Vedanta Excel in Mathematics Book 9 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 51 Vedanta Excel in Mathematics - Book 9 Creative Section 6. a) For each of the following articles, find the discount amount, selling price after discount, VAT amount and selling price with VAT. (i) (ii) (iii) iv) b) The marked price of a bike helmet is Rs 3,000 and 10 % discount is allowed on it. (i) Find the discount amount. (ii) Find the price after discount. (iii) Find its cost with 13 % VAT. c) Salina marks the price of a blanket as Rs 5,500. She allows 20 % discount and adds 13 % VAT. (i) How much discount does she allow? (ii) What is the price of the blanket after discount? (iii) How much does a customer pay for the blanket with value added tax? 7. a) Bipin bought a motorbike for Rs 2,40,000 and fixed its price 20 % above the cost price. Then, he allowed 15 % discount and sold to a customer with 13 % Value Added Tax (VAT). (i) What was the marked price of the bike? (ii) What was the selling price of the bike after discount? (iii) How much did the customer pay for it with VAT? b) Mr. Sharma bought a computer for Rs 50,000 and fixed its price 25% above the cost price. Then, he allowed 10 % discount and sold to a customer with 13% VAT. (i) What was the marked price of the computer? (ii) How much did the customer pay for it with VAT? c) Sabin Electronics Enterprises bought a television for Rs. 16,000 and sold at a profit of 20% to a customer with 13% VAT. How much did the customer pay for the television? d) Mrs. Lama marked the price of a cosmetic item 25% above its cost price. If the cost price of the cosmetic item was Rs 4,400, at what price did she sell it with 13% VAT? 8. a) Mrs. Kandel went to a restaurant with her family. They had three plates of Mo:Mo at Rs 120 per plate, one plate chicken chilly at Rs 220 per plate, and three bottles of cold drink at Rs 40 per bottle. (i) What was the bill with 10% service charge? (ii) How much did she pay to clear the bill with 13% value added tax? M.P. =Rs 25,000 Discount rate = 10% VAT rate = 13% M.P. =Rs 85,000 Discount rate = 14% VAT rate = 13% M.P. =Rs 1,20,000 Discount rate = 14% VAT rate = 13% M.P. =Rs 40,000 Discount rate = 15% VAT rate = 13% Taxation

Vedanta Excel in Mathematics - Book 9 52 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) A group of three friends had two plates of chicken chilly, two plates of french fry, two plates of Mo:Mo and a few glasses of fresh juice in a restaurant. The cost of these items amounts to Rs 900, how much should they pay with 10 % service charge and 13 % VAT to clear the bill? 9. a) A retailer allows 15 % discount on the marked price of an electric fan. If a customer pays Rs 3,842 with 13 % VAT, find: (i) the price of the fan excluding VAT. (ii) the marked price of the fan. (iii) the discount amount. (iv) amount of the value added tax b) After allowing 10 % discount on the marked price of an iPod and levying 13 % value added tax, Mrs. Gurung sold it for Rs. 7,119 to Mrs. Dahal. Find: (i) its price excluding the VAT. (ii) its marked price. (iii) the discount amount (iv) amount of the value added tax c) A tourist paid Rs 5,763 for a carved window made up of wood with a discount of 15% including 13% value added tax (VAT). How much does he get back while leaving Nepal? 10. a) A mobile price is tagged Rs 5,000. When 12% discount is allowed and adding certain percent of VAT the price reaches to Rs 4,972. (i) Find the price of the mobile after allowing discount. (ii) Find out the VAT amount. (iii) Calculate the VAT percent. b) The marked price of a bag is Rs 2,000. The price of the bag becomes Rs 1,921 after 15% discount and adding VAT amount. Find: (i) selling price after discount (ii) the rate of VAT. 11.a) Mrs. Karki purchased a sari for Rs 8,000 and sold it for Rs 11,300 with 13% VAT. (i) Find the selling price excluding VAT. (ii) Find her profit or loss percent. b) A supplier bought a scanner machine for Rs 35,000 and sold it for Rs 47,460 with 13% VAT. (i) Find the selling price excluding VAT. (ii) Find the profit or loss percent of the supplier. Creative section - B 12. a) A wholesaler sold a photocopy machine for Rs 48,000 to a retailer. The retailer spent Rs 2,000 for transportation and Rs 1,500 for the local tax. If the retailer sold it at a profit of Rs 4,500 to a customer, answer the following questions: (i) Calculate the cost price of the machine to the retailer including transportation and local tax. (ii) Find the selling price of the machine. (iii) How much did the customer pay for it with 13% VAT? b) The Buddha supplier sold a digital T-shirt printer for Rs 3,00,000 to Everest supplier. The Everest supplier spent Rs 5,500 for transportation and Rs 2,500 for the local tax and sold at a profit of 10% to a customer. How much did the customer pay for the printer with 13% VAT? Taxation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 53 Vedanta Excel in Mathematics - Book 9 Taxation c) A wholesaler purchased a washing machine for Rs 60,000 and sold it to a retailer at 10% profit. The retailer spent Rs 2,400 for transportation and Rs 1,600 for local tax. Then she sold it to a customer at 12% profit. How much did the customer pay for it with 13% VAT? 13. a) A retailer allowed 4% discount on his goods to make 20% profit and sold a refrigerator for Rs 10,848 with 13% VAT. (i) Find the selling price of refrigerator without VAT. (ii) How much is the marked price of the refrigerator? (iii) Calculate the cost price of the refrigerator to the retailer. (iv) How much is the selling price of the refrigerator if the retailer wants to make only 15% profit? (v) By how much is the discount to be increased so that he can gain only 15%? b) A supplier sold a scanner machine for Rs 41,400 with 15% VAT after allowing 10% discount on it's marked price and gained 20%. By how much is the discount percent to be reduced to increase the profit by 4%? 14. A retailer hired a room in a shopping mall at Rs 45,000 rent per month and started a business of garments. He spent Rs 20,00,000 to purchase different garment items in the first phase and marked the price of each item 30% above the cost price. Then, he allowed 10% discount on each item and sold to customers. His monthly miscellaneous expenditure was Rs 15,000 and the items of worth 10% of the investment remained as stocks after two months. (i) Calculate the investment excluding stocks. (ii) Find the selling price of items after discount then, calculate gross profit. (iii)Find the net profit or loss in two months and express it in percent. Project work and activity section 15. a) Make groups of your friends. Collect different types of goods purchasing bill. Study about the marked price, rate of discount, rate of VAT or other rates of taxes mentioned in the bills. Prepare the reports and present in your class. b) Visit the nearby department store or your local shops. Search and collect the marked price, rate of discount and rate of VAT on different daily using goods. Prepare a report and compare your report with that of your friends. c) Let's become a problem maker and problem solver yourself. Write the values of the variable of your own and find the unknown variables. (i) (ii) (iii) Given: M. P. = .................. Discount= ............. % Discount amount = ? S.P. = ? Given: M. P. = .................. Discount= ............. % Discount amount = ? S.P. = ? Given: M. P. = .................. S.P. = .................. Discount amount = ? Discount Percent = ?

Vedanta Excel in Mathematics - Book 9 54 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (iv) (v) (vi) d) Using the above variables make a word problem of each of (i) to (vi) of your own, related to the real life situations. Then, solve your problems and get the unknown variables. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. Which of the following is the taxable income? (A) CIT (B) Festival expense (C) Remote area allowance (D) Insurance premium 2. The rate of social security tax is (A)1% (B) 10% (C) 20% (D) 30% 3. Which of the following is the tax-exempt income? (A)Salary (B) Festival expense (C) Donation (D) Interest 4. The annual income of an individual is Rs 4,50,000. How much income tax does s/he pay in a year? (A)Rs 4,000 (B) Rs 4,500 (C) Rs 9,000 (D) Rs 9,500 5. A couple is the sole proprietor of a firm, income tax is exempted up to (A)Rs 4,00,000 (B) Rs 4,50,000 (C) Rs 5,00,000 (D) Rs 5,50,000 6. An indirect tax which is charged at the time of consumption of goods and services is (A) VAT (B) Income tax (C) Customs duty (D) Excise duty 7. The current rate of Value Added Tax (VAT) in Nepal is (A) 10% (B) 13% (C) 15% (D) 7% 8. Which of the following formulae is correct? (A) S.P. with VAT = S.P. + VAT% of S.P. (B) S.P. with VAT = S.P. – VAT% of S.P. (C)S.P. with VAT = C.P. + VAT% of C.P. (D) S.P. with VAT = M.P. – VAT% of M.P. 9. The Value Added Tax (VAT) is imposed on (A) M.P. (B) S.P. (C) C.P. (D) Discount 10. If M.P. = Rs x, discount = Rs y and VAT = Rs z, then S.P. with VAT is (A) Rs (x – y + z) (B) Rs (x + y – z) (C) Rs (x + y + z) (D) Rs (x – y – z) Given: S. P. = .................. VAT = .................. S.P. with VAT = ? Given: S. P. = .................. VAT = .................. % S.P. with VAT = ? Given: M. P. = .................. Discount = .............% VAT = ..................% S.P. with VAT = ? Taxation

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 55 Vedanta Excel in Mathematics - Book 9 3.1 Commission – Introduction Let's study the following examples and make the sense of commission. a) Mr. Lama had a piece of land and he wanted to sell it. He announced to provide 2% of selling price to the agent for selling it. Mr. Sharma, a salesperson, came to know about it and searched for the buyers. He sold the land for Rs 50,00,000 and got the amount of 2% of Rs 50,00,000 i.e., Rs 1,00,000 for his service. b) Mrs. Ghale wished to buy a house. She went through an agent. The agent found a house on sale. Mrs. Ghale liked it and bought for Rs 1,20,00,000. She gave 1.5% of the price of the house to the agent. c) Mr. Chaudhary works in a hardware shop. His income of the month, Push is given in the table. Answer the following questions. (i) What is his monthly salary? (ii) What is the total sale of the month? (iii) What is the commission rate? (iv) What amount of money did he get as commission? (v) What is his total income in Push? Commission is the amount of money paid to an agent (or an employee) for performing a business services such as buying and selling goods, property, etc. The commission is usually a percent of the selling price. The percent is called the commission rate. For example, A Publication House pays 5% commission of the total sales of books to its dealer in a year. A real estate company pays 2.5% commission to its agent for selling lands and property. In this way, we calculate the amount of commission as the given percent of the selling price. Month: Push Salary Total sales Commission rate Commission amount Rs 25,000 Rs 8,50,000 1.5% Rs 12,750 Total income Rs 37,750 Unit 3 Commission, Bonus and Dividend

Vedanta Excel in Mathematics - Book 9 56 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember 1. Amount of commission = percent of commission × total selling price. 2. Commission percent = Commission amount S.P. × 100% Worked-out examples Example 1: A real estate company gives 1.5 % commission to its agent on selling a piece of land for Rs 20,00,000 and 2 % commission for additional amount of selling price above the fixed price. If the agent sold the land for Rs 28,50,000, how much commission did the agent receive? Solution: Here, the fixed selling price of the land = Rs 20,00,000 The selling price of the land = Rs 28,50,000 Now, the commission received by the agent = 1.5 % of Rs 20,00,000 + 2 % of (Rs 28,50,000 – Rs 20,00,000) = 1.5 100 × Rs 20,00,000 + 2 100 × Rs 8,50,000 = Rs 30,000 + Rs 17,000 = Rs 47,000 Hence, the agent received the commission of Rs 47,000. Example 2: A plywood factory provides commission to its dealers on the basis of the following monthly transactions. Sales up to Rs 5,00,000, 4 % commission. Sales from Rs 5,00,000 to Rs 7,50,000, 5 % commission. Sales more than Rs 7,50,000, 6 % commission. Calculate the commissions of the following monthly transaction of different dealers. (i) Rs 4,80,000 (ii) Rs 7,20,000 (iii) Rs 8,60,000 Solution: (i) Commission of the sale of Rs 4,80,000 = 4 % of Rs 4,80,000 = Rs 19,200 (ii) Commission of the sale of Rs 7,20,000 = 4 % of Rs 5,00,000 + 5 % of (Rs 7,20,000 – Rs 5,00,000) = Rs 20,000 + Rs 11,000 = Rs 31,000 (iii) Commission of the sale of Rs 8,60,000 = 4% of Rs 5,00,000 + 5% of (Rs 7,50,000 – Rs 5,00,000) + 6% of (Rs 8,60,000 – Rs 7,50,000) = Rs 20,000 + Rs 12,500 + Rs 6,600 = Rs 39,100 Example 3: The monthly salary of an employee in a publication house is Rs 30,000 and 2 % commission is provided when the monthly sales is more than Rs 5,00,000. If the sales of the publication house in a month is Rs 6,75,000, find the income of the employee in the month. Commission, Bonus and Dividend Rs 28,50,000 Rs 20,00,000 Rs 8,50,000 (1.5%) (2%)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 57 Vedanta Excel in Mathematics - Book 9 Commission, Bonus and Dividend Solution: Here, the monthly salary of the employee = Rs 30,000 Commission percent = 2% Sales of the month = Rs 6,75,000 Now, sales for the commission = Rs 6,75,000 – Rs 5,00,000 = Rs 1,75,000 Amount of commission = 2 % of Rs 1,75,000 = 2 100 × Rs 1,75,000 = Rs 3,500 Again, the income of the employee of the month = Rs 30,000 + Rs 3,500 = Rs 33,500 Hence, the income of the employee in the month is Rs 33,500. Example 4: The monthly salary of an employee in a hardware shop is Rs 27,000 and a certain commission is given as per the monthly sales. If the sales of a month is Rs 5,00,000 and the total income of the employee of the month including commission is Rs 32,500, calculate the rate of commission. Solution: Here, the monthly salary of the employee = Rs 27,000 Sales of the month = Rs 5,00,000 Total income of the employee of the month = Rs 32,500 Now, the amount of commission of the month = Rs 32,500 – Rs 27,000 = Rs 5,500 Again, the rate of commission = Amount of commission Sales of the month × 100% = 5,500 5,00,000 × 100% = 1.1% Hence, the required rate of commission is 1.1%. EXERCISE 3.1 General section 1. Find the commission amount received by agent from the table given below. S.N. Property Total sale Rate of commission Commission a) Land Rs 60,50,000 1% ………. b) House Rs 2,40,00,000 1.5% ………. c) Car Rs 60,00,000 2% ………. 2. a) An agent gets 1.5% commission while selling a house for Rs 2 crores 50 lakhs. (i) How much commission will the agent get? (ii) How much amount will the house owner get after paying the commission? b) A real estate company gives 3% commission to it’s agents. An agent sold a piece of land for Rs 12,50,000. (i) How much commission will the agent get? (ii) How much amount will the company receive after paying the commission?

Vedanta Excel in Mathematics - Book 9 58 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. a) An automobile company gives 10% commission to it’s agents for selling second-hand motorbike. If an agent received Rs 16,290 by selling a scooter, at what price did the agent sell the scooter? b) A farmer sold his seasonal vegetables through an agent by giving 2% commission. If the agent got Rs 2,200 commission, at what price did the agent sell the vegetables? c) Mr. Kulman sold a piece of land through a real estate agent. After paying commission at the rate of 3.5% to the real estate agent, he received Rs 1,78,52,500 for selling his land. Calculate the selling price of the land. d) A noodles factory gives 10% commission to a wholesaler for selling its products. After giving the commission to the wholesaler, the company receives Rs 3,15,000 in Baishakh. Find the total sales of noodles by the wholesaler in Baishakh. 4. a) If a sales agent received Rs 30 thousands commission when he sold a second hand taxi for Rs 12 lakhs, what rate of commission did he get? b) A building owner fixed the cost of his building as Rs 27,50,400 and the price above the fixed cost goes to an agent as his commission. If the agent sold it for Rs 29,01,672, find his commission percent. Creative section 5. a) A real estate company gives 5 % commission on selling a piece of land for Rs 10,00,000 and 7 % commission for the additional amount of selling price above the fixed price. If the agent sold the land for Rs 12,99,000, how much commission did he/she receive from the company? b) An insurance company offered 2 % commission for the first 10 lakh and 2.5 % for the rest sum of money collected from new clients by its agents. If an agent is able to collect a sum of Rs 15,30,000 from his new clients, find his total commission. c) A noodle factory provides commission to its sales agents on the basis of the following weekly transactions. Sales Rate of commissions Upto Rs 50,000 2.5 % From 50,000 - Rs 1,00,000 5% Above Rs 1,00,000 6% Now, calculate the commission of the following weekly sales. (i) Rs 48,600 (ii) Rs 72,500 (iii) Rs 1,10,700 6. a) The monthly salary of a salesperson of a subway restaurant is Rs 21,600, and an additional incentive of 1.5 % on the total monthly sale is provided as commission. (i) Calculate his/her total income in a month if he/she makes a total sale of Rs 5,80,000 in that month. (ii) What should be his/her total sale in the next month so that he/she can receive a total income of Rs 31,350 in the month? Commission, Bonus and Dividend

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 59 Vedanta Excel in Mathematics - Book 9 Commission, Bonus and Dividend b) Mr. Bibek is an online salesperson in an online shopping store. His monthly salary is Rs 18,700 and 2% commission is given to him when the monthly sales is more than 5 lakh rupees. If the sales of the store in a month is Rs 7,20,000, calculate his total income of the month. c) Mrs. Nepali draws Rs 19,800 as her monthly salary in a wholesale cosmetic shop and a certain commission is given as per the monthly sales. The sales of a month is Rs 12,00,000 and her total income of the month including commission is Rs 31,800. (i) Find the rate of commission (ii) If the sales of the previous month was Rs 15,00,0000 and the commission was given to her at the same rate, what was her income of previous month? 7. a) Mr. Ghising decides to sell a plot of 10 Aanas of land he owns at the rate of Rs 5,00,000 per Aana. He promises a commission of 3% on the sale of the land to a broker. If he gives a discount of 2% on the original price to the buyer, find his income from the sale of the land, after paying the commission to the broker. b) Mr. Chauhan is a car dealer who purchases a car for Rs 50,00,000. He intends selling the car for Rs 55,00,000. However, he gives a discount of 3% on this price to the buyer and a commission of 2% to a broker. How much profit does he make? 3.2 Bonus Let's study the following examples and get the idea about bonus. a) Mr. Bajracharya is an officer of a corporation. His monthly salary is Rs 40,000. This year, the corporation decided to distribute a bonus amount that is equivalent to his two month’s salary. So, he received Rs 80,000 bonus. b) The board of management of an enterprise had decided to distribute equal amount of bonus to its 50 employees from 10% of its net profit. Bonus is an extra amount of money that is given to an employee as a reward for his/ her good performance. It is an incentive given to an employee besides his/her fixed salary. Bonus is calculated as a certain percent of profit and it is decided by the board of management of any business organization. Facts to remember 1. Total bonus amount = Rate of bonus × Net profit 2. Bonus rate = Total bonus amount Net profit × 100% 3. Bonus amount received by each employee = Total bonus amount Number of employees Example 1: The annual salary of an employee in a business company is Rs 3,02,400. Besides, the company provides 15% bonus from its net profit at the end of each fiscal year. If the company made a profit of Rs 60,50,000 in a fiscal year and decided to distribute bonus to its 30 employees equally, how much income did the employee make in the year?

Vedanta Excel in Mathematics - Book 9 60 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, the annual salary of the employee = Rs 3,02,400 Profit of the company = Rs 60,50,000 The total amount of bonus = 15 % of Rs 60,50,000 = Rs 9,07,500 Now, the amount of bonus received by each employee = Rs 907500 30 = Rs 30,250 Again the annual income of the employee = Rs 3,02,400 + Rs 30,250 = Rs 3,32,650 Hence, the employee made an income of Rs 3,32,650 in the year. Example 2: A cement factory made a net profit of Rs 2,60,00,000 in the last year. The management of the factory decided to distribute 10% bonus from the profit to its 130 employees equally. (i) By what percent should the bonus be increased so that each employee can receive Rs 30,000? (ii) What should be the profit of the factory so that it can provide Rs 44,500 to each employee at 20% bonus? Solution: Here, net profit of the factory = Rs 2,60,00,000, bonus percent = 10% and of employees = 130 (i) Bonus amount received by each employee = Rs 30,000, bonus percent = ? Bonus amount =130× Rs 30,000 = Rs 39,00,000 Bonus percent = Total bonus amount Net profit × 100% = Rs 39,00,000 Rs 2,60,000 × 100% = 15% Hence, the bonus should be increased by 15% - 10% = 5% (ii) Bonus percent= 20%, number of employees = 130, Bonus amount received by each worker = Rs 44,500 Now, bonus amount =130× Rs 44,500 = Rs 57,85,000 Let, net profit of the factory be Rs x Then, bonus amount = bonus % of net profit or, Rs 57,85,000 = 20% of x or, 57,85,000 = 20x 100 ∴ x = Rs 2,89,25,000 Hence, the profit of the factory should be Rs 2,89,25,000. Example 3: When a commercial bank increased its profit from 20% to 25%, the amount of profit increased to Rs 1,25,00,000. If the company decided to distribute 80% bonus to its 50 employees equally from the increased amount of profit, how much bonus will each employee receive? Commission, Bonus and Dividend

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 61 Vedanta Excel in Mathematics - Book 9 Commission, Bonus and Dividend Solution: Let, the annual income of the bank be Rs x. According to question, 25% profit of the income = Rs 1,25,00,000 or, 25 100 × x = Rs 1,25,00,000 ∴ x = Rs 5,00,00,000 When the profit was 20%, profit amount = 20% of Rs Rs 5,00,00,000 = Rs 1,00,00,000 Increased amount of profit = Rs 1,25,00,000 – Rs 1,00,00,000 = Rs 25,00,000 Now, bonus amount = 80% of Rs 25,00,000 = Rs 20,00,000 ∴ Bonus amount received by each employee = Bonus amount Number of employees = Rs 20,00,000 50 = Rs 40,000 Hence, each employee received Rs 40,000 bonus. EXERCISE 3.2 General section 1. Calculate the bonus amount distributed by each of the following companies from the table given below. S.N. Company Net profit Bonus rate Total bonus amount a) Restaurant Rs 50,00,000 10% ………. b) Bike showroom Rs 65,00,000 15% ………. c) Insurance company Rs 90,80,500 20% ………. 2. a) A publication house announced to distribute 10 % bonus equally to its 20 employees from the net profit of Rs 18,36,000 at the end of a fiscal year, find the bonus received by each employee. b) The net profit of a hotel in a fiscal year was Rs 80 lakhs. The board of management of the hotel announced to distribute 25% bonus equally to its 50 employees from the net profit. How much bonus did each employee receive? 3. a) A garment factory announced 20 % bonus to its 25 workers from the net profit at the end of last fiscal year. If every worker received Rs 18,500, how much was the profit of the factory? b) A trading corporation distributed 8% of its net profit equally among its 120 employees last year. If each employee received Rs 24,000 bonus, how much net profit did the corporation make in the last year? 4. a) A business company distributed bonus to its 24 employees from the net profit of Rs 16,48,000. If every employee received Rs 8,240, what was the bonus percent? b) The board of management of an LP gas company announced to distribute Rs 15,500 bonus equally to its 60 employees from its net profit of Rs 46,50,000 in a fiscal year. What percent of its net profit was distributed by the company for the employee bonus?

Vedanta Excel in Mathematics - Book 9 62 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative section 5. a) A garment factory made a net profit of Rs 48,00,000 in the last year. The management of the factory decided to distribute 18 % bonus from the profit to its 25 employees. (i) Find the bonus amount received by each employee. (ii) By what percent should the bonus be increased so that each employee can receive Rs 38,400? (iii) What should be the profit of the company so that it can provide Rs 40,000 to each employee at 20 % bonus? b) A hydro power company made a net profit of Rs 2,50,00,000 in the last year. The board of management of the company decided to distribute 10% bonus from the profit to its 125 employees. (i) Find the bonus amount received by each employee. (ii) By what percent should the bonus be increased so that each employee can receive Rs 24,000? (iii) What should be the profit of the company so that it can provide Rs 25,000 to each employee at 8% bonus? 6. a) When a publication house increased its profit from 20% to 25%, the amount of profit increased to Rs 52,08,000. If the company decided to distribute 60% bonus to its 30 employees equally from the increased amount of profit, how much bonus does each employee receive? b) When cement factory increased its profit from 20% to 25% the amount of profit increased to Rs 87,50,000. If the company decided to distribute 80% bonus to its 50 employees equally from the increased amount of profit, how much bonus will each employee receive? Project work and activity section 7. a) Is your any family member involved in any organisation or corporation. If so, discuss with him/her and learn more about bonus. Prepare a report on bonus and present in the class. b) Make the groups of 5 of your friends. Take the information regarding bonus distribution of any 5 public or private organisations. Present your report in the class. 3.3 Dividend Let's study the following examples and get the idea about dividend. a) At the end of a fiscal year, a bank announced Rs 25 dividend per share to its shareholders. Mr. Wagle had 250 shares of the company and received Rs 6,250 cash dividend. b) A person has bought 1,500 shares out of 1,00,000 shares of Rs 100 per share from a hydroelectricity corporation. If the corporation earned a net profit of Rs 3,50,28,000 in a certain year and it decided to distribute 20% of the net profit to its shareholders, the dividend received by the person as per his/her number of shares is called Rs 1,05,084. Commission, Bonus and Dividend

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 63 Vedanta Excel in Mathematics - Book 9 Commission, Bonus and Dividend A dividend is a payment made by a corporation to its shareholders, usually as a distribution of profits. A dividend is allocated as a fixed amount per share with shareholders receiving a dividend in proportion to their shareholding. Facts to remember 1. Total dividend amount = Rate of dividend × Net profit 2. Dividend rate = Total dividend amount Net profit × 100% 3. Dividend per share = Total dividend amount Total number of shares 4. Dividend for a shareholder with 'n' shares = n × dividend per shares Example 1: Mrs. Bhatta bought 750 shares out of 20,000 shares from a Business Company. If the company earned a net profit of Rs 90,00,000 and it declared to distribute 10% dividend to its shareholders, how much money did Mrs. Bhatta receive? Solution: Here, the net profit of the company = Rs 90,00,000 Now, 10% of Rs 90,00,000 = 10 100 × Rs 90,00,000 = Rs 9,00,000 Again, the profit of 20,000 shares = Rs 9,00,000 The profit of 1 share = Rs 9,00,000 20,000 = Rs 45 The profit of 750 shares = 750 × Rs 45 = Rs 33,750 Hence, she received Rs 33,750 as her dividend. Example 2: A commercial bank sold 30,000 shares. The bank earned a net profit of Rs 2,50,00,000 in a year and distributed a certain percent of profit as dividend. If a shareholder, who has bought 400 shares, received Rs 50,000 dividend, find what percent of profit was distributed as divided by the bank? Solution: Here, the dividend of 400 shares = Rs 50,000 the dividend of 1 share = Rs 50,000 400 = Rs 125 the dividend of 30,000 shares = Rs 30,000 × Rs 125 = Rs 37,50,000 The total dividend = Rs 37,50,000 The net profit = Rs 2,50,00,000 Now, the percent of net profit as dividend = Dividend Net profit × 100% = Rs 37,50,000 Rs 2,50,00,000 × 100% = 15% Hence, 15% of the net profit was distributed as dividend by the bank.

Vedanta Excel in Mathematics - Book 9 64 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Alternative process: The dividend of 400 shares = Rs 50,000 The dividend of 1 share = Rs 50,000 400 = Rs 125 Also, the net profit for 30,000 shares = Rs 2,50,00,000 The net profit for 1 share = Rs 2,50,00,000 30,000 = Rs 2500 3 Now, the percent of profit as dividend = Dividend for 1 share Profit for 1 share × 100% = Rs 125 Rs 2500/3 × 100% = 15% Example 3: Harka Magar bought 500 shares out of 50,000 shares sold by a hydropower company. When the company distributed 25% of its net profit, he received Rs 43,850 as his dividend in a year. Calculate the net profit of the company. Solution: Here, the profit of 500 shares = Rs 43,850 The profit of 1 share = Rs 43,850 500 = Rs 87.70 The profit of 40,000 shares = 50,000 × Rs 87.70 = Rs 43,85,000 Let, the net profit of the bank be Rs x. Now, 25% of Rs x = Rs 43,85,000 or, 25x 100 = Rs 43,85,000 x = Rs 1,75,40,000 Hence, the net profit of the company is Rs 1,75,40,000. EXERCISE 3.3 General section 1. a) Calculate the dividend amount from the table given below. S.N. Company Net profit Dividend rate Dividend (i) Finance Rs 25,00,000 50% ………. (ii) Hydropower Rs 80,00,000 60% ………. (iii) Commercial bank Rs 1,75,40,000 45% ………. b) Find the dividend per share from the table given below. Company Net profit Dividend rate Number of shares Dividend per share (i) A Rs 60,00,000 40% 2,000 ………. (ii) B Rs 98,00,000 45% 9,000 ………. (iii) C Rs 2,45,60,000 70% 30,700 ………. Commission, Bonus and Dividend

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 65 Vedanta Excel in Mathematics - Book 9 Commission, Bonus and Dividend 2. a) A business company makes a net profit of Rs 80,00,000 in a year. The Board of Directors declares 12% cash dividend from the net profit. If the company has sold 1000 shares, answer the following questions. (i) Find the total cash dividend. (ii) Find the dividend for each share. b) A development bank made a net profit of Rs 2,40,00,000 in a year and the management announced to distribute 24% cash dividend from the net profit. If the bank has sold 2500 shares, answer the following questions. (i) Find the total cash dividend. (ii) Find the dividend for each share. Creative section A 3. a) A finance company earned a net profit of Rs 45,20,000 in a year. The company declared to distribute a certain percent of profit as dividend. The total dividend amount was Rs 9,04,000. (i) What percent of net profit was distributed as dividend? (ii) If the net profit of the company was Rs. 50,00,000 in the year, how much dividend amount would be distributed at the same rate? b) A man has 200 shares out of 1000 total shares of a business company. He received Rs 96,000 as dividend in a year which was a certain percent of net profit of Rs 24,00,000. (i) How much was the total dividend amount distributed? (ii) What percent of net profit was distributed as dividend? 4. a) A publication house distributed 21% of dividend to its shareholders from the net profit of a year. If the amount of distributed dividend was Rs 23,52,000, answer the given questions. (i) Calculate the net profit made by the publication house. (ii) If the rate of dividend were only 20%, how much would be the total dividend distributed? b) A small hydropower company distributed the total amount of dividend of Rs 14,58,000 to its shareholders which was 30% of the net profit earned by the company. (i) Find the net profit earned by the company. (ii) If the rate of dividend were only 25%, how much would be total dividend distributed? Creative section B 5. a) Mrs. Rai bought 250 shares out of 10,000 shares from a finance company. The company earned a net profit of Rs 85,20,000 and declared 17% dividend to its shareholders. Calculate the amount of dividend received by Mrs. Rai. b) A Business Company sold 2,500 shares at Rs 1,200 per share. Bishwant bought 450 shares. If the company earned a net profit of Rs 39,00,000 in a year and it announced to distribute 18% dividend from the net profit to its shareholders, find the amount of dividend received by Bishwant.

Vedanta Excel in Mathematics - Book 9 66 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. a) A Cable Car Company sold 3,000 shares to the local people. The company earned a net profit of Rs 1,20,00,000 in a year and distributed a certain percent of profit as dividend. If Anamol, a shareholder who has bought 125 shares received Rs 1,10,000 dividend, answer the following questions. (i) What percent of profit was distributed as dividend? (ii) How much cash dividend would Anamol get at the same percent of net profit if he had 200 shares and the company made a net profit of Rs 1,50,00,000? b) Suntali bought 200 shares out of 5,000 shares sold by a hydropower company to the local people. The company earned a net profit of Rs 75,00,000 in a year and it declared to distribute a certain percent dividend to its shareholders. If she received Rs 81,000, answer the following questions. (i) What percent of the net profit was distributed as dividend? (ii) How much cash dividend would Ravi get if he had 250 shares of the same company, the company made a net profit of Rs 80,00,000 and distributed the dividend at the same percent of profit? 7. a) Mr. Dhurmus bought 500 shares out of 10,000 shares sold by a commercial bank. The bank earned some profit and it distributed 14% of the net profit as dividend in a year. If Dhurmus received Rs 1,03,600 in the year, answer the following questions: (i) What was the net profit of the bank? (ii) How much cash dividend would Pemba get if he had 350 shares of the same bank and the bank distributed 15% of net profit as dividend? b) A Life Insurance Company earned some profit and announced to distribute 40% dividend from its net profit to its shareholders. If a shareholder who bought 300 shares out of 12,000 shares sold by the company received Rs 1,25,400, Answer the following questions. (i) What was the net profit of the company? (ii) If Junu had 400 shares of the same company and the company distributed only 30% of its net profit as dividend, how much cash dividend would she get? Project work and activity section 8. a) Take the information from website or national level of daily news papers or from any business papers and make a report about the share values of any five companies, banks, or any business organisations. Present your report in the class. b) Is your any family member involved in any business organisation as one of the shareholders? If so, discuss with her/him and learn more about bonus, dividend, etc. Then, prepare a report and present in the class. Commission, Bonus and Dividend

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 67 Vedanta Excel in Mathematics - Book 9 OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. Commission is (A) a payment given to the employees based on the sales they made. (B) a payment to the shareholders by the company from the net profit. (C) a payment to the employees by the company from the net profit. (D) a payment to an employee as a reward for good performance. 2. An extra amount of money that is given to an employee as a reward for good performance is called (A) commission (B) bonus (C) dividend (D) tax 3. The payment made by a corporation to its shareholders from certain portion of net profit is called (A) bonus (B) commission (C) allowance (D) dividend 4. Commission is calculated on the basis of (A) cost price (B) marked price (C) selling price (D) net profit 5. Bonus is calculated as a certain percent of (A) profit (B) loss (C) selling price (D) cost price 6. The commission in selling a dozer for Rs 60,00,000 at 0.30% is (A) Rs 18,000 (B) Rs 15,000 (C) Rs 1,800 (D) Rs 1,80,000 7. Mr. Gopal paid a commission to the agent while buying a house for Rs 1,35,00,000 at 0.75% is (A) Rs 11,250 (B) Rs 1,01,250 (C) Rs 1,10,250 (D) Rs 1,12,500 8. A farmer sells the fruits through an agent. If the agent charges 5% commission for selling the fruits worth Rs 45,000; the amount received by the farmer is (A) Rs 2,250 (B) Rs 42,750 (C) Rs 47,750 (D) Rs 40,500 9. The monthly salary of a salesman in a departmental store is Rs 15,000 and additional payment of 1% on the total monthly sale is provided as commission. What is his income in a month if he makes a total sale of Rs 6,60,000 in that month? (A) Rs 15,000 (B) Rs 6,600 (C) Rs 8,400 (D) Rs 2,1600 10. A doctor working in a clinic has monthly salary Rs 40,000. He gets 2% commission on the monthly sales of medicine above 5 lakhs in the clinic. What should be the monthly sales in the clinic so that the monthly income of the doctor in that monthly is Rs 60,000? (A) 8 lakhs (B) 10 lakhs (C) 15 lakhs (D) 20 lakhs

Vedanta Excel in Mathematics - Book 9 68 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 4.1 Introduction Household arithmetic deals with the regular financial activities based on the household purposes. Payments of electricity bill, telephone bill, water bill, etc. are a few examples of household financial activities. Money exchange, calculation of taxi fare are also under the household arithmetic. In this unit, we shall discuss about electricity bill, telephone bill, water bill, and taxi fare. 4.2 Electricity bill We measure the consumption of electricity in the numbers of units. Digital meter and dial meter are two types of electric meters which are used to find the number of units of electricity consumed. 1 unit of electricity = 1 kilowatt hour or 1000 watt hour Thus, 1 unit of electricity means, it is the amount of consumption of electricity by an electric appliance of 1000 watt power in 1 hour. The number of units of electricity consumed is calculated as: Reading of the recent month – Reading of the previous month The meter reading to find the consumption of electricity of every month by a household is performed at the end of each month. The table given below shows the capacity of meter box of the lower voltage limit 230-400 volt connected in our houses and the minimum energy charges. Domestic Consumers Service and Energy Charges (Single Phase) kWh (Monthly) 5 Ampere 15 Ampere 30Ampere 60 Ampere Service Charge (Rs.) Energy Charge (Rs. per unit) Service Charge (Rs.) Energy Charge (Rs.per unit) Service Charge (Rs.) Energy Charge (Rs.per unit) Service Charge (Rs.) Energy Charge (Rs.per unit) 0-20 30.00 0.00 50.00 4.00 75.00 5.00 125.00 6.00 21-30 50.00 6.50 75.00 6.50 100.00 6.50 125.00 6.50 31-50 50.00 8.00 75.00 8.00 100.00 8.00 125.00 8.00 51-100 75.00 9.50 100.00 9.50 125.00 9.50 150.00 9.50 101-250 100.00 9.50 125.00 9.50 150.00 9.50 200.00 9.50 Above 250 150.00 11.00 175.00 11.00 200.00 11.00 250.00 11.00 Unit 4 Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 69 Vedanta Excel in Mathematics - Book 9 Billing method (for 5 Ampere) S.N. Consumption block Charge per unit Billing method 1. 0-20 0.00 Minimum charge: Rs. 30 Energy charge: free Example: 15 units: Rs (30 + 15 × 0) = Total Rs 30 2. 21-30 Rs 6.50 Minimum charge: Rs. 50 Energy charge: Rs. 3 per unit for (0-20) units, Rs. 6.50 per unit for (21-30) units, Example: 26 units: Rs (50 + 20 × 3 + 6 × 6.50) = Total Rs 149 3. 31-50 Rs 8.00 Minimum charge: Rs. 50 Energy charge: Rs. 3 per unit for (0-20) units, Rs 6.50 per unit for (21-30) units, Rs 8.00 per unit for (31-50) units Example: 40 units: Rs (50 +20 × 3 +10 × 6.50+10 × 8) = Total Rs 255 4. 51-100 Rs 9.50 Minimum charge: Rs. 75 Energy charge: Rs. 3 per unit for (0-20) units, Rs 6.50 per unit for (21-30) units, Rs 8.00 per unit for (31-50) units, Rs 9.50 per unit for (51-100) units Example: 80 units: Rs (75+20×3+10×6.50+20×8+30×9.50) = Total Rs 645 5. 101-250 Rs 9.50 Minimum charge: Rs. 100 Energy charge: Rs. 3 per unit for (0-20) units, Rs 6.50 per unit for (21-30) units, Rs 8.00 per unit for (31-50) units, Rs 9.50 per unit for (51-250) units, Example: For 150 units, total charge is Rs(100+20×3+10×6.50+20×8+(50+50)×9.50) = Rs 1,335 6. Above 250 Rs 11.00 Minimum charge: Rs. 150 Energy charge: Rs. 3 per unit for (0-20) units, Rs 6.50 per unit for (21-30) units, Rs 8.00 per unit for (31-50) units, Rs 9.50 per unit for (51-250) units, Rs 11 per unit for more than 250 units Example: For 300 units, total charge is Rs (150+20×3+10×6.50+20×8+(50+150)×9.50+50×11) = Rs 2,885 Household Arithmetic

Vedanta Excel in Mathematics - Book 9 70 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember 1. For the consumers with 5 Ampere meter box, the energy charge is Rs 3 per unit upto 20 units incase of consuming more than 20 units of electricity. 2. The amount of consumption of electricity by an electric appliance of 1000 watt power in 1 hour = 1 kilowatt hour = 1 unit 3. Consumed units = Present reading – Former reading 4. The payment schedule after the meter reading are as follows. (i) Within 7 days of meter reading 2 % rebate is allowed. (ii) From the 8th day to the 15th day, the payment will be according to the bill (iii) From the 16th day to the 30th day, 5 % extra fine (iv) From the 31st day to the 40th day, 10 % extra fine (v) From the 41st day to the 60th day, 25 % extra fine (vi) If the bill is not paid upto the 60th day, the electricity line will be disconnected. Worked-out examples Example 1: The rate of electricity charge up to 20 units is Rs 3 per unit and Rs 6.50 per unit from 21 to 30 units. The meter reading of a household was 01045 units on 1 Baishakh and 01070 units on 1 Jestha. (i) How many units of electricity was consumed in Baishakh? (ii) Find the charge of consumption of electricity with Rs 50 service charge. Solution: (i) Here, consumed electricity = meter reading of 1 Jestha – meter reading of 1 Baishakh = 01070 – 01045 = 25 units Hence, 25 units of electricity was consumed in Baishakh. (ii) Rate of charge up to 20 units = Rs 3 per unit ∴ Charge up to 20 units = 20×Rs 3= Rs 60 Again, The excessive number of units = (25 – 20) units = 5 units Rate of charge from 21 to 30 units = Rs 6.50 per unit So, the charge of 5 units = 5 × Rs 6.50= Rs 32.50 ∴ Total charge of electricity with service charge = Rs 50 + Rs 60 + Rs 32.50 = Rs 142.50 Hence, the required charge of consumption of 25 units is Rs 142.50. Example 2: A 5 A meter box is used in Pushpa’s house. The meter reading of the recent month and the previous month of her house are 1572 and 1492 respectively. Answer the following questions under the given rates and billing system. Units 0-20 21-30 31-50 51-100 101-250 Rate per unit Rs 3 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50 Service charge Rs. 30 Rs. 50 Rs. 50 Rs. 75 Rs. 100 25 units 20 units 5 units (Rs 3 per unit) (Rs 6.50 per unit) Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 71 Vedanta Excel in Mathematics - Book 9 Payment schedule for rebate / fine: Meter reading (days) Within 7 days 8th -15th 16th - 30th 31st - 40th 41st - 60th Rebate/fine 2% rebate - 5% fine 10% fine 25% fine a) Calculate the amount of charge required to pay the electricity bill on the following days. (i) 5th days of meter reading (ii) 14th day of meter reading (iii) 21st day of meter reading (iv) 33rd day of meter reading b) How much more charge should she pay if the payment of the bill was made on 60th day rather than it was paid within 7 days after meter reading? Solution: Here, consumption of electricity = 1572 units – 1492 units = 80 units a) Consumption block No. of units Rate per unit Energy charge Service charge 0-20 21-30 31-50 51-100 20 – 0 = 20 30 – 20 = 10 50 – 30 = 20 80 – 50 = 30 Rs 3 Rs 6.50 Rs 8 Rs 9.50 20×Rs 3= Rs 60 10×Rs 6.50 = Rs 65 20×Rs 8= Rs 160 30×Rs 9.50= Rs 285 Rs 75 [As 80 units lies in (51 – 100) units] Total charge Rs. 570 ∴Total charge = total energy charge + service charge = Rs. 570 + Rs. 75 = Rs. 645 (i) When the payment of bill is made on 5th day (within 7 days) of meter reading, 2% rebate is given. ∴Required amount = Rs 645 – 2% of Rs 645 = Rs 632.10 (ii) When the payment of bill is made on 14th day of meter reading, the payment is as per the bill. ∴Required amount = Rs 645 (iii) When the payment of bill is made on 21st day of meter reading, 5% fine is charged. ∴Required amount = Rs 645 + 5% of Rs 645 = Rs 677.25 (iv) When the payment of bill is made on 33rd day of meter reading, 10% fine is charged. ∴Required amount = Rs 645 + 10% of Rs 645 = Rs 709.50 b) When the payment of bill is made on 60th day of meter reading, 25% fine is charged. ∴Required amount = Rs 645 + 25% of Rs 645 = Rs 806.25 Difference between the payments made on 60th day and within 7 days of the meter reading = Rs 806.25 – Rs 632.10 Rs 174.15 Hence, the required more charge is Rs 174.15. Example 3: Sunayana’s house has a 15A electricity meter box. If she made the payment of Rs 1100 with service charge on the 40th day of meter reading, how many units of electricity was consumed in the month? Calculate it by using the following rates. Household Arithmetic

Vedanta Excel in Mathematics - Book 9 72 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Units 0-20 21-30 31-50 51-100 101-250 Rate per unit Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50 Service charge Rs. 50 Rs. 75 Rs. 75 Rs. 100 Rs. 125 Payment from 31st to 40th day of meter reading: 10% extra fine Solution: Let the consumption of electricity of the month be x units. Consumption block No. of units Rate per unit Energy charge 0-20 21-30 31-50 51-100 101-250 20 – 0 = 20 30 – 20 = 10 50 – 30 = 20 100 – 50 = 50 x – 100 = 150 Rs 4 Rs 6.50 Rs 8 Rs 9.50 Rs 9.50 20 × Rs 4 = Rs 80 10 × Rs 6.50 = Rs 65 20 × Rs 8 = Rs 160 50 × Rs 9.50 = Rs 475 (x – 100) × Rs 9.50 = Rs (9.5x – 950) ∴ Total charge = service charge + total energy charge = Rs 125 + [20 × Rs 4 + 10 × Rs 6.50 + 20 × Rs 8 + 50 × Rs 9.50 + (x – 100) × Rs 9.50] = Rs (125 + 80 + 65 + 160 + 475 + 9.5x – 950) = Rs (9.5x – 45) According to question, Payment of the bill with fine = Rs 1100 or, Rs (9.5x – 45) + 10% of Rs (9.5x – 45) = Rs 1100 or, (9.5x – 45) + 0.1 (9.5x – 45) = 1100 or, (9.5x – 45) + (0.95x– 4.5) = 1100 or, 10.45x = 1149.5 or, x = 110 Hence, 110 units of electricity was consumed in the month. EXERCISE 4.1 General section 1. a) The meter reading of previous month was 3450 units and the meter reading of present month is 3467 units, what is the consumed units of electricity in 1 month? b) The meter readings for the consumption of electricity of a household is 1140 units on 1 Bhadra and 25 units of electricity is found to be consumed in Bhadra, what is the meter reading of 1 Aswin? 2. a) What do you mean by 1 unit of electricity? b) How many units of electricity is consumed when a refrigerator of 100 watts is used for 20 hours? c) The power of a water pump installed at a house is 4 kW. It is used 15 minutes everyday. How many units of electricity is consumed if it is used for 16 days? 3. a) According to the recent electricity tariff rate, the service charge up to 20 units is Rs 30 but energy charge is free; find the charge of consumption of 10 units. b) There is no energy charge but the service charge is Rs 30 up to 20 units; how much bill should be charged by Nepal Electricity Authority (NEA) for the consumption of 18 units? https://www.geogebra.org/m/yph4p67w Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 73 Vedanta Excel in Mathematics - Book 9 Household Arithmetic 4. a) The electricity tariff rate for the consumer using 5 A meter box are given alongside. Answer the following questions. (i) What is the electricity charge for the consumption of 26 units? (ii)What is the electricity charge for the consumption of 28 units? b) The meter reading of Sandhya's house on 1 Baisakh and 1 Jestha are shown in the meter box alongside. Find the electricity charge for the month of Baisakh. c) The meter reading of Kapil's house is shown in the table. Find the electricity charge of the month. Creative section 5. a) The electricity tariff rate fixed by NEA for the consumer using 15 A meter are given alongside. (i) The meter readings of electricity of a household was 3015 units on 1 Magh and 3040 units on 1 Falgun. Find the electricity charge for the month of Magh. (ii) The meter readings of electricity of a household was 6780 units on 1 Chaitra and 6830 units on 1 Baishakh. Find the electricity charge for the month of Chaitra. b) Nepal Electricity Authority has fixed the electricity tariff rates for the consumer using 30 A meter are given alongside. (i) The meter readings of electricity of a household was 1250 units on 1 Kartik and 1300 units on 1 Mansir. Find the electricity charge for the month of Kartik. (ii) The meter readings of electricity of a household was 4040 units on 1 Asar and 4114 units on 1 Saun. Find the electricity charge for the month of Asar. c) The table represents the electricity tariff rates for the consumer using 60 A meter. (i) The meter readings of electricity of a household was 2575 units on 1 Ashoj and 2620 units on 1 Kartik. Find the electricity charge for the month of Ashoj. (ii) The meter readings of electricity of a household was 2002 units on 1 Baishakh and 2080 units on 1 Jeth. Find the electricity charge for the month of Baishakh. Baisakh Jestha 011981 012234 Month Falgun Chaitra Meter reading 2400 2425 Units (kWh) Service charge Energy charge 0-20 Rs 50 Rs 4 21-30 Rs 75 Rs 6.50 31-50 Rs 75 Rs 8 Units (kWh) Service charge Energy charge 0-20 Rs 75 Rs 5 21-30 Rs 100 Rs 6.50 31-50 Rs 100 Rs 8.00 51-100 Rs 125 Rs 9.50 Units (kWh) Service charge Energy charge 0-20 Rs 125 Rs 6 21-30 Rs 125 Rs 6.50 31-50 Rs 125 Rs 8.00 51-100 Rs 150 Rs 9.50 Units (kWh) Service charge Energy charge 0-20 Rs 30 Rs 3 21-30 Rs 50 Rs 6.50

Vedanta Excel in Mathematics - Book 9 74 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. In a household of 5A electricity transmission line, the meter reading of 1 Bhadra was 05724 units and that of 1 Aswin was 05844 units. Answer the following questions under the given rates and billing system. Units 0-20 21-30 31-50 51-100 101-250 Rate per unit Rs 3 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50 Service charge Rs 30 Rs 50 Rs 50 Rs 75 Rs 100 a) Calculate the charge amount required to pay the electricity bill on the following days. (i) 6th days of meter reading (ii) 11th day of meter reading (iii) 22nd day of meter reading (iv) 40th day of meter reading b) How much more amount should be required if the payment is made on 60th day rather than it is within 7 days after meter reading? 7. a) Jasmin’s house has a 5A electricity meter box. She made the payment of Rs 235.20 with service charge on the 3rd day of meter reading. How many units of electricity was consumed in the month? Calculate it by using the following rates. Units 0-20 21-30 31-50 Rate per unit Rs 3 Rs 6.50 Rs 8.00 Service charge Rs 30 Rs 50 Rs 75 Payment within 7 days of meter reading: 2% rebate b) A company has fixed a 15A electricity meter box. It made the payment of Rs 1100 with service charge on the 50th day of meter reading. Apply the rates given in the following table to find the of units of electricity consumed in a month. Units 0-20 21-30 31-50 51-100 101-250 Rate per unit Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50 Service charge Rs 50 Rs 75 Rs 75 Rs 100 Rs 125 Payment from 40th to 60th day of meter reading: 25% extra fine 8. Dinesh, who is using a 5A meter box, recorded the meter reading of first 6 months of last year. Study the following table and answer the questions according to the latest tariff rates of electricity. Month 1 Baishakh 1 Jestha 1 Asar 1 Shrawan 1 Bhadra 1 Aswin Meter reading 5465 5500 5540 5570 5616 5660 a) How many units of electricity was consumed in the month of Asar? b) Which month has maximum consumption? c) How much electricity charge did he pay for the month of (i) Baishakh if the payment was made on 5 Jestha? (ii) Jestha if the payment was made on 20 Asar? d) If he paid the electricity charge of Rs 223 on 10 Kartik for the bill of Aswin, what was the meter reading of 1 Kartik? Payment Rebate/fine Within 7 days 2% rebate 8th – 15th days – 16th – 30th days 5% fine 31st – 40th days 10% fine 41st – 60th days 25% fine Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 75 Vedanta Excel in Mathematics - Book 9 9. a) Mr. Sharma has a 5A meter in his house. He uses 5 CFL bulbs of 15 watt each for 4 hours and an electric heater of 1200 watt for 1 hour everyday. Find the cost of payment of the bill of a month at the rate of Rs 3 per unit up to 20 units, Rs 6.50 per unit from 21 to 30 units and Rs 8 from 31-50 units with Rs 75 service charge, if the payment is made on the 10th day of meter reading. b) Mrs. Bajracharya's house has 15A electricity meter. She uses 5 LED bulbs of 10 watt each for 4 hours, 2 televisions of 60 watt each for 5 hours and a refrigerator of 250 watt for 2 hours everyday. Find the cost of payment of the bill of a month at the rate of Rs 4 per unit up to 20 units, Rs 6.50 per unit from 21 to 30 units and Rs 8 from 31 to 50 units with Rs 75 service charge, if the payment is made on the 35th day of meter reading. Project work and activity section 10. Collect the electricity bills of your house or school of any three successive months. a) Mention the size of meter box fixed in your house or school. b) Calculate the consumption of electricity of each month and the electricity charges. c) Mention a few ways to reduce the electricity charges. d) Describe the online payment system of electricity bill. 4.3 Water bill Nepal Water Supply Corporation (g]kfn vfg]kfgL ;+:yfg) is the concerned authority of the Ministry of Water Supply and Sanitation, the Government of Nepal. A household having water supply facility may have tap with a meter or without a meter. If a meter is not connected in the tap, a customer should pay a lump sum amount fixed by the Nepal Water Supply Corporation. For the payment of a water bill, Nepal Water Supply Corporation has implemented the following rules. Revised tariff rates effective from 1 Falgun 2070: NWSC S.N. Size of pipe (in inch) Minimum consumption (litre) Metered Tap Unmetered Tap Minimum charge (Rs) Additional charge per 1000 litres (Rs) Main Tap Charge (Rs) Branch Tap Charge (Rs) 1. 1 2 " 10,000 110 25 560 200 2. 3 4 " 27,000 1,490 40 3,360 1,600 3. 1" 56,000 3,420 40 9,200 2,700 4. 1.5" 1,55,000 9,600 40 20,700 6,500 5. 2" 3,20,000 21,600 40 42,300 12,000 6. 3" 8,81,000 49,500 40 1,31,400 33,000 7. 4" 18,10,000 97,200 40 2,50,200 62,000 Household Arithmetic

Vedanta Excel in Mathematics - Book 9 76 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revised tariff rates effective from 1 Shrawan 2070: KUKL S. No. Size of Pipe Tap with Meter Tap without Meter (Rs) Minimum Consumption (litre) Minimum Charge (Rs) Additional Consumption per thousand litre (Rs) 1 1 2 " 10,000 100 32 785 2 3 4 " 27,000 1910 71 4595 3 1" 56,000 3960 71 9540 Furthermore, Nepal Water Supply Corporation has implemented the following rules and regulations about the schedule of the payment of the bills. Payment is made after the bill issued Rebate/Fine 1. Within the first and second month 3% rebate 2. Within the third month No rebate and no fine 3. Within the fourth month 10% fine 4. Within the fifth month 20% fine 5. After fifth month 50% fine Facts to remember 1. The consumption of 1,000 litres of water = 1 unit 2. Consumed units = Present reading – Former reading Worked-out examples Example 1: Mr. Regmi has connected a metered tap using half-inch pipe in his house at Biratnagar. The consumption of water in the month of Aswin is 26 units. According to Nepal Water Supply Corporation (NWSC), the minimum charge up to 10 units is Rs 110 and Rs 25 per unit for additional consumption of water, and 50% of the total charge is to be paid as sewerage service charge. Answer the following questions. (i) How many litres of water is consumed in the month of Aswin? (ii) How much sewerage service charge should he pay? (iii)Find the total bill including 50% sewerage service charge. (iv) If he paid only Rs 390 for the consumption of water in Bhadra, how many units of water was consumed in the month of Bhadra? Solution: (i) We know, 1 unit = 1,000 litres of water ∴ 26 units = 26 × 1,000 litres = 26,000 litres Thus, 26,000 litres of water is consumed in Aswin. 26 units 10 units 16 units (Rs 110) (Rs 25 per unit) Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 77 Vedanta Excel in Mathematics - Book 9 (ii) Minimum charge up to 10 units = Rs 110 The additional units of water consumption = (26 – 10) units = 16 units Now, charge of 16 additional units of water = 16 × Rs 25 = Rs 400 ∴ Total charge = Rs 110 + Rs 400 = Rs 510 Also, sewerage service charge = 50% of Rs 510 = Rs 255 (iii) Total bill including sewerage service charge = Rs 510 + Rs 255 = Rs 765 (iv) Let, the total charge of water excluding sewerage service charge be Rs x. Then, x + 50% of x = 390 or, x + 50 100 × x = Rs 390 or, 3x 2 = 390 or, x = 260 Also, The charge for additional units of water = total charge of water – minimum charge = Rs 260 – 110 = Rs 150 ∴The additional units of water = Rs 150 Rs 25 = 6 Total consumed units of water = minimum units + additional units = 10 + 6 =1 6 Thus, the consumption of water in the month of Bhadra is 16 units. Example 2: A hotel in Butwal is using a 3/4'' size of water pipe with meter. The meter reading for the consumption of water of the hotel was 1430 units on 1 Mangsir and 1470 units on 1 Poush. The tariff rates implementing by the NWSC is given below. Size of pipe Minimum consumption Minimum charge Additional charge 3/4" 27,000 litres Rs 1,490 Rs 40 per 1,000 litres a) Find the of units of water consumed in the month of Mansir. b) Calculate the charge to be paid including 50% sewerage service charge if the payment of the bill is made in the following schedule. (i) Within the second month after the bill issued. (i) Within the fourth month after the bill issued. (iii) Within the sixth month after the bill issued. Solution: a) Here, the meter reading on 1 Mangsir = 1430 units The meter reading of 1 Poush = 1470 units ∴Consumed units of water in the month of Mansir = (1470 – 1430) units = 40 units b) According to water tariff provisions of Nepal Water Supply Corporation for using 3/4′′ pipe, The minimum charge of 27,000 litres (i.e., 27 units) = Rs 1,490 The additional units of water consumption = (40 – 27) units = 13 units 40 units 27 units 13 units (Rs 1,490) (Rs 40 per unit) Household Arithmetic

Vedanta Excel in Mathematics - Book 9 78 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Household Arithmetic Now, the charge of additional 13 units of water = 13 × Rs 40 = Rs 520 The total charge = Rs 1490 + Rs 520 = Rs 2,010 ∴ Again, the charge including sewerage service charge = Rs 2,010 + 50% of Rs 2,010 = Rs 3,015 (i) When the payment is made within the second month after the bill issued, 3% rebate is allowed. ∴Required payment = Rs 3,015 – 3% of Rs 3,015 = Rs 2,924.55 (ii) When the payment is made within the fourth month after the bill issued, 10% fine is charged. ∴Required payment = Rs 3,015 + 10 %s of Rs 3,015 = Rs 3,316.50 (iii) When the payment is made within the sixth month after the bill issued, 50% fine is charged. ∴Required payment = Rs 3,015 + 50 % of Rs 3,015 = Rs 4,522.50 EXERCISE 4.2 General section 1. a) How many litres of water is to be consumed to make 1 unit? b) According to water tariff rates of Nepal Water Supply Corporation, what is the minimum charge up to 10 units for the consumption of water by using a half-inch pipe? c) What is the minimum charge for the consumption of water up to 56,000 litres by using a 1′′ pipe according to NWSC? 2. a) The meter reading for the consumption of water in the previous month was 1380 units and the meter reading of present month is 1400 units, find the consumed units of water in 1 month. b) The meter reading of consumption of water in 1 Baishakh was 6699 units and that of 1 Jestha was 7070 units, how many units of water was consumed in the month of Baishakh? c) On 1 Poush, the meter readings of water of a household is 1140 units and the household consumed 15 units of water in Mansir. What was the meter reading of 1 Mansir? 3. a) The water tariff rates fixed by Nepal Water Supply Corporation for the consumers using half-inch pipe are given alongside. Answer the following questions. (i) How much should a household pay for the consumption of 15 units of water in a month? (ii)How much will be charged for a household for the consumption of 24 units of water in a month? Minimum consumption Minimum charge (Rs) Additional charge (Rs) 10 units Rs 110 Rs 25 per unit Sewerage service charge: 50%

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 79 Vedanta Excel in Mathematics - Book 9 Household Arithmetic b) Kathmandu Upatyaka Khanepani Limited (KUKL) has implemented the water tariff rates for the consumers using half-inch pipe are given in the table. Answer the following questions. (i) How much should a household of Kathmandu pay for the consumption of 18 units of water in a month? (ii)How much will be charged for a household of Madhyapur, Thimi for the consumption of 40 units of water in a month? Creative section 4. a) Mr. Upreti has a metered tap using half-inch pipe in his house in Banepa. The meter reading of his house on 1 Baishakh was 1050 units and on 1 Jestha was 1075 units. According to NWSC, the minimum charge up to 10 units is Rs 110 and Rs 25 per unit for additional consumption of water, and 50% of the total charge is to be paid as sewerage service charge. Answer the following questions. (i) How many litres of water is consumed in Baishakh month? (ii) How much sewerage service charge should he pay? (iii) Find the total bill including 50% sewerage service charge. (iv) If he paid only Rs 465 for the consumption of water in Asar, how many units of water was consumed in the month of Asar? b) Mr. Maharjan uses a 3 4 " pipe in his house in Kathmandu. The meter reading of his house on 1 Aswin was 2020 units and on 1 Kartik was 2050 units. According to KUKL, the minimum charge up to 27 units is Rs 1,910 and Rs 71 per unit for additional consumption of water, and 50% of the total charge is to be paid as sewerage service charge. Answer the following questions. (i) How many units of water is consumed in Aswin month? (ii) How much sewerage service charge should he pay? (iii) Find the total bill including 50% sewerage service charge. (iv) If he paid only Rs 3,717 for the consumption of water in Falgun, how many units of water was consumed in the month of Falgun? 5. a) A household of Lahan consumed 32 units of water in a month by using half-inch pipe. Calculate the payment of the bill as per the rules implemented by NWSC including 50% sewerage service charge, if the payment is made within the first month of the bill issued. b) 127 units of water is consumed by using 3 4 " pipe in a hotel in Pokhara. If the payment of the bill is made within the fifth month after the bill issued, how much money is required to clear the bill with 50% sewerage charge? Calculate it as per the tariff rates provisioned by NWSC. Minimum consumption Minimum charge (Rs) Additional charge (Rs) 10 units Rs 100 Rs 32 per unit Sewerage service charge: 50%

Vedanta Excel in Mathematics - Book 9 80 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. a) A hospital at Dhulikhel is using a 3 4 " of water pipe with meter. The meter reading for the consumption of water of the hotel was 1540 units on 1 Kartik and 1600 units on 1 Mansir. The tariff rates implementing by the NWSC is given below. Size of pipe Minimum consumption Minimum charge Additional charge 3/4" 27,000 litres Rs 1,490 Rs 40 per 1,000 litres Calculate the charge to be paid including 50% sewerage service charge if the payment of the bill is made in the following schedule. (i) within the first month after the issue of bill (ii) within the third month after the issue of bill (iii) within the fifth month after the issue of bill (iv) within the seventh month after the issue of bill b) A factory run in Lalitpur is using a 1" of water pipe. The meter reading for the consumption of water in the factory was 3500 units on 1 Falgun and 3580 units on 1 Chaitra. The tariff rates implementing by the KUKL is given below. Size of pipe Minimum consumption Minimum charge Additional charge 1" 56,000 litres Rs 3,960 Rs 71 per 1,000 litres Calculate the charge to be paid including 50% sewerage service charge if the payment of the bill is made in the following schedule. (i) within the second month after the bill issued (ii) within the fourth month after the bill issued (iii) within the sixth month after the bill issued Project work and activity section 7. Collect the water bills of your house or school of any two months. a) Mention the size of pipe fixed in your house or school. b) Calculate the consumption of water of each month and the charges. c) Describe the online payment system of water bill. 4.4 Telephone bill The payment of telephone bill is also a regular financial household activity. Here, we discuss about the telephone bill of Public Switch Telephone Network (PSTN) /Land line. Study the following rules of payment of the telephone bills implemented by Nepal Telecommunications Authority (g]kfn b'/;~rf/ k|flws/0f) 1. Rules for PSTN Telephone Service Tariff Minimum Monthly Rental for Local Telephone (with 175 free calls) is Rs 200 Rate of charge above 175 calls is Re 1 per call (2 minutes per call) Sub-total charge (C) = Minimum charge + charge of additional number of call Telecom Service Charge (TSC) = 13 % of Sub-total charge (C) Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 81 Vedanta Excel in Mathematics - Book 9 Total charge = C + TSC VAT = 13 % of (C + TSC) Grand Total charge = C + TSC + VAT Nepal Telecom Nepal Doorsanchar Company Ltd. Statement For The Month: 13% 13% No. : Date : Local call, STD call and ISD call are three types of telephone calls. The rates of charge of these calls are different. 2. Rules for Nepal Telecom GSM Postpaid Service Tariff S.N. Service Type Call charge 1. Call Inside NT Network: On-Net Call Rs 1.50 per minute Outside NT Network: Off-Net Call Rs 2 per minute 2. Short Message Service (SMS) • Re 1 per SMS (On-Net) • Rs 1.25 per SMS (Off-Net) • Rs 5 per SMS (International-Net) 3. Friends and Family (FNS) Call Service Rs 0.70 per minute (Maximum 5 people) 3. Rules for Nepal Telecom GSM Pre-paid Service Tariff S.N. Service Type Charge Peak hour (6 AM-6 PM) Off-peak hour (6 PM-6 AM) Monthly rental 1. Call On-Net Call Re 1 per minute Rs 0.55 per minute Rs 300 Off-Net Call Rs 1.5 per minute Rs 1.5 per minute 2. SMS (a) On-Net: Re 1 per SMS (b) International-Net: Rs 5 per SMS 3. FNS Rs 0.55 per minute (Maximum 6 people) 4. Video call Rs 2 per minute Household Arithmetic

Vedanta Excel in Mathematics - Book 9 82 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember 1. 13% TSC and 13% VAT have been applying as per Government rule since 1 Shrawan 2075. 2. 2% Ownership Tax (OT) has been applying on every GSM and CDMA Postpaid mobile recharge as per Government rule since 1 Shrawan 2077. Worked-out examples Example 1: An office made a total of 1175 calls in the month of Baishakh. According to Nepal Telecommunication Authority (NTA), the minimum charge up to 175 calls is Rs 200 and Re 1 for each additional call. 13% TSC and 13% VAT are applied as per Government rule. a) How much monthly rental charge is fixed? b) What is the sub total amount of telephone bill? c) What is the telephone charge with TSC? d) How much amount should the office pay with VAT to clear the bill? Solution: a) Here, the minimum charge up to 175 calls = Rs 200 So, the monthly rental charge is Rs 200. b) Now, the additional number of calls = 1175 – 175 = 1000 Charge for additional number of calls = 1000 × Re 1 = Rs 1000 ∴Sub-total charge (C) = Charge for 1175 calls = Minimum charge + Additional charge = Rs 200 + Rs 1000 = Rs 1,200 c) Also, TSC = 13% of sub-total charge (C) = 13% of Rs 1,200 = Rs 156 ∴ Total charge (TC) = Sub-total charge (C) + TSC = Rs 1200 + Rs 156 = Rs 1,356 d) Again, VAT = 13% of total amount = 13% of Rs 1,356 = Rs 176.28 ∴Grand total amount = C + TSC + VAT = Rs 1,356 + Rs 176.28 = Rs 1,532.28 Hence, the office should pay Rs 1,532.28 to clear the telephone bill. 1175 calls 175 calls 1000 calls (Rs 200) (Re 1 per call) Alternate process c) Total charge (TC) = C + 13% TSC of C = Rs 1200+13% of Rs1200 = Rs 1,356 d) Grand total charge = TC + 13% of TC = Rs 1,356+13% of Rs1356 = Rs 1,532.28 Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 83 Vedanta Excel in Mathematics - Book 9 Example 2: The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each extra call of 2 minutes duration is Re 1. If a person paid Rs 893.83 with 13 % TSC and 13 % VAT for his telephone bill, answer the following questions. a) Find the charge of telephone calls without VAT. b) Find the charge of telephone calls without TSC. c) Find the total number of calls made. Solution: a) Let the charge of telephone calls without VAT be Rs x. Then, x + 13 % of x = Rs 893.83 or, 113x 100 = Rs 893.83 or, 1.13 x = Rs 893.83 or, x = Rs 791 So, the charge of telephone calls without VAT is Rs 791. b) Let the charge of telephone calls without TSC be Rs y. Then, y + 13 % of y = Rs 791 or, 113x 100 = Rs 791 or, 1.13 x = Rs 791 or, x = Rs 700 So, the charge of telephone calls without TSC is Rs 700. c) The minimum charge up to 175 calls = Rs 200 The charge for the extra calls = Rs 700 – Rs 200 = Rs 500 Now, the number of extra calls = Rs 500 Re 1 = 500 Hence, the total number of calls is 175 + 500 = 675 calls. Example 3: Mr. Shrestha uses a postpaid NTC mobile. The monthly rental charge for Nepal Telecom GSM postpaid mobile service is Rs 300. On-net call charge is Rs 1.5 per minute, 13% TSC is added and 13% VAT is levied on it then 2% ownership tax (OT) is charged. a) How much minimum charge should he pay in a month? b) How long can he make calls on a mobile with recharge of Rs 200? Solution: a) Here, monthly rental charge = Rs 300 Charge with 13% TSC = 113% of Rs 300 = Rs 339 Charge with 13% VAT = 113% of Rs 339 = Rs 383.07 Charge with 2% OT = 102% of Rs 383.07 = Rs 390.73 Hence, he should pay minimum charge of Rs 390.73 Alternate process a) Let, charge without VAT be Rs x. Then, 113% x = Rs 893.83 or, 1.13x = Rs 893.83 ∴ x = Rs 791 b) Let, charge without TSC be Rs y. Then, 113% y = Rs 791 or, 1.13 y = Rs 791 ∴ y = Rs 700 Answer checking: When sub-total charge = Rs 700 ∴ Grand total = 113% of 113% of Rs 700 = Rs 893.83, which is given in the question. Alternate process a) Minimum charge = 113% of 113% of 102% of Rs 300 = Rs 390.73 Household Arithmetic

Vedanta Excel in Mathematics - Book 9 84 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) Let the call charge without OT be Rs x. Then, x + 2 % of x = Rs 200 or, 102x 100 = Rs 200 or, 1.02 x = Rs 200 or, x = Rs 196.08 So, the charge of calls without OT is Rs 196.08 Also, let the call charge without VAT be Rs y. Then, y + 13 % of y = Rs 196.08 or, 113y 100 = Rs 196.08 or, 1.13 y = Rs 196.08 or, y = Rs 173.52 So, the charge of calls without VAT is Rs 173.52 Again, let the call charge without TSC be Rs z. Then, z + 13 % of z = Rs 173.52 or, 113z 100 = Rs 173.52 or, z = Rs 153.56 By the recharge of Rs 200, only Rs 153.56 is used for talk time. Now, on-net call charge = Re 1.5 per minute. ∴Talk time for Rs 153.56 = 153.56 1.5 minutes = 102.37 minutes = 60 minutes + 42.37 minutes = 1 hour + 42 minutes + 0.37 × 60 seconds = 1 hour + 42 minutes + 22.2 seconds = 1 hour 42 minutes 22.2 seconds EXERCISE 4.3 General section 1. a) The telephone reading of the former month was 5500 calls and the call reading of current month is 6000 calls, how many telephone calls is made in 1 month? b) The call reading of Baishakh was 8800 calls and that of Jestha was 9900. If the minimum charge upto 175 calls is Rs 200 and Re 1 for each additional call, calculate the subtotal amount of telephone bill. 2. a) The subtotal amount of a telephone bill is Rs 1,000. Find the grand total amount of the bill with 13% TSC and 13% VAT. b) What will be the grand total amount of a telephone bill with 13% TSC and 13% VAT if the subtotal amount is Rs 2,000? Alternate process b) Let, the amount for talk time be Rs x. Then, 113% of 113% of 102% of x = Rs 200 or, 1.13 × 1.13 × 1.02x = 200 ∴ x = Rs 153.56 Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 85 Vedanta Excel in Mathematics - Book 9 Creative section 3. a) A school made a total of 2175 calls in the month of Baishakh. According to Nepal Telecommunication Authority (NTA), the minimum charge up to 175 calls is Rs 200 and Re 1 for each additional call. 13% TSC and 13% VAT are applied as per the Government rule. (i) How much monthly rental charge is fixed? (ii) What is the sub total amount of telephone bill? (iii) Find the TSC and VAT amounts. (iv) How much total amount should the school pay to clear the bill? b) Nepal Telecommunication Authority (NTA) has determined the rental charge of Rs 200 with 175 free calls, Re 1 for each extra call and, 13% TSC and 13% VAT. In a hospital, the telephone reading of the months of 1 Aswin and 1 Kartik was 7438 calls and 9601 calls respectively. (i) How many calls were made in Aswin? (ii) What is the sub total amount of telephone bill? (iii) Find the TSC and VAT amounts. (iv) How much amount should the hospital pay to clear the bill? c) The minimum charge up to 175 calls is Rs 200. If the charge for each additional call is Re 1, how much is the charge for 475 calls with 13 % TSC and 13 % VAT? d) The charge for STD call from Bhadrapur to Surkhet for 1 minute is Rs 1.25. If Mr. Rajbanshi makes a call for 10 minutes, how much should he pay with 13 % TSC and 13 % VAT? e) The charge of ISD call from Bharatpur to Melbourne, Australia is Rs 20 per minute. If Nirmal made a call for 5 minutes, calculate the cost paid by him with 13 % TSC and 13 % VAT. 4. a) The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each extra call of 2 minutes duration is Re 1. If Mr. Malla paid Rs 510.76 with 13 % TSC and 13 % VAT for his telephone bill, answer the following questions. (i) Find the charge of telephone calls without VAT. (ii) Find the charge of telephone calls without TSC. (iii) Find the total number of calls that he made. b) Mrs. Gurung has a shop of cultural dresses. She uses a landline phone. She paid Rs 1,276.90 with 13 % TSC and 13 % VAT for her telephone bill for the month of Poush. Given that the minimum charge of telephone calls up to 175 calls is Rs 200 and the charge for each extra call is Re 1. (i) Find the charge of telephone calls without VAT. (ii) Find the charge of telephone calls without TSC. (iii) Find the number of total calls made in Poush. Household Arithmetic

Vedanta Excel in Mathematics - Book 9 86 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5. a) Mr. Regmi uses a postpaid NTC mobile. The monthly rental charge for Nepal Telecom GSM postpaid mobile service is Rs 300. If onnet call charge is Rs 1.50 per minute, 13% TSC is added and 13% VAT is levied on it, then 2% ownership tax (OT) is charged. (i) How much minimum charge should he pay in a month? (ii) How long can he make calls on a mobile with recharge of Rs 100? b) Mrs. Magar uses a postpaid NTC mobile. The monthly rental charge for Nepal Telecom GSM postpaid mobile service is Rs 300. If on-net call charge is Rs 1.50 per minute, 13% TSC is added and 13% VAT is levied on it, then 2% ownership tax (OT) is charged. (i) How much minimum charge should he pay in a month? (ii) How long can he make calls on a mobile recharge of Rs 500? Project work and activity section 6. a) Collect the telephone bills of your house or school of any two months and bring to your class. Discuss about the number of calls of the former month and current month, sub-total charge, total charge and grand total charge. b) Ask to your parents about the mobile recharge card for postpaid sim. Discuss about the talk time of a mobile topup of Rs 100. 4.5 Calculation of taxi fare in a taximeter Nowadays, with the increasing facilities of taxi services, especially in the urban area, the payment of taxi fare is also becoming a regular financial activity. A device used in taxis that automatically records the distance travelled and the fare payable is called taximeter. Nepal Bureau of Standards and Metrology (NBSM) -g]kfn u'0f:t/ tyf gfktf}n laefu_ is responsible to implement the rules, regulations, and the rates of fare and monitoring that it is carried out fairly or correctly. Here, we shall discuss about the calculation of fare in a taximeter. Following are the rates of fare in taximeter implemented by NBSM department: Time Minimum fare Fare of per 200 meters 6: 00 am to 9:00 pm Rs 50 Rs 10 9: 00 pm to 6:00 am Rs 75 Rs 15 Remember that taxi fare of 9:00 pm to 6:00 am is 1.5 times of the fair of 6:00 am to 9:00 pm. While travelling between 6:00 am to 9:00 pm, Rs 50 appears in the monitor of the taximeter immediately when it is flagged down. Then, the fare goes on at the rate of Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 87 Vedanta Excel in Mathematics - Book 9 Rs 10 per 200 m. However, in the case of travelling between 9:00 pm to 6:00 am, when the taximeter is flagged down Rs 75 appears in the monitor and the fare goes on at the rate of Rs 15 per 200 m. If the taxi is asked to wait by a passenger, the additional waiting charges are applied according to the rules given in the table. Worked-out examples Example 1: The latest rates of taxi fare implemented by NBSM are given in the table below. Time Minimum fare Fare of per 200 meters Waiting charge 6: 00 am - 9:00 pm Rs 50 Rs 10 Rs 10 per 2 minutes 9: 00 pm - 6:00 am Rs 75 Rs 15 Rs 15 per 2 minutes Sunayana called a metered taxi at 7:00 a.m. and asked to wait for 6 minutes. Then, she travelled from Banasthali to Tribhuvan International Airport (TIA) to drop her bother at airport. If the distance between Banasthali to airport is 8.3 km, answer the following questions. (i) What was the minimum fare appeared immediately after the meter was flagged down? (ii) What was the waiting charge? (iii) Calculate the total fare paid by Sunayana. (iv) If she again hired a taxi and returned back to Banasthali through the same route, how much more or less fare would she pay? Solution: (i) Here, 7:00 a.m. lies between 6: a.m. and 9:00 p.m. Thus, the minimum fare is Rs 50. (ii) Now, the waiting charge for 2 minutes = Rs 10 The waiting charge for 1 minute = Rs 10 2 = Rs 5 ∴ The waiting charge for 6 minutes = Rs 5 × 6 = Rs 30 (iii) Distance travelled = 8.3 km = 8300 m Now, the fare of 200 m = Rs 10 The fare of 1 m = Rs 10 200 The fare of 8300 m = Rs 10 200 × 8300 = Rs 415 ∴ The total fare = Rs 50 + Rs 415 + Rs 30 = Rs 495 She paid Rs 495 to travel Airport from Banasthali. (iv) While returning back to Banasthali, taxi was not asked to wait. So, the fare = Rs 50 + Rs 415 = Rs 465 Now, the difference between two fare = Rs 495 – Rs 465 = Rs 30 Hence, she paid Rs 30 less while returning back to Banasthali. Time Waiting charge 6: 00 am to 9:00 pm Rs 10 per 2 minutes 9: 00 pm to 6:00 am Rs 15 per 2 minutes Household Arithmetic

Vedanta Excel in Mathematics - Book 9 88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: Shashwat hired a taxi and travelled a certain distance at 10:30 pm. He paid the total fare of Rs 555 including the additional waiting charge for 14 minutes. If the minimum fare is Rs 75, the fare per 200 m is Rs 15 and the waiting charge is Rs 15 per 2 minutes, answer the following questions. (i) Which department is responsible for implementation of the rates of taxi fare? (ii) How much was the waiting charge? (iii) Find the distance travelled by him. Solution: (i) Nepal Bureau of Standards and Metrology (NBSM) department is responsible for implementation and monitoring the rates of taxi fare. (ii) The minimum charge = Rs 75 The waiting charge for 2 minutes = Rs 15 The waiting charge for 1 minute = Rs 15 2 ∴ The waiting charge for 14 minutes = Rs 15 2 × 14 = Rs 105 (iii) Now, the taxi fare excluding the minimum fare and waiting charge = Rs 555 – Rs 75 – Rs 105 = Rs 375 Again, Rs 15 is the fare of 200 metres Re 1 is the fare of 200 15 metres Rs 375 is the fare of 200 15 × 375 meters = 5000 meters = 5 km Hence, he travelled 5 km. EXERCISE 4.4 General section 1. a) Answer the following questions. (i) Which department is responsible for implementation of the rates of taxi fare? (ii) What is the minimum fare of taxi from 6:00 a.m. to 9:00 p.m.? (iii) What is the taxi fare for each 200 m from 6:00 a.m. to 9:00 p.m.? (iv) What is the minimum fare of taxi from 9:00 p.m. to 6:00 a.m.? (iv) What is the taxi fare for each 200 m from 9:00 p.m. to 6:00 a.m.? b) According to rates of taxi fare implemented by Nepal Bureau of Standards and Metrology (NBSM), answer the following questions. (i) What is the waiting charge for 2 minutes from 6:00 a.m. to 9:00 p.m.? (ii) What is the waiting charge for 2 minutes from 9:00 pm to 6:00 a.m.? 2. a) Suntali hired a taxi and travelled 6 km at 3:30 p.m. If the minimum fare is Rs 50 and the fare goes on at the rate of Rs 10 per 200 metres, calculate the total fare paid by her. b) Madan Bahadur hired a taxi and travelled 9 km at 8:30 a.m. If the minimum fare is Rs 50 and the fare goes on at the rate of Rs 10 per 200 metres, calculate the total fare paid by him. Household Arithmetic

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 89 Vedanta Excel in Mathematics - Book 9 c) Mr. Sampang hired a taxi and travelled 15 km at 10: 00 pm. If the minimum fare is Rs 75 and the fare goes on at the rate of Rs 15 per 200 metres, calculate the total fare paid by him. Creative section 3. The table given below shows the latest rates of taxi fare implemented by NBSM. Use the rates to solve the following questions. Time Minimum fare Fare of per 200 meters Waiting charge 6: 00 a.m. - 9:00 p.m. Rs 50 Rs 10 Rs 10 per 2 minutes 9: 00 p.m. - 6:00 a.m. Rs 75 Rs 15 Rs 15 per 2 minutes a) Mrs. Kandel and her family wished to visit zoo at Jawalakhel which is at a distance of 4.6 km from her home located in Kalanki. For this, she hired a metered taxi and asked it to wait for 6 minutes, then they travelled to zoo at 10:00 a.m. (i) What was the minimum fare appeared immediately after the meter was flagged down? (ii) What was the waiting charge? (iii) Calculate the total fare paid by her. (iv) If she hired another taxi and returned back to her home through the same route, how much more or less fare did she pay? b) At 2:45 a.m., Mr. Poudel had to go hospital which was at a distance of 15 km from his home. He hired a metered taxi and asked it to wait for 4 minutes. (i) What was the minimum fare appeared immediately after the meter was flagged down? (ii) What was the waiting charge? (iii) Find the taxi fare paid by him. (iv) If he hired another taxi and returned back to his home at 7:00 a.m. through the same route, how much more or less fare would he pay than travelling to hospital? 4. a) Bishwant hired a taxi and travelled a certain distance at 6:15 a.m. He paid the total fare of Rs 450 including the additional waiting charge for 20 minutes. If the minimum fare is Rs 50, the fare per 200 m is Rs 10 and the waiting charge is Rs 10 per 2 minutes, answer the following questions. (i) How much was the waiting charge? (ii) Calculate the fare excluding the waiting charge and minimum fare. (iii) Find the distance travelled by him. b) Dipika hired a taxi and travelled a certain distance at 9:40 p.m. She paid the total fare of Rs 675 including the additional waiting charge for 10 minutes. If the minimum fare is Rs 75, the fare per 200 m is Rs 15 and the waiting charge is Rs 15 per 2 minutes, answer the following questions. (i) How much was the waiting charge? (ii) Find the distance travelled by her. Household Arithmetic

Vedanta Excel in Mathematics - Book 9 90 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. What does 1 unit of electricity mean? (A) Consumed amount of electricity by electric appliance of 1 watt power in 1 hour. (B) Consumed amount of electricity by electric appliance of 10 watt power in 1 hour. (C) Consumed amount of electricity by electric appliance of 100 watt power in 1 hour (D) Consumed amount of electricity by electric appliance of 1000 watt power in 1 hour. 2. The consumption of electricity of recent month is calculated by (A) Meter reading of recent month – Meter reading of previous month (B) Meter reading of following month – Meter reading of recent month (C) Meter reading of previous month – Meter reading of recent month (D) Meter reading of following month – Meter reading of previous month 3. The meter readings of the month Baishakh, Jesth and Asar are 1127, 1155 and 1180 respectively. The consumption of the electricity in the month Jestha is (A) 28 units (B) 25 units (C) 53 unts (D) 38 units 4. According to payment schedule of consumption of electricity, if the payment is made within 7 days of meter reading then there is provision of (A) 3% of rebate is allowed. (B) 2% of rebate is allowed. (C) 5% extra fine (D) No rebate no fine 5. What is the grand total in the telephone bill if the sub-total is Rs 400? (A) Rs 440 (B) Rs 452 (C) Rs 497.20 (D) Rs 406.80 6. According to Nepal Telecommunication Authority, the minimum charge for 175 calls is (A) Rs 100 (B) Rs 200 (C) Rs 175 (D) Rs 350 7. The compulsory provision of sewerage service charge in water bill is (A) 20% of water consumption charge (B) 40% of water consumption charge (C) 50% of water consumption charge (D) 100% of water consumption charge 8. When the water bill is paid within the third month of bill issued, there will be (A) 3% rebate in bill (B) no rebate no fine (C) 10% fine (D) 20% fine 9. The consumer gets 3% rebate in water bill if the payment is made within (A) 1st month after the bill issued (B) 2nd month after the bill issued (C) 1st or 2nd month after the bill issued (D) 3rd month after the bill issued 10. According to Nepal Bureau of Standards and Metrology (NBSM), what is the minimum taxi fare from 6:00 a.m. to 9:00 p.m.? (A) Rs 7.20 (B) Rs 10.80 (C) Rs 14 (D) Rs 21

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 91 Vedanta Excel in Mathematics - Book 9 Assessment -II 1. Mrs. Bista is a secondary level English teacher. Her monthly basic salary is Rs 45,990. She earns monthly Rs 2,000 as dearness allowance, one month’s basic salary once in a year as the festival expenses, Rs. 10,000 as clothing allowance. She has contributed 10% of her basic salary in the Employee’s Provident Fund (EPF) and the same amount is deposited by the government in the fund. For unmarried people For married people Tax rates Annual income (Rs) Annual income (Rs) Up to 5 lakh Up to 6 lakh 1% 5-7 lakh 6-8 lakh 10% 7-10 lakh 8-11 lakh 20% Based on the above information, answer the following questions. (a) Define income tax. (b) Find her annual taxable income. (c) How much income tax should she pay annually if she pays Rs 60,000 annual premium of life insurance? 2. The marked price of a carved window made of wood is Rs 5,500. A Canadian tourist bought it with 13% VAT after allowing 20% discount. (a) What is the selling price of the window after discount? (b) By what percent is the selling price of the window with VAT more or less than marked price? 3. Mrs. Magar bought 500 shares out of 50,000 shares sold by a hydro power company. When the company distributed 15% of its net profit, she received Rs 22,500 as cash dividend in a year. (a) Find the cash dividend of each share. (b) Find the net profit of the company. (c) How much cash dividend would Mrs. Magar receive if she had 300 shares of the same company and the company distributed 18% of net profit as dividend? 4. Janasewa Secondary School has connected with 5A electricity meter, 0.5 inch pipe and Nepal telephone landline phone. Answer the following questions according to the given rate of charges. Electricity charge Rate of water consumption per month (0.5 inch pipe) Telephone charge rate Units (Landline) Service charge Energy charge 0-20 Rs 30 Rs 3 • Min. charge up to 10 units: Rs 100 • Extra charge= Rs 32 per unit • Min. charge up to 175 calls = Rs 200 • Charge for each extra call= Rs Re 1 21-30 Rs 50 Rs 6.50 31-50 Rs 50 Rs 8.00 51-100 Rs 75 Rs 9.50 (a) How much telephone charge does it pay if 300 calls is made in Baisakh? (b) Calculate the amount required to pay the electricity bill of the month of Baisakh if the meter reading of Baisakh 1 is 04560 and that of Jeth 1 is 04610. (c) If 40 units of water was consumed in the month Baisakh, find the water bill of the month Baisakh.

Vedanta Excel in Mathematics - Book 9 92 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 5.1 Area – Looking back Classwork-Exercise 1. Fill in the blanks with correct answers as quickly as possible. a) The formula to calculate the area of rectangle is ………………… b) The formula to find the area of square is ………………… c) The formula to find the area of parallelogram is ………………… d) The formula to calculate the area of triangle is ………………… 2. Calculate the areas mentally and write in the blank space as quickly as possible. a) The length of a rectangular carpet is 3 m and its breadth is 2 m. Its area is …….. b) A squared garden is 30 feet long. Its area is ….…. c) The base of a parallelogram is 15 cm and its height is 8 cm. Its area is ….…. d) The area of triangle whose base is 20 inch and height is 9 inch is ….…. e) A right angled triangle has base 26 cm and perpendicular 10 cm. Its area is ..…... 5.2 Mensuration Mensuration is the branch of mathematics which studies the measurement of the geometric figures. It deals with length, area, and volume of different types of 2-dimensional and 3-dimensional shapes. Mensuration is used in the field of architecture, medicine, construction, etc. It is necessary for everyone to learn formulae used to find the perimeter and area of 2-dimensional figures as well as the surface area and volume of 3-dimensional solids in day to day life. 5.3 Perimeter and area of plane figures - review Perimeter of a plane figure is the total of the length of its boundary. The unit of perimeter is the same as the unit of length, for example centimeter (cm), metre (m), etc. are the units of perimeter. Perimeter of a triangle = a + b + c, where, a, b and c are the length of sides of a triangle. Perimeter of a rectangle = 2(l + b), where, l is the length and b is the breadth of a rectangle. Perimeter of a square = 4l, where, l is the length of each side of a square. Perimeter of a circle = 2πr, where, r is the radius of a circle. Unit 5 Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 93 Vedanta Excel in Mathematics - Book 9 Area of a plane figure is the measure of the plane surface enclosed by its boundary line. The unit of area is square centimetre (cm2 ) square metre (m2 ), etc. It is noted that ‘square metre’ and ‘metre square’ have different meanings. 9 square metre means the area is 9 m2 . However, 9 metre square means a square each of whose side is 9 m long. 5.4 Area of triangle We can find the area of different types of triangles by using special formulae which are generalised on the basis of the given measurements of various parts of the triangles. (i) Area of a triangle in terms of its base and height (altitude) In the adjoining D ABC, AD ⊥ BC. So, AD is the height and BC is the base of the triangle ABC. Here, area of D ABC = 1 2 base × height = 1 2 BC × AD = 1 2 b × h (ii) Area of a right-angled triangle In the given right-angled D ABC right angled at B, the perpendicular AB is the height of the triangle and BC is its base. Area of right-angled D ABC = 1 2 base × height = 1 2 base × perpendicular = 1 2 BC × AB (iii) Area of an equilateral triangle In the given equilateral triangle ABC, AB = BC = CA = a (suppose). AD perpendicular to BC is drawn. In an equilateral triangle, perpendicular drawn from a vertex to its opposite side bisects the side. ∴ BD = DC = a 2 By using Pythagoras Theorem in right-angled D ABD, AD = AB2 – BD2 = a2 – a2 4 = 3 2 a Now, area of D ABC = 1 2 base × height = 1 2 BC × AD = 1 2 a × 3 2 a = 3 4 a2 Thus, area of equilateral triangle = 3 4 (side)2 (iv) Area of an isosceles triangle In the given isosceles triangle ABC, AB = AC = a and BC = b (suppose). AD perpendicular to BC is drawn. A C D B A B C h D b A B C p b A h a B D C a a a 2 a 2 Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur We know that, in an isosceles triangle, perpendicular drawn from the vertical angle to the base bisects the base. Therefore, BD = DC = b 2 By using Pythagoras Theorem in right-angled D ABD, AD = AB2 – BD2 = a2 – b2 4 = 4a2 – b2 2 Now, area of D ABC = 1 2 base × height = 1 2 BC × AD = 1 2 b × 4a2 – b2 2 = 1 4 b 4a2 – b2 Thus, area of an isosceles triangle = 1 4 b 4a2 – b2 (v) Area of scalene triangle in terms of its three sides (Heron's formula) Heron of Alexandria [A.D. 10 – A.D. 70 (CE)] was a Greek mathematician and engineer. He lived in the city of Alexandria in Egypt, and is one of the greatest “experimenter” of antiquity. He wrote books on mathematics, mechanics and physics. In mathematics, he is mostly remembered for Heron’s formula that allows us to calculate the area of a triangle using only the lengths of its sides. In the adjoining scalene triangle ABC, suppose the sides BC = a, CA = b and AB = c. Let's draw AD⊥BC. Therefore, AD = h is the height of DABC. Let the length of DC be x cm. ∴ The length of BD = (a – x) cm Here, the perimeter of D ABC = 2s = BC + CA + AB = a + b + c ∴ 2s = a + b + c or, s = a + b + c 2 which is called the semi-perimeter of the triangle. Now, in right-angled D ACD, by using Pythagoras Theorem, AD2 = AC2 – DC2 or, h2 = b2 – x2 .................(i) Also, in right-angled D ABD, AD2 = AB2 – BD2 or, h2 = c2 – (a – x)2 .................(ii) From equations (i) and (ii), we get, c2 – (a – x)2 = b2 – x2 or, c2 – a2 + 2ax – x2 = b2 – x2 or, x = a2 + b2 – c2 2a ................. (iii) A a a B b D C 2 b 2 A h b a B a – x D x C c https://www.geogebra.org/m/ejsvzhgy Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 95 Vedanta Excel in Mathematics - Book 9 Again, from the equations (i) and (iii), we get, h2 = b2 – a2 + b2 – c2 2a 2 = b + a2 + b2 – c2 2a b – a2 + b2 – c2 2a = 2ab + a2 + b2 – c2 2a 2ab – a2 – b2 + c2 2a = (a + b)2 – c2 2a c2 – (a – b)2 2a = (a + b + c) (a + b – c) 2a (c + a – b) (c – a + b) 2a = (a + b + c) (a + b + c – 2c) (a + b + c – 2b) (a + b + c – 2a) 4a2 Now, substituting, a + b + c = 2s, we get, h2 = 2s (2s – 2c) (2s – 2b) (2s – 2a) 4a2 = 16s (s – a) (s – b) (s – c) 4a2 h = 2 a s(s – a) (s – b) (s – c) Then, area of D ABC = 1 2 base × height = 1 2 BC × h = 1 2 a × 2 a s(s – a) (s – b) (s – c) = s(s – a) (s – b) (s – c) Thus, when three sides of a triangle are given, its area A = s(s – a) (s – b) (s – c) Worked-out examples Example 1: Find the area of the triangle and the quadrilateral given below. Solution: a) In ∆ABC, s = 14 + 15 + 13 2 cm = 21 cm ∴ Area of ∆ABC= s (s – a) (s – b) (s – c) = 21 (21 – 15) (21 – 13) (21 – 14) = 21 × 6 × 8 × 7 = 7 × 3 × 3 × 2 × 2 × 4 × 7 = 7 × 3 × 2 × 2 = 84 cm2 . b) In ∆ABD, by Pythagoras theorem, we have, BD = AB2 + AD2 = 42 + 32 = 5 cm a) 14 cm A B C 15 cm 13 cm 13 cm A 12 cm C D B 3 cm 4 cm b) Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Now, area of ∆ABD = 1 2 × AB × AD = 1 2 × 4 × 3 = 6 cm2 In ∆BDC, s = 13 + 12 + 5 2 cm = 15 cm ∴ Area of ∆BDC = s (s – a) (s – b) (s – c) = 15 (15 – 13) (15 – 12) (15 – 5) = 15 × 2 × 3 × 10 = 30 sq. cm. Hence, area of quad. ABCD = Area of (∆ABD + ∆BDC) = 6 sq. cm. + 30 sq. cm = 36 sq. cm. Example 2: Find the area of the given parallelogram, in which AB = DC = 41 cm and AD = BC = 14 cm and diagonal BD = 28 cm. Solution: The semi-perimeter of ∆ABD is s = 41 cm + 28 cm + 15 cm 2 = 42 cm Now, area of ∆ABD = s (s – a) (s – b) (s – c) = 42 (42 – 41) (42 – 28) (42 – 15) = 42 × 1 × 14 × 27 = 126 cm2 ∴ Area of parm. ABCD = 2 × area of D ABD = 2 × 126 cm2 = 252 cm2 Example 3: The traffic signal board is an equilateral triangle with perimeter 180 cm, find its area. Solution: Here, the perimeter of equilateral traffic signal board = 180 cm or, 3a = 180 cm or, a = 60 cm ∴Semi-perimeter (s) = 180 cm 2 = 90 cm Now, area (A) = s (s – a) (s – b) (s – c) = 90 (90 – 60) (90 – 60) (90 – 60) = 90 × 30 × 30 × 30 = 2430000 = 1558.85 cm2 Hence, the area of the signal board is 1,558.85 cm2 Example 4: Find the area of an isosceles triangle whose base is 10 cm and length of each of the equal side is 13 cm Solution: Here, in an isosceles triangle, base (b) = 10 cm and equal sides (a) = 13 cm ∴Semi-perimeter (s) = 13 cm + 13 cm + 10 cm 2 = 18 cm Now, area of the triangle (A) = s (s – a) (s – b) (s – c) = 18 (18 – 13) (18 – 13) (18 – 10) = 18 × 5 × 5 × 8 = 3600 = 60 cm2 D 15 cm 28 cm 41 cm C A B Alternate process Area of the equilateral triangular board (A) = 3 4 a2 = 3 4 × (60 cm)2 =1558.85 cm2 10 cm 13 cm 13 cm Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 97 Vedanta Excel in Mathematics - Book 9 Alternatively, Area of isosceles triangle (A)= b 4 4a2 – b2 =10 4 4(13)2 – 102 =5 2 676 – 100 =5 2 ×24=60cm2 Hence, the area of isosceles triangle is 60 cm2 . Example 5: The perimeter of a triangle is 48 cm. If the ratios of its sides are 4 : 5 : 3, find: (i) the lengths of its sides (ii) its area Solution: Here, the perimeter of the triangle = 48 cm ∴ The semi-perimeter of the triangle = 48 2 cm = 24 cm Let the sides of the triangle are 4x cm, 5x cm and 3x cm. (i) Now,4x + 5x + 3x = 48 cm or, 12x = 48 cm or, x = 48 12 cm = 4 cm. Again, 4x = 4 × 4 cm = 16 cm, 5x = 5 × 4 cm = 20 cm and 3x = 3 × 4 cm = 12 cm. ∴ a = 16 cm, b = 20 cm and c = 12 cm. (ii) Area of the triangle = s (s – a) (s – b) (s – c) = 24 (24 – 16) (24 – 20) (24 – 12) = 24 × 8 × 4 × 12= 96 cm2 . Hence, the required area of the triangle is 96 cm2 . Example 6: The sides of a triangular garden are in the ratio 13:14:15. If the area of the garden is 336 m2 , find its perimeter. Solution: Let, the sides of the garden be a = 13x m, b =14x m and c = 15x m. Now, semi-perimeter (s) = 13x + 14x + 15x 2 = 42x 2 = 21x Also, Area of the garden = s (s – a) (s – b) (s – c) or, 336 m2 = 21x(21x – 13x)(21x – 14x)(21x – 15x) or, 336 = 21x × 8x × 7x × 6x or, 336 = 7056x4 or, 336 = 84x2 or, x2 = 4 or, x = 2 Thus, the sides of the garden are; a = 13x = 13 × 2 m = 26 m, b= 14x = 14 × 2m = 28m and c = 15x = 15 × 2m = 30m Hence, the perimeter of the garden is a+ b + c = 26 m + 28m + 30m = 84 m. Example 7: The umbrella given alongside is made by stitching 8 triangular pieces of cloth, each piece measuring 61 cm, 61 cm and 22 cm. (i) How much cloth is required for the umbrella? (ii) If the cost of 1 cm2 is Rs 0.25, find the total cost of cloth. Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 98 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, the sides of a triangular piece of cloth are a = 61 cm, b = 61 cm and c = 22 cm (i) Now, semi-perimeter (s) = 61 cm + 61 cm + 22 cm 2 = 144 cm 2 = 72m ∴ Area of a triangular piece = s (s – a) (s – b) (s – c) = 72(72 – 61)(72 – 61)(72 – 22) = 72 × 11 × 11 × 50 = 435600 = 660 cm2 Also, the area of cloths required to make the umbrella (A) = 8×660 cm2 = 5,280cm2 (ii) Again, the total cost of cloth (T) = Area (A) ×Rate (R) = 5280 ×Rs 0.25 = Rs 1,320 Hence, the total cost of cloth required to make the umbrella is Rs 1,320. Example 8: Mr. Pattel is a farmer in Janakpur. He has a field in the shape of a rhombus. The perimeter of the field is 800 m and one of its diagonal is 240 m. (1 hectare = 10,000 m2 ) (i) Find the length of each side of the field. (ii) Find the area of the field in square meter. (iii) Find the area of the field in hectare. Solution: Here, perimeter of the field in the shape of rhombus ABCD = 800m (i) Now, perimeter of field = 800 m or, 4× length of each side (l) = 800 m or, l = 200 m ∴AB = BC = CD = AD = 200 m (ii) In ∆BCD, semi-perimeter (s) = 200 m + 200 m + 240 m 2 = 640 m 2 = 320 m ∴ Area of ∆BCD = s (s – a) (s – b) (s – c) = 320(320 – 200)(320 – 200)(320 – 240) = 320 × 120 × 120 × 80 = 368640000 = 19,200 m2 Thus, the area of the field ABCD = 2 × Area of ∆BCD = 2 × 19,200 m2 =38,400 m2 (iii) Again, 10,000 m2 = 1 hectare ∴ 38,400m2 = 38400 10000 hectare = 3.84 hectares Hence, the area of his field is 3.84 hectares. EXERCISE 5.1 General section 1. a) What is the area of triangle whose base is x cm and height is y cm? b) Write down the area of right angled triangle in which base is m cm and perpendicular is n cm. A B C D 240 m Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 99 Vedanta Excel in Mathematics - Book 9 c) What is the area of equilateral triangle whose side is ‘x’ cm? d) If the length of each equal sides of an isosceles triangle is p cm and base is q cm, what is its area? e) The sides of a triangle are x, y and z units respectively, what is the semi-perimeter of the triangle? f) The semi-perimeter of triangle having side lengths a cm, b cm and c cm is s cm. Find its area. g) What is the area of triangle whose sides are p cm, q cm and r cm? 2. a) Find the area of the following triangles. (i) (ii) (iii) (iv) b) The edges of a triangular kitchen garden are 20 ft, 21 ft and 29 ft long. Calculate the area of the garden. c) The sides of the triangular park are 193 m, 194 m and 195 m. Find its area. d) Find the area an isosceles triangle in which each of equal sides is 25 cm and the base is 14 cm. e) Find the area of an equilateral triangle whose perimeter is 12 cm. Creative section - A 3. Find the area of the following quadrilaterals. a) b) c) d) e) f) 4. a) If the ratio of the sides of a triangle is 3: 4: 5 and its perimeter is 36 cm, find: (i) the measure of its sides (ii) its area b) The perimeter of a triangular park is 84 m and the ratio of its sides is 13: 14: 15, find: (i) the measure of its sides (ii) its area 13 cm 15 cm 14 cm P Q R A B C 10 cm 8 cm 6 cm 41 cm 41 cm 18 cm L N M 25 cm 12 cm X Y Z 17 cm 15 cm A B D C 14 cm 13 cm 48 cm 25 cm P Q R S 12 cm 25 cm 17 cm 10 cm 9 cm D A B C 30 cm 16 cm 17 cm W X Y Z D B C A 21 cm 12 cm 16 cm 13 cm A B C D 13 cm 21 cm 11 cm 20 cm Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) The ratio of the sides of a triangular field is 12:5:13 and its area is 120 m2 , find the perimeter of the field. d) The area of triangular kitchen garden is 324 sq. ft and its edges are in the ratio 9:10:17. Find the perimeter of the garden. Creative section - B 5. a) The shape of a piece of land is a parallelogram whose adjacent sides are 12 m and 17 m and the corresponding diagonal is 25 m. Find the area of the land. b) A park is in the shape of parallelogram. The lengths of two adjacent edges are 51m and 52 m and the corresponding diagonal is 53 m, what is the area of the park? c) A garden is in the shape of a rhombus whose each side is 15 m and its one of two diagonals is 24 m. Find the area of the garden. d) A farmer has a field in the shape of a rhombus. The perimeter of the field is 400m and one of its diagonals is 120m. Find the area of the field in hectare. (1 hectare = 10,000 m2 ) 6. a) An umbrella is made by stitching 8 triangular pieces of cloth of different colors as shown in the figure. Each piece measures 50 cm, 50 cm and 28 cm. (i) How much cloth of each color is required for the umbrella? (ii) If the rate of cost of the cloth is Rs 0.15 per sq. cm, find the cost of the cloths required to make the umbrella. b) Mrs. Limbu made a rangoli which has 8 identical triangles. Each triangle measures 20 cm, 13 cm and 11 cm. (i) How much surface of the floor is covered with the design? (ii) If the rate of cost of the colour is Rs 2.50 per sq. cm, find the cost of the colours required to make the design. c) A floral design on a floor of a room is made up of 16 triangular pieces of tiles. The sides of each piece of tile are 28 cm, 15 cm and 41 cm. (i) Find the area of the design. (ii) Estimate the cost of polishing the tiles at Rs 1.50 per sq. cm. 5.5 Local Land Measurement Units in Nepal In Nepal, people from different parts of the country have different culture, custom, life style and languages. Similarly, the people use different customary units to measure the area of land in different regions though the metric system has been the official standard since 1968 AD. In Terai region, Bigha-Kattha-Dhur measurements are common to measure the area of land while Ropani-Aana measurements are common in hilly and mountainous regions. 20cm 11cm 13cm Mensuration (I): Area