Vedanta Excel in mathematics Book 10 for TARA SIR (1)-1-106_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 51 Vedanta Excel in Mathematics - Book 10 Compound interest Example 16: The yearly compound interest on a sum of money for two successive years are Rs 320 and Rs 336 respectively. (a) Calculate the rate of interest (b) Find the original sum. Solution: Here, difference between the C. I. of two successive years= Rs 336 – Rs 320 = Rs 16 ∴ Rs 16 is the simple interest of one year on Rs 320. (a) Rate of interest= I × 100 P × T = 16 × 100 320 × 1 = 5% p.a. (b) Again, the interest of the original sum (P) for the first year = Rs 320 P 1 + R 100 T – 1 = Rs 320 or, 1 + 5 100 1 – 1 = Rs 320 or, P 20 = Rs 320 or, P = Rs 6,400 Hence, the rate of interest is 5% p.a. and the original sum is Rs 6,400. Example 17: The compound interest calculated yearly on a certain sum of money for the second year is Rs 1,650 and for the third year is Rs 1,815. (a) Calculate the rate of interest. (b) Find the original sum of money. Solution: Here, the difference between the C. I. of two successive years = Rs 1,815 – Rs 1,650 = Rs 165 ∴ Rs 165 is the interest on Rs 1,650. (a) Now, rate of interest (R) = I × 100 P × T = 165 × 100 1650 × 1 = 10% p.a. (b) C.A. for the first year = P 1 + R 100 T = P 1 + 10 100 1 = 11P 10 So, 11P 10 is the principal for the second year. ∴ C. I. in the second year = 11P 10 1 + R 100 T – 1 = 11P 10 1 + 10 100 1 – 1 = 11P 100 From the question, 11P 100 = Rs 1,650 or, P = Rs 15,000 Hence, the rate of interest is 10% p.a. and the original sum is Rs 15,000. EXERCISE 2.1 General section 1. a) If C.A. and C.I. are the compound amount and compound interest of a sum P in T years at R% p.a. respectively, write the relationships among the following variables. (i) P, T, R, and C.A. (Compounded annually) (ii) P, T, R, and C.I. (Compounded annually)

Vedanta Excel in Mathematics - Book 10 52 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest (iii) P, T, R, and C.A. (Compounded semi-annually) (iv) P, T, R, and C.I. (Compounded semi-annually) (v) P, T, R, and C.A. (Compounded quarterly) (vi) P, T, R, and C.I. (Compounded quarterly) b) The compound amount on a sum P in T years M months at R % p.a. is C.A. Write the relation among P, T, M, R, and C.A. compounded annually. c) If C.I. is the compound interest of a sum P in T years M months at R % p.a., write the relation among P, T, M, R and C.I. compounded annually. d) If C.A. is the compound amount of a sum P at the different rates R1 %, R2 %, and R3 % in the first, the second and the third years respectively. Write the relation among C.A., R1 , R2 , and R3 (compounded annually). 2. Calculate the compound interest without using the formula. a) P = Rs 5,000, T = 2 years, R = 4% b) P = Rs 15,000, T = 3 years, R = 10% 3. a) Find the interest compounded annually and the simple interest of the sum of Rs 2,000 in 1 year at 5 % p.a. Are both types of interest same? Write your conclusion. b) Find the amount compounded annually. (i) Principal (P) = Rs 4,000, time (T) = 2 years, rate (R) = 5 % p.a. (ii) Principal (P) = Rs 7,500, time (T) = 2 years, rate (R) = 12 % p.a. (iii) Principal (P) = Rs 12,000 , time (T) = 1 year 4 months, rate (R) = 6% p.a. (iv) Principal (P) = Rs 3,125, time (T) = 11 2 years, rate (R) = 8 % p.a. c) Find the interest compounded annually. (i) Principal (P) = Rs 8,000, time (T) = 3 years, rate (R) = 5 % p.a. (ii) Principal (P) = Rs 16,000, time (T) = 2 years, rate (R) = 10 % p.a. (iii) Principal(P) = Rs 6,000, time(T) = 2 years 6 months, rate(R) = 10 % p.a. (iv) Principal (P) = Rs 3,125, time (T) = 11 2 years, rate (R) = 8 % p.a. d) Find the amount compounded half-yearly (semi-annually). (i) Principal (P) = Rs 5,000, time (T) = 1 year, rate (R) =4 % p.a. (ii) Principal (P) = Rs 16,000, time (T) = 11 2 year, rate (R) = 10 % p.a. e) Find the interest compounded semi-annually. (i) Principal (P) = Rs 2,500, time (T) = 1 year, rate (R) = 8 % p.a. (ii) Principal (P) = Rs 40,000 time (T) = 11 2 years, rate (R) = 10 % p.a. f) Find the amount compounded quarterly. (i) Principal (P) = Rs 20,000, time (T) = 1 year, rate (R) =12 % p.a. (ii) Principal (P) = Rs 16,000, time (T) = 6 months, rate (R) = 10 % p.a. g) Find the interest compounded quarterly. (i) Principal (P) = Rs 30,000, time (T) = 1 year, rate (R) =12 % p.a. (ii) Principal (P) = Rs 75,000, time (T) = 9 months, rate (R) = 10 % p.a. www.geogebra.org/classroom/ra7tqqry Classroom code: RA7T QQRY Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 53 Vedanta Excel in Mathematics - Book 10 Compound interest Creative section-A 4. a) Mrs. Tharu deposited Rs 65,000 in Lumbini Bank at the rate of 10% p.a. for 1 year. How much amount will she get under the following cases? (i) The bank pays the interest compounding yearly. (ii) The bank pays the interest compounding half-yearly. (iii) The bank pays the interest compounding quarterly. b) Minakshi invested Rs 85,000 for 1 year in Goodwill Finance at the rate of 8% per annum. (i) How much interest will she receive if it is compounded in each year? (ii) How much interest will she receive if it is compounded in each 6 month? (iii) How much interest will she receive if it is compounded in each 3 month? 5. a) Samriddhi has deposited Rs 40,000 in Lumbini Bank at 10% p.a. for 1 year. (i) Write the formula to find the yearly compound amount on a sum P for T years at R% per annum. (ii) Calculate the yearly compound amount. (iii) Find the amount if the bank pays the interest compounding half-yearly. (iv) Find the difference between the amounts compounded yearly and quarterly. b) Bishwant took a loan of Rs 50,000 at the rate of 6% p.a. for 1 year from Agriculture Development Bank to promote his vegetable farming. (i) What does P represent in the usual notation, C.A.= P 1 + R 100 T ? (ii) Find the annual compound amount that he has to pay to the bank. (iii) How much amount should he pay to the bank according to the semi-annual compound interest system? (iv) Find the difference between the amounts compounded semi-annually and quarterly. 6. a) You are going to deposit Rs 1,00,000 for 2 years in a bank. You have got the following information about the rate of interest given by two banks. Bank- A Yearly compound interest rate: 9% p.a. Bank- B Half-yearly compound interest rate: 8% p.a. (i) Which formula should be used to find the interest on a sum P for T years at R% per annum compounded annually? (ii) Calculate the interest earned by saving the money in bank-A. (iii) In which bank would you deposit the money and why? Write with reason. b) Your sister is going to deposit Rs 2,00,000 for 2 years in a bank. She has got the following information about the rate of interest given by two banks. Bank- P Yearly compound interest rate: 5% p.a. Bank- Q Half-yearly compound interest rate: 4% p.a. (i) Write the formula which is used to find the interest compounded half-yearly. (ii) Find the interest that she earned by saving the money in bank-P. (iii) Which bank would you suggest her to deposit the money and why? Write with reason.

Vedanta Excel in Mathematics - Book 10 54 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest 7. a) Balen wants to deposit Rs 90,000 in a bank for 2 years. The bank offers 10% per annum compound interest with three options (annually compound interest, semi-annually compound interest and quarterly compound interest). (i) Among the three options, identify the best one from which he gets more interest. (ii) How much amount will he receive after 2 years according to the interest compounding semi-annually? (iii) At the end of 1 year, if Balen withdraws the amount received according to semiannual compound interest and deposits it for the rest of the period to receive quarterly compound interest, how much amount will he get at the end? b) A retired service man deposited Rs 50,000 in his fixed deposit account in a commercial bank for 2 years at 10% per annum half yearly compound interest. (i) How many times is the interest calculated in the case of half yearly compound interest in 2 years? (ii) Calculate the compound interest received by the man at the end of 2 years. (iii) In how many years will the compound amount of Rs 50,000 be Rs 66,550 at the same rate of yearly compound interest? 8. a) Mr. Waiba borrowed a loan of Rs 80,000 from Mr. Pariyar at the rate of 10% p.a. simple interest for 2 years. He immediately deposited the same sum at the same rate of yearly compound interest for the same duration of time in a bank. (i) Define compound interest. (ii) Find the compound interest received by Mr. Waiba after 2 years. (iii) After paying the principal and interest to Mr. Pariyar, how much money is left with Mr. Waiba? b) Sunayana borrowed a sum of Rs 25,000 from Shashwat at 12 % p.a. yearly compound interest for 1 year 6 months and lent to Bishwant at the same rate of compound interest compounded half-yearly for the same interval of time. (i) Calculate the interest that Sunayana has to pay to Shashwat. (ii) How much profit did she make? (iii) By how much more would she gain if Bishwant paid the interest compounding quarterly rather than compounding semi-annually at the same rate? 9. a) Januka lends Rs 75,000 to Manisha for 2 years. They make an agreement that the rate of interest for the first year is 8% and for the second year is 10%. (i) Write the formula to find the amount compounded annually for 2 years when the rates of interest of the first year is R1 % and the second year is R2 %? (ii) How much amount will Manisha pay to Januka to clear the debt? (iii) How much more or less amount would Manisha have paid to Januka if the the rate of interest was 9% p.a. for 2 years? b) Swornim deposited Rs 1,00,000 for 3 years in a bank. The bank pays the interest at 4 %, 5 % and 7 % in the first, second and the third years respectively. (i) Write the formula to find the amount compounded annually for 3 years when the rates of interest of the first, second and third years are R1 %, R2 % and R3 % respectively. (ii) How much amount will he withdraw after 3 years? (iii) How much more or less amount would he receive if the bank paid him the interest compounded yearly at 6% p.a. for 3 years?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 55 Vedanta Excel in Mathematics - Book 10 Compound interest 10. a) The simple interest on a sum of money for 2 years at 8% p.a. is Rs 8,000. (i) Find the principal. (ii) Calculate the compound interest on the same sum and at the same rate for the same duration, if the interest is reckoned half-yearly. b) A person paid Rs 6,200 compound interest on a certain sum of money that he borrowed from a bank at 10% p.a. compounded annually for 11 2 years. (i) Calculate the sum. (ii) Find the compound interest on the same sum and at the same rate for the same period of time, if the interest had to be paid semi-annually. 11. a) The compound interest on the sum of money at 8 % p.a. for 2 years is more than the simple interest on the same sum at the same rate for the same time by Rs 76.80. (i) Find the sum. (ii) Find the compound interest compounded annually. b) Kaji Basnet borrowed a certain sum of money from Roshan Shrestha at the rate of 9 % p.a. simple interest for 2 years. He immediately lent this money to Raju Jha at the same rate of compound interest for the same period of time. The interest paid by Kaji is Rs 243 less than the interest paid by Raju. (i) Calculate the sum. (ii) How much interest did Raju Jha pay to Kaji Basnet? 12. a) The difference between the annual and semi-annual compound interest at the rate of 10% per annum for 2 years is Rs 881. (i) Write the formula to find the semi-annual compound interest at the given rate for 1 year. (ii) Find the principal. (iii) Find the interest of the first year compounded quarterly. b) The difference between the interest compounded quarterly and semi-annually at the rate of 20% p.a. for 1 year is Rs 2643. (i) What type of interest is calculated by the formula, C.I. = P 1 + R 400 4T – 1 ? (ii) Find the principal. (iii) Find the interest compounded annually on the same sum at the same rate and time. 13. a) Saleen deposits Rs 5,000 in a bank at 12% p.a. interest compounded annually. (i) In how many years will Rs 5,000 amount to Rs 6,272 at the given rate of interest? (ii) What will be the amount compounded half-yearly on Rs 5,000 at the same rate for the same time? b) Mrs. Dhital deposited Rs 10,000 in her fixed deposit account of a bank at 10% p.a. compound interest. (i) In how many years would she withdraw a compound amount of Rs 12,100? (ii) Find the amount compounded half-yearly on Rs 10,000 at the same rate for the same time? 14. a) Rajesh took a loan of Rs 16,000 from his friend Gopal at the rate of 5 paisa per rupee per year interest compounded annually. (i) Find the rate of interest in percent per year. (ii) In how many years would he pay a compound interest of Rs 2,522 to Gopal?

Vedanta Excel in Mathematics - Book 10 56 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Compound interest b) Amrita borrows a loan of Rs 55,000 from the Agriculture bank. The rate of compound interest is 4 paisa per rupee per year. (i) Find the rate of interest in percent per year. (ii) In how many years does she pay the compound interest of Rs 4,488? 15. a) Rita is a school teacher. She deposited Rs 80,000 in her fixed deposite account at compound interest compounded annually for 2 years for the expenses of her son's study and the compound amount at the end of one year is Rs 88,000. (i) Write the formula to find yearly compound amount for principal Rs 'P', time 'T' years and rate of interest 'R' % p.a. (ii) Find the annual rate of compound interest. (iii) How much compound amount does she get at the end of 2 years? b) Mrs. Bajracharya invests Rs 8,000 for 3 years at a certain rate of interest compounded annually. At the end of one year the sum amounts to Rs 8,800. (i) Find the rate of interest. (ii) Calculate the amount at the end of second year. (iii) Find the amount at the end of third year. c) (i) At what rate percent per annum compound interest does Rs 3,125 amount to Rs 3,380 in 2 years? (ii) What will be the quarterly compound amount on Rs 50,000 at the same rate during the same time? 16. a) If the compound amounts of a sum of money in 2 years and 3 years are Rs 17,640 and Rs 18,522 respectively, find: (i) the rate of interest (ii) Find the sum. b) If a sum becomes Rs 6,655 in 3 years and Rs 7,320.50 in 4 years interest being compounded annually, find: (i) the rate of interest (ii) Find the sum. 17. a) The compound interest of a sum of money in 1 year and 2 years are Rs 450 and Rs 945 respectively. (i) Find the rate of interest compounded yearly. (ii) Find the sum. b) The compound interest of a sum of money in 1 year is Rs 350 and in 2 years is Rs 724.50. (i) Find the rate of interest compounded yearly. (ii) Find the sum. 18. a) Roshan borrowed Rs 5,00,000 from a commercial bank at the rate of 10% p.a. compounded annually for 2 years. After one year, the bank changed its policy to pay the interest compounded half-yearly at the same rate. (i) How much interest should he have to pay in the first year? (ii) How much interest should he have to pay in the second year? (iii) How much less interest would he have to pay if the interest rate was not changed? b) Gita borrowed Rs 85,000 from a bank at the rate of 12% p.a. compounded semiannually for 2 years. After one year, the bank changed its policy to charge the interest compounded quarterly at the same rate. (i) How much interest will she have to pay in the first year? (ii) How much interest will she have to pay in the second year? (iii) How much less interest would she have to pay if the interest rate was not changed?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 57 Vedanta Excel in Mathematics - Book 10 Compound interest 19. a) Mr. Lamichhane earned Rs 4,00,000 as commission from the sale of Mrs. Shah's land plot. Immediately, he deposited the commission amount in the fixed deposit account in Nepal Bank for two years at the rate of 10% per annum interest being compounded semi-annually. The bank charges 5% tax on the interest. After one year, the bank changed its policy and decided to calculate the interest quarterly. (i) Name the system of calculating interest in each 6 months on the amount of just previous 6 months as principal. (ii) Find the net interest that Mr. Lamichhane received in the first year after deducting the tax. (iii) Compare the net interests of the first and second years. b) Mrs. Lama got Rs 8,00,000 by selling her second hand car. She deposited the whole money in the fixed deposit account in a bank for two years at the rate of 8% per annum interest being compounded half yearly. But after one year the bank changed its policy and decided to calculate the interest quarterly. The bank charges 5% tax on the interest. (i) What type of interest is calculated in each 4 months by taking the amount of just previous 4 months as principal? (ii) Calculate the net interest received by Mrs. Lama in the first year after deducting the tax. (iii) Compare the net interests of the first and second years. 20. a) Mr. Ghising took a loan of Rs 2,50,000 from a bank for 2 years at the rate of 10% annual compound interest. After a year, he paid back Rs 1,00,000 in order to reduce the debt. (i) How much did he pay at the end of second year to clear his debt? (ii) Find the total interest paid by him in two years. (iii) If he had paid the loan only at the end of the second year, how much less or more interest would he have to pay? b) Mrs. Dahal borrowed Rs 3,00,000 from a Finance Company for 2 years at the rate of 12% annual compound interest. After a year, she paid back Rs 2,00,000 in order to reduce the debt. (i) How much did she pay at the end of second year to clear her debt? (ii) Find the total interest paid by her in two years. (iii) If she had paid the loan only at the end of the second year, how much less or more interest would she have to pay? Creative section-B 21. a) If a sum of money becomes Rs 1,40,450 in 1 year and Rs 1,48,877 in 11 2 years, interest being compounded semi-annually, calculate the rate of interest and the sum. b) According to half-yearly compound interest system, the interest of a sum of money in 6 months is Rs 2,200 and in 1 year is Rs 4,510 respectively. Find the rate of interest and the sum. 22. a) The yearly compound interest on a sum of money for two successive years are Rs 400 and 440 respectively. Calculate the rate of interest and the original sum. b) The compound interest calculated yearly on a certain sum of money for the second year is Rs 1,320 and for the third year is Rs 1,452. Calculate the rate of interest and the original sum of money.

Vedanta Excel in Mathematics - Book 10 58 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 23. a) Krishna lend altogether Rs 10,000 to Gopal and Radha for 2 years. Gopal agrees to pay simple interest at 12 % p.a. and Radha agrees to pay compound interest at the rate of 9 % p.a. If Radha paid Rs 596.70 more than Gopal as the interest, find how much did Krishna lend to each. b) Suntali deposited Rs 9,000 altogether in her saving account and fixed deposit account in a bank. Saving account gives her 5% p.a. interest compounded annually and fixed deposit account gives 10% p.a. interest compounded halfyearly. If she got Rs 160 more interest from fixed deposit account at the end of one year, find how much money did she deposit in her each account? 24. a) A bank has fixed the rate of interest 10% p.a. semi-annually compound interest in account X and 12% p.a., annually compound interest in account Y. If you are going to deposit Rs 50,000 for 2 years, in which account do you deposit and why? Give your reason with calculation. b) A business person borrowed Rs 1,20,000 from a commercial bank at the rate of 10% p.a. compounded annually for 2 years. After one year the bank changed it’s policy to pay the interest compounded semi-annually at the same rate. What is the percentage difference between the interest of the first year and second year? Give reason with calculation. c) Sujita deposited Rs 4,00,000 in fixed deposit in a commercial bank for 2 years at 10% p.a. compounded half yearly. After 1 year the bank changed its policy and decided to give compound interest compounded quarterly at the same rate. The bank charged 5% tax on the interest as per government’s rule. What is the percentage difference between the interest of the first and second year after paying tax. Multiple Choice Section 25. Tick the correct alternative. (a) The interest compounded semi-annually on Rs. P for T years at R% p.a. is (A) CI = P 1 + R 100 T –1 (B) CI = P 1 + R 200 2T –1 (C) CI = P 1 + R 300 3T–1 (D) CI = P 1 + R 400 4T –1 (b) If a sum P is placed at the rate of R% compound interest for T years, then the relation C.A. = P 1 + R 400 4T is applicable to calculate (A) amount compounded annually (B) amount compounded semi-annually (C) amount compounded quarterly (D) amount compounded monthly (c) Which of the following statements is true for a sum at a certain rate for a year? (A) The interest compounded annually is equal to simple interest. (B) The interest compounded annually is equal to the interest compounded semiannually. (C) The interest compounded annually is equal to the interest compounded quarterly. (D) The interest compounded semi-annually is equal to the interest compounded quarterly. (d) What will be compound amount when Rs. 5,000 is invested for 2 years at 10% p.a.? (A) Rs 6,655 (B) Rs 6,050 (C) Rs 5,500 (D) Rs 6,077.53 (e) How long will it take to get interest Rs 1,155 compounded annually on Rs 5,500 at the rate of 10% per annum? (A) 1 year (B) 2 years (C) 18 months (D) 3 years

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 59 Vedanta Excel in Mathematics - Book 10 3.1 Population growth - introduction Let’s discuss upon the following questions in order to make the sense on population growth. 1. a) The population of a municipality in the beginning of B.S. 2081 is 1,00,000 and it is estimated that the population of the city increases every year by 10%. (i) What will be the population of the city in the beginning of B. S. 2082? (ii) What will be the population of the city in the beginning of B. S. 2083? b) Can we calculate the population of the municipality by using the formula of compound amount? 2. According to Nepal Census 2078 B.S., the population of Nepal is 2,91,92,480 and the annual growth rate is 0.93%. What is the estimated population of Nepal of 2079 B.S.? Population growth is the increase in the number of individuals in a population. Many of the world’s countries have seen a sharp rise in population. However, it is found to remain nearly constant or slightly decreasing in some other countries. The way of increasing population is exactly similar to the way of increasing compound amount because every year population increases from the previously increased population. Population of the previous date (P) Population after T years (Pt ) To find the population of a place after certain time, we apply the formulae under the following cases. Case I: If the population of a country or a place at a certain time = P, the rate of growth of population = R % p.a., the population of the country or place after T years = Pt , Then, PT = P T 1 + R 100 Similarly, the increased population in T years = PT – P = P T 1 + R 100 – P = P 1 + R 100 T – 1 Unit 3 Population Growth and Depreciation

Vedanta Excel in Mathematics - Book 10 60 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation Case II: The growth of population is also affected by the number of deaths, in-migrants and out-migrants at the end of the given period of time. If Min and Mout are the number of in-migrants and out-migrants at the end of the given period of time respectively, ∴The actual population in T years, (PT) = P T 1+ R 100 – no. of deaths+Min– Mout Case III: If the rate of growth of population every year is different, i.e., R1 , R2 , ..., RT are the rates of growth of population in the first year, second year, and ... Tth years then the population after T years is calculated by using the following formula, PT = P 1 + R1 100 1 + R2 100 ... 1 + RT 100 Case IV: If the population of place is P and it decreases every year by R%, then, the population of the place after T years, PT = P T 1 – R 100 Facts to remember 1. Population growth rate (R) = birth rate – death rate 2. Population growth rate (R) = birth rate + rate of in-migrants during the given time Worked-out Examples Example 1: The population of a municipality before 2 years was 60,000 and the rate of annual growth of population is 2%. (a) If the initial population of a place is P, population growth rate is R% p.a. and the population after T year is PT , write the relationship among PT , P, R and T. (b) Find the present population of municipality. (c) If the numbers of in-migrants and out-migrants at the end of 2 years were 750 and 410 respectively and 620 people died within this time interval, what change would be in present population of the municipality? Solution: Here, the population of the municipality before 2 years (P) = 60,000 The rate of growth of population (R) = 2% (a) The relationship among PT , P, R and T is, PT = P T 1 + R 100 (b) The present population, PT = P T 1 + R 100 = 60,000 2 1 + 2 100 = 62,424 (c) The number of in-migrants (Min) = 750 The number of out-migrants (Mout) = 410 The number of deaths (D) = 620

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 61 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation Now, the present population of the municipality = 62424 – D + Min – Mout = 62424 – 620 + 750 – 410 = 62,144 Also, difference = 62,424 – 62,144 = 280 Hence, the present population of the municipality would be less by 280. Example 2: The present population of a Metropolitan city is 16,87,500. The rate of growth of population of the city is 4% per year. (a) Write the formula to find the increased population in T years? (b) What will be the increased population of the city in 2 years? (c) What will be increased population at the end of third year if the population growth rate after 2 years is 5% per annum? Solution: Here, the present population of the city (P) = 16,87,500 The rate of growth of population (R) = 4% p.a. Time (T) = 2 years (a) The formula to find the increased population is, IP = P 1 + R 100 – 1 T (b) Now, the increased population of the city in 2 years = 16,87,500 1 + 4 100 – 1 2 = 1,37,700 Hence, the increased population of the Metropolitan city in 2 years is 1,37,700. (c) For the third year; initial population (P) = 16,87,500 + 1,37,700 = 18,25,200 Population growth rate (R) = 5% and time (T) = 1 year Again, the increased population of the city in 3rd year = P 1 + R 100 – 1 T = 18,25,200 1 + 5 100 – 1 1 = 91,260 Hence, the population of the Metropolitan city will be increased by 91,260 at the end of the third year. Example 3: Two years ago the population of a village was 40,000. If the birth rate is 4.5% p.a. and death rate is 2% p.a., answer the following questions. (a) Find the annual population growth rate of the village. (b) Find the present population of the village. Solution: Here, the population of the village before 2 years (P) = 40,000 (a) The annual growth rate of the population (R) = 4.5% – 2% = 2.5% (b) Now, the present population (PT) = P T 1 + R 100 = 40,000 2 1 + 2.5 100 = 42,025 Hence, the present population of the village is 42,025. Example 4: The population of a village increases every year by 5%. At the end of two years, if 1,025 people migrated to other places and the population of the village remained 10,000, what was the population of the village in the beginning? Solution: Here, the total population including the migrated number = 10,000 + 1,025 = 11,025 We know that PT = P T 1 + R 100

Vedanta Excel in Mathematics - Book 10 62 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation or, 11,025 = P 2 1 + 5 100 or, 11,025 = 1.1025 P or, P = 11025 1.1025 = 10,000 Hence, the population of the village in the beginning was 10,000. Example 5: The population of a town before 3 years was 1,50,000. If the annual growth rates of the population in the last 3 years were 2%, 4% and 5% respectively every year, find the population of the town at the end of 3 years. Solution: Here, the population of the town before 3 years (P) = 1,50,000 The annual growth rates in every year were R1 = 2%, R2 = 4% and R3 = 5% Now, the population of the town at the end of 3 years is PT = P 1 + R1 100 1 + R2 100 1 + R3 100 = 1,50,000 1 + 2 100 1 + 4 100 1 + 5 100 = 1,50,000 × 1.02 × 1.04 × 1.05 = 1,67,076 Hence, the population of the town at the end of 3 years is 1,67,076. Example 6: In the beginning of 2021 A.D., the population of a town in Turkey was 50,000. The population growth rate of the town is 2% p.a. In the beginning of 2023 A.D. 3,020 people died due to the earth quake. Find the population of the village in the beginning of 2024 A.D. Solution: Here, Population of the town in the Turkey in the beginning of 2021 A.D. (P) = 50,000 Time (T) = 2 years Population of the town at the end of 2 years (PT) = ? We have PT = P T 1 + R 100 = 50,000 2 1 + 2 100 = 50,000 (1.02)2 = 52,020 After 2 years, 3,020 people died. So, the population of the town in the beginning of 2023 A.D. = 52,020 – 3,020 = 49,000 Again, the population in the beginning of 2024 A.D. = P T 1 + R 100 = 49,000 1 1 + 2 100 = 49,000 × 1.02 = 49,980 Hence, the population of the town in the beginning of 2024 A.D. is 49,980. Example 7: In the beginning of 2078 B.S. the population of a city was 5,00,000 and it was 6,65,500 at the end of 2080 B. S. (a) Find the annual growth rate of population. (b) What will be the population of the city at the end of 2082 B.S. at the same rate of growth of the population?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 63 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation Solution: Here, the population of the city in the beginning of 2078 B.S. (P) = 5,00,000 Population of the city at the end of 2080 B.S. (Pt ) = 6,65,500 Time (T) = 3 years The rate of growth of population (R) = ? (a) We have PT = P T 1 + R 100 = 5,00,000 3 1 + R 100 or, 6,65,500 = 5,00,000 100 + R 3 100 or, 100 + R 3 100 = 665500 500000 = 11 3 10 or, 100 + R 100 = 11 10 or, R = 10% (b) Again, the population of the city at the end of 2080 B.S. (P)= 6,65,500 Population of the city at the end of 2082 B.S. (Pt ) = ? Time (T) = 2 years and Rate (R) = 10% p.a. Then,Pt = P T 1 + R 100 = 665500 2 1 + 10 100 = 665500 × 1.21= 8,05,255 Hence, the population of the city at the end of 2082 B.S. will be 8,05,255. EXERCISE 3.1 General section 1. a) If the initial population of a place is P and the rate of population growth is R% p.a., write the formula to find the population of the place after T years. b) If the initial population of a region is P and the rate of growth of population is R% p.a., write the formula to find the increased population in T years. c) The rates of annual growth of population of a town in 3 years are R1 %, R2 %, and R3 %. If the present population of the town is P, write the formula to find the population after 3 years. 2. a) The population of a town in the beginning of 2080 B.S. was 32,400. Within a year the population increased 3% by birth and 2% by immigration. Find the population of the village in the beginning of 2081 B.S. b) The present population of a village is 30,000. If it is increased at the rate of 10 % per annum, what will be the population after 2 years? c) After two years the population of a town will be 33,620 at the population growth rate of 2.5 % p.a., find the present population of the town. d) A city has population 1,75,760. If its population growth rate is 4 % p.a., find its population before 2 years. 3. a) The rate of growth of a plant is 5% every month. If the height of the plant in the beginning of Baisakh 2081 is 20 cm, find its height at the end of Asar 2081.

Vedanta Excel in Mathematics - Book 10 64 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation b) A house owner made an agreement to increase the house rent by 10% every year. If the rent of the house this year is Rs 16,000, find the house rent after 3 years. c) The growth rate of a certain type of useful bacteria is 10% per day. During a research work in a laboratory, if the bacteria are grown upto the number of 2.662 × 1012 in 3 days, how many bacteria were there 3 days ago? d) Due to the annual increment of price of land in a rapidly growing area by 25%, the present value of a piece of land is Rs 12,50,000 per ropani. How much was the value of the land before 4 years? 4. a) In how many years will the population of a town be 2,09,475 from 1,90,000 at the growth rate of 5% per annum? b) The population of a town is 96,250. In how many years would it be 1,04,104 if the population increases at the rate of 4% every year? 5. a) The population of a village increased from 10,000 to 11,000 in one year. Find the rate of growth of population. b) One year ago, the population of a village was 10,000. If the present population of the village is 10,210, find the population growth rate. Creative section A 6. a) The present population of a rural municipality is 30,000. The annual population growth rate of the place is 10%. (i) What does P denote in the population of a place after T year, PT= P T 1 + R 100 ? (ii) What will be the population of the village after 2 years? (iii) If the numbers of in-migrants and out-migrants at the end of 2 years were 50 and 80 respectively and 30 people died within this time interval, what change would be in population after 2 years in the rural municipality? b) 3 years ago, the population of Janakpur Sub-metropolis was 2,00,000 and the rate of annual growth of population is 5% per annum. (i) If the initial population of a place is P, population growth rate is R% p.a. and the population after T year is PT , write the relationship among PT , P, R and T. (ii) Find the present population of Janakpur Sub-metropolis. (iii) If the numbers of in-migrants and out-migrants at the end of 3 years were 850 and 540 respectively and 700 people died within this time interval, what change would be in present population of the Janakpur Sub-metropolis? 7. a) 2 years ago, the population of Bharatpur Metropolitan was 3,75,000. The rate of population growth of that Metropolitan is 4%. (i) Write the formula to find the increased population of a place after T years. (ii) Find the increased population of the metropolitan in two years. (iv) Find the difference between the population of the beginning of the last year and beginning of this year of the metropolitan. b) According to the students enrollment statistics published in 2079 B.S., the number of students admitted in 2078 B.S. in a region was 3,20,000. The annual rate of growth of the number of students is 5%. (i) Identify whether the population growth trend is compound or simple. (ii) How many new students were admitted by 2081 B.S.? (iii) How many more students were admitted in 2081 B.S. than in 2080 B.S.?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 65 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation 8. a) The present population of a Tansen municipality is 50,000. It is estimated that the population of the municipality increases by 4% and 5% respectively in next 2 years. (i) If the population growth rate of a certain place for first two years be R1 % and R2 % respectively, write the formula to find the population after 2 years. (ii) What will be the population of the municipality after two years? (iii) If the population of the municipality were increased at the rate of 5% per annum during the 2 years, by how many would the population of the municipality be more or less? b) The present population of Lumbini province is approximately 52,00,000. The population increases 1.25% by birth and 0.25% by immigration in the province. (i) Find the annual population growth rate. (ii) What will be the population of the province after 2 years? (iii) If the population was increased by 0.75% by immigration and 1.25% by birth during 2 years in the province, how many more people would be increased? 9. a) The present population of a village is 2,500 and its annual growth rate is 2%. (i) What does PT = P T 1 + R 100 represent in usual notation? (ii) In how many years, will the population of the village become 2,601? (iii) What will be the population of a town after the same period of time if the present population of the town is 1,10,000 and population growth rate is 3% p.a.? b) The present population of Chandragiri municipality is approximately 1,40,000 and the population increases at the rate of 4% every year. (i) In how many years, will the population of the municipality become 1,51,424? (ii) If the present population of Godawari municipality is approximately 1,00,000 and its population increases by 2% every year, what will be the population of the municipality after the same period of time? 10. a) At present, the population of Neelakantha municipality is 77,500 and the population of the municipality is estimated to be 83,824 in two years. (i) What is the population growth rate? (ii) Find the population of the municipality after one year at the same growth rate. b) The population of a municipality in the beginning of 2078 B.S. was 1,80,000 and at the end of 2080 B.S. was 2,39,580. (i) Find the rate of growth of population per year. (ii) Find the population of the municipality in the beginning of 2079 B.S. at the same growth rate. c) Nabin returned back from South Korea and deposited 60 lakh rupees in a bank at the rate of 5% p.a. annual compound interest for 2 years. At the end of 2 years, he withdrew the whole amount received from the bank and bought a plot of land. (i) How much amount did he withdraw at the end of 2 years? (ii) If the value of the land became Rs 88,04,565 in 3 years, find the annual growth rate of value of plot of the land. Creative section-B 11. a) The population of a village increases every year by 5%. At the end of two years, 460 people were migrated to other village and the population of the village remained 26,000. What was the population of the village in the beginning?

Vedanta Excel in Mathematics - Book 10 66 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation b) The population of a town increases every year by 10%. At the end of two years, if 5,800 people were added by migration and the total population of the town became 30,000, what was the population of the town in the beginning? 12. a) In the beginning of 2078 B.S., the population of a town was 1,00,000 and the rate of growth of population is 2% every year. If 8,000 people migrated there from different places in the beginning of 2079 B.S., what is the population of the town in the beginning of 2081 B.S.? b) In the beginning of the year 2020, the population of a village was 25,000 and the rate of growth of population is 10% every year. In the beginning of 2021, if 500 people migrated to other places, what was the population of the town in the beginning of 2024? 13. a) In the beginning of 2021 A.D. the population of a city was 2,50,000 and it was 3,02,500 at the end of 2022 A.D. What will be the population of the city at the end of 2024 A.D. at the same rate of growth of the population? b) In the beginning of 2079 B.S. the population of a municipality was 15,625 and it was 16,900 at the end of 2080 B. S. What will be the population of the municipality at the end of 2081 B.S. at the same rate of growth of the population? Project Work and Activity Section 14. a) Make the different groups of your friends. Visit different parts of your Ward and collect the statistics of the present population of your Ward. Visit to the concerned Ward Administration Office and get the statistics of the population of the Ward in the last census report. Calculate the rate of growth of population of your Ward. b) How many students were there in your school before 2 years and how many students are there in this year? Collect these data from your school administration office and calculate the growth rate of students in your school. c) Visit the available website and search the rate of growth of population of various countries in the world. Compare these rates with the rate of growth of population of our country. Multiple Choice Questions Section 15. Tick (√) the correct option (a) If the population of a municipality increases 2% by birth and 3% by immigration every year, what is the rate of growth of population? (A) 5% p.a. (B) 6% p.a. (C) 1% p.a. (D) 4% p.a. (b) The present population of a place is P and the annual population growth rate is R%, what will be the population of the place after T? (A) P T 1 + R 100 (B) P 2T 1 + R 200 (C) P T 1 – R 100 (D) P 2T 1 – R 200 (c) The initial population of a place is Po and the population growth rate is R% per annum, what will be the increased population of that place after T years? (A) Po T 1 + R 100 –1 (B) Po [ T 1 + R 100 –1] (C) Po [1– T 1 + R 100 ] (D) Po T 1+ R 100 +1] (d) The time duration from the beginning of B.S. 2078 to the end of B.S. 2080 is … (A) 1 years (B) 2 year (D) 3 years (D) 2.5 years (e) If the population growth rate of a certain place for first two years be R1 % and R2 % respectively then what is the formula to calculate the population after 2 years? (A) P 1 + R1 100 1 + R2 100 (B) P 1 – R1 100 1 – R2 100 (C) P 1 + R1 – R2 100 (D) P 1 + R1 + R2 100

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 67 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation 3.2 Depreciation When any asset such as furniture, vehicles, machinery items, etc. are being used for some time, their values are decreased. The reduction of value of an item due to its constant use is known as depreciation. Depreciation may be simple or compound. In simple depreciation, the amount of reduction is constant every year from every depreciated value. For example, suppose the original cost of a machine is Rs 60,000 and every year it is depreciated by Rs 6,000. Then, After 1 year, its price = Rs 60,000 – Rs 6,000 = Rs 54,000 After 2 years, its price = Rs 54,000 – Rs 6,000 = Rs 48,000 After 3 years, its price = Rs 48,000 – Rs 6,000 = Rs 42,000 and so on. On the other hand, in case of compound depreciation, the amount of reduction is calculated every year from every depreciated value. For example, suppose Mrs. Thapa bought a VESPA scooter for RS. 3,75,000 and its value depreciates every year by 10%. Then, After 1 year, its value = Rs 3,75,000 – 10% of Rs 3,75,000 =Rs 3,37,500 After 2 year, its value = Rs 3,37,500 – 10% of Rs 3,37,500 =Rs 3,03,750 After 3 year, its value = Rs 3,03,750 – 10% of Rs 3,03,750 =Rs 2,73,375 and so on. The compound depreciation is calculated in the similar way of calculation of compound interest, but here the value is gradually decreasing. Case – I: If V is the original price, R is the rate of depreciation, T is the time period for depreciation and VT is the depreciated value after T years, then VT = V T 1 – R 100 Also, the depreciated amount = V – VT = V – V T 1 – R 100 = V 1 – T 1 – R 100 Case – II: If R1 , R2 , R3 , ... be the rates of depreciation in the first, second, third, ... years respectively, then the value after T years (VT ) = V 1 – R1 100 1 – R2 100 1 – R3 100 ... The reduction in the value of article at a certain rate over a period of time is called compound depreciation.

Vedanta Excel in Mathematics - Book 10 68 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation Worked-out Examples Example 1: Sangita bought a new scooter for Rs 2,10,000. After 2 years, she sold it to her friend Anjali at 10% per year compound depreciation. (a) What does R indicate in the formula VT = V T 1 – R 100 (b) How much did Anjali pay for the scooter? (c) Find the total amount of depreciation in the price of scooter? Solution: Here, the initial cost of the scooter (V) = Rs 2,10,000 The time period for depreciation (T) = 2 years The rate of compound depreciation (R) = 10% per year (a) In the formula VT = V T 1 – R 100 ; R indicates the rate of depreciation. (b) Now, the price after 2 years (VT) = V T 1 – R 100 = Rs 2,10,000 2 1 – 10 100 = Rs 2,10,000 (0.9)2 = Rs 1,70,100 Hence, Anjali paid Rs 1,70,100 for the scooter. (c) Total amount of depreciation = V – VT = Rs 2,10,000 – Rs 1,70,100 = Rs 39,900 Hence the total amount of depreciation is Rs 39,900. Example 2: Mr. Dhurmus purchased a taxi for Rs 18,00,000. He earned a profit of Rs 5,40,000 in 3 years. He sold it after 3 years at 5% p.a. compound depreciation. (a) Find the selling price of the taxi. (b) Find his profit or loss. Solution: Here, the original cost of the taxi (V) = Rs 18,00,000. The time period for depreciation (T) = 3 years The rate of compound depreciation (R) = 5% p.a. (a) Now, the cost after depreciation (VT) = P T 1 – R 100 = Rs 18,00,000 3 1 – 5 100 = Rs 18,00,000 (0.95)3 = Rs 15,43,275 ∴ The taxi was sold for Rs 15,43,275. (b) Again, the profit in 3 years = Rs 5,40,000. ∴ The value of the taxi with profit = Rs 15,43,275 + Rs 5,40,000 = Rs 20,83,275 But the original cost of the taxi = Rs 18,00,000 ∴ Profit = Rs 20,83,275 – Rs 18,00,000 = Rs 2,83,275 Hence, his profit is Rs 2,83,275. www.geogebra.org/classroom/jxwe387n Classroom code: JXWE 387N Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 69 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation Example 3: The compound depreciation of shares of a hydropower company for 2 years is at the rate of 2% p.a. Ritwik bought a certain number of shares of the company and sold it for Rs 24,010 after 2 years. (a) Find the value of shares before 2 years. (b) How many shares at Rs 100 per share did he sell? (c) If he borrowed the sum required to buy the shares from a bank at compound interest rate of 10% p.a., how much would be insufficient to clear the debt after two years. Solution: (a) Here the value of shares before 2 years = Rs V. The rate of depreciation for 2 years = 2% p.a. Now, the present value of the shares (VT) = Rs 24,010 or, V T 1 – R 100 = Rs 24,010 or, V 2 1 – 2 100 = Rs 24,010 or, V × 49 50 × 49 50 = Rs 24,010 or, V = Rs 24,010 × 50 × 50 49 × 49 = Rs 25,000 (b) Also, the number of shares = Total value Rate of cost = 25,000 100 = 250 So, the required number of shares sold before 2 years were 250. (c) Again, principal (P) = Rs 25,000, Time (T) = 2 years, Rate (R ) = 10% p.a. Compound amount = ? We have, CA = P T 1 + R 100 = Rs 25,000 2 1 + 10 100 = Rs 30,250 Insufficient amount = Rs 30,250 – Rs 24,010 = Rs 6,240 Thus, Rs 6,240 would be insufficient if he bought the shares by borrowing the sum from the bank. Example 4: Bishwant purchased a mountain bike for Rs 32,000. If he used it for 3 years and sold to Sunayana for Rs 19,652, find the rate of compound depreciation. Solution: Here, the initial cost of the bike (V) = Rs 32,000 The cost of bike after 3 years (VT) = Rs 19,652 Time (T) = 3 years We have, VT = V T 1 – R 100 or, 19652 = 32,000 100 – R 3 100

Vedanta Excel in Mathematics - Book 10 70 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation or, 19652 32000 = 100 – R 3 100 or, 17 3 20 = 100 – R 3 100 or, 17 20 = 100 – R 100 or, 1700 = 2,000 – 20R or, R = 15% Hence, the rate of compound depreciation is 15% p.a. Example 5: Nishchal bought a Royal Enfield Bullet bike. The value of the bike depreciates each year at a certain rate. The value of the bike becomes Rs 4,45,500 in 2 years and Rs 4,00,950 in 3 years. (a) Find the rate of annual compound depreciation. (b) Find the original cost of the bike. Solution: (a) Let, the cost price of the bike be Rs V and the annual rate of depreciation be R. Here, the value of the bike after 2 years (V2 ) = Rs 4,45,500 or, V 2 1 – R 100 = Rs 4,45,500 …(i) Also, the value of the bike after 3 years (P3 ) = Rs 4,00,950 or, V 3 1 – R 100 = Rs 4,00,950 …(ii) Hence, the annual rate of depreciation is 10% p.a. (b) Now, dividing equation (ii) by equation (i), we get V 3 1 – R 100 V 2 1 – R 100 = 400950 445500 or, 1 – R 100 = 9 10 or, 100 – R 100 = 9 10 or, R = 10% Again, substituting the value of R in equation (i), we get V 2 1 – R 100 = Rs 4,45,500 or, V × 81 100 = Rs 4,45,500 or, V = Rs 5,50,000 Hence, the original cost of the bike is Rs 5,50,000.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 71 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation EXERCISE 3.2 General section 1. a) If the original value of a printing machine is V and the rate of compound depreciation is R% p.a., write the formula to calculate the depreciated value of the machine after T years. b) Write the formula to calculate the amount of deprecation of a motorbike in T year, if the initial price of the bike is V and the rate of compound depreciation is R% p.a. c) The rates of compound depreciation of a photocopy machine in 3 years are R1 %, R2 %, and R3 %. If the initial value of the machine is V, write the formula to calculate its value after 3 years. 2. a) The value of a machine is Rs 4,50,000. If its value is depreciated by 7% every year, find its value after 1 year. b) The price of a microbus is Rs 24,30,000. What will be its value after 1 year if it depreciates at the rate of 10% per annum? 3. a) If the value of a laptop which was bought for Rs 61,200 depreciates at 10% annually, find its value after 2 years. b) A set of office furniture costing Rs 96,000 is depreciated at the rate of 5% per year. What will be its cost after 3 years? 4. a) If the cost is depreciated at the rate of 12% per annum, the cost of a photocopy machine becomes Rs 69,696 after 2 years. Find the original price of the machine. b) If the cost is depreciated at the rate of 20% per annum, the cost of a taxi after 3 years becomes Rs 9,21,600. Calculate the original price of the taxi. 5. a) The original value of a computer is Rs 96,000. If it is depreciated by 15% every year with the compound depreciation, by how much is its value depreciated in 3 years? b) Mr. Mahato bought a building for Rs 25,20,000. He sold it after 3 years at the rate of compound depreciation of 10% p.a. By how much was the value of the building depreciated? 6. a) A mobile costing Rs 6,000 is depreciated per year and after 2 years its price becomes Rs 5,415. Find the rate of compound depreciation. b) If the value of an electric item depreciated from Rs 18,000 to Rs 14,580 in two years, find the rate of compound depreciation. 7. a) The original price of furniture is Rs 40,000. If the rate of depreciation is 5% per annum, after how many years will its price become Rs 36,100? b) A few years ago a man bought a computer for Rs 1,25,000. In the recent year, he sold it for Rs 64,000 at 20% annual rate of depreciation. How long did he use the computer? Creative section – A 8. a) Sajan bought a laptop for Rs 75,000. The price of laptop depreciates at the rate of 10% p.a. (i) What does R represent in the price after T years (VT) = V T 1 – R 100 ? (ii) What will be the price of laptop after 2 years? (iii)If he sold the laptop after 3 years at the same rate of compound depreciation, how much less amount would he get than it was sold after 2 years? b) Rina bought an i-Phone for Rs 1,50,000. She used it for 3 years. Due to household problem, she sold it at the rate of 20% p.a. compound depreciation.

Vedanta Excel in Mathematics - Book 10 72 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation (i) Write the formula to find the price of an article costing Rs P after T years at R% per annum compound depreciation. (ii) At what cost did she sell the i-Phone? (iii) How much more amount would she get if she sold the i-Phone at the same rate of depreciation after 2 years than it was sold after 3 years? 9. a) Mr. Magar bought a tripper for Rs 35,00,000. He used it for transporting construction materials and earned Rs 15,00,000 in 2 years. He sold it at 15% p.a. compound depreciation. (i) Define compound depreciation. (ii) Calculate the price of the tripper after 2 years. (iii) Find his total profit or loss percent during 2 years. b) Mrs. Thapa left the hotel business and she started a cow farming with an investment of Rs 20,00,000 two years ago. During 2 years, she earned Rs 5,00,000 from the farm. However, it was depreciated at the rate of 10% per annum due to Lumpy Skin Disease (LSD) in cows and she sold it. (i) What do you mean by the rate of compound depreciation is 10% per annum? (ii) Calculate the present price of her cow farming. (iii) Find her total profit or loss percent during 2 years. 10. a) If the cost is depreciated at the rate of 12% per annum, the cost of a photocopy machine becomes Rs 61,952 after 2 years. (i) If V = initial price, T= time, R = rate of depreciation and VT = price after T years, write the relationship among V, T, R and VT . (ii) Find the original price of the machine. (iii) If the rate of compound depreciation of the first and second years were 10% and 15% respectively, what would be the price of the photocopy machine after 2 years? b) Three years ago, Mrs. Kandel bought a scooter. Now, she has sold the scooter for Rs 1,12,999 in the system of compound depreciation at the rate of 15% p.a. (i) How much did she pay for it by that time? (ii) If the rate of compound depreciation of the first, second and third years were 10% and 15% and 20% respectively, what would be the present value of the scooter? c) The rate of compound depreciation of shares of a life insurance company for 2 years is 5% p.a. Ramu bought a certain number of shares of the company and sold it for Rs 36,100 after 2 years. (i) Find the price of the shares before 2 years. (ii) How many shares of Rs 100 did Ramu sell? (iii) If Ramu borrowed the sum required to buy the shares from a bank at a compound interest rate of 10% p.a., how much would be insufficient to him to clear the debt?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 73 Vedanta Excel in Mathematics - Book 10 Population Growth and Depreciation 11. a) The original value of a washing machine is Rs 60,000. The value of the washing machine is depreciated by 15% every year. (i) What does the usual notation VT = V 1 – T 1 – R 100 stand for? (ii) By how much is the value of the machine depreciated in 3 years? (iii)If the value of the washing machine was depreciated by Rs 16,260 in 3 years, what would be the rate of compound depreciation? b) Mr. Mahato bought an electric car for Rs 40,00,000. He sold it after 3 years at the rate of compound depreciation of 10% p.a. (i) Write the formula to find the amount of deprecation after T years at R% p.a. (ii) By how much was the value of the electric car depreciated? (iii)If the value of the electric car was depreciated by Rs 15,43,500 in 3 years, what would be the rate of compound depreciation? 12. a) Maya Limbu bought a television for Rs 64,000. In the recent year, she sold it for Rs 46,240 at 15% annual rate of compound depreciation. (i) How long did she use the television? (ii) If she bought a refrigerator for Rs 40,000 and used it as long as she used the television and sold at the rate of 20% p.a. compound depreciation, how much money would she receive from the refrigerator? b) A few years ago, Binaya bought a computer for Rs 1,25,000. In the recent year, he sold it for Rs 64,000 at 20% annual rate of depreciation. (i) How long did he use the computer? (ii) If he bought an i-Pad for Rs 48,000, used it as long as he used the computer and sold at the rate of 20% p.a. compound depreciation, how much money would he receive from the i-Pad? 13. a) Gita purchased a scooter for Rs 2,40,000 and some gold for Rs 5,00,000. For 2 years, the price of the scooter depreciated at a compound rate of 10% per annum, while the price of gold has been increasing at 5% p.a. compound rate of growth. (i) Write the formula to find the compound growth. (ii) What will be the price of her gold after 2 years? (iii)Will the total price of gold and scooter after 2 years be Rs 7,45,650? Write with calculation. b) In B.S. 2078, Arjun purchased a car for Rs. 30,00,000 and land for Rs. 70,00,000. The price of the car depreciated at a compound rate of 5% per annum, while the price of land increased at a certain compound rate till B.S. 2080. (i) How long was the value of car depreciated? (ii) What was the price of car in B.S. 2080? (iii)In B.S. 2080, if the total price of land and car was Rs 1,05,72,700 then what was the rate of compound growth in the price of land? Creative section –B 14. a) Mr. Chhetri bought a scanner machine 3 years before for Rs 40,000. If the value of the machine is depreciated by 5 %, 8 % and 10 % in the first, the second and the third year respectively. At what price did he sell the machine at the end of 3 years?

Vedanta Excel in Mathematics - Book 10 74 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Population Growth and Depreciation b) The present value of an asset is Rs 7,20,000. If its value is deprecated by 10% in the first 2 years and 15% in the third year, find its value after 3 years. 15. a) At the annual rate of compound depreciation, if the value of a computer becomes to Rs 40,500 in 2 years and Rs 36,450 in 3 years, find the rate of depreciation. Also, find its original cost. b) Phurba Lama bought a car and used it for 2 years. He depreciated the value of car at a rate and sold to Tilak Gurung for Rs 36,10,000. After using it 1 year, Tilak also sold the car to Dipendra Jha for Rs 34,29,500 at the same rate of annual depreciation. Find the annual rate of compound depreciation in the valuation of the car. Find the cost price of the car for Phurba. Project Work and Activity Section 16. a) Make the different groups of your friends. Visit different showrooms or shops where the second-hand vehicles are sold. Also visit different companies and organizations where the second-hand electronic items, electrical machines, vehicles, etc. are in sale. Ask the original price and the second-hand price of those items and calculate the rate of depreciation. Also calculate the profit or loss percent by selling those items. b) Visit the available website and search the rates of depreciation of different types of materials such as machinery items, vehicles, furniture, buildings, etc. Do these materials have the same rate of deprecation? Multiple Choice Questions Section 17. Tick (√) the correct option (a) The present value of a machine is Rs. P and its value depreciates every year by R%. What will be the value of the machine after T years? (A) P T 1 – R 100 (B) P T 1 + R 100 (C) P T 1 + R 200 (D) P 2T 1 – R 200 (b) The initial value of an article is P. What will be its value after 3 years if the rates of depreciation in three successive are R1 %, R2 % and R3 % respectively? (A) P 1 + R1 100 1 + R2 100 1 + R3 100 (B) P 1 – R1 100 1 – R2 100 1 – R3 100 (C) P 1 – R1 100 1 + R2 100 1 – R3 100 (D) P 1 + R1 100 1 – R2 100 1 – R3 100 (c) The value of a house at present is Rs P and annual rate of compound depreciation is R%, what be amount of depreciation after T years? (A) Dv = P 1 + T 1 – R 100 (B) Dv = P T 1 – R 100 – 1 (C) Dv = P 1 – R T 100 – 1 (D) Dv = P 1 – T 1 – R 100 (d) The original value of a car is Rs P, annual depreciation rate is R% and its value after T year will be Rs PT , then the relation between P and PT is … (A) PT > P (B) PT ≥ P (C) PT < P (D) PT ≤ P

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 75 Vedanta Excel in Mathematics - Book 10 4.1 Currency exchange - introduction Let’s study the following examples about foreign currency exchange. 1. Ms. Sampang exchanged Nepali rupees into Australian Dollars from a bank before flying to Australia for her higher study. 2. Mr. Sharma returned back from USA with some dollars. He exchanged his US Dollars into Nepali rupees from a money exchange centre. 3. Mr. Chaudhary is in Canada. He planned the family tour to some European countries. So, he exchanged his some Canadian dollars with European Euro. Let’s study the Nepal Foreign Exchange (forex) data provided by the Central Bank of Nepal - Nepal Rastra Bank on 2079-12-03 and answer the following questions. Foreign Exchange Rates Fixed By Nepal Rastra Bank Currency Code Symbol Unit Buying/Rs. Selling/ Rs. Indian Rupee INR ₹ 100 160.00 160.15 Open Market Exchange Rates (For the purpose of Nepal Rastra Bank) U.S. Dollar USD $ 1 131.86 132.46 European Euro EUR € 1 140.06 140.70 UK Pound Sterling GBP £ 1 159.16 159.88 Swiss Franc CHF Fr 1 143.05 143.70 Australian Dollar AUD $ 1 87.56 87.96 Canadian Dollar CAD $ 1 95.96 96.40 Singapore Dollar SGD $ 1 90.40 90.84 Japanese Yen JPY ¥ 10 9.87 9.92 Chinese Yuan CNY ¥‎ 1 19.10 19.18 Saudi Arabian Riyal SAR ر.س 1 35.11 35.27 Qatari Riyal QAR ر.ق 1 36.11 36.27 Thai Baht THB ฿ 1 3.80 3.82 UAE Dirham AED د.إ 1 35.90 36.07 Malaysian Ringgit MYR RM 1 29.41 29.54 South Korean Won KRW ₩ 100 10.02 10.06 Swedish Kroner SEK Kr 1 12.48 12.54 Danish Kroner DKK Kr 1 18.81 18.90 Hong Kong Dollar HKD $ 1 16.80 16.88 Kuwaiti Dinar KWD د.ك 1 429.74 431.69 Bahrain Dinar BHD د.ب 1 349.80 351.39 Note: Under the present system the open market exchange rates quoted by different banks may differ. Unit 4 Currency and Currency Exchange

Vedanta Excel in Mathematics - Book 10 76 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Currency and Currency Exchange a) Which bank determines the money exchange rates everyday in Nepal? b) What are the buying and selling rates of US $ 1 in Nepali rupees? c) What are the buying and selling rates of South Korean Won 100 in Nepali rupees? d) Which one is the highest value of currency according to the above table? e) Which currency has the lowest value according to the above forex? f) What is the difference between the selling and buying rates of Kuwaity Dinar? A currency exchange is a licensed business that allows customers to exchange one currency for another. Currency exchange of physical money is usually done over a counter at a teller station. The currencies can be exchanged through banks, money exchanged centres, authorised forex dealers, hotel and the international airport. International trade requires an organised system for exchanging money. The exchanging rate between two currencies is used for this purpose. The exchanging rate is either the bank selling or the bank buying rate. An exchanging rate is the price of one nation’s currency in terms of another nation’s currency. The buying rate is the rate at which banks or money dealers will buy foreign currency, and the selling rate is the rate at which they will sell that currency. Most of the world’s currencies, including the euro (EUR), the US dollar (USD), the Canadian dollar (CAD), the Australian dollar (AUD), and the British pound (GBP), are floating, or variable. This means their values and their exchanging rates depend on the international money market. Facts to remember 1. When an individual needs the foreign currency, the bank or money exchange centre sells him/her it by using selling rate. 2. When an individual has foreign currency and he/she needs exchanging it into the currency of his/her own country, the bank or money exchange centre buys his/her the foreign currency by using buying rate. 4.2 Revaluation and Devaluation of currency Revaluation in currency is a significant rise in a county’s official exchange rates in comparison to a foreign currency. The process of revaluation can only be done by the central bank of the revaluing country. When the value of Nepali rupees is increased in the market relative to foreign currency, the amount of rupees required to exchange with a unit of foreign currency declines. For example:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 77 Vedanta Excel in Mathematics - Book 10 Currency and Currency Exchange If the exchange rate of 1 USD was NPR 130. 40 last week and this week Nepalese currency is revaluated by 2% in comparison with USD, then new exchange rate is calculated as 1 USD = NPR 130.40 – 2% of NPR 130.40 = NPR 127.79 On the other hand, devaluation of a currency is a deliberate lowering of an official exchange rate of a country in comparison with a foreign currency. The nation’s monetary policy results the devaluation of own currency. For example; If the exchange rate of 1 GBP was NPR 155.25 and after 10 days, Nepali currency is devaluated by 4% in comparison with Pound Sterling, then new exchange rate is calculated as 1 GBP= NPR 155.25 + 4% of NPR 155.25 = NPR 161.46 Facts to remember 1. If 1 Foreign Currency (F.C.) = NPR x and Nepali currency is devaluated by d% in comparison with the Foreign Currency (F.C.) then new exchange rate is 1 F.C. = NPR (x + d% of x). 2. If 1 Foreign Currency (F.C.) = NPR x and Nepali currency is revaluated by r% in comparison with the foreign currency (F.C.) then new exchange rate is 1 F.C. = NPR (x – r% of x). Worked-out Examples Example 1: Find the difference of selling and buying rates of US $ 500 in Nepali rupee. (Buying rate of $1 = 131.86 and selling rate of $1 = Rs 132.46) Solution: The buying rate of US$ 1 = Rs 131.86 The buying rate of US$ 500 = 500 × Rs 131.86 = Rs 65,930 Also, the selling rate of US$ 1 = Rs 132.46 The selling rate of US$ 500 = 500 × Rs 132.46 = Rs 66,230 The difference due to the selling and buying rates = Rs 66,230 – Rs 65,930 = Rs 300 Example 2: The buying and selling rates of GBP (£) 1 fixed by Nepal Rastra Bank on a certain day are NPR 159.16 and NPR 159.88 respectively. (a) How much Nepali rupees can you exchange for £ 2,500? (b) How many Pounds can you exchange with NPR 2,38,740? Solution: Here, buying rate: GBP £ 1 = NPR 159.16 and selling rate: GBP £ 1 = NPR 159.88 (a) The bank buys the pounds by using buying rate. Buying rate of GBP £ 1 = NPR 159.16 ∴ GBP £ 2500 = 2500 × NPR 159.16 = NPR 3,97,900 (b) The bank sells the pounds by using selling rate.

Vedanta Excel in Mathematics - Book 10 78 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Currency and Currency Exchange Selling rate of GBP £ 1 = NPR 159.88 or, NPR 159.88 = £ 1 or, NPR 1 = £ 1 159.88 ∴ NPR 2,38,740 = £ 1 159.88 × NPR 2,38,740 = £ 1,500 Example 3: Study the following exchange rate table: Country Currency Exchange rate United Kingdom Pounds (£) 1 £ = NPR 158.50 United States Dollars ($) 1 $ = NPR 130.60 In Nepal, the cost of a television is Rs 1,61,000. In England the same television costs £ 1,050 and in the USA $ 1,220. In which country is the television cheaper? Solution: The cost of the television in Nepal = NPR 1,61,000 The cost of the television in England is NPR = 1,050 × NPR 158.50 = NPR 1,66,425 The cost of the television in USA is NPR = 1,220 × NPR 130.60 = NPR 1,59,332 Hence, the television is cheaper in United States. Example 4: The buying and selling rates of AUS ($) 1 in Dolphin Money Exchange Centre are NPR 87.56 and NPR 87.96 respectively. (a) How many Australian dollars should it buy and sell to make a profit of NPR 1,760 on the day? (b) Pemba needs AUS ($) 5,000 for the higher study in Australia and the money exchange centre charges 1.5% as commission, how much Nepali rupees does he require if he exchanges the money from the same centre on the same day? Solution: Here, buying rate: AUS ($) 1 = NPR 87.56 and selling rate: AUS ($) 1 = NPR 87.96 (a) Let, the required Australian dollars to be bought and sold be AUS ($) x. Then, C.P. of AUS ($) x = NPR 87.56 x and S.P. of AUS ($) x = NPR 87.96 x Now, profit (P) = NPR 1,760 or, 87.96x – 87.56x = 1760 or, 0.4x = 1760 or, x = 4400 Hence, the money centre should buy and sell AUS ($) 4,400. (b) The money exchange centre sells the Australian dollar to Pemba. Exchange rate of AUS ($) 1 = NPR 87.96 ∴ AUS ($) 5,000 = 5,000 × NPR 87.96 = NPR 4,39,800 Commission = 1.5% of NPR 4,39,800 = NPR 6,597 ∴The total money required for Pemba = NPR 4,39,800 + NPR 6,597 = NPR 4,46,397

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 79 Vedanta Excel in Mathematics - Book 10 Currency and Currency Exchange Example 5: Mr. Rajiv Sharma bought some EURO (€) for NPR 3,35,000 at the rate of EURO (€) 1 = NPR 134 to visit a few European countries. Unfortunately, because of his Visa problem, he cancelled his trip. Within a week Nepali rupee is devaluated by 5%. He again exchanged his EURO to Nepali rupee after a week. (a) How much Euro did he get? (b) What is the new exchange rate after devaluation in Nepali rupee? (c) How much did he gain or lose? Solution: (a) Here, NPR 134 = Euro (€) 1 ∴NPR 3,35,000 = € 335000 134 = € 2,500 (b) After the devaluation of Nepali rupee by 5%, EUR (€) 1 = NPR 134 + 5% of NPR 134 = NPR 140.70 (c) We have, EUR (€) 1 = NPR 140.70 ∴ EUR (€) 2,500 = 2,500 × NPR 140.70 = NPR 3,51,750 Hence, his gain = NPR 3,51,750 – NPR 3,35,000 = NPR 16,750 Example 6: A trader purchased 500 pieces of Nepali carpet at Rs 6,500 per piece. He exported them to United Kingdom with 5% export tax. If he sold them at £ 65 per piece in UK, calculate his profit or loss percent. (£ 1 = NPR 159.88) Solution: Here, C.P. of 500 piece of carpets = 500 × Rs 6,500 = Rs. 32,50,000 Also, C.P. with 5% export tax = Rs 32,50,000 + 5% of Rs 32,50,000 = Rs 34,12,500 Again, S.P. of 500 pieces of carpet in UK = 500 × £ 65 = £ 32,500 Now, £ 1 = Rs 159.88 £ 32,500 = 32,500 × Rs 159.88 = Rs 51,96,100 ∴ Profit = Rs 51,96,100 – Rs 34,12,500 = Rs 17,83,600 And, profit percent = Actual profit C.P. × 100% = Rs 17,83,600 Rs 34,12,500 × 100% = 52.27% Hence, his profit percent is 52.27%. 4.3 Money exchange by using chain rule Chain rule is an alternative method of unitary method and ratio and proportion method to find the value of unknown variable in a compound proportion. Study the following examples and learn about this method. If A = B and B = C, then C = A ∴ A × B × C = B × C × A

Vedanta Excel in Mathematics - Book 10 80 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Currency and Currency Exchange Example 7: If US $ 1 = NPR 130.00 and GBP £ 1 = NPR 155.00, convert $ 775 into pound. Solution: Here, $ 1 = Rs 130 Rs 155 = £ 1 £ x = $ 775 Hence, $ 775 is equal to £ 650 according to the given exchange rate. Example 8: Mr. Ram Baral has a flight to Japan for the business purpose. He has some US dollars and he wants to exchange US $ 5,300 into Japanese Yen. The exchange rate of 1 US dollar is NPR 120 and the exchange rate of 10 Japanese Yen is NPR 10.60 (a) How much Japanese Yen can he exchange with US $ 5,300? (b) If he exchanges the dollar on the day when the Nepali rupee is devaluated by 10% relative to Dollar and the exchange rate of Japanese Yen is unchanged, how much Yen can he exchange with US $ 5,300? (c) If the bank charges 1% commission after devaluation of Nepali rupee, how much Japanese Yen can he exchange with US $ 5,300? Solution: (a) Here, UD $ 1 = NPR 120 NPR 10.60 = JPY ¥ 10 Suppose, ¥ x = US $ 5300 By using chain rule, we get 1 × 10.60 × x = 120 × 10 × 5300 or, 10.6x = 6360000 or, x = 600000 Thus, he can exchange 600,000 Yen with 5,300 dollars. (b) After 10% devaluation on Nepali rupee in comparison with US dollar the new exchange rate becomes, 1 US dollar = NPR 120 + 10% of NPR 120 = NPR 132 Also, UD $ 1 = NPR 132 NPR 10.60 = JPY ¥ 10 Suppose, ¥ x = US $ 5,300 By using chain rule, we get 1 × 10.6 × x = 132 × 10 × 5,300 or, 10.6 x = 6996000 or, x = 660000 \ US $ 5,300 = ¥ 660,000 So, he can exchange 660,000 Yen with 5,300 dollars when the Nepali rupee is devaluated by 10% as compared to the US dollar. Now, using chain rule, we get, 1 × 155 × x = 130 × 1 × 775 or, x = 130 × 775 155 = £ 650

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 81 Vedanta Excel in Mathematics - Book 10 Currency and Currency Exchange c) Again, the bank sells ¥ 660,000 for 5,300 US dollars. Rate of commission = 1% ∴Commission amount = 1% of ¥ 660,000 = ¥ 6,600 Change after deduction of commission amount = ¥ 660000 – ¥ 6600 = ¥ 653,400 Thus, he can exchange 653,400 Yen with 5,300 dollars when the bank charges 1% commission after devaluation of Nepali rupee in comparison to the US dollar. Example 9: A merchant imports laptops from China and sells them in Nepal. He buys each laptop for ¥ 2,500 Chinese Yuan (CYN) with custom duty charge. He makes 20% profit and sells each laptop with 13% value added tax. (a) How much Nepali rupee does the merchant pay for each laptop? (b) If Rajiv buys the laptop with the merchant, how much should he pay for it? (c) If Rajiv uses the laptop for 2 years and sells it at the rate of 15% p.a. compound depreciation, at what cost does he sells it? (d) If Rajiv had sold it at the rate of 10% p.a. compound depreciation than 15% p.a., how much more or less amount would he receive? (CYN 1 = NPR 18) Solution: (a) Here, C.P. of each laptop = CYN 2,500 = 2500 × NPR 18 = NPR 45,000 So, the merchant pays Rs 45,000 for each laptop. (b) Now, S.P. of each laptop=C.P.+20% of C.P. = Rs 45,000+20% of Rs 45,000 = Rs 54,000 ∴S.P. with VAT = S.P. + 13% of S.P. = Rs 54,000 + 13% of Rs 54,000 = Rs 61,020 Hence, Rajiv buys a laptop for Rs 61,020 from the merchant. (c) Again, rate of compound depreciation (R) = 15% p.a., time (T) = 2 years We have, the price of laptop after 2 years (VT) = V T 1 – R 100 = Rs 61,020 P 2 1 – 15 100 = Rs 44,086.95 Hence, Rajiv sells the laptop for Rs 44,086.95 after 2 years. (d) Rate of compound depreciation (R) = 10% p.a., time (T) = 2 years The price of laptop after 2 years (VT) = V T 1 – R 100 = Rs 61,020 P 2 1 – 10 100 = Rs 49,426.20 Difference = Rs 49,426.20 – Rs 44,086.95 = Rs 5,339.25 Hence, he would receive Rs 5,339.25 more if he had sold the laptop at the rate of 10% compound depreciation rather than it was sold at 15% compound depreciation.

Vedanta Excel in Mathematics - Book 10 82 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Currency and Currency Exchange EXERCISE 4.1 General section 1. a) Name the bank that determines the money exchange rates on each day in Nepal. b) If Kusum needs to exchange US $ 500 in to Nepali rupees from a bank, does the bank use buying or selling rate to exchange her dollar? c) The buying and selling rates of 1 European Euro (€) are Rs 140.06 and Rs 140.70 respectively. If Sandeep needs € 3,500 for his Europe tour, which rate does the bank use while exchanging his rupees into Euro (€)? 2. a) Define the meaning of devaluation of Nepali currency in comparison to US Dollar. b) What do you mean by revaluation of Nepali currency relative to Pound Sterling? 3. The exchange rates of foreign currencies on 16 March 2023 are given below in the table. Use the exchange rates and solve the following problems. Country Currency Exchange rate India Rupee (₹) ₹ 100 = NPR 160 United States Dollar ($) US $ 1 = NPR 132.46 European Union Euro (€) EUR € 1 = NPR 140.70 United Kingdom Pound Sterling (GBP £) GBP £ 1 = NPR 159.88 Australia AUD ($) AUD ($) 1 = NPR 87.96 Canada CAD ($) CAD ($) 1 = NPR 96.40 Singapore SGD ($) SGD ($) 1 = NPR 98.21 Japan Japanese Yen (¥) JPY (¥) 10 = NPR 9.92 Saudi Arab Riyal (SAR) SAR 1 = NPR 35.27 Qatar Qatari Riyal (QAR) QAR 1 = NPR 36.27 Malaysia Ringgit (MYR) MYR 1 = NPR 29.54 South Korea Won (KRW ₩) ₩ 100 = NPR 10.06 a) Convert the following currencies into Nepali Currency (NPR) (i) INR (₹)17,550 (ii) US $ 990 (iii) £ 630 (iv) SAR 3,240 (v) QAR 2,880 (vi) ₩ 50,000 (vii) MYR 2,460 (viii) AUD 1,230 b) Convert NPR 1,80,000 into (i) INR (ii) CAD (iii) GBP (iv) JPY 4. a) Hari K.C. works as a supervisor in Malaysia. His monthly salary is RM 2,000. If the exchange rate is MYR 1 = NPR 29.41, how much is his monthly salary in Nepali currency? b) Bishu Rai is a pre-primary teacher in Australia. She earns AUD$ 58 per hour. She teaches 5 hours per day except Sunday. The exchange rate of AU$ 1 = NPR 87.56. (i) Find her earning per week in Nepali rupees. (ii) How much Nepali rupee does she earn in 4 weeks? 5. a) The buying and selling rates of 1 USD ($) fixed by Nepal Rastra Bank on a date are NPR 131.86 and NPR 132.46 respectively. (i) How much Nepali rupees can be exchanged with US $ 800? (ii) How many dollars can be exchanged with NPR 6,62,300? b) The purchasing and selling rates of 1 Euro (€) on a day are NPR 140.06 and NPR 140.70 respectively.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 83 Vedanta Excel in Mathematics - Book 10 Currency and Currency Exchange (i) How many Nepali rupees can be exchanged with EUR (€) 2,400? (ii) How many European Euros (€) can be bought by NPR 3,37,680? 6. a) How many Pound Sterling should a money exchange centre buy and sell to get Rs 1,800 profit. (Buying rate: 1 GBP = NPR 159.16 and selling rate: 1 GBP = NPR 159.88). b) The buying and selling rates of 1 Canadian dollar in Annapurna Money Exchange Centre are NRs 95.96 and NRs 96.40. How many Canadian dollars should it buy and sell to make a profit of NRs 4,400? 7. a) Manoj Regmi is going to visit Thiland for his family trip. He estimated to exchange US $ 4,000 in a bank. If the bank charges 1.5% commission for exchanging the money, how many Nepali rupees is required for him? (US$ 1 = NPR 132.40) b) Shashwat needs 10,50,000 Japanese Yen for the higher study in Japan. How many Nepali rupees should he manage to exchange at the rate of JPY 10 = NPR 9.48 if the bank charges 2% commission? 8. a) If 260 dollars can be exchanged for 200 pounds and 1 pound is exchanged for Rs 148, how many dollars can be exchanged for Rs 59,200? b) If $ 5 = £ 4 and Rs 342 = $ 3, how many pounds (£) will be equal to Rs 8,550? c) The exchange rate of 10 Japanese Yen on a certain day was Rs 10.25. On the same day, if $ 3 was exchanged for Rs 345, how many Yens could be exchanged for $ 123? d) If the exchange rate of £ 1 is Rs 147.00 and the exchange rate of US $ 1 is Rs 113.00, how many dollars can be exchanged for 100 sterling pounds? e) If Rs 2,280 = $ 20 and Rs 2,190 = £ 15, how many dollars can be exchanged for £ 50? f) If QAR 20 =$ 5.40 and $ 5 = Rs 560, how many QAR can be exchanged for Rs 50,000? g) If the exchange rate of MYR 1 is Rs 28 and the exchange rate of $ 1 is Rs 113, find the exchange rate of dollars for MYR 100. 9. a) In Nepal the cost of a mobile set is Rs 1,20,000. In England the same set costs £ 950 and in America US$ 1,200. In which country is the mobile set cheaper? (£ 1 = NPR 159.88, $ 1 = NPR 132.46) b) Mrs. Magar wants to buy a book online. She finds a publisher in London selling the book for £ 15. This publisher is offering free transportation on the product. She then, finds the same book from a publisher in New York for $17 excluding a transportation fee of $2. Which publisher should she buy the book from? (Exchange rate: £ 1 = NPR 159.88, $ 1 = NPR 132.46) Creative section-A 10. a) Nirmal Shrestha bought some American dollars for NPR 5,40,000 at the exchange rate of 1 USD ($) = NPR 120 to visit USA. Unfortunately, because of his Visa problem, he cancelled his trip. Within a week Nepali rupee is devaluated by 5%. He again exchanged his dollars to Nepali rupee after a week. (i) How many dollars did he get with Rs 5,40,000? (ii) What is the new exchange rate after devaluation in Nepali rupee? (iii) How much did he gain or lose?

Vedanta Excel in Mathematics - Book 10 84 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Currency and Currency Exchange b) A businessman exchanged Rs 6,60,000 into Pounds at the rate of $ 1 = NPR 150. After a few days, Nepali currency was revaluated by 10% relative to Pounds and he exchanged his pounds into Nepali currency again. (i) How many pounds did he get with Rs 6,60,000? (ii) Calculate the exchange rate after the revaluation of Nepali rupee. (iii) Find his profit or loss. c) Suresh went to a commercial bank to exchange US dollar to visit Switzerland. On that day, the buying rate of US dollar was Rs 132.20 and selling rate was Rs 133. (i) How many US dollar did Suresh receive by exchanging Rs 3,32,500 with US dollars? (ii) On the same day, if his friend who returned back to Nepal from USA, exchanged $ 1,800, how much Nepali rupees did the friend receive? (iii) After a week, if the selling rate of US dollar was increased to Rs 134.33, by what percent was the Nepali currency devaluated? d) Pradeep exchanged some Euros from a bank in order to visit Germany. According to the currency exchange rate of that day, the buying rate of 1 Euro was Rs 142.05 and selling rate was Rs 142.60. (i) How many Euros did he exchange with Rs 3,56,500? (ii) If he could not go to Germany due to his health problem, how many Nepali rupees did he get back by exchanging Euros at the same buying and selling rates? (iii) How much profit or loss did he have from this transaction? Write with reason. 11. a) Surya Gurung bought 10 Tola gold in Hong Kong for HKD$ 3,900 per Tola at the rate of HKD$ 1 = NPR 16.88. and bought to Nepal. (i) Find the cost of gold in NPR if 25% custom duty was charged. (ii) Find the selling price of gold in NPR if 20% profit was made. (iii) Find the selling price of the gold with 13% VAT in Nepal. b) A merchant purchased 500 pieces of ‘Nepali Pasmina’ at Rs 2,700 per piece and exported to USA. ($ 1 = NPR 132.00) (i) What is the cost of ‘Pasminas’ with 5% export tax? Find in rupees. (ii) If he sold them at $35 per piece in USA, what was his profit or loss percent? 12. a) Mr. Menyambo is a British Army. He has a flight to USA. He wants to exchange £ 6,500 into US dollar. The exchange rates are 1 GBP (£) = NPR 156.60 and 1 US $ = NPR 130. (i) How many dollars can he exchange with £ 6,500? (ii) If he exchanges the dollar on the day when the Nepali rupee is revaluated by 10% relative to US Dollar and the exchange rate of Pound is same, how many dollars can he exchange with £ 6,500? (iii) If the bank charges 1% commission after revaluation of Nepali rupee, how many dollars can he exchange with £ 6,500?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 85 Vedanta Excel in Mathematics - Book 10 Currency and Currency Exchange b) A businessman arrived Nepal from Kuwait with some Kuwaiti Dinars and he booked a ticket immediately for China. He exchanged 111,000 Kuwaiti Dinar into Chinese Yuan. (Given exchange rates are 1 KWD = NPR 420 and 1 CNY = NPR 18.50) (i) How many Chinese Yuan did he exchange? (ii) Had the Nepali rupee been devaluated by 2.5% relative to the Dinar and the exchange rate of Chinese Yuan is unchanged, how many Yuan would have been exchanged? (iii) Had the bank charged 1.5% commission after devaluation of Nepali rupee, how many Yuan would he have been exchanged? 13. a) Pujan has a mobile showroom in Tamrakar Complex, Kathmandu. He imports i-Phones from U.S.A. and sells in Nepal. He imports each i-Phone for US$ 800 with custom duty charge. He makes 25% profit in each i-Phone and sells with 13% value added tax. (US$ 1 = NPR 132) (i) How much Nepali rupee does he pay for each i-Phone? (ii) If Melina buys an i-Phone set from Pujan's showroom, how much should she pay for it? (iii) Due to financial problem, Melina sells the i-Phone to Ritesh after using it for 2 years at the rate of 40% p.a. compound depreciation, at what cost does Ritesh buy the i-Phone? (iv) If Melina had sold it at the rate of 25% p.a. compound depreciation than 40% p.a., how much more or less amount would she receive? b) A merchant exports Gurkha Stainless Steel Watch in UK from Nepal. He purchases each watch for NPR 21,700 with export tax. He makes 20% profit and sells each watch with 10% VAT. (GBP 1 = NPR 155) (i) Find the cost of each watch in Pounds. (ii) If Mr. Harry buys a watch from him, how much should he pay for it? (iii) If Mr. Harry uses the watch for 2 years and sells it at the rate of 5% p.a. compound depreciation, at what cost does he sell it? (iv) If Mr. Harry had sold it at the rate of 10% p.a. compound depreciation than 5% p.a., how much more or less amount would he get? Creative section-B 14. Prabin exchanged NPR 3,45,200 into some Canadian Dollar at the rate of CAD $1 = NPR 86.30. After 4 days Nepali currency was devaluated by 10% compared to Canadian dollar and he exchanged his Canadian Dollars into Nepali currency. How much profit or loss did he make? 15. Anita Pandey imports the electrical items from Japan. For this purpose, she exchanged some Japanese Yen for NPR 6,60,000 at the rate of ¥ 10 = NPR 8.80. Next day, Nepali currency was revaluated by 5%. She decided not to bring the electrical items from Japan and exchanged her Yen back to Nepali rupees. How much profit or loss did she make?

Vedanta Excel in Mathematics - Book 10 86 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Currency and Currency Exchange 16. Prabesh processed his visa for security guard in UAE and took a loan of Rs 2,00,000 from Annapura Finance for 2 years at 12% per annum compound interest rate. After 1 year, he sent 5000 UAR Dirhams to his home to pay off the loan. At the rate of 1 UAE Dirham = NPR 35.60, his wife exchanged all the Dirham and reduced the loan. He cleared the remaining loan at the same interest rate after 1 year. How much total money did he pay to clear all the loan to the finance? Project Work and Activity Section 17. a) Collect the information from the daily newspapers or from internet about today's exchange rates of different currencies in terms of Nepali currency. Then find the exchange rate between the following currencies: (i) US ($) and GBP (£) (ii) US ($) and AUD ($) (iii) GBP (£) and EUR ( ) (iv) EUR ( ) and US ($) (v) AUD ($) and GBP (£) (vi) US ($) and CAD ($) b) Collect the information from the daily newspapers or from internet about the cost of oil per barrel and cost of gold per 10 gram in the international market. Calculate the selling price of these materials in Nepali market according to the present rates of custom duty, VAT and other rates of taxes. Multiple Choice Section 18. Tick (√) the correct options. (a) Which of the following symbols is used for representing Indian Rupees (INR)? (A) $ (B) ₹ (C) € (D) £ (b) If the buying and selling rates of 1 USD are NPR 131.86 and NPR 132.46 respectively, how much rupees can be exchanged with US $ 300? (A) NPR 39,585 (B) NPR 39,558 (C) NPR 39,738 (D) NPR 35,985 (c) If the buying and selling rates of 1 UAE Dirham are NPR 35.90 and NRP 36.07 respectively, how many Dirhams can be exchanged with NPR 3,60,700? (A) AED 10,000 (B) AED 10,047.35 (C) AED 9,000 (D) AED 10,500 (d) The exchange rate of 1 Euro is NPR 130.40, what will be the new rate if the Nepali currency is devaluated by 5% in comparison to Euro? (A) NPR 123.88 (B) NPR 117.36 (C) NPR 143.44 (D) NPR 136.92 (e) How many Danish Kroner can be exchanged with NPR 74,460 at 1 Kr = NPR 18.25 if the bank charges 2% commission? (A) Kr 4,000 (B) Kr 3,998.40 (C) Kr 4,161.60 (D) Kr 4,163.27

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 87 Vedanta Excel in Mathematics - Book 10 Currency and Currency Exchange Assessment -II 1. Suppose you are going to deposit Rs 50,000 in Nepal Bank. The information attached on the notice-board is given below. Notice! Dear valued customer, Fixed Deposits earn you a higher return on your savings than an ordinary savings account. We provide you two types of fixed deposit plans to suit your needs. Please! Deposit your money in the Fixed Deposit Account on time. Fixed Deposit Account- A Fixed Deposit Account- B Yearly compound interest at 10% p.a. Half-yearly compound interest at 8% p.a. Answer the following questions based on the above information. (a) Write the formula to find the yearly compound interest. [1K] (b) How much interest is earned on account-A in 2 years? [1U] (c) In which account will you deposit and why? Write with reason. [2HA] 2. Mr. Magar borrowed Rs 3,00,000 from a commercial bank at the rate of 10% p.a. compounded half yearly for 2 years. After one year, the bank changed its policy to pay the interest compounded quarterly at the same rate. (a) How much interest should he have to pay in the first year? [2U] (b) How much interest should he have to pay in the second year? [1A] (c) How much less interest would he pay if the bank had charged interest compounded annually for 2 years? [2HA] 3. Mr. Subba, working in UAE, sends 20,000 Dirham to his wife in Nepal. His wife exchanges the Dirham into Nepali rupees at the exchange rate of 1 Dirham = NPR 35. She spends Rs 50,000 for household expenses and invests the remaining money in her account in a bank for 2 years at the rate of 10% p.a. interest being compounded annually. However, the bank changes its policy after 1 year and pays the interest compounding half-yearly at the same rate. (a) How much Nepali rupee will she receive while exchanging the Dihram? [1U] (b) What will be her balance in the beginning of second year? [2A] (c) What will be her balance at the end of second year? [1A] 4. Mr. Sanjiv bought some Euro (€) for NPR 5,50,000 at the exchange rate of Euro (€) 1 = NPR 125 to visit a few European countries. Unfortunately, because of Visa problem, they cancelled the trip. Within a week Nepali rupee is devaluated by 3%. They again exchanged their Euro to Nepali rupee after a week. (a) How many Euro did he buy? [1U] (b) What is the new exchange rate after devaluation in Nepali rupee? [1A] (c) How much did he gain or lose? [1A] (d) After devaluation in Nepali rupee, if the bank charged 2.5% commission, how many rupees would he gain or lose? [2A] 5. On 2080-11-20, a man bought a plot of land for Rs. 8,00,000 in a rural municipality and immediately he invested Rs 27,00,000 and built a house on it. Given that the value of land increases every year by 20% and that of the house decreases every year by 20%, in how many years will the values of the land and house be equal? [2A]

Vedanta Excel in Mathematics - Book 10 88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Unit 5 Mensuration (I): Pyramid 5.1 Pyramid –introduction Let’s observe the following objects and answer the following questions. Rubik's cube Glue stick Vastu Yantra Volleyball Birthday cap (a) Identify and name the shapes of the above objects. (b) Which of the above objects have only flat faces or only curved surface? (c) Name the objects which have both the flat and curved surfaces. (d) Which of the above objects have polygonal base, triangular faces and vertex? Let’s observe a few real life examples of pyramids. A pyramid is a solid having a polygonal base and triangular faces. The triangular faces meet at a common point called vertex (or apex). The polygonal base of a pyramid may be a triangle, rectangle, square, pentagon, hexagon, etc. On the basis of the types of polygonal bases, there are different types of pyramids. Its base is a triangle. A triangular pyramid or a tetrahedron Its base is a rectangle. A rectangular pyramid Its base is a square. A square-based pyramid apex face Base

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 89 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid Facts to remember 1. A pyramid has three main parts – a polygonal base, triangular faces and apex. 2. The triangular faces of pyramid except the base are called lateral faces. 3. The pyramid in which the base is a regular polygon and perpendicular drawn from the vertex (apex) passes through the centre of its base is called a right pyramid. 4. In a right pyramid, the triangular faces are congruent and isosceles. 5. The line segments created by the intersecting faces are called edges. 5.2 Surface area of square based pyramid The pyramid given alongside is a square-based pyramid. All the triangular faces of a square-based pyramid are congruent. The perpendicular drawn from the vertex to the base is the height of the pyramid. OP = h is the height of the pyramid. The perpendicular drawn from the vertex to the opposite side of each triangular face is called the slant height of the pyramid. OQ = l is the slant height of the pyramid. Squared based pyramid Net Component faces The surface area of a square based pyramid can easily be calculated by representing the three dimensional (3D) figure into a two dimensional (2D) net. P D A B a l C A B D C a a a l l l l a O O O O P Q O Height Slant height Vertex l a h

Vedanta Excel in Mathematics - Book 10 90 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid 1. Area of base = Area of square ABCD = (side)2 = a2 2. Area of ∆AOB = ∆BOC = ∆COD = ∆AOD = 1 2 a × l = 1 2 a l 3. The lateral surface area (L.S.A.) of the pyramid = sum of area of triangular faces = 4 × area of a triangle = 4 × 1 2 a l = 2al 4. The total surface area (T.S.A.) of the pyramid = Area of base + Area of triangular faces = a2 + 2al 5.3 Volume of square based pyramid Let’s make six identical square based pyramids of base side 2a units and height a units. Then, area of base of each pyramid = (2a) 2 sq. units. Let’s reassemble all six pyramids as shown in the figure given below to form a cube having each side 2a units. Let, the volume of each pyramid is V. Now, volume of 6 identical pyramids = Volume of a cube having each side 2a units or, 6V = (2a) 3 or, 6V = (2a) 2 × 2a or, V = 1 6 × (2a) 2 × 2a or, V = 1 3 × (2a) 2 × a ∴Volume of pyramid = 1 3 × Area of base (A) × Height (h) In the adjoining square-based pyramid with length of base a, OP is the perpendicular drawn from the vertex to the surface of the base. So, OP is the height (or altitude) of the pyramid. The volume of a square-based pyramid = 1 3 ×Area of the base×height (h) = 1 3 ×a2 × h = 1 3 a2 h www.geogebra.org/classroom/qtx4agmz Classroom code: QTX4 AGMZ Vedanta ICT Corner Please! Scan this QR code or browse the link given below: P a O a h

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 91 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid Facts to remember 1. In right angled ∆POQ; OQ = half of length of base = a 2 , PO = height of the pyramid = h and PQ = slant height = l. By using Pythagoras theorem, OQ2 + OP2 = PQ2 i.e., a 2 2 + h2 = l 2 2. In right angled ∆PQC; QC = half of length of base =a 2 , PQ = slant height= l and PC = length of slanted edge = e. By using Pythagoras theorem, QC2 + PQ2 = PC2 i.e., a 2 2 + l 2 = e2 O P A B C Q D a h l e Worked-out Examples Example 1: Given figure is a square-based pyramid in which the vertical height is 8 cm and the length of the base is 12 cm. (a) How many triangular faces are there in a square based pyramid? (b) Find the volume of the pyramid. (c) Calculate the lateral surface area of the pyramid. (d) Find the total cost of painting all the faces of the pyramid at Rs 5 per 8 sq. cm. Solution: Here, length of the base of pyramid (a) = 12cm vertical height of the pyramid (h) = 8cm (a) There are 4 triangular faces in a square based pyramid. (b) Now, volume of the pyramid (V) = 1 3 a2 h = 1 3 × 12 × 12 × 8 cm3 = 384 cm3 (c) Also, PQ = 1 2 × RS = 1 2 × 12 cm = 6 cm In right angled ∆OPQ, OQ = OP2 + PQ2 = 82 + 62 = 10 cm ∴Slant height (l) = 10 cm Lateral surface area of the pyramid = 2al = 2 × 12 cm × 10 cm = 240 cm2 (d) Total surface area of the pyramid = a2 + 2al = (12cm)2 + 240 cm2 = 384 cm2 Again, the rate of cost of painting the pyramid = Rs 5 per 8 sq. cm = Rs 5 8 per cm2 ∴ The total cost of painting the pyramid = Area × Rate = 384 × Rs 5 8 = Rs 240 O Q P R S 8 cm 12 cm8 cm 12 cm

Vedanta Excel in Mathematics - Book 10 92 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid Example 2 : The solid alongside is a square-based pyramid in which the length of its each edge is 13 cm. (a) What does l represent in the formula, e2 = (a 2) 2 + l2 ? (b) Find the slant height of the pyramid. (c) Find the total surface area of the pyramid. Solution: Here, length of the base (a) = 10 cm length of the slant edge (e) = 13 cm (a) In the formula, e2 = (a 2) 2 + l 2 ; l represents the slant height of the pyramid. (b) Now, EB = 1 2 × AB = 1 2 × 10 cm = 5 cm Then, in right angled D BEO, OE = OB2 – EB2 = 132 – 52 = 12 cm ∴ Slant height (l) = 12 cm (c) Now, the total surface area of the pyramid = a2 + 2al = (10 cm)2 + 2×10×12 cm2 = 340 cm2 Hence, the required total surface area of the pyramid is 340 cm2 . Example 3: In the given figure, the vertical height (AB) and the length of the slant edge (AC) of the square-based pyramid are 24cm and 25cm respectively. Find the volume of the pyramid. Solution: Here, vertical height (AB) = h = 24cm Length of the slant edge (AC) = 25 cm In right angled D ABC, BC = 252 – 242 = 7cm ∴ DC (d) = 2 × BC = 2 × 7cm = 14cm Now, area of base = 1 2 d2 = 1 2 × 142 cm2 = 98 cm2 Again, the volume of the pyramid = 1 3 area of base × height = (1 3 × 98 × 24) cm3 = 784 cm3 Hence, the required volume of the pyramid is 784 cm3 . A 10 cm B C D O 13 cm A 5 cm B C D O E 13 cm Alternatively e2 = a 2 2 + l 2 or, 132 = 10 2 2 + l 2 or, 144 = l 2 or, l = 12 cm P C B D A 25 cm 24 cm Area of a square = 1 2 × (diagonal)2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 93 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid Example 4: In the given figure, the length of a side of the base of the pyramid having square base is 14 cm and the volume of the pyramid is 1,568 cm3 , find the total surface area of the pyramid. Solution: Here, the length of the base of the pyramid (a) = 14 cm The volume of the pyramid = 1,568 cm3 or, 1 3 × a2 × h = 1,568 cm3 or, 1 3 × 14 × 14 × h = 1,568 cm3 or, h = 24 cm In the figure, slant height (OQ) = l = OP2 + PQ2 = 242 + 72 = 25 cm Now, the total surface area of the pyramid = a2 + 2al = (14cm)2 + 2 × 14 × 25 cm2 = 896 cm2 Hence, the required total surface area of the pyramid is 896 cm2 . Example 5: In a square-based pyramid, the ratio of vertical height and slant height is 4:5. The total surface area of the pyramid is 2,400 cm2 . (a) Write the formula for finding the total surface area of the square based pyramid. (b) Find the length of base of the pyramid. (c) If the pyramid is a hollow pyramidal vessel made up of tin sheet, how many litres of water can be filled in the vessel? Solution: (a) The total surface area of square based pyramid = a2 + 2al (b) Let, the vertical height of a square-based pyramid (h) = 4x cm and slant height (l) = 5x cm. Now, In right angled ∆OPQ, PQ = OQ2 – OP2 = (5x)2 – (4x)2 = 9x2 = 3x cm ∴AB (a) = 2 × PQ = 2 × 3x cm = 6x cm Also, total surface area of the pyramid = a2 + 2al or, 2,400 cm2 = (6x)2 + 2 × 6x × 5x or, 2,400 = 36x2 + 60x2 or, 2,400 = 96x2 or, x2 = 25 or, x = 5 ∴ The length of base (a) = 6x = 6 × 5 cm = 30 cm, the vertical height (h) = 4x = 4 × 5 cm = 20 cm and slant height (l) = 5x cm = 5 × 5 cm = 25 cm (c) Again, volume of the pyramid (V) = 1 3 a2 h = 1 3 × 30 × 30 × 20 cm3 = 6,000 cm3 We know, 1000 cm3 = 1 l ∴ Capacity of the hollow pyramidal vessel = 6,000 1,000 l = 6 l 14 cm Q O 14 cm 7cm24 cm Q P O 4x cm 5x cm D C A B P Q O

Vedanta Excel in Mathematics - Book 10 94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid Example 6: In a community, people want to make a temple in the public place. The roof of the temple is in the shape of square-based pyramid of side length 20 ft. and slant height 15 ft. The roof of the temple is covered with zinc sheet. (a) How much zinc sheet is required for the roof? (b) What would be the cost for the zinc sheet required to cover the roof of the temple at the rate of Rs 125 per sq. ft? Solution: Here, side of the base of pyramid-shaped roof (a) = 20 ft. slant height of the roof (l) = 15 ft. (a) Now, lateral surface area of the roof (L.S.A.) = 2al = 2 × 20 ft. × 15 ft. = 600 sq. ft. (b) Again, rate of zinc sheet (R) = Rs 125 per sq. ft. ∴ Total cost of zinc sheet = Area × Rate = 600 × Rs 125 = Rs 75,000 Hence, the required area of zinc sheet is 600 sq. ft. and the cost of the zinc sheet is Rs 75,000. Example 7: 30 tourists are coming from America to visit Mt. Everest. They have planned to stay at Everest base camp for 4 days. For this purpose, they ordered some square-based pyramid tents in Nepal. A tent can hold 2 people and each person has 6 ft × 3 ft space on the ground with 24 cu. ft of air to breathe. (a) Find the side length of base of each tent. (b) Find the height of each tent. (c) Find the surface area of required tents. (d) Find the total cost of all tents at the rate of Rs 110 per sq. ft. Solution: Here, total number of tourists = 30 Number of tourists who can stay in each square-based pyramid tent = 2 ∴Required number of tents = 30 2 =15 (a) Now, the space for 1 person on the ground of the tent = 6 ft × 3 ft ∴ The space for 2 people on the ground of 1 tent = 2 × (6 ft × 3 ft) = 36 sq. ft. But, area of squared base of each tent = the space required for 2 people or, a2 = 36 sq. ft or, a2 = (6 ft.)2 \ a = 6 ft. Hence, the length of side of base of each tent is 6 ft. (b) Also, volume of air required to breathe for 1 person = 24 cu. ft. ∴ Volume of air required to breathe for 2 people = 2 × 24 cu. ft. = 48 cu. ft. But, volume of each tent = the volume of air required for 2 people or, 1 3 × a2 h = 48 sq. ft or, 62 × h = 144 \ h = 4 ft. Hence, the height of each tent is 4 ft. We have, slant height of each tent (l) = a 2 2 + h2 = 6 2 2 + 42 = 9 + 16 = 5 ft. (c) Again, surface area of 1 tent = 2al = 2 × 6 ft. × 5 ft. = 60 sq. ft. ∴The total of the surface area of 15 tents = 15 × 60 sq. ft. = 900 sq. ft. (d) Total cost of tents = Area × Rate = 900 × Rs 110 = Rs 99,000 Hence, the total cost of the tents is Rs 99,000.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 95 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid EXERCISE 5.1 General section 1. a) In the adjoining square-based pyramid, the length of side = a, vertical height = h and the slant height = l. Write the formula to find: (i) area of base (ii) lateral surface area (iii) total surface area (iv) volume b) If the length of the base of a square-based pyramid is x unit and slant height is y unit, write the formulae to find the lateral surface area of the pyramid. c) The length of the base of a square-based right pyramid is p unit and slant height is q unit, write the formula to find the total surface area of the pyramid. d) The length of the base of a square base pyramid is m unit and height of pyramid is n unit. Write the formula to find its volume. 2. a) Find the area of triangular faces (L.S.A.) of the following square-based pyramids. (i) (ii) (iii) (iv) b) The length of the base of a square-based pyramid is 30 cm and its slant height is 17 cm. Find the lateral surface area of the pyramid. c) The length of the base of a square-based pyramid is 12 cm and its slant height is 15 cm. Find its total surface area. 3. a) Calculate the volume of the following square-based pyramids. (i) (ii) (iii) (iv) b) The height of a square-based pyramid having the area of its base 81 sq. metres is 18 metres. Find the volume of the pyramid. Creative section-A 4. a) The area of the base of a square based pyramid is 100 cm2 and its height is 12 cm. (i) Write the formula to find the area of the triangular surfaces of a square based pyramid. (ii) Find the length of side of the base of the pyramid. (iii) Find the total surface area of the pyramid. b) The length of base of a right pyramid having square base is 24 cm and its volume is 3,072 cm3 . (i) Write the formula to find the volume of a square based pyramid. (ii) Find the height of the pyramid. (iii) Calculate the total surface area of the pyramid. h l P a 6cm 6cm 5cm 10cm 13cm 10cm 14cm 25cm 20cm26cm 20cm 12cm 12cm 8cm 14cm 14cm24cm 7 ft. 7 ft. 6 ft. 9 ft. 9 ft. 14 ft.

Vedanta Excel in Mathematics - Book 10 96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid 5. a) In the square based pyramid given alongside; the length of base is 16 cm and height is 6 cm. (i) How many triangular faces are there in a square based pyramid? (ii) Calculate the slant height. (iii) Find the lateral surface area of the pyramid. (iv) Find the total cost of painting all the faces of the pyramid at Rs 5 per 20 square centimeter. b) A square based solid pyramid has length of base 18 cm and height 12 cm. (i) What does PQ represent in the pyramid? (ii) What is the length of slant height? (iii) Calculate the area of triangular faces. (iv) Find the total cost of painting all the faces of the pyramid at Rs 3 per 16 sq. cm. c) The height and slant height of a square based pyramid made up of paper are 12 cm and 13 cm respectively. (i) Write the relation among the length of base (a), height (h) and slant height (l). (ii) Find the length of base of pyramid. (iii) Find the lateral surface area of the pyramid. (iv) Find the total cost of paper required to make the pyramid at Rs 3 per 10 cm2 . d) Based on the given square based pyramid, answer the following questions. (i) Find the length of base of the pyramid. (ii) Compare between the area of base and area of triangular faces. 6. a) A metalic square based pyramid is given alongside. (i) Define square based pyramid. (ii) Find the vertical height of the given pyramid. (iii) Calculate the volume of the pyramid. (iv) If 1 cm3 of the metal weighs 5 g, find the mass of the pyramid. b) The given square based pyramid is made up of wood. (i) Name the line segment that joins the vertex to the centre of square base of the pyramid. (ii) Find the height of the pyramid. (iii) Calculate the volume of the pyramid. (iv) If 3.08 g of wood occupies 1 cm3 of air in the space, find the mass of the pyramid. O Q 6cm C P D A 16 cm B 18 cm O Q 12cm C P D A B A 16cm B C O D 10cm 13cm 10cm O N 13cm C M D 12cm A B24cm 25cm O D A P B C

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 97 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid c) A hollow square based pyramid made up of glass is given aside. (i) Find the length of base of the pyramid. (ii) How many litres of water can it hold? d) The square based pyramidal vessel given in the figure is hollow and made up of metal sheet. (i) What is the measure of height of the pyramid? (ii) Estimate the capacity of the vessel. 7. a) In the given square-based pyramid, the volume of the pyramid is 1568 cm3 and half of the length of the side of the base (OM) = 7 cm. (i) Find the height of the pyramid. (ii) Find the slant height of the pyramid. (iii) Calculate the area of triangular faces of the pyramid. b) In the given figure, if the length of a side of the base of the pyramid having square base is 12cm and the volume of the pyramid is 384 cm3 . (i) Find the height of the pyramid. (ii) Find the slant height of the pyramid. (iii) Find the total surface area of the pyramid. 8. a) In the given figure, the total surface area of the square-based pyramid is 96 cm2 and the side of the square base is 6 cm. (i) Find the slant height of the pyramid. (ii) Find the height of the pyramid. (iii) Find the volume of the pyramid. b) The adjoining figure is a square-based pyramid. The slant height of the pyramid is 13 cm and the total surface area is 360 square cm. (i) Find the length of each side of its base. (ii) Find its vertical height. (iii) Find the volume of the pyramid. Creative section-B 9. a) The ratio of length of side and slant height of a square based pyramid is 6 : 5 and its total surface area is 384 cm2 . (i) Write the relation of the area of base (A), height (h) and volume (V) of the pyramid. (ii) Compare the area of base and the area of triangular surfaces of the pyramid. (iii) Find the volume of the pyramid. O s Q R P T A 17 cm 15 cm M N 17cm S O P Q R 16cm O D C P A B M 7cm 12 cm D C E A 6 cm B 13 cm

Vedanta Excel in Mathematics - Book 10 98 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid b) In 1485, Leonardo Da Vinci’s sketched a parachute designs in the shape of a square based pyramids. For a parachute design, he sketched it with square base having side lengths of 6 feet each and height of 7 feet. (i) How much air would be captured in the pyramid once it is deployed fully? (ii) How much canvas would require to make a parachute? 10. a) Mr. Pandey has built a house for the purpose of art gallery in which roof is in the shape of a square-based pyramid. The length of each side of base of roof is 24 m and its slant height 13 m. The roof is covered with zinc sheet. (i) How many square meters of zinc sheet are required for the roof? (ii) What would be the cost for the zinc at the rate of Rs 120 per sq. meter? b) 40 tourists came to visit Mt. Annapurna from Canada. They planned to stay at Annapurna base camp for 4 days. For this purpose, they ordered some square-based pyramid tents in Nepal. A tent can hold 8 people and each person has 6 ft × 3 ft space on the ground with 48 cu. ft of air to breathe. (i) Find the length of side of base of each tent. (ii) Find the total cost of all tents at the rate of Rs 450 per sq. ft. Project Work and Activity Section 11. Make a square based pyramid by using cardboard paper. Measure its length of each side of base and height. Calculate its volume and surface area. Then, present in the classroom. 5.4 Cone –introduction Let’s observe a few real life examples of cone. Ice-cream cone Birthday cap Traffic cone Top of stupa and temple A cone is like a pyramid with a circular base. A pyramid is the special case of a cone where the base is a polygon. A cone can be thought of as a pyramid with an infinite number of faces. A cone is a three-dimensional geometrical shape that narrows smoothly from a flat circular base to a point called the apex or vertex. P (vertex) O A

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 99 Vedanta Excel in Mathematics - Book 10 Mensuration (I): Pyramid Activity Let’s take a circular piece of paper with centre O. Cut off the sector APB and join the edges AO and BO to form a cone. P A B O P A B O B P A B O A cone has a curved surface and a circular base. O is the centre of the circular base, OA is the radius, OP is the vertical height of the cone and PA is the slant height. Facts to remember 1. In the given right cone, OQ = radius of base = r, PO = vertical height = h and PQ = slant height = l. 2. In right angled ∆POQ, By using Pythagoras theorem, OP2 + OQ2 = PQ2 i.e., r2 + h2 = l 2 h r l P O Q 5.5 Surface area of cone Activity Step 1: Take a hollow cone made up of with a paper. Step 2: Cut it along the slant height and unroll it into a sector of circle such that the total arc length of sector is equal to the circumference of the circle i.e.,2πr. . Step 3: Divide the sector into a few smaller sectors of same size. Step 4: Assemble the smaller sectors in the reverse order to form a parallelogram with base πr and height l as shown in the figure. Step 5: Cut a smaller sector into two halves and re-assemble them to have a rectangle. www.geogebra.org/classroom/gvyrnbdh Classroom code: GVYR NBDH Vedanta ICT Corner Please! Scan this QR code or browse the link given below: P O A r l l l 2pr l l × pr pr l l pr pr

Vedanta Excel in Mathematics - Book 10 100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Mensuration (I): Pyramid Now, the curved surface area of the cone = area of the rectangle with length πr and bread l = πr × l = πrl Thus, C.S.A. of cone =πrl Again, the net of a solid cone consists of a small circle and a sector of a larger circle. ∴ Total surface area of the cone = Area of base + C.S.A. = πr2 + πrl = πr (r + l) Thus, T.S.A. of cone = πr (r + l) 1.6 Volume of cone Activity Step 1: Take a hollow right-circular cone of a known height h and base radius r. Also, take a hollow right-circular cylinder of the same height h and base radius r such that it is open at the top and closed at the bottom. Step 2: Fill the cone with water (or sand) up to the brim and pour this water into the cylinder. We can see that it only fills one third part of the cylinder. Step 3: Fill the same cone with water up to the brim for the second time and empty it into the same cylinder. We see that the cylinder is still not completely filled. Step 4: Fill the same cone with water up to the brim for the third time and empty it into the same cylinder. It is now observed that the cylinder is completely filled up to the brim. From the above experiment, we observed that the volume of the cylinder is equal to three times the volume of a cone. Thus, volume of 3 cones = volume of a cylinder = Area of base (A) × Height (h) = πr2 h ∴ Volume of a cone = 1 3 πr2 h Facts to remember In a cone having radius of base r units and height h units; 1. Circumference of base = 2πr 2. Area of base = πr2 3. C.S.A. = πrl 4. T.S.A. = πr (r + l) 5. Volume of cone = 1 3 × Area of base × height= 1 3 πr2 h l r h l l r www.geogebra.org/classroom/rwdzc7mf Classroom code: RWDZ C7MF Vedanta ICT Corner Please! Scan this QR code or browse the link given below: