of Rajan and Devan who share profits and losses 2:1, calculate interest on capital at 6% p.a. for the year ending 31^st December 2018.
Balance sheet as on 31^st December 2018

On 1^st April, 2018, Rajan introduced an additional capital of ₹ 40,000 and on 1^st September, 2018, Devan introduced ₹ 30,000. Drawings of Rajan and Devan during the year were ₹ 20,000 and ₹ 10,000 respectively. Profit earned during the year was ₹ 70,000.
Answer:

Question 10.
Ahamad and Basheer contribute ₹ 60,000 and ₹ 40,000 respectively as capital. Their respective share of profit is 2 : 1 and the profit before interest on capital for the year is ₹ 5,000. Compute the amount of interest on capital in each of the following situations:

1. if the partnership deed is silent as to the interest on capital
2. if the interest on capital @ 4% is allowed as per the partnership deed
3. if the partnership deed allows interest on capital @ 6% per annum.

Answer:
1. No Interest on capital is allowed.

2. Since the profit is sufficient, Interest on capital will be provided.
Ahamad:
60, 000 × \(\frac { 4 }{ 100 }\) = ₹ 2, 400
Basheer:
40, 000 × \(\frac { 4 }{ 100 }\) = ₹ 1, 600

3. Since the profit is insufficient, Interest on capital will be provided.
Ahamad:
60, 000 × \(\frac { 6 }{ 100 }\) = ₹ 3, 600
Basheer:
40, 000 × \(\frac { 6 }{ 100 }\) = ₹ 2, 400
Profit of 5,000 will be distributed to the partners in their capital ratio of 3:2.

Question 11.
Mani is a partner, who withdrew ₹ 30,000 on 1^st September, 2018. Interest on drawings is charged at 6% per annum. Calculate interest on drawings on 31^st December, 2018 and show the journal entries by assuming that the fluctuating capital method is followed.
Answer:
Mani:
30, 000 × \(\frac { 6 }{ 100 }\) = ₹ 600
Interest on drawings of Mani = ₹ 600.

Question 12.
Santhosh is a partner in a partnership firm. As per the partnership deed, interest on drawings is charged at 6% per annum. During the year ended 31^st December, 2018 he withdrew as follows:

Calculate the amount of interest on drawings.
Interest on Drawings = Amount x Rate of Interest x Period
Feb 1 ⇒ 2,000 x \(\frac { 6 }{ 100 }\) = ₹ 600
May 1 ⇒ 10, 000 x \(\frac { 6 }{ 100 }\) x \(\frac { 8 }{ 12 }\) = ₹ 400
July 1 ⇒ 4,000 x \(\frac { 6 }{ 100 }\) x \(\frac { 6 }{ 12 }\) = ₹ 120

Question 13.
Kumar is a partner in a partnership firm. As per the partnership deed, interest on drawings is charged at 6% per annum. During the year ended 31^st December, 2018 he withdrew as follows:

Interest on Drawings = Amount x Rate of Interest x Period
March 1 ⇒ 4, 000 x \(\frac { 6 }{ 100 }\) = ₹ 200
June 1 ⇒ 4, 000 x \(\frac { 6 }{ 100 }\) x \(\frac { 7 }{ 12 }\) = ₹ 140
Sep 1 ⇒ 4,000 x \(\frac { 6 }{ 100 }\) x \(\frac { 4 }{ 12 }\) = ₹ 80

Question 14.
Mathew is a partner who withdrew ₹ 20,000 during the year 2018. Interest on drawings is charged at 10% per annum. Calculate interest on drawings on 31^st December 2018.
Mathew:
20 000 x \(\frac { 10 }{ 100 }\) x \(\frac { 6 }{ 12 }\) = ₹ 1, 000

Question 15.
Santhosh is a partner in a partnership firm. As per the partnership deed, interest on drawings is charged at 6% per annum. During the year ended 31^st December, 2018 he withdrew as follows:

Calculate the amount of interest on drawings by using product method.

Interest on drawings = Product x Rate of interest x \(\frac { 1 }{ 12 }\)
= 1,44,000 x \(\frac { 6 }{ 100 }\) x \(\frac { 1 }{ 12 }\) = ₹ 720

Question 16.
Kavitha is a partner in a firm. She withdraws ₹ 2,500 p.m. regularly. Interest on drawings is charged @ 4% p.a. Calculate the interest on drawings using average period, if she draws

1. At the beginning of every month
2. In the middle of every month
3. At the end of every month

Answer:
1. At the beginning of every month = 2, 500 x 12 x \(\frac { 4 }{ 100 }\) x \(\frac { 13 }{ 24 }\) = ₹ 650
2. In the middle of every month = 2,500 x 12 x \(\frac { 4 }{ 100 }\) x \(\frac { 12 }{ 24 }\) = ₹ 600
3. At the end of every month = 2, 500 x 12 x \(\frac { 4 }{ 100 }\) x \(\frac { 11 }{ 24 }\) = ₹ 550

Question 17.
Kevin and Francis are partners. Kevin draws ₹ 5,000 at the end of each quarter. Interest on drawings is chargeable at 6% p.a. Calculate interest on drawings for the year ending 31^st March 2019 using the average period.
Answer:
Calculation of interest on drawings of Kevin.
Total amount of drawings = 5000 x 4 = 20,000

= 20,000 x \(\frac { 6 }{ 100 }\) x \(\frac { 4.5 }{ 12 }\) = ₹ 450

Question 18.
Ram and Shyam were partners. Ram withdrew ₹ 18,000 at the beginning of each half year. Interest on drawings is chargeable @ 10% p.a. Calculate interest on the drawings for the year ending 31^st December 2018 using the average period.
Answer:
Total amount of drawing: 18, 000 x 2 = 36,000
Interest on drawings = Amount x Rate of Interest x
= 36,000 x \(\frac { 10 }{ 100 }\) x \(\frac { 9 }{ 12 }\) = ₹ 2700

Question 19.
Janani, Kamali and Lakshmi are partners in firm sharing profits and losses equally. As per the terms of the partnership deed, Kamali is allowed a monthly salary of ₹ 10,000 and Lakshmi is allowed a commission of ₹ 40,000 per annum for their contribution to the business of the firm. You are required to pass the necessary journal entry. Assume that their capitals are fluctuating.
Answer:

Question 20.
Sibi and Manoj are partners in a firm. Sibi is to get a commission of 20% of net profit before charging any commission. Manoj is to get a commission of 20% on net profit after charging all commission. Net profit for the year ended 31^st December 2018 before charging any commission was ₹ 60,000. Find the commission of Sibi and Manoj. Also, show the distribution of profit.
Answer:

Question 21.
Anand and Narayanan are partners in firm sharing profits and losses in the ratio of 5 : 3. On 1^st January 2018, their capitals were ₹ 50,000 and ₹ 30,000, respectively. The partnership deed specifies the following:

1. Interest on capital is to be allowed at 6% per annum.
2. Interest on drawings charged to Anand and Narayanan are ₹ 1,000 and ₹ 800, respectively.
3. The net profit of the firm before considering interest on capital and interest on drawings amounted to ₹ 35,000.

Give necessary journal entries and prepare profit and loss appropriation account as of 31^st December 2018. Assume that the capitals are fluctuating.
Answer:
Profit and Loss Appropriation Account

Journal Entries

Question 22.
Dinesh and Sugumar entered into a partnership agreement on 1^st January 2018, Dinesh contributing ₹ 1,50,000 and Sugumar ₹ 1,20,000 as capital. The agreement provided that:

1. Profits and losses to be shared in the ratio 2 : 1 as between Dinesh and Sugumar.
2. Partners to be entitled to interest on capital @ 4% p.a.
3. Interest on drawings to be charged Dinesh: ₹ 3,600 and Sugumar: ₹ 2,200
4. Dinesh to receive a salary of ₹ 60,000 for the year, and
5. Sugumar to receive a commission of ₹ 80,000

During the year ended on 31^st December 2018, the firm made a profit of ₹ 2,20,000 before adjustment of interest, salary, and commission.
Prepare the Profit and loss appropriation account.
Profit and Loss Appropriation Account

Question 23.
Antony and Ranjith started a business on 1st April 2018 with capitals of ₹ 4,00,000 and ₹ 3,00,000 respectively. According to the Partnership Deed, Antony is to get the salary of ₹ 90,000 per annum, Ranjith is to get 25% commission on profit after allowing salary to Antony and interest on capital @ 5% p.a. but before charging such commission. Profit-sharing ratio between the two partners is 1:1. During the year, the firm earned a profit of ₹ 3,65,000.
Answer:
Prepare profit and loss appropriation account. The firm closes its accounts on 31^st March every year.

Samacheer Kalvi 12th Accountancy Accounts of Partnership Firms-Fundamentals Additional Questions and Answers

I. Choose the correct answer

Question 1.
Under the fixed capital system, the capitals of the partners ……………. year after year.
(a) keep changing
(b) remain fixed
(c) Both are possible
Answer:
(b) remain fixed

Question 2.
Under fluctuating capital system, the capitals of the partners year after years …………….
(a) keep changing
(b) Remain fixed
(c) Both are possible
Answer:
(a) keep changing

Question 3.
Under the fluctuating capital system, the partners ……………. accounts are opened.
(a) current
(b) drawing
(c) capital
Answer:
(c) capital

Question 4.
Under the fixed capital system, the profits and losses of partners will be transferred to their ……………. accounts.
(a) current
(b) drawings
(c) Both
Answer:
(a) current

Question 5.
Interest capital is calculated as the …………….
(a) opening capital
(b) closing capital
(c) Both
Answer:
(a) opening capital

Question 6.
……………. is an extension of profit and loss account
(a) Balance sheet
(b) Profit and loss appropriation account
(c) Both
Answer:
(b) Profit and loss appropriation account

Question 7.
The persons who have entered into partnership are collectively known as …………….
(a) partnership
(b) partners
(c) firm
Answer:
(c) firm

Question 8.
Name the method of calculating interest on drawings of the partner if different amounts are withdrawn are different dates …………….
(a) Direct method
(b) Product method
(c) Average period method
Answer:
(A) Product method

Question 9.
Which of the following items, does not appear in the profit and loss appropriation account?
(a) Salaries to partners
(b) Interest on capital
(c) Interest on drawings
(d) Drawings
Answer:
(d) Drawings

II. Fill in the Blanks

Question 10.
The partners share the ……………. of the business.
Answer:
Profit and losses.

Question 11.
The ……………. accounts of partners may show credit or debit balance.
Answer:
Current.

Question 12.
Interest on partners drawings is changed to their respective …………….
Answer:
Capital A/c.

Question 13.
Salary to the partner will be paid only if it is allowed by the …………….
Answer:
Agreement

Question 14.
The capital accounts of partners may be ……………. or fluctuating.
Answer:
Fixed

Question 15.
Interest Drawings for the regular interval is calculated by the formula …………….
Answer:
Total Drawings x \(\frac { Rate }{ 100 }\) x \(\frac { 16 }{ 12 }\)

Question 16.
In the case of fluctuating capital, where will you record drawings, interest on drawing …………….
Answer:
Dr. Side of Capital A/c

III. Match the following

Question 17.

Answer:
(ii) 4 3 2 1

IV. Short-answer Questions

Question 1.
What is a Current Account?
Answer:
Under the fixed capital method, Capital Account and Current accounts are maintained. In current a/c all adjustments relating to partners are recorded on the credit side of the current account viz. Interest on capital, the share of profits, salary and commission etc., are recorded. On the debit side, drawings, interest on drawings, share of loss are recorded. Current accounts sometimes shows credit balance or debit balance.

Question 2.
Interest on Capital: Mr. A and B started a business on 1.4.2014, with a capital of Rs. 60,000 and 50,000, respectively. On 1^st July 2014. Mr. A withdrew Rs 8,000 from his capital. Mr. B introduced additional capital of Rs 10,000 on 30.9.2014. Calculate interest on capital @ 5% p.a. for year ending 31.3.2015.
Solution:

Question 3.
Prince, Queen, and Kings had a capital of Rs. 1,60,000, Rs. 1,20,000 and Rs. 80,000 respectively, on 1.4.2010. Queen withdraw Rs. 16,000 on 30.9.2010. King introduced additional capital Rs. 24,000 on 31.12.10. Calculate interest on capital @ 6% p.a. on 31.3.2011.
Solution:

Question 4.
A and B are partners, sharing profits and losses in an equal ratio with Capital of Rs. 50,000 and Rs 40,000 on 1.4.2017. On 1^st July 2017, A introduced Rs 10,000 as his additional capital, where B introduced only Rs 1,000. Interest 10% p.a. Calculate interest on capital.
Solution:

Question 5.
Interest on Drawings:
Sundar and Shanmugam are two partners equally. Sundar drew regularly Rs. 4,000 end of every month. Shanmugam draws Rs. 8,000 regularly beginning of every month. Calculate interest on their drawings @10%.
Solution:
Interest on Drawings of Sundar:
Total drawings x \(\frac { Rate }{ 100 }\) x \(\frac { 5.5 }{ 12 }\)
4000 x 12 x \(\frac { 10 }{ 100 }\) x \(\frac { 5.5 }{ 2 }\)
= ₹ 2, 200
Interest on Drawings of Shanmugam:
Total drawing x \(\frac { Rate }{ 100 }\) x \(\frac { 6.5 }{ 12 }\)
8000 x 12 x \(\frac { 10 }{ 100 }\) x \(\frac { 6.5 }{ 12 }\)
= ₹ 5, 200

Question 6.
Priya and Kala are partners Priya draws Rs. 8,000 at end of each quarter. Interest on drawings @6% p.a. Kala draws Rs. 2000/- per month at the end of the month.
Solution:
Interest of drawings:
Priya: 8,000 x 4 = 32,000

V. Exercise

Question 1.
Arun and Arora were partners sharing profits and losses in the ratio 5:3. Their fixed capitals on 1.4.2016 were Arun Rs. 60,000; Arora Rs. 80,000. Interest on capital @ Rs. 12%. Interest on drawings @15% p.a. Profit for the year ended 31.3.2017 before all above adjustment was Rs. 12,600. Drawings: Arun Rs. 2,000. Arora Rs. 4,000 during the year. Prepare profit and loss appropriation account.
Solution:

Interest on Drawings:
Arun: 2000 x \(\frac { 15 }{ 100 }\) x \(\frac { 6 }{ 12 }\) = ₹ 150
Arora: 4000 x \(\frac { 15 }{ 100 }\) x \(\frac { 6}{ 12 }\) = ₹ 300

Question 2.
Write up the capital accounts and current accounts of the partners Kaviya and Divya from the following:

Solution:
Capital Account

Current Account

Question 3.
Distribution of Profits: Anitha, Ramita were partners sharing profit and losses in the ratio of 7:3. Their capitals were Rs. 80,000 and Rs. 60,000, respectively.

1. Interest on capital @10% p.a.
2. Interest on drawings @12% p.a.
3. Both to get a salary of Rs. 10,000 each per annum.
4. Anitha to get a commission of 10% on the net profit before charging such commission.

The profit for the year Rs. 60,000. Drawings were Anitha Rs. 12,000 and Ramita Rs. 8,000. Show profit and loss appropriation account and the capital A/c.
Solution:
Profit and Loss Appropriation Account

Capital Accounts

Students can Download Computer Science Chapter 1 Function Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science 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 Computer Science Solutions Chapter 1 Function

Samacheer Kalvi 12th Computer Science Function Text Book Back Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
The small sections of code that are used to perform a particular task is called ……………………….
(a) Subroutines
(b) Files
(c) Pseudo code
(d) Modules
Answer:
(a) Subroutines

Question 2.
Which of the following is a unit of code that is often defined within a greater code structure?
(a) Subroutines
(b) Function
(c) Files
(d) Modules
Answer:
(b) Function

Question 3.
Which of the following is a distinct syntactic block?
(a) Subroutines
(b) Function
(c) Definition
(d) Modules
Answer:
(c) Definition

Question 4.
The variables in a function definition are called as ……………………….
(a) Subroutines
(b) Function
(c) Definition
(d) Parameters
Answer:
(d) Parameters

Question 5.
The values which are passed to a function definition are called ……………………….
(a) Arguments
(b) Subroutines
(c) Function
(d) Definition
Answer:
(a) Arguments

Question 6.
Which of the following are mandatory to write the type annotations in the function definition?
(a) Curly braces
(b) Parentheses
(c) Square brackets
(d) Indentations
Answer:
(b) Parentheses

Question 7.
Which of the following defines what an object can do?
(a) Operating System
(b) Compiler
(c) Interface
(d) Interpreter
Answer:
(c) Interface

Question 8.
Which of the following carries out the instructions defined in the interface?
(a) Operating System
(b) Compiler
(c) Implementation
(d) Interpreter
Answer:
(c) Implementation

Question 9.
The functions which will give exact result when same arguments are passed are called ……………………….
(a) Impure functions
(b) Partial Functions
(c) Dynamic Functions
(d) Pure functions
Answer:
(d) Pure functions

Question 10.
The functions which cause side effects to the arguments passed are called ……………………….
(a) Impure functions
(b) Partial Functions
(c) Dynamic Functions
(d) Pure functions
Answer:
(a) Impure functions

PART – II
II. Answer The Following Questions

Question 1.
What is a subroutine?
Answer:
Subroutines are the basic building blocks of computer programs. Subroutines are small sections of code that are used to perform a particular task that can be used repeatedly. In Programming languages these subroutines are called as Functions.

Question 2.
Define Function with respect to Programming language?
Answer:
A function is a unit of code that is often defined within a greater code structure. Specifically, a function contains a set of code that works on many kinds of inputs, like variants, expressions and produces a concrete output.

Question 3.
Write the inference you get from X: = (78)?
Answer:
Value 78 being bound to the name X.

Question 4.
Differentiate interface and implementation?
Answer:
Interface:
Interface just defines what an object can do, but won’t actually do it.

Implementation:
Implementation carries out the instructions defined in the interface.

Question 5.
Which of the following is a normal function definition and which is recursive function definition?
Answer:
(I) Let Recursive sum x y:
return x + y

(II) let disp:
print ‘welcome’

(III) let Recursive sum num:
if (num! = 0) then return num + sum (num – 1) else
return num

1. Recursive function
2. Normal function
3. Recursive function

PART – III
III. Answer The Following Questions

Question 1.
Mention the characteristics of Interface?
Answer:
Characteristics of interface:

1. The class template specifies the interfaces to enable an object to be created and operated properly.
2. An object’s attributes and behaviour is controlled by sending functions to the object.

Question 2.
Why strlen is called pure function?
Answer:
strlen (s) is called each time and strlen needs to iterate over the whole of ‘s’. If the compiler is smart enough to work out that strlen is a pure function and that ‘s’ is not updated in the lbop, then it can remove the redundant extra calls to strlen and make the loop to execute only one time. This function reads external memory but does not change it, and the value returned derives from the external memory accessed.

Question 3.
What is the side effect of impure function. Give example?
Answer:
Impure Function:

• The return value of the impure functions does not solely depend on its arguments passed. Hence, if you call the impure functions with the same set of arguments, you might get the different return values. For example, random( ), Date( ).
• They may modify the arguments which are passed to them.

Question 4.
Differentiate pure and impure function?
Answer:
Pure Function:

1. The return value of the pure functions solely depends on its arguments passed.
2. If you call the pure functions with the same set of arguments, you will always get the same return values.
3. They do not have any side effects.
4. They do not modify the arguments which are passed to them.

Impure Function:

1. The return value of the impure functions does not solely depend on its arguments passed.
2. If you call the impure functions with the same set of arguments, you might get the different return values. For example, random( ), Date( ).
3. They have side effects.
4. They may modify the arguments which are passed to them.

Question 5.
What happens if you modify a variable outside the function? Give an example?
Answer:
When a function depends on variables or functions outside of its definition block, you can never be sure that the function will behave the same every time it’s called.
For example let y: = 0
(int) inc (int) x
y: = y + x;
return (y)
In the above example the value of y get changed inside the function defintion due to which the result will change each time. The side effect of the inc ( ) function is it is changing the data ‘ of the external visible variable ‘y’.

PART – IV
IV. Answer The Following Questions

Question 1.
What are called Parameters and write a note on?
Answer:

1. Parameter without Type
2. Parameter with Type Parameters (and arguments)

Parameters are the variables in a function definition and arguments are the values which are passed to a function definition.

(I) Parameter without Type
Let us see an example of a function definition:
(requires: b> = 0)
(returns: a to the power of b)
let rec pow a b: =
if b = 0 then 1
else a * pow a (b – 1)
In the above function definition variable ‘b’ is the parameter and the value which is passed to the variable ‘b’ is the argument. The precondition (requires) and postcondition (returns) of the function is given. Note we have not mentioned any types: (data types). Some language compiler solves this type (data type) inference problem algorithmically, but some require the type to be mentioned.

In the above function definition if expression can return 1 in the then branch, by the typing rule the entire if expression has type int. Since the if expression has type ‘int ’, the function’s return type also be ‘inf. ‘b ’is compared to 0 with the equality operator, so ‘b ’is also a type of ‘int. Since a is multiplied with another expression using the * operator, ‘a’ must be an int.

(II) Parameter with Type
Now let us write the same function definition with types for some reason:
(requires: b > 0)
(returns: a to the power of b)
let rec pow (a: int) (b: int): int : =
if b = 0 then 1
else a * pow b (a – 1)
When we write the type annotations for ‘a ’ and ‘b ’ the parentheses are mandatory. Generally we can leave out these annotations, because it’s simpler to let the compiler infer them. There are times we may want to explicitly write down types. This is useful on times when you get a type error from the compiler that doesn’t make sense. Explicitly annotating the types can help with debugging such an error message.

The syntax to define functions is close to the mathematical usage: the definition is introduced by the keyword let, followed by the name of the function and its arguments; then the formula that computes the image of the argument is written after an = sign. If you want to define a recursive function: use “let rec ” instead of “let
Syntax: The syntax for function definitions:

let rec fnal a2 … an : = k
Here the fn is a variable indicating an identifier being used as a function name. The names ‘al ’ to ‘an ’ are variables indicating the identifiers used as parameters. The keyword ‘rec ’ is required if fn ’ is to be a recursive function; otherwise it may be omitted.
For example: let us see an example to check whether the entered number is even or odd.
(requires: x> = 0)
let rec even x : = x = 0 || odd (x – 1)
return ‘even’
(requires: x> = 0)
let odd x : =
x< >0 && even (x – 1)
return ‘odd’
The syntax for function types:
x → y
x₁ → x₂ → y
x₁ → … → x_n → y
The ‘x’ and ‘y’ are variables indicating types. The type x → y is the type of a function that gets an input of type ‘x’ and returns an output of type ‘y’. Whereas x₁ → x₂ → y is a type of a function that takes two inputs, the first input is of type ‘x1 ’ and the second input of type ‘x2’, and returns an output of type ‘y’. Likewise x1 → … → xn → y has type ‘x’ as input of n arguments and ‘y’ type as output.

Question 2.
Identify in the following program
Answer:
let rec gcd a b : =
if b < > 0 then gcd b (a mod b) else return a
(I) Name of the function
gcd

(II) Identify the statement which tells it is a recursive function
let rec

(III) Name of the argument variable
a, b

(IV) Statement which invoke the function recursively
gcd b(a mod b) [when b < > 0]

(V) Statement which terminates the recursion
return a (when b becomes 0).

Question 3.
Explain with example Pure and impure functions?
Answer:
Pure functions:
Pure functions are functions which will give exact result when the same arguments are passed. For example the mathematical function sin (0) always results 0. This means that every time you call the function with the same arguments, you will always get the same result. A function can be a pure function provided it should not have any external variable which will alter the behaviour of that variable.
Let us see an example
let square x
return: x * x
The above function square is a pure function because it will not give different results for same input. There are various theoretical advantages of having pure functions. One advantage is that if a function is pure, then if it is called several times with the same arguments, the compiler only needs to actually call the function once. Let’s see an example let i: = 0;
if i < strlen (s) then – Do something which doesn’t affect s ++ i If it is compiled, strlen (s) is called each time and strlen needs to iterate over the whole of ‘s’.

If the compiler is smart enough to work out that strlen is a pure function and that ‘s’ is not updated in the loop, then it can remove the redundant extra calls to strlen and make the #loop to execute only one time. From these what we can understand, strlen is a pure function because the function takes one variable as a parameter, and accesses it to find its length.

This function reads external memory but does not change it, and the value returned derives from the external memory accessed. Impure functions: The variables used inside the function may cause side effects through the functions which are not passed with any arguments.

In such cases the function is called impure function. When a function depends on variables or functions outside of its definition block, you can never be sure that the function will behave the same every time it’s called. For example the mathematical function random Q will give different outputs for the same function call, let Random number let a : = random( ) if a > 10 then
return: a
else
return: 10
Flere the function Random is impure as it is not sure what will be the result when we call the function.

Question 4.
Explain with an example interface and implementation?
Answer:
Interface Vs Implementation:
An interface is a set of action that an object can do. For example when you press a light switch, the light goes on, you may not have cared how it splashed the light. In Object Oriented Programming language, an Interface is a description of all functions that a class must have in order to be a new interface.

In our example, anything that “ACTSLIKE” a light, should have function defnitions like turn on ( ) and a turn off ( ). The purpose of interfaces is to allow the computer to enforce the properties of the class of TYPE T (whatever the interface is) must have functions called X, Y, Z, etc.

A class declaration combines the external interface (its local state) with an implementation of that interface (the code that carries out the behaviour). An object is an instance created from the class. The interface defines an object’s visibility to the outside world.

The difference between interface and implementation is:

Interface:
Interface just defines what an object can do, but won’t actually do it. Implementation carries out the instructions defined in the interface.

Implementation:
Implementation carries out the instructions defined in the interface.
In object oriented programs classes are the interface and how the object is processed and executed is the implementation.

Characteristics of interface

1. The class template specifies the interfaces to enable an object to be created and operated properly.
2. An object’s attributes and behaviour is controlled by sending functions to the object.

For example, let’s take the example of increasing a car’s speed.

The person who drives the car doesn’t care about the internal working. To increase the speed of the car he just presses the accelerator to get the desired behaviour. Here the accelerator is the interface between the driver (the calling / invoking object) and the engine (the called object). In this case, the function call would be Speed (70): This is the interface.

Internally, the engine of the car is doing all the things. It’s where fuel, air, pressure, and electricity come together to create the power to move the vehicle. All of these actions are separated from the driver, who just wants to go faster.

Let us see a simple example, consider the following implementation of a function that finds the minimum of its three arguments:
let min 3 x y z : =
if x < y then
if x < z then x else z
else
if y < z then y else z

Practice Programs
Question 1.
Write algorithmic function definition to find the minimum among 3 numbers?
Answer:
let min 3 x y z : =
if x < y then
if x < z then x else z
else
if y < z then y else z

Question 2.
Write algorithmic recursive function definition to find the sum of n natural numbers?
Answer:
let rec sum num:
if (num! = 0) then return num + sum (num – 1)
else
return num

Samacheer kalvi 12th Computer Science Function Additional Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
……………………… are expressed using statements of a programming language.
(a) Algorithm
(b) Procedure
(c) Specification
(d) Abstraction
Answer:
(a) Algorithm

Question 2.
……………………… are the basic building blocks of a computer programs.
(a) Code
(b) Subroutines
(c) Modules
(d) Variables
Answer:
(b) Subroutines

Question 3.
In programming languages, subroutines are called as …………………………..
(a) Functions
(b) Task
(c) Modules
(d) Code
Answer:
(a) Functions

Question 4.
Find the correct statement from the following.
(a) a : = (24) has an expression
(b) (24) is an expression
Answer:
(a) a : = (24) has an expression

Question 5.
……………………… binds values to names.
(a) Algorithms
(b) Variables
(c) Interface
(d) Definitions
Answer:
(d) Definitions

Question 6.
Identify the statement which is wrong.
(a) Definitions are expressions
(b) Definitions are distinct syntactic blocks.
(c) Definitions can have expressions, nested inside them.
Answer:
(a) Definitions are expressions

Question 7.
The name of the function in let rec pow ab : = is …………………………
(a) Let
(b) Rec
(c) Pow
(d) a b
Answer:
(c) Pow

Question 8.
In function definition pre condition is given by ……………………….
(a) Needed
(b) Let
(c) Returns
(d) Requires
Answer:
(d) Requires

Question 9.
In function definition post condition is given by …………………………
(a) Needed
(b) Let
(c) Returns
(d) Requires
Answer:
(c) Returns

Question 10.
In b = 0, = is ……………………….. operator
(a) Assignment
(b) Equality
(c) Logical
(d) Not equal
Answer:
(b) Equality

Question 11.
The formula should be written after ……………………….. sign
(a) +
(b) –
(c) =
(d) ++
Answer:
(c) =

Question 12.
To define a recursive function, …………………………. is used.
(a) Let
(b) Let r
(c) Let rfn
(d) Let rec
Answer:
(d) Let rec

Question 13.
Find which is false.
(a) All function definitions are static
(b) All function definitions are dynamic
Answer:
(b) All function definitions are dynamic

Question 14.
A ……………………….. combines the external interface with an implementation of that interface.
Answer:
class declaration

Question 15.
An …………………………. is an instance created from the class.
(a) Object
(b) Functions
(c) Subroutines
(d) Definitions
Answer:
(a) Object

Question 16.
Find the statement which is not true.
(a) The interface defines an objects visibility to the outside world
(b) Interface defines what an object can do.
(c) In object oriented programs, objects are interfaces
Answer:
(c) In object oriented programs, objects are interfaces

Question 17.
An ………………………… attributes and behaviour is controlled by sending functions to the object.
Answer:
Objects

Question 18.
The class template specifies the ………………………. to enable an object to be created and operated properly.
Answer:
Interfaces

Question 19.
The accelerator is the …………………………… between the driver and the engine.
(a) Interface
(b) Object
(c) Instruction
(d) Code
Answer:
(a) Interface

Question 20.
sin (0) = 0 is an example for ………………………. function.
(a) Impure
(b) Pure
(c) Interface
(d) Instruction
Answer:
(b) Pure

Question 21.
Find the impure function from the following.
(a) Sin (0)
(b) Square x
(c) Strlen (s)
(d) None of these
Answer:
(d) None of these

Question 22.
The function random ( ) is an example for …………………….. functions.
Answer:
Impure

Question 23.
Why is the function random( ) is a impure function?
(a) It gives different outputs for same function call
(b) It gives different outputs when 0 is given
(c) It will not give different output
Answer:
(a) It gives different outputs for same function call

Question 24.
Which function definition, doesn’t modify the arguments passed to them?
(a) Pure function
(b) Impure function
(c) Object
(d) Interface
Answer:
(a) Pure function

Question 25.
How many parameters are defined in the function let rec gcd a b : =
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question 26.
In the function definition, the keyword let is followed by …………………………
(a) Function name
(b) Arguments
(c) Parameters
(d) Implementations
Answer:
(a) Function name

Question 27.
Find the correct statement from the following function definitions. let rec p on a b : =
(a) data type of the parameters are given
(b) data type of the parameters are not mentioned
Answer:
(b) data type of the parameters are not mentioned

Question 28.
If a function is not a recursive one, then ……………………………. is used
(a) abc
(b) gcd
(c) let
(d) let rec
Answer:
(c) let

Question 29.
Find the name of the function,
let rec even x : =
(a) Let
(b) Rec
(c) Even
(d) x
Answer:
(c) Even

Question 30.
Match the following function definitions with their terms.
let rec odd xy : =

1. Keyword – (i) Xy
2. Recursion – (ii) Odd
3. Function name – (iii) Rec
4. Parameters – (iv) let

(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)
(b) 1 – (i) 2 – (ii) 3 – (iii) 4 – (iv)
(c) 1 – (iv) 2 – (i) 3 – (ii) 4 – (iii)
(d) 1 – (i) 2 – (iv) 3 – (ii) 4 – (iii)
Answer:
(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)

Question 31.
In object oriented programming language, an is a description of all functions that a class must have
(a) Object
(b) Class
(c) Interface
(d) Code
Answer:
(c) Interface

Question 32.
The ……………………… defines an object’s visibility to the outside world.
(a) Object
(b) Interface
(c) Pure function
(d) Impure function
Answer:
(b) Interface

Question 33.
Find the correct statement.
(i) Evaluation of pure function causes side effects to its output.
(ii) Evaluation of Impure function causes side effects to its output.
Answer:
(ii) Evaluation of Impure function causes side effects to its output.

PART – II
II. Answer The Following Questions

Question 1.
What are the two types of parameter passing?
Answer:

1. Parameter without type
2. Parameter with type

Question 2.
In the function definition
let rec pow a b : = Is it recursive function. If so Explain. Why?
Answer:
Yes it is a recursive function. It is given in the function definition as rec which indicates recursive function.

Question 3.
Write the syntax for the function definitions?
Answer:
let rec f_n a₁ a₂ … a_n : = k
f_n : Function name
a₁ … a_n – variable
rec: recursion

Question 4.
Define recursive functions: How will you define it?
Answer:
A function definition which calls itself is called recursive functions. It is given by let rec.

PART – III
III. Answer The Following Questions

Question 1.
Write note on Definitions?
Answer:
Definitions bind values to names, Definitions are not expressions, Definitions are distinct syntactic blocks. Definitions can have expressions nested inside them, and vice – versa.

Question 2.
Write the pre condition and post condition for the function?
Answer:
(requires: b > 0)
– (returns: a to the power of b) let rec pow(a : int) (b : int): int: =

1. Pre condition : b > 0
2. Post condition : a to the power of b.

Question 3.
Give function definition for the Chameleons of Chromeland problem?
Answer:
let rec monochromatize abc : =
if a > 0 then
a, b, c : = a – 1, b – 1, c + 2
else
a: = 0, b: = 0, c: = a + b + c
return c

Question 4.
Give the flow chart for Chameleons of Chromeland problem?
Answer:
The algorithm is depicted in the flowchart as below:

Question 5.
Give the example function definition for parameter with type?
Answer:
Parameter with Type:
Now let us write the same function definition with types for some reason:
(requires: b> 0)
(returns: a to the power of b ) let rec pow (a: int) (b: int): int : =
if b = 0 then 1
else a * pow b (a – 1)

Students can Download Economics Chapter 12 Mathematical Methods for Economics Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Economics 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 11th Economics Solutions Chapter 12 Mathematical Methods for Economics

Samacheer Kalvi 11th Economics Mathematical Methods for Economic Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
Mathematical Economics is the integration of ____________
(a) Mathematics and Economics
(b) Economics and Statistics
(c) Economics and Equations
(d) Graphs and Economics
Answer:
(a) Mathematics and Economics

Question 2.
The construction of demand line or supply line is the result of using ____________
(a) Matrices
(b) Calculus
(c) Algebra
(d) Analytical Geometry
Answer:
(d) Analytical Geometry

Question 3.
The first person used the mathematics in Economics is ____________
(a) Sir William Petty
(b) Giovanni Ceva
(c) Adam Smith
(d) Irving Fisher
Answer:
(a) Sir William Petty

Question 4.
Function with single independent variable is known as ____________
(a) Multivariate Function
(b) Bivariate Function
(c) Univariate Function
(d) Polvnomial Function
Answer:
(c) Univariate Function

Question 5.
A statement of equality between two quantities is called ____________
(a) Inequality
(b) Equality
(c) Equations
(d) Functions
Answer:
(c) Equations

Question 6.
An incremental change in dependent variable with respect to change in independent variable is known as ____________
(a) Slope
(b) Intercept
(c) Variant
(d) Constant
Answer:
(a) Slope

Question 7.
(y – y₁) = m (x – x₁) gives the ____________
(a) Slope
(b) Straight line
(c) Constant
(d) Curve
Answer:
(b) Straight line

Question 8.
Suppose D = 50 – 5P. When D is zero then ____________
(a) P is 10
(b) P is 20
(c) P is 5
(d) Pis-10
Answer:
(a) Pis 10

Question 9.
Suppose D = 150 – 50P. Then, the slope is
(a) -5
(b) 50
(c) 5
(d) – 50
Answer:
(d) – 50

Question 10.
Suppose determinant of a matrix A = 0, then the solution ____________
(a) Exists
(b) Does not exist
(c) Is infinity
(d) Is zero
Answer:
(b) Does not exist

Question 11.
State of rest is a point termed as ____________
(a) Equilibrium
(b) Non – Equilibrium
(c) Minimum Point
(d) Maximum Point
Answer:
(a) Equilibrium

Question 12.
Differentiation of constant term gives ____________
(a) one
(b) zero
(c) infinity
(d) non – infinity
Answer:
(b) zero

Question 13.
Differentiation of xⁿ is ____________
(a) nx ^{(n – 1)}
(b) nx ^{(n + 1)}
(c) zero
(d) one
Answer:
(a) nx ^{(n – 1)}

Question 14.
Fixed Cost is the ____________ term in cost function represented in mathematical form.
(a) Middle
(b) Price
(c) Quantity
(d) Constant
Answer:
(d) Constant

Question 15.
The first differentiation of Total Revenue function gives ____________
(a) Average Revenue
(b) Profit
(c) Marginal Revenue
(d) Zero
Answer:
(c) Marginal Revenue

Question 16.
The elasticity of demand is the ratio of ____________
(a) Marginal demand function and Revenue function
(b) Marginal demand function to Average demand function
(c) Fixed and variable revenues
(d) Marginal Demand function and Total demand function
Answer:
(b) Marginal demand function to Average demand function

Question 17.
If x + y = 5 and x – y = 3 then, Value of x ____________
(a) 4
(b) 3
(c) 16
(d) 8
Answer:
(a) 4

Question 18.
Integration is the reverse process of
(a) Difference
(b) Mixing
(c) Amalgamation
(d) Differentiation
Answer:
(d) Differentiation

Question 19.
Data processing is done by ____________
(a) PC alone
(b) Calculator alone
(c) Both PC and Calculator
(d) Pen drive
Answer:
(c) Both PC and Calculator

Question 20.
The command Ctrl + M is applied for ____________
(a) Saving
(b) Copying
(c) getting new slide
(d) deleting a slide
Answer:
(c) getting new slide

Part – B
Answer the following questions in one or two sentences

Question 21.
If 62 = 34 + 4x  what is x?
solution:
If 62 = 34 + 4x what is x
Given:
62 = 34 + 4x
34 + 4x = 62
4x = 62 – 34
4x = 28
x = \(\frac { 28 }{ 4 }\)
x = 7

Question 22.
Given the demand function q = 150 – 3p, derive a function for MR.
Solution:
Given:
q = 150 – 3p
\(\frac { dq }{ dq }\) = 0 – 3 (1)
\(\frac { dq }{ dq }\) = – 3
Revenue = price × quantity = p (150 – 3p)
= p (150 – 3p)
= 150p – 3p²
MR = \(\frac { dq }{ dq }\) = 150 \(\frac { dq }{ dq }\) -6p \(\frac { dq }{ dq }\)
= 150 (\(\frac { – 1 }{ 3 }\)) – 6p (\(\frac { – 1 }{ 3 }\))
= – 50 + 2p
= – 50 = 2 (\(\frac { 150 – q }{ 3 }\))
= 50 + 100 + \(\frac { – 2 q }{ 3 }\)
= 50 – \(\frac { 2 q }{ 3 }\)
= 50 – \(\frac { 2 }{ 3 }\) (150 – 3p)
MR = 2p – 50

Question 23.
Find the average cost function where TC = 60 + 10x +15x²
Solution:
Given:
Formula = \(\frac{TC}{x}\)
Average cost function = \(\frac{60}{x}\) + \(\frac{10x}{x}\) + \(\frac { 15x^{ 2 } }{ x } \)
Average cost = \(\frac{60}{x}\) + 10 + 15x

Question 24.
The demand function is given by x = 20 – 2p – p² where p and x are the price and the quantity respectively. Find the elasticity of demand for p = 2.5.
Solution:
Given:
x = 20 – 2p – p²
p = 2.5
e_d = (\(\frac { p }{ x }\)) (\(\frac { dx }{ dp }\))
If p = 2.5 then x,
x = 20 – 2 (2.5) – (2.5)²
= 20 – 5 – (6.25)
= 15 – 6.25
x = 8.75
Here x = 20 – 2p – p²
\(\frac { dx }{ dp }\) = 0 – 2 (1) – 2p
= -2 – 2p
If P = 2.5 then
\(\frac { dx }{ dp }\) = -2 – 2 (2.5)
= -2 – 5
\(\frac { dx }{ dp }\) = – 7
e_d = \(\frac { -2.5 }{ 8.75 }\) × – 7
= 0.2857 × – 7
e_d = 2

Question 25.
Suppose the price p and quantity q of a commodity are related by the equation q = 30 – 4p – p²
find :

1. e_d at p = 2
2. MR

Solution:
Given:
q = 30 – 4p – p²
1. e_d = (\(\frac { p }{ x }\))(\(\frac { dx }{ dp }\))
If P = 2 then
x = 30 – 4 (2) – 2²
= 30 – 8 – 4
=20 – 12
x = 18
\(\frac { dx }{ dp }\) =30 – 4p – p²
= 0 – 4 – 2p
= 0 – 4 – 2 (2)
= -4 -4
= -8
e_d = \(\frac { 2 }{ 18 }\) (-8)
x = – \(\frac { 16 }{ 18 }\)
e_d = – 0.88

2. MR
TR = p ×q
TR = p (30 – 4p – p²)
= 30p – 4p² – p³
MR = \(\frac { d(TR) }{ dp }\)
= 30 – 8p – 3p²
If p = 2
MR = 30 – 8 (2) – 3 (2)²
= 30 – 16 12
MR = 2

Question 26.
What is the formula for elasticity of supply if you know the supply function?
Solution:
Elasticity of supply = (\(\frac { p }{ q }\)) (\(\frac { dq }{ dp }\))

Question 27.
What are the main menus of MS Word?
Answer:

1. Home menu
2. Insert
3. Page layout
4. Reference
5. Review
6. View
7. Print layout
8. Outline
9. Task pane
10. Toolbars
11. Header and footer
12. Footnotes

Part – C
Answer the following questions in One Paragraph

Question 28.
Illustrate the uses of Mathematical Methods in Economics.
Answer:

1. Mathematical Methods help to present the economic problems in a more precise form.
2. Mathematical Methods help to explain economic concepts.
3. Mathematical Methods help to use a large number of variables in economic analyses.
4. Mathematical Methods help to quantify the impact or effect of any economic activity implemented by the Government or anybody. There are of course many other uses.

Question 29.
Solve for A: quantity demanded if 16x – 4 = 68 + 7x.
Solution:
Given:
16x  – 4 = 68 + 1x
16x – 7x = 68 + 4
9x = 72
x = (\(\frac { 72 }{ 9 }\))
x = 8

Question 30.
A firm has the revenue function R = 600q – 0.03q² and the cost function is C = 150q + 60,000, where q is the number of units produced. Find AR, AC, MR and MC.
Solution:
Given:
(I) Average Revenue = \(\frac{R}{q}\)
\(\frac { 600q-0.03q^{ 2 } }{ q } \) = \(\frac{600q}{q}\) – \(\frac { 0.03q^{ 2 } }{ q } \)
AR = 600 – 0.03q

(II) Average Cost = \(\frac{C}{q}\)
= \(\frac{150q}{q}\) + \(\frac{60000}{q}\)
AC = 150 + \(\frac{60000}{q}\)

(III) Marginal Revenue = \(\frac{dR}{dq}\)
R = 600q – 0.03q²
\(\frac{dR}{dq}\) = 600 – (0.03) (2q)
MR = 600 – 0.06q

(IV) Marginal Cost = \(\frac{dC}{dq}\)
C = 150q + 60000
MC = 150

Question 31.
Solve the following linear equations by using Cramer’s rule.
x₁ – x₂ + x₃ = 2
– x₁ – x₂ – x₃ = 0
– x₁ – x₂ – x₃ = – 6
Solution:

= 1 (- 1 – 1) + 1 (- 1 – 1) + 1 (- 1 + 1)
= 1(- 2) + 1 (0)
∆ = – 2 – 2
∆ = – 2 – 2
∆ = – 4

= 2 (- 1 – 1) + 1 (0 + 6) + 1 (0 – 6)
= 2 (- 2) + 1(6) + 1 (- 6)
= – 4 + 6 – 6
∆ = – 4

= 1 (0 – 6) – 2 (- 1 – 1)+1(- 6 + 0)
= 1 (- 6) – 2 (- 2) + 1 (- 6)
= – 6 + 4 – 6
= – 12 + 4
∆x₃ = -8

1 (- 6 + 0) + 1 (- 6 + 0) + 2 (- 1 + 1)
= 1 (- 6) + 1 (- 6) + 2 (0)
= – 6 – 6
∆x₃ = – 12

Question 32.
If a firm faces the total cost function TC = 5+ x² where x is output, what is TC when x is 10?
Solution:
Given:
TC = 5 + x²
If x = 10
TC = 5 + 10²
= 5 + 100
TC = 105

Question 33.
If TC = 2.5q³ – 13q² + 50q + 12 derive the MC function and AC function.
solution:
Given:
TC= 2.5q³ – 13q² + 50q + 12
Marginal cost function
MC = \(\frac { d(TC) }{ dQ }\)
= \(\frac { d }{ dq }\) [2.5q³ – 13q² + 50q + 12]
= 2.5 (3q²) – 13 (2q) + 50 (1) + 0
MC = 7.5q – 26q + 50
Average cost AC =\(\frac { TC }{ q }\)
= \(\frac{2.5 q^{3}-13 q^{2}+50 q+12}{q}\)
= \(\frac{2.5 q^{3}}{q}\) – \(\frac{13 q^{2}}{q}\) + \(\frac { 50q }{ q }\) + \(\frac { 12 }{ q }\)
AC = 2.5q – 13q + 50 + \(\frac { 12 }{ q }\)

Question 34.
What are the steps involved in executing an MS Excel Sheet?
Answer:
MS Excel is used in data analysis by using formulas. A spreadsheet is a large sheet of paper which contains rows and columns.
To start an excel there are various options.
Click Start → Program → Microsoft Excel
Double click the MS Excel Icon from the Desktop.

Part – D
Answer the following questions in about a page

Question 35.
A Research scholar researching the market for fresh cow milk assumes that Qt = f (Pt, Y, A, N, Pc) where Qt is the quantity of milk demanded, Pt is the price of fresh cow milk, Y is average household income, A is advertising expenditure on processed pocket milk, N is population and Pc is the price of processed pocket milk.

1. What does Qt= f (Pt, Y,A,N, Pc) mean in words?
2. Identify the independent variables.
3. Make up a specific form for this function. (Use your knowledge of Economics to deduce whether the coefficients of the different independent variables should be positive or negative.)

Solution:
1. Qt is the function of Pt, Y, A, N, Pc
The determinants of demand are
Pt = Price of fresh cow milk.
Y = Average household income
A = Advertising expenditure on processed pocket milk
N = Population
Pc = Price of processed pocket milk.

2. Y and N are independent variables

3. Average household income and population are directly proportional to quantity demanded of cow’s milk (ie if Y and N increases Qt also increase)
Price of fresh cow milk, advertising expenditure on pocket milk and price of processed pocket milk are inversely proportional to Qt.
There
Qt = – aPt + by – CA + dN – ePc

Question 36.
Calculate the elasticity of demand for the demand schedule by using differential calculus method P = 60 – 0.2 Q where price is

1. zero
2. Rs.20
3. Rs.40

Solution:
P = 60 – 0.2 Q
0.2 Q = 60 – P
(0.2 Q) 10 = (60 – p) 10
2 Q = 600 – 10p
Q = 300 – 5p
Given demand function Q = 300 – 5p
1. If P = 0 then Q = 300 – 5 (0)
= 300
\(\frac { dq }{ dp }\) = 300 – 5p
= 0 – 5
\(\frac { dq }{ dp }\) = – 5
e_d = \(\frac { p }{ p }\)(\(\frac { dq }{ dp }\))
= \(\frac { 0 }{ 300 }\)(-5)
e_d = 0

2. If p = 20 then Q = 300 = 5 (20)
= 200
e_d = \(\frac { 20 }{ 200 }\) (-5)
= – \(\frac { 100 }{ 200 }\)
e_d = – 0.5

3. If = 40 then Q = 300 – 5 (40)
= 300 – 200
Q = 100
e_d = \(\frac { p }{ q }\)(\(\frac { dq }{ dp }\))
= \(\frac { 40 }{ 100 }\) (-5)
e_d = – \(\frac { 200 }{ 100 }\)
e_d = -2

Question 37.
The demand and supply functions are pd=1600 – x² and ps = 2x² + 400 respectively. Find the consumer’s surplus and producer’s surplus at an equilibrium point.
Answer:
At equilibrium p_d = p_s
1600 – x² = 2x² + 400
– x² – 2x² = 400 – 1600
– 3x² = – 1200
x² = \(\frac { 1200 }{ 3 }\)
x² = 400
x = 20
If x = 20 p_o = 1600 – (20)²
= 1600 – 400
p_o = 1200
p_o x_o = 1200 × 20
= 24000

= 8000 – \(\frac { 8000 }{ 3 }\)
= \(\frac { 24000 – 8000 }{ 3 }\)
= \(\frac { 16000 }{ 3 }\)
C.S = 5333.3

Question 4.
What are the ideas of information and communication technology used in economics?
Answer:
Information and Communication Technology (ICT) is the infrastructure that enables computing faster and accurate

In economics the uses of mathematical and statistical tools need the support of ICT for data compiling, editing, manipulating and presenting the results.

Samacheer Kalvi 11th Economics Introduction To Micro-Economics Additional Questions and Answers

Match the following and choose the answer using the codes given below

Question 1.

(a) 1 2 3 4
(b) 2 3 4 1
(c) 3 4 2 1
(d) 4 3 1 2
Answer:
(c) 3 4 2 1

Question 2.

(a) 4 3 1 2
(b) 2 3 4 1
(c) 1 2 3 4
(d) 3 2 1 4
Answer:
(a) 4 3 1 2

Question 3.

(a) 1 2 3 4
(b) 3 4 1 2
(c) 4 3 1 2
(d) 2 3 4 1
Answer:
(b) 3 4 1 2

Choose the incorrect statement

Question 4.
(a) Differentiation of constant is zero
(b) Differentiation of xn is x
(c) Constant value is known as fixed cost
(d) x° = 1 when x = 0
Answer:
(d) x° = 1 when x = 0

Question 5.
(a) The population density of T.N is 480 as per 2011 census.
(b) There are 30 national highways in Tamil Nadu.
(c) Tamil Nadu has the highest installed wind energy capacity in India.
(d) The headquarters of southern railway is at Trichy
Answer:
(b) There are 30 national highways in Tamil Nadu.

Pick the odd one out

Question 6.
(a) Arithmetic
(b) Comparison
(c) Text concatenation
(d) Foot note
Answer:
(d) Foot note

Question 7.
(a) X = independent variable
(b) Y = dependent variable
(c) X = f (y)
(d) M – slope
Answer:
(c) X = f (y)

Choose the correct statement

Question 8.
(a) X + Y = 6 and X – Y = 2 then X = 4
(b) If D = 60 – 6P when D is 0 then P is 20
(c) In the equation 0.1 X² + 1 OX + 100 constant = 10
(d) If Y = 6X³ then slope is 9X:
Answer:
(a) X + Y = 6 and X – Y = 2 then X = 4

Choose the incorrect pair

Question 9.

Answer:
(c) \(\frac { d(c) }{ dx }\) = 0 where c is a variable

Question 10.

Answer:
(a) ∫f(x) dx – (i) F (x) + C

Fill in the blanks with suitable option given below

Question 11.
__________ is the first to use mathematics in economics.
(a) Giovanni Ceva
(b) William petty
(c) J.M.Keynes
(d) None
Answer:
(b) William petty

Question 12.
The formula for constructing a straight line is __________
(a) \(\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\)
(b) y = mx
(c) \(\frac { Change in x }{ Change in y }\)
(d) (y -y₁) = m (x – x₁)
Answer:
(d) (y -y₁) = m (x – x₁)

Question 13.
If the determinant ∆ = ?, then solution does not exist
(a) 1
(b) 0
(c) 2
(d) -1
Answer:
(b) 0

Question 14.
Power rule ∫xdx ______
(a) \(\frac{x^{2}}{n+1}\) + C
(b) \(\frac { pq }{ n + 1 }\) + C
(c) \(\frac{\mathrm{x}^{(\mathrm{n}+1)}}{\mathrm{n}+1}\) + C
(d) None
Answer:
(c) \(\frac{\mathrm{x}^{(\mathrm{n}+1)}}{\mathrm{n}+1}\) + C

Question 15.
______ is a rectangular array of numbers systematically arranged in rows and columns within brackets.
(a) Matrix
(b) Determinants
(c) Function
(d) None
Answer:
(a) Matrix

Choose the best option

Question 16.
__________ provides the solution of a system of linear equations with ‘n’ variables and ‘n’ equations
(a) Demand function
(b) Consumer’s surplus
(c) Equilibrium
(d) Cramer’s rule
Answer:
(d) Cramer’s rule

Question 17.
The first known writer to apply the mathematical method to economic problems was __________
(a) Giovanni Ceva
(b) William petty
(c) Keynes
(d) None
Answer:
(a) Giovanni ceva

Question 18.
In y = f (x) which is the independent variable
(a) f
(b) y
(c) x
(d) None
Answer:
(c) x

Question 19.
Formula for marginal cost is
(a) MC = \(\frac { d(TC) }{ dQ }\)
(b) MC = \(\frac { d(TC) }{ dP }\)
(c) MC = \(\frac { d(TC) }{ dQ }\)
(d) None
Answer:
(a) MC = \(\frac { d(TC) }{ dQ }\)

Question 20.
Function with a single independent variable are called _____
(a) Multivariate function
(b) Consumption function
(c) Simple univariate function
(d) None
Answer:
(c) Simple univariate function

Part – B
Answer the following questions in one or two sentences

Question 1.
What is economic analysis?
Answer:
Economic analysis is a systematic approach to determine the optimum use of scare resources and choose available alternatives and select the best alternative to achieve a particular objective.

Question 2.
What is the slope of a straight line ?
Answer:
Slope or gradient of the line represents the ratio of the changes in vertical and horizontal lines

Question 3.
State the phases of evolution of ICT?
Answer:

1. Computer
2. PC
3. Micro Processor
4. Internet and
5. Wireless links.

Question 4.
What are determinants?
Answer:
The determinant is an arrangement of the same elements of the corresponding matrix into rows and columns by enclosing vertical lines.

Question 5.
State the application of differential calculus.
Answer:

1. To find the rate of change in demand with respect to price.
2. To find the rate of change in income with respect to the investment.

Question 6.
What is ICT?
Answer:
Information and Communication Technology (ICT) is the infrastructure that enables computing faster and accurate.

Question 7.
What is a worksheet?
Answer:
A worksheet is a table like a document containing rows and columns with data and formula

Part – C
Answer the following questions in One Paragraph

Question 1.
The marginal cost function for producing x units is y = 23 + 16.x – 3.x2 and the total cost for producing zero unit is Rs. 40. Obtain the total cost function and the average cost function.
Solution:
Given Marginal cost functions
y = 23 + 16x – 3x²
C = 40

Average cost function = \(\frac { TC }{ x }\)
= \(\frac{23 x+8 x^{2}-x^{3}+40}{x}\)
AC = 23 + 8x – x + \(\frac { 40 }{ x }\)

Question 2.
Find the solution of the system of equations
5x₁ + 3x₂ = 30
6x₁ – 2X₂ = 8
Solution :

Question 3.
If the demand function is p = 35 – 2x – x² and the demand x_o is 3, what will be the consmer’s surplus
Solution:
Given:
p = 35 – 2x – x²
If x = 3 then
p = 35-2 (3) – 3²
= 35 – 6 – 9
p = 20

= 35 (3) – 9 – 9 – 60
= 105 – 9 – 9 – 60
CS = 27 units

Question 4.
Given the total cost function TC = 15 + 3Q2 + 7Q3 drive the marginal cost function.
Solution:
Given:
TC = 15 + 3Q² + 7Q²
MC = \(\frac { d(15) }{ dQ }\) + \(\frac{\mathrm{d}\left(3 \mathrm{Q}^{2}\right)}{\mathrm{d} \mathrm{Q}}\) + \(\frac{\mathrm{d}\left(7 \mathrm{Q}^{3}\right)}{\mathrm{d} \mathrm{Q}}\)
= 0+ 3 (2) Q + 7 (3) Q + (3) Q²
MC = 6Q + 21 Q²

Question 5.
Find the value of the determinant of the matrix
Solution:
Given:

= 3 (-5) -4 (-19) +7 (-3)
= -15 + 76 – 21
A = 40

Question 6.
Find the value of the determinant of the matrix A =.
Solution:
Given:

Part – D
Answer the following questions in about a page

Question 1.
Find the equation of a straight line which passes through two points (2,2) and (4, -8) which are (x₁, y₁) and (x₂, y₂) respectively.
Note: For drawing a straight line, at least two points are required. Many straight lines can pass through a single point.
Solution:
Here x₁ = 2 y_1- = 2 and x₂ = 4 y₂ = – 8
Formula for constructing a straight line.

Slope (m) = -5
y intercept = 12
Constant = C
This is of this from y = mx + C
y = 12 – 5x
When x = 0
y = 12
When y = 0
x = \(\frac { 12 }{ 5 }\) = 2.4

Question 2.
Find the solution of the equation system
Solution:
7x₁ – x₂ – x₃ = 0
10x₁ – 2x₂ + x₃ = 8
6x₁ + 3x₂ – 2x₃ = 7

= 7 (4 – 3) + 1 (-20 -6) -1 (30 + 12)
= 7 (1) + 1 (-26) -1 (42)
+ 7 -26 – 42
∆ = – 61

= 0 (4 -3) + 1 (-16 -7) – 1 (24+14)
= 0 + 1 (-23) – 1 (38)
= -23 – 38
= -61

= 7 (-16 – 7) – 0 (- 20 – 6) – 1 (70 – 48)
= 7 ( -23) + 0 (-20 – 6) – 1(70 – 48)
= – 161 – 22
= – 183

= 7(-14 -24) + 1 (70 – 48) + 1 (30 + 12)
= 7 (-38) + 1 (22) + 0
= – 266 + 22
= – 244