Super 20 for Class 12 Tamil Nadu Board 2022 Commerce Stream

Note: 2015-16 Percentage of share holder fund 5, 40,000 = 9,00,000 × 100 = 60% 2, 70, 000 Percentage of Non-current liabilities = 9,00,000 × 100 = 30% Percentage of current liabilities 90, 000 = 9,00,000 × 100 = 10% Percentage of Non-current assets 7, 20,000 = 9,00,000 × 100 = 80% Percentage of current assets 1, 80, 000 = 9,00,000 × 100 = 20% Note: 2016-17 6, 00, 000 Percentage of share holder fund = 10,00,000 × 100 = 60% 2, 50, 000 Percentage of Non-current liabilities = 10,00,000 × 100 = 25% Percentage of current liabilities 1, 50, 000 = 10,00,000 × 100 = 15% Percentage of Non-current assets = 8, 00, 000 × 100 = 80% Percentage of current assets 10, 00, 000 2, 00, 000 = 10,00,000 × 100 = 20% [OR] (b) Calculate turnover ratios from the following information from ‘A’ Ltd. Particulars As on 31. 03. 18 As on 31. 03. 19 Inventory 3,60,000 4,40,000 Trade receivable 7,40,000 6,60,000 Trade payable 1,90,000 2,30,000 Fixed Assets 6,00,000 8,00,000 Additional information: Revenue from operation for the year 35,00,000 Purchases for the year 21,00,000 Cost of revenue from operations 16,00,000 Assume that sales and purchases are for credit. Ans. (i) Inventory turnover ratio = Cost of revenue from operation Average Inventory Sample Paper - 1    E-23

Cost of revenue from operations ` 16,00,000 Average Inventory = Opening inventory + Closing inventory 2 = 3,60,000 + 4, 40,000 = 8, 00, 000 =` 4,00,000 2 2 16, 00, 000 ∴ Inventory turnover ratio = 4,00,000 = 4 times (ii) Trade receivable turnover ratio = Credit revenuefrom operations Average trade receivable Credit revenue from operations = ` 15,00,000 Average trade receivable = Opening trade receivable + closing trade receivable 2 = 7, 40,000 + 6, 60, 000 = 14, 00, 000 = ` 7,00,000 2 2 35, 00, 000 ∴ Trade receivable turnover ratio = 7,00,000 = 5 times (iii) Trade payable turnover ratio = Net credit purchase Average trade payable Average trade payable = Opening trade payable + Closing trade payable 2 = 1,90,000 + 2,30,000 4, 20,000 = ` 2,10,000 2= 2 21, 00, 000 ∴ Trade payable turnover ratio = 2,10,000 = 10 times Revenue from operation (iv) Fixed assets turnover ratio = Average fixed assets Revenue from operations = ` 35,00,000 A verage fixed assets = Operationalfixed assets + Closing fixed assets 2 14, 00, 000 = 6, 00, 000 + 8,00,000 = 2 = ` 7,00,000 2 ∴ Fixed assets turnover ratio = 35, 00, 000 = 5 times 7, 00, 000 E-24    Accountancy – XII

4SAMPLE PAPER – Time: 2 ½ Hours (UNSOLVED) Maximum Marks: 90 PART - I Answer all the questions. Choose the correct answer. [20×1 = 20] (d) Profit 1. The excess of assets over liabilities is .................... . (a) Loss (b) Cash (c) Capital 2. Receipts and payments account is a ....................... . (a) Nominal A/c (b) Real A/c (c) Personal A/c (d) Representative personal A/c 3. Statement of affairs is a ........................ . (a) Statement of income and expenditure (b) Statement of assets and liabilities (c) Summary of cash transactions (d) Summary of credit transactions 4. Receipts and payments account records receipts and payment of ...................... . (a) Revenue nature only (b) Capital nature only (c) Both revenue and capital nature (d) None of these 5. In the absence of an agreement partners are entitled to ...................... . (a) Salary (b) Commission (c) Interest on loan (d) Interest on capital 6. The profit or loss arising from the partnership business is shared by the partners in the ...... . (a) Old ratio (b) New ratio (c) agreed ratio (d) Sacrificing ratio 7. Super profit is the difference between ........... . (a) Capital employed and average profit (b) Assets and liabilities (c) Average profit and normal profit (d) Current year’s profit and average profit 8. Book profit of 2017 is ` 35,000. Non- recurring income included in the profit is ` 1,000 and abnormal loss charged in the year 2017 was ` 2,000 then the adjusted profit is ....................... . (b) ` 35,000 (c) ` 38,000 (d) ` 34,000 (a) ` 36,000 9. The profit or loss or revaluation of assets and liabilities is transferred to the capital account of ......................... . (a) The old partners (b) The new partners (c) All the partners (d) The sacrificing partners 10. A and B are sharing profit and losses in the ratio of 5:3. They admit R as a partner giving him 1/5 share of profits. Find out the sacrificing ratio ........................ . (a) 1:3 (b) 3:1 (c) 5:3 (d) 3:5 11. At the time of retirement of a partners determination of gaining ratio required ................... . (a) To transfer revaluation profit or loss (b) To distribute accumulated profits and losses (c) To adjust goodwill (d) None of these E-70

12. The final amount due to a retiring partner is not paid immediately, it is transferred to ...................... . (a) Bank A/c (b) Retiring partner’s capital A/c (c) Retiring partner’s loan A/c (d) Other partners’ capital A/c 13. After the forfeited shares are reissued, the balance in the forfeited shares account should be transferred to ...................... . (a) General Reserve account (b) Capital Reserve account (c) Securities premium account (d) Surplus account 14. The amount received over and above the par value is credited to ..................... . (a) Securities premium account (b) Calls in advance account (c) Share capital account (d) Forfeited shares account 15. Balance sheet provides information about the financial position of a business concern ................. . (a) Over a period of time (b) As on a particular date (c) For a period of time (d) For the accounting period 16. Which of the following tools of financial statement analysis is suitable when data relating to several years are to be analysed? (a) Cash flow statement (b) Common size statement (c) Comparative statement (d) Trend analysis 17. Current ratio indicates .................... . (a) Ability to meet short term obligation (b) Efficiency of management (c) Profitability (d) Long Term Solvancy 18. Debt equity ratio is a measure of ...................... . (a) Short-Term Solvancy (b) Long-Term Solvancy (c) Profitability (d) Efficiency 19. Which sub menu displays groups ledgers and vouchers types in Tally? (a) Inventory voucher (b) Accounting voucher (c) Company Info (d) Accounting Info 20. Which are the predefined ledgers in Tally? (i) Cash (ii) Profit and loss A/c (iii) Capital A/c (a) Only (i) (b) Only (ii) (c) Both (i) and (ii) (d) Both (ii) and (iii) PART - II [7×2 = 14] Answer any seven questions in which question No. 30 is compulsory. 21. What is conversion method? 22. Give four examples for capital receipts of not-for-profit organisation. 23. Sundar drew regularly ` 4,000 at their end of every month. Calculate interest on drawings at 10 % P.A. 24. From the following information, calculate the value of goodwill on the basis of 3 years purchase of average profits of last four years 2015 profit - 5,000; 2016 profit - ` 8,000; 2017 loss - ` 3,000; 2018 profit - ` 6,000. Sample Paper - 4    E-71

25. A and B were partners of the firm ratio 7:5 on 1.4.2018 the firm’s book showed a reserve fund ` 1,20,000 on the above date they decided to admit C into the firm pass entry. 26. Mani, Nagappan, Ulaganathan are partners Sharing profit with ratio of 4:3:3. Ulaganathan retires and share is taken by Mani and Nagappan in the ratio of 8:2. Calculate new ratio. 27. Abdul Ltd issues 50,000 shares of ` 10 each payable fully on application pass the entry shares are issued at. 28. What is meant by “Financial Analysis”? 29. Give any two differences between current ratio & Quick ratio. 30. What are accounting reports? PART - III [7 x 3 = 21] Answer any seven questions in which question No. 40 is compulsory. 31. From the following details find out total sales made during the year. ` Debtors on 1st April 2018 50,000 Cash received from debtors during the year 1,50,000 Returns inward 15,000 Bad debts 5,000 Debtors on 31st March 2019 70,000 Cash Sales 1,40,000 32. Compute income from subscription for the year 2018 from the following particulars relating to a club. Particulars 1.1.2018 (`) 31.12.2018 (`) Outstanding subscription 3,000 5,000 Subscription received in advance 4,000 7,000 Subscription received during the year 2018: ` 45,000. 33. From the following information prepare capital accounts of partners, Shanthi and Sumathi when their capitals are fixed. Particulars Shanthi Sumathi Capital on 1st January 2018 1,00,000 80,000 Current account on 1st January 2018 (cr) 5,000 3,000 Additional capital introduced 1st June 2018 10,000 20,000 Drawings during 2018 20,000 13,000 Interest on drawings 500 Share of profits 2018 10,000 300 Interest on capital 6,300 8,000 Salary 9,000 5,400 Commission Nil Nil 1200 E-72    Accountancy – XII

34. For the purpose of admitting a new partner a firm has decided to value goodwill at 3 years purchase of average profits of the last 4 years using weighted average method. Profits of the past 4 years and the respective weight. Year 2015 2016 2017 2018 Profit 20,000 22,000 24,000 28,000 weight 12 3 4 Compute the value of goodwill. 35. Sri Ram and Raj are partners sharing ratio 2:1. Nelson joins as a partner on 1st April 2017. The following adjustments are to made (i) Increase the value of stock by ` 5,000 (ii) Bring into record investment of ` 7,000 which had not been recorded in the books of the firm (iii) Reduce the value of office equipment by ` 10,000 (iv) A provision would also be made for outstanding wages for ` 9,500. Give journal entries and prepare revaluation A/c. 36. A, B and C are partners in a firm sharing profit and losses in the ratio of 5:3:2 on 31st March 2018. ‘C’ retires from the firm. Goodwill appeared in the books of a firm ` 1,20,000 by assuming fluctuating capital account. Pass the journal entries. (a) Write off the entire amount of existing goodwill (b) Write off half of the existing goodwill 37. From the following information relating to a partnership firm, find out the value of its goodwill based on 3 years purchase of average profits of the last 4 years: (a) Profits of the years 2015, 2016, 2017 and 2018 are ₹ 10,000, ₹ 12,500, ₹ 12,000 and ₹ 11,500, respectively. (b) The business was looked after by a partner and his fair remuneration amounts to ₹ 1,500 per year. This amount was not considered in the calculation of the above profits. 38. From the following particulars of Mani Ltd and Kani Ltd, prepare a common size income statement for the year ended 31st March 2019. Particulars Mani Ltd ` Kani Ltd ` Revenue from operations 2,00,000 2,50,000 Other income 30,000 25,000 Expenses 1,10,000 1,25,000 39. Calculate gross profit ratio, net profit ratio, stock turnover ratio, opening stock ` 1,00,000; Purchases ` 3,50,000; Gross profit ` 1,50,000; Net profit 90,000; Revenue from operations ` 4,50,000; closing stock ` 1,50,000. 40. What is meant by voucher and what are its types? Answer all the questions. PART - IV [7×5 = 35] 41(a) From the following details you are required to calculate credit sales and credit purchases by preparing total debtors account, total creditors account, bills receivable account and bills payable account. Particulars ` Particulars ` Balances as on 1st April 2018 Balances as on 31st March 2019 Sundry debtors 2,40,000 Sundry debtors 2,20,000 Sample Paper - 4    E-73

Bills receivable 30,000 Sundry creditors 1,50,000 Sundry creditors 1,20,000 Bills receivable 8,000 Bills payable 10,000 Bills payable 20,000 Other information: ` ` Cash received from debtors 6,00,000 Payments against bill payable 30,000 Discount allowed to customers 25,000 Cash received for bills receivable 60,000 Cash paid to creditors 3,20,000 Bills receivable dishonoured 4,000 Discount allowed by suppliers 10,000 Bad debts 16,000 [OR] (b) From the following balance sheets of Brindha and Praveena who share profits and losses in the ratio of 3:4, calculate interest on capital at 6% p.a. for the year ending 31st December 2017. Balance sheet as on 31st December 2017 Liabilities ` Assets ` Capital accounts: Sundry assets 80,000  Brindha 30,000   Praveena 40,000 Profit and loss appropriation A/c 10,000   80,000   80,000 On 1st July 2017, Brindha introduced an additional capital of ` 6,000 and on 1st October 2017, Praveena introduced ` 10,000. Drawings of Brindha and Praveena during the year were ` 5,000 and ` 7,000, respectively. Profit earned during the year was ` 31,000. 42(a) From the following receipts and payments of friends football club for the year ending 31st march 2017. Prepare income and expenditure for the year ending 31st March 2017. In the Books of Friends Foot Ball Club Dr. Receipts and payments Account for the year ended 31.3.17Cr. Receipts `` Payments `` To Balance b/d By furniture 7,000 Cash 1,000 By sports materials purchased 800 Bank 10,000 11,000 By special dinner expenses 1,500 To subscription 5,000 By electricity changes 900 To legacies 6,000 By balance c/d To collection for 2,000 Cash in hand 1,800 special dinner Cash at bank 12,000 13,800 24,000 24,000 E-74    Accountancy – XII

Additoinal information: (i) The club had furniture of ` 12,000 on 1st April 2016. Ignore depreciation on furniture. (ii) Subscription outstanding for 2016-17 - ` 600 (iii) Stock of sports materials on 31.3.2017 - ` 100 (iv) Capital fund as on 1st April 2016 was ` 23,000 [OR] (b) The following particulars are available in respect of the business carried on by a partnership firm. (i) Profits earned 2016 - ` 25,000; 2017 - ` 23,000; 2018 - ` 26,000 (ii) Profit of 2016 includes a non - recurring income of ` 2,500. (iii) Profit of 2017 is reduced by ` 3,500 due to stock destroyed by fire. (iv) The stock was not insured. But it is decided to insure the stock in future. The insurance premium is estimated to be ` 250 per annum. You are required to calculate the value of goodwill of the firm on the basis of 2 years purchase of average profits of the last three years. 43(a) The following particular calculate the goodwill on the (i) Basis of two years purchase of super profits earned on average basis during the above mentioned three years by (ii) Capitalisation method. The firm earned net profit for 3 years 2013 - ` 1,90,000; 2014 - ` 2,20,000; 2015 - ` 2,50,000. The capital employed in the firm ` 4,00,000, risk involved 15% is considered to be a fair return on the capital. The remuneration of the partners during the period ` 1,00,000 P.a. [OR] (b) Vetri and Ranjit are partners sharing profit in the ratio 3:2. Their balance sheet as on 31st December 2017 is as under. Liabilities `` Assets ` Capital Accounts Furniture 25,000 Vetri 30,000 Stock 20,000 Ranjit 20,000 50,000 Debtors 10,000 Reserve fund 5,000 Cash in hand 35,000 Sundry creditors 45,000 Profit and loss A/c loss 10,000 1,00,000 1,00,000 On 1.1.2018 they admit suriya into their firm as a partner on the following arrangements. (i) Suriya brings ` 10,000 as capital for 1/4% share of profit (ii) Stock to be depreciated by 10% (iii) Debtors to be revalued at ` 7,500. (iv) Furniture to be revalued at ` 40,000. (v) There ia an outstanding wages ` 4,500 not yet recorded. Prepare revaluation account, partners capital account and the balance sheet of the firm after admission. Sample Paper - 4    E-75

4 4(a) Mani, Rama, Devan are partners in a firm sharing profit and losses in the ratio of 4:3:3. Their balance sheet as on 31st March 2019. Liabilities `` Assets ` Capital Accounts Buildings 80,000 Mani 50,000 Stock 20,000 Rama 50,000 Furniture 70,000 Devan 50,000 1,50,000 Debtors 20,000 Sundry creditors 20,000 Cash in hand 10,000 Profit and loss A/c 30,000 2,00,000 2,00,000 Mani retired from the partnership on 31.03.2019. subject to the following adjustment. (i) Stock to be depreciated ` 5,000. (ii) Provision for doubtful debts to be created for ` 1,000. (iii) Buildings to be appreciated by ` 16,000. The final amount due to Mani is not paid immediately prepare revaluation A/c, Capital Account. [OR] (b) Roja, Neela and Kanaga are partners sharing profits and losses in the ratio of 4:3:3. On 1st April 2017, Roja retires and on retirement, the following adjustments are agreed upon: (i) Increase the value of building by ` 30,000. (ii) Depreciate stock by ` 5,000 and furniture by ` 12,000. (iii) Provide an outstanding liability of ` 1,000 Pass journal entries and prepare revaluation account. 45(a) Keerthiga company issued share of ` 10 each at 10% premium. Payable ` 2 on application, ` 3 on allotment (including premium), ` 3 on first call and ` 3 on second and final call. Journalise the transactions relating to forfeiture of shares for the following situations. (i) Mohan who holds 50 shares failed to pay the second and final call and his shares were forfeited. (ii) Mohan who holds 50 shares failed to pay the allotment money first call and second and final call money and his shares were forfeited. (iii) Mohan who holds 50 shares failed to pay the allotment money and first call and his shares were forfeited after the first call. [OR] (b) Prepare comparative statement of profit and loss of Ahmed Ltd with the helps of following information. Particulars 31.03.2013 31.03.2012 Revenue from operation 8,00,000 5,00,000 Purchase of stock in trade 4,20,000 2,40,000 Charges in inventories 80,000 60,000 Employee benefit expenses 20,000 40,000 E-76    Accountancy – XII

Other expenses 40,000 30,000 Income Tax 50% 40% Employees benefit expenses = Salaries wages, PF, staff welfare 46 (a) From the following particulars, prepare comparative statement of financial position of Kala Ltd. Particulars 31st March, 2017 31st March, 2018 `` I EQUITY AND LIABILITIES   1. Shareholders’ Fund 3,00,000 3,60,000    (a) Share capital 50,000 50,000    (b) Reserves and surplus   2. Non-current liabilities 50,000 40,000     Long-term borrowings   3. Current liabilities 20,000 12,000     Trade payables  Total 4,20,000 4,62,000 II ASSETS   1. Non-current assets   (a) Fixed assets 2,50,000 2,90,000   (b) Non-current investments 50,000 40,000   2. Current assets   Inventories 80,000 1,00,000   Cash and cash equivalents 40,000 32,000 Total 4,20,000 4,62,000 [OR] (b) From the following information calculate the following ratio. (i) Net Profit Ratio (ii) Debtors Equity Ratio (iii) Quick Ratio Paid up capital 20,00,000 Indirect expenses 4,00,000 Capital Reserve 2,00,000 Current liability 3,00,000 9% debentures 8,00,000 Opening inventory 50,000 Net Revenue from 14,00,000 Closing Inventory 20% operation more than opening inventory Gross Profit 8,00,000 47(a) What are the Salient Feature of tally ERP.9? [OR] (b) Find out total purchases and total sales from the following details by preparing necessary accounts. Sample Paper - 4    E-77

One-mark & Two-mark Answers for Unsolved Sample Papers Sample Paper- 4 1. (c) Capital PART - II 2. (b) Real A/c 3. (b) Statement of assets and liabilities 11. (c) To adjust goodwill 4. (c) Both revenue and capital nature 12. (c) Retiring partner loan A/c 5. (c) Interest on loan 13. (b) Capital Reserve account 6. (c) agreed Ratio 14. (a) Securities premium account 7. (c) Average profit and normal profit 15. (b) As on a particular date 8. (c) ` 38,000 16. (d) Trend Analysis 9. (a) The old partners 17. (a) Ability to meet short term obligation 10. (c) 5:3 18. (b) Long Term Solvancy 19. (d) Accouting Info 20. (c) Both (i) and (ii) PART - II 21. It is one of the methods to find out profit or loss of a business when accounts are kept under single entry system. If it is desired to calculate the profit or loss by preparing trading and profit and loss account under single entry system. Then it is called conversion method. It is conversion of single entry into double entry. 22. (i) Life membership fees (ii) Legacies (iii) Sale of fixed assets (iv) Specific donations 23. Interest on drawings of Sundar = Total drawings × Rate × Average period 100 12 = 4,000 × 12 × 10 × 5.5 100 12 = ` 2,200 24. Goodwill = Average profit × No. of years of purchase Total profit Average profit = Number of years = 5,000 + 8,000 − 3,000 + 6,000 = 16, 000 = ` 4,000 Good will 4 4 = Average profit × No. of years of purchase = ` 4,000 × 3 Good will = ` 12,000 E-134

25. Journal Entry Date Particulars L.F. Dr. (`) Cr. (`) 1,20,000 Reserve Fund A/c Dr. 70,000 50,000 To A’s capital A/c To B’s capital A/c (Reserve fund transferred to old partners capital accounts in old ratio) 26. Calculation of new ratio: Mani Nagappan Ulaganathan Old ratio 4/10 : 3/10 : 3/10 Gaining ratio = 3 : 2 Gain = 3 × 3 : 2× 3 5 10 5 10 New share = 9 : 6 50 50 = 4 + 9 : 3+6 10 50 10 50 = 20 + 9 : 15 + 6 50 50 = 29 : 21 New ratio 50 50 = 29 : 21 27. In the books of Abdul Ltd Journal entry Date Particulars L.F. Debit Credit 5,00,000 Bank A/c Dr. 5,00,000 5,00,000 To Share Application (Application money receive) Share Application A/c Dr. 5,00,000 To share capital A/c (Application money transfer to share capital) 28. Financial Analysis is a systematic process of classifying the data into simple groups and making a comparison of various groups with one another to pin-point the strong points and weakness of the business. 29. Current Ratio Quick Ratio It indicates whether the firm is in a position It indicates whether the firm is in a position to pay its current liabilities with in a year. to pay its current liabilities immediately with in a month. Ideal Standard Ratio is 2:1 Quick Ratio Standard is 1:1 Answers    E-135

30. Accounting report is a compilation of accounting information that are derived from the accounting records of a business concern. Accounting reports may be classified as routine reports and special purpose reports. E-136 Accountancy – XII

BUSINESS MATHEMATICS & STATISTICS QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to Total be Answered Marks Part-I 1 20 20 Multiple Choice Questions Part-II 2 7 14 (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily) Part-III 3 7 21 (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily) Part-IV 57 35 Total Marks 90 S.No. Weightage of Marks Weightage 1. 30% 2. Purpose 40% 3. Knowledge 20% 4. Understanding 10% Application Skill/Creativity F-1

1SAMPLE PAPER – Time: 3 Hours (SOLVED) Maximum Marks: 90 PART-I I. Answer all the questions. Choose the most suitable answer from the given four alternatives and write the option code with the corresponding answer. [Answers are in bold] [20 × 1 = 20] 1   1. If A =  2  , then the rank of AAT is ....................... . (a) 0  3 (b) 1 (c) 2 (d) 3 2. For the system of equations x + 2y + 3z = 1, 2x + y + 3z = 2, 5x + 5y + 9z = 4 ..................... . (a) there is only one solution (b) there exists infinitely many solutions (c) there is no solution (d) none of these 3. ∫ log x dx , x > 0 is .................... . x (a) 1 (log x)2+ c (b) – 1 (log x)2 2 2 22 (c) x2 −c (d) x2 +c 11 1 1 4. = If ∫ f (x)dx 1,=∫ x f (x)dx a and ∫ x2 f (x)dx = a2 , then ∫ (a − x)2 f (x) dx is ..................... . 00 0 0 (a) 4 a2 (b) 0 (c) 2 a2 (d) 1 5. The marginal revenue and marginal cost functions of a company are MR = 30 − 6x and MC = −24 + 3x, where x is the product, then the profit function is .............. . (a) 9x2 + 54x (b) 9x2 – 54x (c) 54x – 9x2 (d) 54x – 9x2 + k 2 2 6. Area bounded by the curve y = 1 between the limits 1 and 2 is .................. . x (a) log 2 sq.units (b) log 5 sq.units (c) log 3 sq.units (d) log 4 sq.units 7. The order and degree of the differential equation dd=x2 2y dy + 5 are ..................... respectively. dx (a) 2 and 3 (b) 3 and 2 (c) 2 and 1 (d) 2 and 2 8. The differential equation formed by eliminating a and b from y = aex + be− x is ..................... . (a) d2 y − y =0 (b) d2y − dy =0 (c) d2y = 0 (d) d2y − x =0 dx 2 dx2 dx dx2 dx2 F-3

( ) 9. If h = 1, then ∆ x2 = ................. . (c) 2x + 1 (d) 1 (a) 2x (b) 2x – 1 10. If f (x) = x2 + 2x + 2 and the interval of differencing is unity, then ∆ f (x) ................ . (a) 2x – 3 (b) 2x + 3 (c) x + 3 (d) x – 3 11. A discrete probability distribution may be represented by ...................... . (a) table (b) graph (c) mathematical equation (d) all of these 12. Probability which explains x is equal to or less than particular value is classified as ............ . (a) discrete probability (b) cumulative probability (c) marginal probability (d) continuous probability 13. An experiment succeeds twice as often as it fails. The chance that in the next six trials, there shall be at least four successes is ................. . (a) 240/729 (b) 489/729 (c) 496/729 (d) 251/729 14. The average percentage of failure in a certain examination is 40. The probability that out of a group of 6 candidates atleast 4 passed in the examination are ............. . (a) 0.5443 (b) 0.4543 (c) 0.5543 (d) 0.4573 15. Which one of the following is probability sampling? (a) purposive sampling (b) judgement sampling (c) simple random sampling (d) convenience sampling 16. An estimator is a sample statistic used to estimate a .................... . (a) population parameter (b) biased estimate (c) sample size (d) census 17. While computing a weighted index, the current period quantities are used in the ............. . (a) Laspeyre’s method (b) Paasche’s method (c) Marshall Edgeworth method (d) Fisher’s ideal method 18. A time series consists of ................ . (b) Four components (a) Five components (d) Two components (c) Three components 19. If number of source is not equal to number of destinations, the assignment problem is called ................. . (a) balanced (b) unsymmetric (c) symmetric (d) unbalanced 20. Decision theory is concerned with .......... . (a) analysis of information that is available (b) decision making under certainty (c) selecting optimal decisions in sequential problem (d) All of the above F- 4 Business Mathematics and statistics –XII

PART-II [7 × 2 = 14] II. Answer any seven questions. Question No. 30 is compulsory. 1 2 3   21. Find the rank of A =  2 3 4  3 5 7 1 2 3   Ans. Given A=  2 3 4   3 5 7  123 Consider 2 3 4 = 1 (21 – 20) – 2 (14 – 12) + 3 (10 – 9) = 1 – 4 + 3 = 0 357 Since third order minor equals zero, ρ(A) < 3 12 Consider 2 3 = 3 – 4 = –1 ≠ 0 There is a minor of order 2 which is not zero. Hence ρ (A) = 2 22. Integrate the (3 + x) (2 – 5x) with respect to x. Ans. (3 + x)(2 − 5x) = 6 – 15 x + 2x – 5x2 = 6 – 13 x – 5x2 (3 + x) (2 – 5x) dx = (6 – 13x – 5x2) dx = 6 dx – 13 x dx – 5x2 dx = 6x − 13x2 − 5x3 + c 23 23. Find the area bounded by the line y = x, the x-axis and the ordinates x = 1, x = 2. Ans. Given lines are y = x, x – axis, x = 1, x = 2  y Required area = ∫ ∫2 2  x2 2 2− 1 y=x  2 = 2 ydx= xdx=  1 1 1 = 3 sq.units 0 x 2 x=1 x=2 24. Find the differential equation: y = cx + c – c3 Ans. y = cx + c – c3  ..........(1) Here c is a constant which has to be eliminated. Differentiating w.r.t x , dy = c ..........(2) Using (2) in (1) we get,=y  dy  dx equation.  dx  dy  dy 3 x + dx −  dx  which is the required differential Sample Paper -1 F-5

25. If f (x) = x2 + 3x, then show that ∆ f (x) = 2 x + 4. Ans. f (x) = x2 + 3 x ∆ f (x) = f (x + h ) – f ( x ) = (x + h )2 + 3 (x + h ) – x 2 – 3x = x2 + 2 xh + h2 + 3x + 3h – x2 – 3x = 2 xh + 3 h + h2 Put h = 1,∆ f (x) = 2 x + 4 26. Two coins are tossed simultaneously. Getting a head is termed as success. Find the probability distribution of the number of successes. Ans. Let X be the number of observed heads. The sample space S = {(H H), (H T), (T H), (T T)} X takes the values 2,1,1,0. Hence the probability distribution of the number of successes is given by X = xi 0 12 P (X = xi ) 1 11 4 24 27. Define Bernoulli trials. Ans. A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q, is called a Bernoulli trial. Example 1, Tossing of a coin (Head or Tail) Example 2, Writing an exam (Pass or Fail) 28. A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Tamil Nadu State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. Calculate standard error of mean. Ans. Given n = 1000, X = 119, σ = 30 S.E. = σ = 30 = 30 = 0.9487 n 1000 31.623 29. Mention the components of the time series. Ans. Components of Time Series There are four types of components in a time series. They are as follows: (i) Secular Trend (ii) Seasonal variations (iii) Cyclic variations (iv) Irregular variations 30. What is feasible solution and non degenerate solution in transportation problem? Ans. Feasible Solution: A feasible solution to a transportation problem is a set of non-negative values xij (i = 1, 2,.., m, j = 1, 2, ...n) that satisfies the constraints. Non degenerate basic feasible Solution: If a basic feasible solution to a transportation problem contains exactly m + n – 1 allocations in independent positions, it is called a Non degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem. F- 6 Business Mathematics and statistics –XII

PART-III III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]  1 2 −1 3    31. Find the rank of the following matrices.  2 4 1 −2   1 2 −1 3   3 6 3 −7    Ans. Let A =  2 4 1 −2  3 6 3 −7 Let us transform the matrix A to an echelon form by using elementary transformations.  1 2 −1 3    A=  2 4 1 −2  3 6 3 −7  1 2 −1 3    ~  0 0 3 −8  r2 → r2 – 2r1  0 0 6 −16  r3 → r3 – 3r1  1 2 −1 3   3 −8 r3 → r3 – 2r2 ~  0 0  0 0 0 0  The above matrix is in echelon form. The number of non zero rows is 2 ⇒ ρ (A) = 2. 32. Integrate the  2x − 1 2 with respect to x.  2x    1 2 2  1  1  2x  =  2x  2x − 2 2 ( ) ( )Ans. ∫ 2x + 2x − 2x ( ) = 2x − 2 + 1 2x 2 So ∫  2x − 1  dx = ∫ 2xdx −∫ 2dx + ∫ 1 dx + c  2x 2x = x2 − 2x + 1 log x +c 2 33. Calculate the producer’s surplus at x = 5 for the supply function p = 7 + x. Ans. Given supply function is p = 7 + x, x0 = 5 p0 = 7 + x0 = 7 + 5 = 12 x0 5 ∫ Producer’s surplus PS = x0 p0 − ∫ p(x) dx = 5(12) − (7 + x)dx 00 x2 5  2 0 =60 − 35 – 25 = 25 = 60 −  7x + 2 2 Hence the producer’s surplus is 25 units. 2 Sample Paper -1 F-7

34. Solve: y (1 – x) – x dy =0 Ans. y (1 – x) – x dx dy = 0 dx Separating the variables, y (1 – x) = x dy dx (1− x) dx = dy y x Integrating both the sides,  1x −1 dx = dy y log x – x = log y + c is the general solution. 35. If h = 1, then prove that (E − 1 ∆) x3 = 3x2 − 3x +1. Ans. h = 1 To prove (E −1 ∆) x 3 = 3x2 − 3x +1 L.H.S (E −1 ∆) x 3 = E −1 (∆ x 3 ) = E −1[ ( x + h)3 – x 3 ] = E −1( x + h)3 – E −1 (x 3) = (x – h + h)3 – (x – h)3 = x3 – (x – h)3 But given h = 1 So (E −1 ∆) x 3 = x3 – ( x – 1)3 = x3 – [ x3 – 3x2 + 3x – 1] = 3 x2 – 3 x + 1 = RHS 36. A person tosses a coin and is to receive ` 4 for a head and is to pay ` 2 for a tail. Find the expectation and variance of his gains. Ans. Let X be the discrete random variable which denotes the gain of the person. The probability distribution of X is given by X=x 4 –2 P (X) 1/2 1/2 (Here, since a coin is tossed the probability is equal for the outcomes head or tail) E(X) = 4 1 + (– 2)  1  = 2 – 1 = 1  2   2  E(X2) = (4)2 1 + (– 2)2 1  2   2  = 16 1 +4  1  = 8 + 2 = 10  2   2  Var(X) = E(X2) – [E(X)]2 = 10 – 1 = 9 Thus the expectation of his gains is 1 and variance of his gains is 9. F- 8 Business Mathematics and statistics –XII

37. Write down the conditions for which the binomial distribution can be used. Ans. The binomial distribution can be used under the following conditions: (a) The number of trials (or) observations ‘n’ is fixed (finite). (b) Each observation is independent of each other. (c) In every trial, there are only two possible outcomes - success or failure. (d) The probability of success ‘p’ is the same for each outcome. 38. State any three merits of stratified random sampling. Ans. (i) A random stratified sample is superior to a simple random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled. (ii) A stratified random sample can be kept small in size without losing its accuracy. (iii) It is easy to administer,if the population under study is sub-divided. 39. Define Time Reversal Test. Ans. It is an important test for testing the consistency of a good index number. This test maintains time consistency by working both forward and backward with respect to time (here time refers to base year and current year). Symbolically the following relationship should be satisfied, P01 × P10 = 1 Fisher’s index number formula satisfies the above relationship P0F1   p1q0  p1q1  p0q0  p0q1 When the base year and current year are interchanged, we get,   P1F0 p0q1  p0q0 P0F1 × P1F0 = 1 p1q1  p1q0 40. Given the following pay-off matrix (in rupees) for three strategies and two states of nature. Strategy States-of-nature S1 E1 E2 40 60 S2 10 – 20 S3 – 40 150 Select a strategy using each of the following rule (i) Maximin (ii) Minimax Ans. (i) Maximin principle Strategy States - of - nature Minimum payoff E1 E2 S1 40 60 40 S2 10 –20 – 20 S3 – 40 150 – 40 Sample Paper -1 F-9

Max (40, –20, – 40) = 40. Since the maximum pay- off is Rs. 40, the best strategy is S1 according to maximin rule. (ii) Minimax principle Strategy States - of - nature Maximum payoff E1 E2 S1 40 60 60 S2 10 –20 10 S3 – 40 150 150 Min (60,10,150) = 10. Since the minimum pay- off is Rs.10, the best strategy is S2 according to minimax rule. PART - IV IV. Answer all the questions. [7 × 5 = 35] 41. (a) Two products A and B currently share the market with shares 50% and 50% each respectively. Each week some brand switching takes place. Of those who bought A the previous week, 60% buy it again whereas 40% switch over to B. Of those who bought B the previous week, 80% buy it again where as 20% switch over to A. Find their shares after one week and after two weeks. If the price war continues, when is the equilibrium reached? Ans. Given that two products A and B have shares 50 % and 50% respectively. (A A) ⇒ those who bought A the previous week, will buy it again = 60 % = 0.6 (A B) ⇒ those who bought A the previous week, will buy B now = 40 % = 0.4 (B A) ⇒ those who bought B the previous week, will switch to A = 20 % = 0.2 (B B) ⇒ those who bought B will again buy B = 80 % = 0.8 AB The transition probability matrix is given by T = A  0.6 0.4   0.8  B  0.2 AB The current position of A and B in the market is ( 0.5 0.5) After one week A B A  A B B  0.6 0.4  The shares of A and B are given by ( 0.5 0.5)  0.8  0.2 AB = ( 0.3 + 0.1 0.2 + 0.4) AB = ( 0.4 0.6) So after one week the market share of A is 0.4 ×100 =40% and that of B is 0.6 ×100 =60% 100 100 F- 10 Business Mathematics and statistics –XII

After two weeks AB AB The shares of A and B are given by ( 0.4 0.6) A  0.6 0.4   B  0.2 0.8  AB AB = (0.24 + 0.12 0.16 + 0.48) = (0.36 0.64) Thus after two weeks, A will have 36% of shares and B will have 64% of shares. As the time goes, equilibrium will be reached in the long run. At this point A + B = 1 We have (A B)  0.6 0.4  = (A B)  0.8   0.2 By matrix multiplication, (0.6 A + 0.2 B 0.4A + 0.8 B) = (A B) Equating the corresponding elements, 0.6 A + 0.2 B = A 0.6 A + 0.2 (1 – A) = A (using A + B = 1) 0.6 A + 0.2 – 0.2 A = A 0.2 = A – 0.4 A A = =0.2 0=.33 33% 0.6 B = 1 – 0.33 = 0.67 = 67% Thus the equilibrium is reached when the share of A is 33% and share of B is 67%. [OR] (b) Evaluate ∫ ( x + 1)2 log x dx Ans. ∫ (x +1)2 log x dx We use integration by parts method. Let u = log x ⇒ du = 1 dx x dv = ( x + 1)2 dx ⇒ v = ( x +1)3 3 We have ( x + 1)2 log x dx = ( x + 1)3 log x − ∫ ( x + 1)3  1  dx  x  3 3 ( ) = ( x +1)3 log 3 x − 1 ∫ x3 + 3x2 + 3x +1 dx 3 x = ( x +1)3 log x − 1 ∫  x2 + 3x + 3 + 1  dx 3  x  3 = 13 ( x +1)3 log x − x3 − 3x2 − 3x − log x  +c 3 2   Sample Paper -1 F-11

42. (a) The demand equation for a product is x = 100 − p and the supply equation is x = p − 10 .Determine the consumer’s surplus and producer’s surplus, under 2 market equilibrium. Ans. Given demand equation is x = 100 − p and supply equation is x = p −10 2 So the demand law is x2 = 100 – p ⇒ pd = 100 – x2 Supply law is given by x + 10 = p 2 ⇒ ps = 2(x + 10) Under equilibrium pd = ps ⇒ 100 – x2 = 2(x + 10) 100 – x2 = 2 x + 20 x2 + 2x – 80 = 0 (x + 10) (x – 8) = 0 x = – 10,8 The value of x cannot be negative, So x = 8 When x0 = 8, p0 = 100 – 82 = 100 – 64 = 36 8 ( ) CS = ∫ 100 − x2 dx − (8)(36) 0 − x3 8 − 288 3 0  = 100x = 800 − 512 − 288 = 1024 3 3 So Consumer Surplus = 1024 units 38 PS = 8 (36) – ∫ 2(x +10)dx 0 8  x2 + 10 x  = 288 – 2   2 0 = 288 – 2  64 + 80  2 = 288 – 2[112] = 64 So Producer Surplus = 64 units [OR] F- 12 Business Mathematics and statistics –XII

(b) Elasticity of a function Ey is given by Ey = −7 x Find the function Ex Ex (1 − 2x)(2 + 3x) 3 when x = 2, y = 8 . Ans. Given η = Ey = −7 x ⇒ x dy = −7 x Ex (1− 2x)(2 + 3x) y dx (1− 2x)(2 + 3x) dy = (1 − −7 x + 3x) dx y 2x)(2 x dy = 7∫ (2x dx + 2) ....(1)1) y − 1)(3x 1 = A+ B (2x −1)(3x + 2) (2x −1) (3x + 2) 1 = A ( 3x + 2 ) + B ( 2x – 1 ) Let x=1 ⇒ 1 = A  3 + 2  2  2  1 = A  7  ⇒ A = 2  2  7 Let x = −2 ⇒ 1 = B 2  −2  − 1 3  3  1 = B −4 − 1 3 1 = B −7  ⇒ B =−3 3  7 Using these values in (1) we get dy 23 y = 7∫ 7 1 dx −7∫ 7 2 dx 2x − 3x + dy = ∫ 2dx −∫ 3dx y 2x −1 3x + 2 log y = log (2x – 1) – log (3x + 2) + log k y =  2x −1  k  3x + 2  When x = 2, y = 3 ⇒ 3 = 3 k 8 8 8 ⇒ k = 1 Hence the function is y = 2x −1 3x + 2 Sample Paper -1 F-13

43. (a) The slope of the tangent to a curve at any point (x, y) on it is given by (y3 − 2yx2)dx + (2xy2 − x3) dy = 0 and the curve passes through (1, 2). Find the equation of the curve. Ans. Given the slope is (y3 − 2yx2)dx + (2xy2 − x3) dy = 0 (y3 − 2yx2)dx = – (2xy2 − x3) dy dy = y3 − 2 yx2 dx x3 − 2xy2 This is a homogeneous differential equation Put y = vx and dy = v + x dv dx dx v + x dv = v3x3 − 2vx3 = v3 − 2v dx x3 − 2x3v2 1 − 2v2 x dv = v3 − 2v – v dx 1 − 2v2 = v3 − 2v − v + 2v3 = 3v3 − 3v 1 − 2v2 1 − 2v2 Separating the variables, 1 − 2v2 dx 3(v3 − v) dv = x ∫1 1− 2v2 dv = dx x 3 v(v2 −1) Using partial fraction method, 1 − 2v2 = 1 − 2v2 =A + B + C v(v2 −1) v(v +1)(v −1) v v +1 v −1 1 – 2 v2 = A(v +1)(v –1) + B v (v –1) + C v (v +1) Put, v = 0 ⇒ A = -1 v = 1 ⇒ C = − 1 v = –1 ⇒ B = − 1 22 1 ∫ −1 dv − ∫ 1 1) dv − 1 ∫ v 1 1 dv = dx 3 v 2(v + 2 −  x  13 − log v − 1 log(v + 1) − 1 log(v − 1) = log x + log c 2 2 – 1 log v + log v +1 + log v −1 = log cx 3 − 1 log v v2 −1 = log cx 3 ⇒ 1 = cx 1 ( )v v2 −1 3 F- 14 Business Mathematics and statistics –XII

1 = c3 x3 y v v2 −1 x ( ) Replace, v = 1 = c3 x3 y2 y x2 −1 x 1 = c3 x3 ⇒ x2 = c3 x3 y y2 − x2 y2 − x2 y x x2 xy y2 − x2 = 1 = 1 , k = c3 c3x kx The curve passes through (1,2). (i.e) x = 1, y = 2 2 3 =1 k Thus , xy y2 − x2 = 2 3 (or) xy y2 − x2 = 2 3 is the required solution. [OR] x (b) The area A of circle of diameter ‘d’ is given for the following values. D 80 85 90 95 100 A 5026 5674 6362 7088 7854 Find the approximate values for the areas of circles of diameter 82 and 91 respectively. Ans. Let diameter be x and area be y We have to find value of y when (a) x = 82 and (b) x = 91 We first find the difference as given below x y ∆y ∆2y ∆3y ∆4y 80 5026 648 85 5674 40 Forward interpolation 688 -2 90 6362 38 4 726 2 95 7088 40 Backward interpolation 766 100 7854 Sample Paper -1 F-15

Now x0 + nh = 82 x0 = 80, h = 5 80 + 5 n = 82 n = 82 − 80 = 0.4 5 ( 0.4) ( 0.4) ( 0.4 ) ( 0.4 − 1) ( 2) y (82) = 5026 + (648) + (0.4 + 1) (40) + 0.4 − (−2) 1! 2! 3! + (0.4)(0.4 −1)(0.4 − 2)(0.4 − 3) (4) 4! y(82) = 5026 + 259.2 – 4.8 – 0.128 – 0.1664 = 5280.1056 (b) x = 91, xn= 100, h = 5 We use backward difference formula here 91 = 100 + 5 n n = 91 − 100 = −1.8 5 Using the backward differences from the table. We get, y (91) = 7854 + (−1.8) (766) + (−1.8)(−1.8 +1) (40) + (−1.8)(−1.8 +1)(−1.8 + 2) (2) 1! 2! 3! + (−1.8)(−1.8 +1)(−1.8 + 2)(−1.8 + 3) (4) 4! ( −1.8) ( −1.8) ( −1.8) ( −0.8) ( 0.2 ) = 7854 − 1378.8 + (−0.8)(40) + (−0.8)(0.2)(2) + (1.2)(4) 2 6 24 = 7854 – 1378.8 + 28.8 + 0.096 + 0.0576 = 6504.15 Hence the area of circle when diameter is 82 is 5281, area of circle when diameter is 91 is 6504. 44. (a) The probability density function of a continuous random variable X is f (x) =  a + bx 2 , 0 ≤ x ≤ 1  0, otherwise 3 where a and b are some constants. Find (i) a and b if E(X) = 5 (ii) Var(X). Ans. Given that X is a continuous random variable and f (x) is density function. ∞1 So ∫ f (x)dx = 1 ∫⇒ (a + bx2 ) dx = 1 −∞ 0 1  ax + bx3  = 1 ⇒ a + b = 1 ..........(1)  3 0 3  Again it is given that E (X) = 3 5 F- 16 Business Mathematics and statistics –XII

∞ 1 3 0 5 Now E (X) = x f (x) dx ⇒ −∞ ∫ ∫ x(a + bx2 ) dx = ∫ 1 = 3 (ax + bx3 ) dx 05  ax2 + bx4 1 = 3  4  5  2 0 a + b =3 .........(2) 245 [OR] We solve (1) and (2) to find the values of a and b. a+b =1 3 a + b = 6 25 Subtract b  1 − 1 = 1− 6  3 2  5 −1 b = −1 ⇒ b = 6 65 5 From (1) ⇒ a = 1 − b =1 − 1  6  =1 − 2 =3 3 3  5  5 5 Thus a = 3 , b = 6 55 ( ii) Var (X) = E (X2) – [E(X)]2 ∞ 1 x2  3 6 x2  0  5 5  E(X 2) = x2 f =( x) dx −∞ ∫ ∫ + dx ∫ = 13 x2 + 6 x4  dx 0  5 5  =  x3 + 6x5 1 =1 + 6 =11  25  5 25 25  5 0 Var (X) = 11 −  3 2 = 11 − 9 = 2 25  5  25 25 25 (b) Derive the mean and variance of poisson distribution. Ans. Let X be a poisson random variable with parameter λ. The p.m.f is given by P(x, λ) = P(X= x) = e−λλx x! ∞ Mean E (X) = ∑ x p(x,λ) x=0 Sample Paper -1 F-17

∑= ∞ x e−λλx x=0 x! ∑=  ∞ λx−1  λe−λ   x=1 ( x − 1)! = λe−λ 1 + λ + λ2 + λ3 + ... = λe−λ (eλ ) =λ  1! 2! 3!  Variance (X) = E (X2) – [E (X)]2 ∞ ∞ x 2 e−λλx x 2 p (x,λ) = ∑ ∑E (X2) = x=0 x=0 x! ∞ [x (x–1) + x] e−λλx = ∑ x=0 x! ∑ ∑ = ∞ x (x – 1) e−λλx + ∞ x e−λλx x=0 x! x=0 x! ∞ e−λ λ x ∞ λx−2 + λ x−2 x=2 x−2 ! ∑ ∑ ( )( ) = + E(X) = e−λ λ2 x=2 ! = e−λ λ2 1 + λ+ λ2 + ...... + λ = e–λ λ2 (eλ) + λ = λ2 + λ  1! 2!  Var(X) = λ2 + λ - λ2 = λ Thus the mean and variance of Poisson distribution are both equal to λ. 45. (a) A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain (a) no more than 2 rejects? (b) at least 2 rejects? Ans. L et X be the binomial random variable denoting the number of metal pistons. Let p be the probability of rejections. Given that p = 12% = 12 = 0.12, q = 0.88, n = 10. 100 So X ∼ B (0.12,10). Hence the p.m.f of X is given by P (X = x ) = 10Cx (0.12)x (0.88)10−x (a) P (X ≤ 2 ) = P (X = 0) + P (X = 1) + P (X = 2) = 10C0 (0.12)0 (0.88)10 + 10C1 (0.12)1 (0.88)9 + 10C2 (0.12)2 (0.88)8 = (0.88)10 + 10(0.12) (0.88)9 + 45(0.12)2 (0.88)8 = (0.88)8 + [(0.88)2 + (1.2) (0.88) + 45 (0.12)2] = (0.35963) [0.7744 + 1.056 + 0.648] = (0.35963) (2.4784) = 0.89131 Thus out of a batch of 10 pistons, the probability of no more than 2 rejects is 0.89131 (b) P(2 ≤ C) = 1 - P(C < 2) = 1 - [P(C = 0) + P(C = 1)] = 1 - [(0.88)10 + 10 (0.12) (0.88)9] = 1 - [(0.88)9 [0.88 + 1.2] = 1 - [ (0.88)9 (2.08)] F- 18 Business Mathematics and statistics –XII

To find (0.88)9 using calculator. (0.88)9 = [(0.88)3]3 = [0.681472]3 = 0.316478 ⇒ P(2 ≤ C) = 1 - [(0.316478) (2.08)] = 1 - 0.65827 = 0.34173 Thus out of 10 pistons, the probability that atleast 2 will be rejected is 0.34173. [OR] (b) Explain the types of sampling. Ans. The different types of sampling are (i) simple random sampling, (ii) Stratified random sampling and (iii) Systematic sampling. (i) In simple random sampling, every item of the population has equal chance for being selected. The sampling can be done with replacement (or) without replacement. A random sampling from a finite population with replacement is equivalent to a sampling from an infinite population without replacement. This technique will give useful results only if the population is homogeneous. The following are some of the methods of selecting a random sample. (a) Use of an unbiased die or coin: If we have to choose between two alternatives, a coin is tossed and depending on head or tail course of action is taken. A die can be employed if there are six different alternatives. (b) Lottery sampling: Here a random sample is selected by identifying each element of the population by means of a card of a pack of uniform cards or (by writing the number on pieces of paper) and to select required number of cards after a thorough mixing of the cards. (c) Random numbers: Random numbers are formed of ‘random digits’ and arranged in the form of a table having a number of rows and columns. Tippett’s numbers form one such table wherein 40,000 digits were selected at random from census reports and combined by groups of four into 10,000 numbers. (ii) In stratified random sampling, a population of ‘n’ units is divided into L sub-populations of N1, N2, ............NL. The sub-populations being non-overlapping and mutually exhaustive so that N = N1 + N2 + ............+ NL. Each sub-populations is known as stratum. If we select n1, n2, ............nl items, respectively, from these strata, we get a stratified sample. If a simple random sample is taken from each stratum, the whole procedure is referred to as stratified random sampling. (iii) Systematic sampling is a form of restricted random selection which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in a serial order and every ith element, starting from any of the first i items is chosen. For example, suppose we require a 5% sample of students from a college where there are 2000 students, we select a random number from 1 to 20. If it is 12, then our sample consists of students with numbers 12,32,52,72,......1992. Sample Paper -1 F-19

46. (a) Calculate the Laspeyre’s, Paasche’s and Fisher’s price index number for the following data. Interpret on the data. Commodities Base Year Current Year Price Quantity Price Quantity A 170 562 72 632 B 192 535 70 756 C 195 639 95 926 D 187 128 92 255 E 185 542 92 632 F 150 217 180 314 Ans: Base Year Current Year Commodities Price Quantity Price Quantity p0q0 p0q1 p1q0 p1q1 (p0) (q0) (p1) (q1) 95540 107440 40464 45504 A 170 562 72 632 102720 145152 37450 52920 124605 180570 60705 87970 B 192 535 70 756 23936 47685 11776 23460 100270 116920 49864 58144 C 195 639 95 926 32550 47100 39060 56520 479621 644867 239319 324518 D 187 128 92 255 E 185 542 92 632 F 150 217 180 314 Total Laspeyre’s price index number  P0L1 p1q0 100 ⇒ P0L1  239319 100 = 49.9 p0q0 479621 Paasche’s price index number  P0P1 p1q1 100 ⇒ P0P1  324518 100 = 50.32 p0q1 644867 Fisher’s price index number  P0F1 = p1q0  p1q1 × 100 p0q0 p0q1 P0F1 = 239319 × 324518 ×100 479621× 644867 P0F1 = 0.2511 ×100 = 50.1 From the index numbers, we conclude that for the same quantity, the price has reduced by 50.5% in the current year compared to the base year according to Laspeyre’s index. By the Paasche’s index number we see that the price has reduced by 49.68% in the current year, and according to Fisher’s index number it has reduced by 49.9% in the current year. [OR] F- 20 Business Mathematics and statistics –XII

(b) The Annual production of a commodity is given as follows: Year 1995 1996 1997 1998 1999 2000 2001 162 171 182 158 180 178 Production 155 (in tones) Fit a straight line trend by the method of least squares. Ans. Computation of trend values by the method of least squares. (ODD years) Production X2 Trend values Year (x) (in tonnes) (Y) X = x – 1998 XY (Yt) 1995 155 – 3 9 – 465 159.57 1996 162 – 2 4 – 324 162.86 1997 171 – 1 1 – 171 166.14 1998 182 0 0 0 169.43 1999 158 1 1 158 172.72 2000 180 2 4 360 176.00 2001 178 3 9 534 179.29 N=7 ΣY = 1186 ΣX = 0 ΣX2 = 28 ΣXY = 92 ΣYt = 1186.01 a= Y  1186 = 169.429 b=  XY  92 = 3.286 7  X2 28 N Therefore, the required equation of the straight line trend is given by Y = a + b X (i.e) Y = 169.429 + 3.286 X (or) Y = 169.429 + 3.286 (x – 1998) The trends values are obtained by When x = 1995, Yt = 169.429 + 3.286 (1995 – 1998) = 159.57 When x = 1996, Yt = 169.429 + 3.286 (1996 – 1998) = 162.86 When x = 1997, Yt = 169.429 + 3.286 (1997 – 1998) = 166.14 When x = 1998, Yt = 169.429 + 3.286 (1998 – 1998) = 169.43 When x = 1999, Yt = 169.429 + 3.286 (1999 – 1998) = 172.72 When x = 2000, Yt = 169.429 + 3.286 (2000 – 1998) = 176.00 When x = 2001, Yt = 169.429 + 3.286 (2001 – 1998) = 179.29 47. (a) Consider the following transportation problem. D 1 D 2 D 3 D 4 Availability O1 5 8 3 6 30 O2 4 5 7 4 50 O3 6 2 4 6 20 Requirement 30 40 20 10 Determine initial basic feasible solution by VAM. Sample Paper -1 F-21

Ans. Let ‘ai’ denote the availability and ‘bj’ denote the requirement ∑ai = 30 + 50 + 20 = 100 and ∑bj = 30 + 40 + 20 + 10 = 100 ∑ai = ∑bj So the given problem is balanced transportation problem. Hence there exists a feasible solution to the given problem. For VAM, we first find the penalties for rows and columns. We allocate units to the maximum penalty column (or) row with least cost. F irst allocation: D1 D2 D3 D4 a Penalty O1 5 8 3 6 30 (2) O2 4 5 7 4 50 (1) O3 (20) 4 6 20/0 (2) 62 a Penalty b 30 40/20 20 10 30/10 (2) j (2) Penalty (1) (3) (1) L argest penalty = 3. ∴ Allocate min (40,20) to (O3, D2 ) S econd allocation: D1 D2 D3 D4 O1 5 (20) 6 83 O2 4 5 7 4 50 (1) bj 30 20 20/0 10 Penalty (1) (3) (4) (2) L argest penalty = 4. ∴ Allocate min (20,30) to (O1, D3 ) a Penalty Third allocation: (1) D1 D2 D4 O1 5 8 6 10 O2 (20) 4 50/30 (1) 45 bj 30 20/0 10 Penalty (1) (3) (2) Largest penalty is 3 ∴ Allocate min (20,50) to (O2, D2 ) F- 22 Business Mathematics and statistics –XII

Fourth allocation: D1 D4 a Penalty O1 5 6 10 (1) O2 4 (10) 30/20 (0) 4 bj 30 Penalty (1) 10/0 (2) L argest penalty is 2 ∴ Allocate min (10,30) to (O2, D4 ) Fifth allocation: D1 a Penalty (10) O1 5 10/0 − O2 (20) 20/0 − 4 bj 30/10/0 Penalty (1) Largest penalty is 1 ∴ Allocate min (30,20) to (O2, D1 ). Balance 10 units we allot to (O1, D1) Thus we have the following allocations: D1 D2 D3 D4 a O1 (10) (20) 6 30 5 83 O2 (20) (20) (10) 50 4 5 74 O3 (20) 4 6 20 62 b 30 40 20 10 j Transportation schedule : O1→ D1, O1→ D3, O2→ D1, O2→ D2, O2→ D4, O3→ D2 ( i.e) x11 = 10, x13 = 20, x21 = 20, x22 = 20, x24 = 10, x32 = 20 Total cost = (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2) = 50 + 60 + 80 + 100 + 40 + 40 = 370 Thus the least cost by VAM is Rs.370. [OR] Sample Paper -1 F-23

(b) The following table summarizes the supply, demand and cost information for four factors S1, S2, S3, S4 shipping goods to three warehouses D1, D2, D3. D 1 D 2 D 3 Supply S1 2 7 14 5 S2 3 3 1 8 S3 5 4 7 7 S4 162 14 Demand 7 9 18 Find an initial solution by using north west corner rule. What is the total cost for this solution? Ans. Let ‘ai’ denote the supply and bj denote the demand. Then total supply = 5 + 8 + 7+14 = 34 and Total demand = 7 + 9 + 18 = 34 ∑ai = ∑bj. So the problem is balanced transportation problem and we can find a basic feasible solution, by North west corner rule. First allocation: D 1 D2 D 3 Supply ( ai ) S1 (5) 2 7 14 5/0 3 S2 3 4 18 6 77 S3 5 9 2 14 18 S4 1 Demand bj 7/2 Second allocation: D1 D2 D3 ai (2) 8/6 S2 3 3 1 S3 5 4 7 7 S4 1 62 14 bj 2/0 9 18 F- 24 Business Mathematics and statistics –XII

Third allocation: D2 D3 (ai ) S2 (6) 1 6/0 3 S3 4 7 7 S4 6 2 14 bj 9/3 18 Fourth allocation: D2 D3 ai S3 (3) 7 7/4 4 S4 6 2 14 Fifth allocation: bj 3/0 18 D 3 ai S3 (4) 4/0 7 S4 (14) 14/0 2 bj 18/14/0 We first allot 4 units to cell (S3, D3) and then the balance 14 units to cell (S4, D3). Thus we get the following allocations: D1 D2 D 3 Supply S1 (5) 2 7 14 5 S2 (2) 3 (6) 3 18 S3 5 (3) 4 (4) 7 7 S4 1 6 (14) 2 14 Demand 7 9 18 The transportation schedule : S1→ D1, S2→ D1, S2→ D2, S3→ D2, S3→ D3, S4→ D3 (i.e) x11 = 5, x21 = 2, x22 = 6, x32 = 3, x33 = 4, x43 = 14 Total cost = (5 × 2) + (2 × 3) + (6 × 3) + (3 × 4) + (4 × 7) + (14 × 2) = 10 + 6 + 18 + 12 + 28 + 28 = 102 Thus the initial basic solution is got by NWC method and minimum cost is Rs.102. Sample Paper -1 F-25

4SAMPLE PAPER – Time: 3 Hours (UNSOLVED) Maximum Marks: 90 PART-I I. Answer all the questions. Choose the most suitable answer from the given four alternatives and write the option code with the corresponding answer. [20 × 1 = 20] AB 1. If T = A  0.7 0.3 is a transition probability matrix, then the value of x is ................... .   B  0.6 x  (a) 0.2 (b) 0.3 (c) 0.4 (d) 0.7 2 . If the number of variables in a non - homogeneous system AX = B is n, then the system possesses a unique solution only when ................... . (a) ρ (A) = ρ (A,B) > n (b) ρ (A) = ρ (A,B) = n (c) ρ (A) = ρ (A,B) < n (d) none of these 3. ∫ 2x dx is ................... . (c) 2x + c (d) log 2 + c log 2 2x (a) 2x log 2+c (b) 2x +c ∞ ∫4. e−2x dx is ................... . 0 (b) 1 (c) 2 (d) 1 (a) 0 2 5. The profit of a function p(x) is maximum when ................... . (a) MC – MR = 0 (b) MC = 0 (c) MR = 0 (d) MC + MR = 0 6. The demand and supply functions are given by D(x) = 16 − x2 and S(x) = 2x2 + 4 are under perfect competition, then the equilibrium price x is ................... . (a) 2 (b) 3 (c) 4 (d) 5 7 . Solution of dx + Px = 0 (c) x = py + c (d) x = cy dy (a) x = ce py (b) x = ce – py 8. The differential equation formed by eliminating A and B from y = e−2x (Acos x + B sin x) is .................. . (a) y2 − 4y1 + 5 = 0 (b) y2 + 4y – 5 = 0 (c) y2 − 4y1 – 5 = 0 (d) y2 + 4y1 + 5 = 0 9 . If c is a constant then ∆c = .................. . (a) c (b) ∆ (c) ∆2 (d) 0 10. If ‘n’ is a positive integer ∆n [∆– n f (x)] .............. . (a) f (2x) (b) f (x + h) (c) f (x) (d) ∆ f (x) F- 71

1 1. Given E(X) = 5 and E(Y) = – 2, then E(X – Y) is ............. . (a) 3 (b) 5 (c) 7 (d) –2 1 2. Which of the following is not possible in probability distribution? (a) ∑ p(x) ≥ 0 (b) ∑ p(x) = 1 (c) ∑ xp(x) = 2 (d) p(x) = – 0.5 13. Forty percent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. For a full flight, what is the mean of the number of passengers who do not check in any luggage? (a) 6.00 (b) 6.45 (c) 7.20 (d) 7.50 1 4. Which of the following cannot generate a Poisson distribution? (a) The number of telephone calls received in a ten-minute interval (b) The number of customers arriving at a petrol station (c) The number of bacteria found in a cubic feet of soil (d) The number of misprints per page 1 5. The standard error of sample mean is ..................... . (a) σ (b) σ (c) σ (d) σ2 2n n nn 1 6. Errors in sampling are of ...................... . (a) Two types (b) three types (c) four types (d) five types 17. The components of a time series which is attached to short term fluctuation is .................... . (a) Secular trend (b) Seasonal variations (c) Cyclic variation (d) Irregular variation 1 8. Laspeyre’s index = 110, Paasche’s index = 108, then Fisher’s Ideal index is equal to .......... . (a) 110 (b) 108 (c) 100 (d) 109 19. A type of decision - making environment is ............. . (a) certainty (b) uncertainty (c) risk (d) all of the above 20. Which of the following methods is used to verify the optimally of the current solution of the transportation problem? (a) Least cost method (b) Vogel’s method (c) North west corner rule (d) None of these PART-II [7 × 2 = 14] II. Answer any seven questions. Question No. 30 is compulsory. 21. Find the rank of the matrix A= 4 5 2 2  3 2 1 6  4 4 8 0 22. Integrate the 3x + 5 with respect to x. 23. Find the area bounded by the curve y = x2 and the line y = 4. 24. Solve: (1− x) dy − (1+ y) dx = 0. 25. Prove that ∆ ∇ = ∆ –∇ 26. Define random variable. F- 72 Business Mathematics and Statistics–XII

27. If the probability of success is 0.09, how many trials are needed to have a probability of atleast one success as 1/3 or more? 28. A random sample of size 50 with mean 67.9 is drawn from a normal population. If the S.E of the sample mean is 0.7 , find 95% confidence interval for the population mean. 29. Define Laspeyre’s price index number. 30. Find an initial basic feasible solution by the North West Corner Rule (NWC). Supply 9 15 12 10 6 8 13 23 Demand 9 3 11 27 21 14 25 PART-III III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21] 31. Find the rank of the matrix by reducing to echelon form. 1 1 1 1  12  A =  1 3 −2 2 0 −3 32. Integrate the 3x + 2 with respect to x. ( x − 2)( x − 3) 33. Find the area of the region lying in the first quadrant bounded by the region y = 4x2, x = 0, y = 0 and y = 4. 34. Find the differential equation of the family of parabola with foci at the origin and axis along the x-axis. 35. Following are the population of a district. Year(x) 1881 1891 1901 1911 1921 1931 Population(y) 363 391 421 - 467 501 (Thousands) Find the population of the year 1911. 36. The distribution of a continuous random variable X in range (–3, 3) is given by p.d.f. 1 (3 + x)2 , −3 ≤ x ≤ −1 16  1 f=( x)  16 (6 − 2x2 ), −1 ≤ x ≤ 1   1  16 (3 − x)2 , 1≤ x ≤ 3 Verify that the area under the curve is unity. Sample Paper - 4 F- 73

37. Assume that a drug causes a serious side effect at a rate of three patients per one hundred. What is the probability that atleast one person will have side effects in a random sample of ten patients taking the drug? 38. A random sample of 500 apples was taken from large consignment and 45 of them were found to be bad. Find the limits at which the bad apples lie at 99% confidence level. 39. Calculate by a suitable method, the index number of price from the following data: Commodity 2002 2012 Price Quantity Price Quantity A 10 20 16 10 B 12 34 18 42 C 15 30 20 26 40. Find the investment option using Maximin rule for the following: Economy Alternatives Growing Stable Declining Bonds 40 45 5 Stocks 70 30 –13 Mutual Funds 53 45 –5 PART-IV IV. Answer all the questions. [7 × 5 = 35] 41. (a) Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method. [OR] (b) Solve the following homogeneous differential equations. x dy − y= x2 + y2 dx 42. (a) The demand and supply functions under perfect competition are pd = 1600 − x2 and ps = 2x2 + 400, respectively. Find the producer’s surplus. [OR] (b) The population of a city in a census taken once in 10 years is given below. Estimate the population in the year 1955. Year 1951 1961 1971 1981 Population in lakhs 35 42 58 84 ∫ 43. (a) Evaluate dx [OR] ex + 6 + 5e−x F- 74 Business Mathematics and Statistics–XII

(b) Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function.  0,for x ≤ 0   x , for 0 ≤ x < 1 2  =F( x)  1 , for1 ≤ x < 2  2  x , for 2 ≤ x < 4  4  1,for x ≥ 4 (a ) Is the distribution function continuous? If so, give its probability density function? (b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes? 44. (a) Out of 750 families with 4 children each, how many families would be expected to have (i) atleast one boy (ii) atmost 2 girls (iii) and children of both sexes? Assume equal probabilities for boys and girls. [OR] (b) A simple random sample of size 100 has mean (i) 15, the population variance being 25. Find an interval estimate of the population mean with a confidence level of 95% and 99%. (ii) If the population variance is not given, what should be done to find out the required estimates? 45. (a) The following table shows the number of salesmen working for a certain concern. Year 1992 1993 1994 1995 1996 No. of salesmen 46 48 42 56 52 Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997. [OR] (b) Solve (i) dy = ae y (ii) 1+ x2 = xy dy dx 1+ y dx 46. (a) Compute (i) Laspeyre’s (ii) Paasche’s (iii) Fisher’s Index numbers for the 2010 from the following data. Commodity Price Quantity 2000 2010 2000 2010 A 12 14 18 16 B 15 16 20 15 C 14 15 24 20 D 12 12 29 23 [OR] Sample Paper - 4 F- 75

(b) The sales of a commodity in tones varied from January 2010 to December 2010 as follows: In Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 2010 Sales (in 280 240 270 300 280 290 210 200 230 200 230 210 tones) Fit a trend line by the method of semi-average. 14000 47. (a) The marginal cost function of a commodity is given by MC = 7x + 4 and the fixed cost is ₹18,000. Find the total cost and average cost. [OR] (b) Find the trend of production by the method of a five-yearly period of moving average for the following data: Year 126 1979 123 1980 Production 117 1981 ('000) 128 1982 125 1983 124 1984 130 1985 114 1986 122 1987 129 1988 118 1989 123 1990 F- 76 Business Mathematics and Statistics–XII

ONE-MARK & TWO-MARK ANSWERS FOR UNSOLVED SAMPLE PAPERS SAMPLE PAPER – 4 1. (c) 0.4 2. (b) ρ (A) = ρ (A,B) = n 3. (c) 2x + c 4. (d) 12 log 2 5. (a) MC – MR = 0 6. (a) 2 7. (b) x = ce – py 8. (d) y2 + 4y1 + 5 = 0 9. (d) 0 10. (c) f (x) 11. (c) 7 12. (d) p(x) = – 0.5 13. (a) 6.00 14. (b) The number of customers arriving at a petrol station σ 15. (c) n 16. (a) Two types 17. (d) Irregular variation 18.(d) 109 19. (d) all of the above 20. (a) Least cost method 4 5 2 2  3 2 1 6  21. Given A = 4 4 8 0 Order of A is 3 × 4. So ρ(A) ≤ 3 Consider the third order minor, 452 3 2 1 = 4 (16 – 4) – 5(24 – 4) + 2 (12 – 8) 4 4 8 = 48 – 100 + 8 = – 44 ≠ 0 There is a minor of order 3 which is not zero∴ρ(A) = 3 1 22. 3x + 5 = (3x + 5)2 This is of the form (ax+b)n where a = 3, b = 5, n = 1 2 (ax + b)n+1 + c ∫ (ax + b=)n dx a(n +1) So (3x + 1 dx = (3x + )5 1 +1 + c 1 2 5)2 3 2 + 1 33 = (=33x+325) 2 (3x + 5)2 3 +c = 2 ( 3x + 5) 2 +c 9 9 2 F- 109

23. Given the parabola is y = x 2 and line y = 4 y The parabola is symmetrical about the y – axis. So required area = 2 [Area in the first quadrant y=4 between limits y = 0 and y = 4] y = x2 44 = 2∫ xdy = 2∫ y dy 00 0x  2 3 4 4=.8 32 sq.units = 2  y 2=  3 0 3 3 24. (1− x) dy − (1+ y) dx = 0 Separating the variables, (1 – x ) dy = (1+ y ) dx dy = dx 1+ y 1− x I ntegrating, we get ∫ 1 y dy = ∫ 1 x dx 1+ 1− log(y + 1) = – log(1– x) + log c log(y + 1) + log(1– x) = log c ⇒ ( y + 1) (1 – x) = c 25. LHS ∆ ∇ = (E – 1)  E −1 = (E – 1) 1 − 1   E  E  =E–1–1+ 1 = (E – 1) – 1 − 1  E E  = (E – 1) –  E− 1  = ∆ –∇ = RHS  E  26. A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as X,Y and Z etc. If X and Y are r.v’s then X+Y is also a r.v. 27. Given p = 0.09 (success) q = 0.91 (failure) We have to find number of trials ‘n’. According to the problem, P(X ≥ 1 ) > 1 3 (We must have atleast one success) 1 1 2 1 – P (X < 1) > 3 1 – P (X = 0) > 3 (or) P (X = 0) < 3 Using p.m.f, we have, n C0 (0.09)0 (0.91)n < 2 ⇒ (0.91)n < 2 3 3 We can use log tables to calculate ‘n’ (or) by trial method try for n = 1,2, ..... using calculator. We observe that (0.91)5 < 2 . Thus we need minimum 5 trial or more. 3 F- 110 Business mathematics and statistics–XII

28. n = 50, x = 67.9, 9 5% confidence limits for population mean are µ are x – Zα/2   ≤μ≤ x + Zα/2    n   n  σ is given as n 0.7 and Zα/2 = 1.96 ⇒ 67.9 – (1.96) ( 0.7 ) ≤ μ ≤ 67.9 + (1.96) 0.7 66.26 ≤ μ ≤ 69.54 Thus the 95% confidence intervals for estimating μ is given by (66.26, 69.54) 29. The weighted aggregate index number using base period weights is called Laspeyre’s price index number. P0L1   p1q0 100  p0q0 Where p1 is current year price p0 is base year price q0 is base year quantity 30. Total demand = Total supply = 60 Supply (10) 15 12 10/0 9 (11) (12) 13 23/12/0 68 (2) (25) 9 3 11 27/25/0 Demand 21/11/0 14/2/0 25/0 Total cost = (10 × 9) + (11 × 6) + (12 × 8) + (2 × 3) + (25 × 11) = 90 + 66 + 96 + 6 + 275 = 533 Answers F- 111

COMPUTER APPLICATIONS QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to be Total Marks answered Multiple Choice Questions 1 15 15 Very short answers: 2 6 12 (Totally 9 questions will 3 6 18 be given. Any one question should be answered compulsorily.) Short answers: (Totally 9 questions will be given. Any one question should be answered compulsorily.) Essay type 55 25 Total 70 Practical Marks 20 Internal Assessment 10 100 Total Marks S.No. Weightage of Marks Weightage 1. 30% 2. Purpose 40% 3. Knowledge 20% 4. Understanding 10% Application Skill/Creativity G-1

Time: 3 Hours 1SAMPLE PAPER – (SOLVED) Maximum Marks: 70 PART - I [15 × 1 = 15] Choose the correct answer. Answer all the questions  [Answers are in Bold] 1. RTF file format was introduced by ......................... . (a) TCS (b) Microsoft (c) Apple (d) IBM 2. Place option is present in .................... menu. (a) File (b) Edit (c) Layout (d) Window 3. The ................. diagram gives a logical structure of the database graphically? (a) Entity-Relationship (b) Entity (c) Architectural Representation (d) Database 4. Which of the below symbols is a newline character? (a) \\r (b) \\n (c) /n (d) /r 5. What will be the output of the following PHP code ? <?php if (-100) print “hi” ; else print “how are u”; ?> (a) how are u (b) hi (c) error (d) no output 6. PHP supports which types of looping techniques? (a) for loop (b) while loop (c) foreach loop (d) all the above 7. Wi-Fi is short name for ........................... . (a) Wireless Fidelity (b) Wired fidelity (c) Wired fiber optic (d) Wireless fiber optic 8. Communication over ................... is be made up of voice, data, images and text messages. (a) Social media (b) mobile network (c) whatsapp (d) software 9. How many parameter are required for MYSQLi connect function in PHP? (a) 2 (b) 3 (c) 4 (d) 5 10. Pick the odd one out from the following. (a) node (b) label (c) domain (d) server 11. ARPANET stands for ................................... . (a) American Research Project Agency Network (b) Advanced Research Project Area Network (c) Advanced Research Project Agency Network (d) American Research Programs And Network G-3

12. SME stands for .............................. . (a) Small and medium sized enterprises (b) Simple and medium enterprises (c) Sound messaging enterprises (d) Short messaging enterprises 13. Match the following List A List B A1. First Digit B1. Account number A2. 9th to 15th Digit B2. MII Code A3. First 6 Digits B3. BIN Code A4. Last Digit B4. Check digit A1 A2 A3 A4 (a) B4 B3 B2 B1 (b) B2 B1 B3 B4 (c) B2 B3 B4 B1 (d) B2 B4 B3 B1 14. The security authentication technology does not include ( i) Digital Signatures (ii) Digital Time Stamps (iii) Digital Technology ( iv) Digital Certificates (a) (i), (ii) & (iv) (b) (ii) & (iii) (c) (i), (ii) & (iii) (d) all the above 15. EDIFACT stands for .......................................... . (a) EDI for Admissible Commercial Transport (b) EDI for Advisory Committee and Transport (c) EDI for Administration, Commerce and Transport (d) EDI for Admissible Commerce and Trade PART - II A nswer any six questions. Question No. 21 is compulsory. [6 × 2 = 12] 16. List out image file formats. TIFF (Tagged Image File Format), BMP (Bitmap), DIB (Device Independent Bitmap), GIF (Graphics Interchange Format), JPEG (Joint Photographic Experts Group), TGA (Tagra), PNG (Portable Network Graphics). 17. What are the ACID properties? A CID Properties – The acronym stands for Atomicity, Consistency, Isolation and Durability. Atomicity follows the thumb rule “All or Nothing”, while updating the data in database for the user performing the update operation. Consistency ensures that the changes in data value to be constant at any given instance. Isolation property is needed during concurrent transaction. Durability is defied as the system’s ability to recover all committed transactions during the failure of storage or the system. G-4 Computer Applications – XII

18. Define Looping Structure in PHP. ▪ Loop structures in PHP is an iterative control structures that involves executive the same block of code a specified num ber of times. ▪ Loops that iterate for fixed no of times is also called as Bounded loops. ▪ PHP supports four types of loops. • for Loop • While Loop • foreach Loop • Do While Loop 19. Define Computer Network. ▪ A set of computers connected together for the purpose of sharing resources is called as computer networks. ▪ At present, Internet is the most common resource shared everywhere. 20. List any four domain names. Domain Name Meaning com Commercial Organisation edu Educational Institutions gov Government (US) mil Military groups 21. Explain the History of open source software. H istory of open source software: ▪ In the early day of computing, programmers and developers shared software in order to learn from each other ▪ Eventually, it moved to the way side of commercialization of software in 1970-1980 ▪ The Netscape communication corporation released their popular Netscape Communicator Internet Suite as free software. This made others to look into how to bring the free software ideas. ▪ The Open Source Initiative was founded in Feb 1998 to encourage use of the new term open-source. 22. Define E-Commerce. E-Commerce can be described as the process of buying or selling products, services or information via computer networks. 23. Define non-repudiation. Non-repudiation: prevention against violation agreement after the deal. Repudiation refers to any act of relinquishing responsibility for a message. Non-repudiation ensures that the signer who digitally signed the document cannot deny having signed it. 24. What is meant by directories in EDIFACT? The versions of EDIFACT are also called as directories. These EDIFACT directories will be revised twice a year; on 1st April and 1st October to include new or update existing EDIFACT messages. EDIFACT directories have names like D.18B ( D stands for Directory, 18 is the year and A/B indicates the month of release). Sample Paper - 1 G-5