Oasis School Mathematics-10 47 22. A man took a loan of Rs. 2,50,000 at the rate of 8% p.a. compounded annually. (a) what would be his loan at the end of first year? (b) If he paid Rs. 1,00,000 as first installment at the end of first year, how much loan is left? (c) In beginning of second year if the bank changed the interest rate 6% p.a. compounded semi annually, how much loan is left to pay after paying Rs. 1,00,0000 as second installment at the end of second year? (d) how much should be paid at the end of third year to clear the loan? 23. Saleem took a loan of Rs. 3,00,000 from the bank at the rate of 10% p.a. compounded annually. (a) how much would it be amounted at the end of first year? (b) how much loan is left to pay if he paid Rs. 1,00,000 as the first installment at the end of first year? (c) how much loan is left if he paid Rs. 1,00,000 as the second installment at the end of second year? (d) at the beginning of third year, bank changed the interest rate 8% p.a. compounded semi annually, find how much should be paid at the end of third year to clear the loan? 24. Aadhya deposited Rs. 20,000 in a bank at the rate of 10% C.I. compounded annually. (a) how much interest does she get at the end of second year? (b) If she withdrew Rs. 5,000 at the second year, how much interest is collected in next 2 years? (c) If she deposited Rs. 15,000 at the beginning of 5th year, what would be the interest in last year? (d) Find the total interest received by her in 5 years. 25. Mingma deposited Rs. 50,000 in a bank at the rate of 6% p.a. compounded half yearly. (a) what would be the compound interest at the end of first year? (b) what would be the compound interest in the second year if bank changed the interest rate 10% p.a. compounded yearly. (c) In the beginning of third year, if he deposited Rs. 25,000 what would be the interest in the third year? (d) What is the total interest in 3 years?
48 Oasis School Mathematics-10 Answers 1. Consult your teacher 2. (a) Rs. 3025, Rs. 525 (b) Rs. 3785.60, Rs. 285.60 (c) Rs. 63672.48,Rs. 3672.48 3. (a) Rs. 1399.68, Rs. 199.68 (b) Rs. 3993, Rs. 993 4. (a) Rs. 1092.73, Rs. 92.73 (b) Rs. 5624.32, Rs. 624.32 5. (a) (i) Rs. 960, (ii) Rs. 1008 (iii) 5% (b) (i) Rs. 3600, (ii) Rs. 3783, (iii) 5.08% (c) Rs.2,42, 400, (ii) Rs. 2,40,000, (iii) 1% 6. (a) Rs. 77.5 (b) Rs. 405 7. (a) Rs. 113.08 (b) Rs. 19.09 8. (a) Second scheme (b) First scheme (c) Account N 9. (a) Rs. 8000 (b) Rs. 2000 10. (a) Rs. 930 (b) Rs. 5132.16 11. (a) 2 years (b) 1 1 2 years (c) 3 years (d) 3 years 12. (a) 10% (b) 10% (c) 5 paisa 13. (a) Rs. 24000 (b) Rs. 4400 14. (a) 20,000 (b) Rs. 16,000 15. (a) 10%, Rs. 6,000 (b) 4%, Rs. 3125 (c) 10%, Rs. 362.81 16. (a) (i) Rs. 2000, (ii) Rs. 400 (b) (i) Rs. 4250, (ii) Rs. 525.30, (iii) Rs. 15.30 (c) (i) 10,000, (ii) Rs. 2155.06, (iii) Rs. 55.06 17. (a) Rs. 2200, 2.2 years (b) Rs. 420, 10.5% 18. (a) Ram: Rs. 3600, Shyam: Rs. 3,000 (b) First: Rs. 2,000, Second Rs. 2,500 19. (a) 8 1 3 %, Rs. 792 (b) Rs. 8,000, 10% 20. (b) 5%, Rs. 4,000 (c) Consult your teacher. 21. (b) 10%, Rs. 5,000 (c) Consult your teacher. 22. (a) Rs. 2,70,000, (b) Rs. 1,70,000, (c) Rs. 80,353, (d) Rs. 85,246.50 23. (a) Rs. 3,30,000, (b) Rs. 2,30,000, (c) Rs. 1,53,000, (d) 1,65,484.80 24. (a) Rs. 4200, (b) Rs. 4032, (c) Rs. 3,823.20, (d) Rs. 12055.20 25. (a) Rs. 3045, (b) Rs. 5,304.50, (c) Rs. 8,334.95, (d) Rs. 16,684.45 26. (a) Rs. 1,25,000 (b) Rs. 1, 37, 500 (c) Rs. 37, 500 (d) Rs.15,000 more. Project Work Get into a small group of 4/5 students. As a group, visit different financial institutions (Banks, cooperatives, etc.) of your locality. Find out the financial institution with a better scheme of taking a loan or depositing money. 26. A man took a loan of Rs. 2,50,000 from a bank for 2 years at the rate of 10% p.a compound yearly. (a) If he paid Rs. 1, 50,000 at the end of first year, how much loan is left at the end of first year? (b) How much should he have to pay at the end of second year to clear the loan? (c) Find the total interest paid by him in 2 years. (d) If he paid all the loan at the end of second year, how much more or less interest should have to be paid.
Oasis School Mathematics-10 49 3.2 Population Growth While calculating the compound amount and the compound interest, there is growth of money at a certain rate after some time. There may be similar type of situation in the growth in population also. Therefore, using the similar type of formulae, we can obtain the population after a certain period of time. If, P0 = Initial population PT = Population after T years R = Rate of population growth per annum T = Time period then, Population after T years PT = P R T 0 1 100 + ....................... (i) But the population of a country is also affected by the migration and the death of people. ∴ Actual population after T years PT = P R T 0 1 100 + – D + Min – Mout....................... (ii) Where, D = Number of dead people Min = Number of in-migrants Mout = Number of out-migrants If there is no migration and only deaths then, Unit 3 Growth and Compound Depreciation Sunayana kept Rs. 1200 in a bank for 2 years at the rate of 4% p.a. How much will he get after 2 years if the interest is compounded annually? The present population of a village is 1200. If it is growing at the rate of 4% p.a., what will be the population of the village after 2 years? What are the similarities in these two questions? Can we calculate the population of a village after 2 years by using the same formula of compound amount? The first question is about the growth of money whereas the second one concerns growth of population. 3.1 Warm-up Activities Discuss the following in your class and draw a conclusion.
50 Oasis School Mathematics-10 PT = P R T 0 1 100 + – D....................... (iii) If there is no death, only migration then, PT = P R T 0 1 100 + + Min – Mout ....................... (iv) If there is no death and no out migrants, only in-migrants then, PT = P R T 0 1 100 + + Min ....................... (v) If the rate of population growth is different in different years, population after T years PT = P R R R 0 1 2 3 1 100 1 100 1 100 + + + ....................... (vi) If the population decreases with the rate R% p.a., then, population after T years PT = P R T 0 1 100 − ....................... (vii) There are many situations in our daily life where growth taken place. • Income of a family • Economic growth of country • Production growth • Growth in the number of students. What happens if growth rate is different in different year? Then, PT = P0 (1 + R1 100 )(1 + R2 100 ) (1 + R3 100 )......... Where, R1 , R2 , R3 ..........are the growth rate in first year, second year, third year, ............... Note • If both the birth rate and death rate are given, then R = birth rate – death rate. • If the rate of out-migrants and in-migrants are given, R = Rate of Min– Rate of Mout . • If the birth rate, death rate, rate of in migration and out migration are given, on such condition Growth rate = Birth rate – Death rate +In migration rate – Out migration rate Remember ! P0 Present population Population before T years Population after T years Present population PT
Oasis School Mathematics-10 51 Here, Population of the village (P0 ) = 5,000 Growth rate (R) = (2-1)% = 1% Time (T) = 1 year We have, PT = P0 (1+ R 100)T = 5,000 (1+ 1 100)1 = 5,000 × 1.01 = 5,050 Alternative method . Worked Out examples example 1 the population of a village was 5000. Within a year, 2% people migrated in and 1% people died due to different causes. What would be the population of the village after one year? Solution: Here, Population of a village = 5000 Number of migrants entered = 2% of 5,000 = 2 100 × 5,000 = 100 Number of dead people = 1% of 5,000 = 1 100 × 5,000 = 50 ∴ Number of people after 1 year = (5,000 + 100 – 50) = 5,050 example 2 two years ago, the population of a village was 12000. It is increasing at the rate of 5% every year. If 150 people entered and 175 people left the village, calculate the present population of the village. Solution: Here, Population before 2 years (P0 ) = 12,000 Growth rate (R) = 5% Min = 150 Mout = 175 T = 2 years We have, PT = P R T 0 1 100 + + Min – Mout = 12000 1 5 100 2 + + 150 – 175 = 12000 1 1 20 25 2 + – = 12000 21 20 25 2 – = 12000 441 400 × – 25 = 30 × 441 – 25 = 13230 – 25 = 13205 ∴ Present population = 13,205
52 Oasis School Mathematics-10 example 3 Population of a municipality is a 60,000.In the first year. The growth rate of population is 2%, in second year, growth rate is is 3% and in the third year, the population decreased by 1%. Find the population of the municipality after 3 years. What percent of population is increased in 3 years? Solution: Here, growth rate of population is different in 3 different years. So, R1 = 2%, R2 = 3% and R3 = 1% Since, the population is decreased in third year, PT = P0 (1 + R1 100 )(1 + R2 100 ) (1 – R3 100 ) = 60,000(1 + 2 100 )(1 + 3 100 ) (1 – 1 100 ) = 60,000 (1.02) (1.03) (0.99) = 62405 (approx.) Now, the total increment of the population in 3 years is 62,405 – 60,000 = 2,405. Increment in 3 years = 2405 60,000 × 100% = 4% example 4 the population of a town increases every year by 10%. At the end of 2 years, the total population of the town was 30,000. If 5,800 people were added by migration, what was the population of the town at the beginning? Solution: Here, Rate (R) = 10% per year. Time (T) = 2 years Population after 2 yrs. = 30,000 Migrated population = 5,800 PT = 30,000 – 5,800 = 24,200. We have, PT = P R 1 100 2 + 24,200 = P0 2 1 10 100 + or, 24,200 = P0 2 11 10 We have, PT = P0 (1 + R 100) T + Min or, 30,000 = P0 (1 + R 100) T + 5800 or, 30,000–5800 = P0 (110 100) 2 or, 24,200 = P0 ( 11 10 ) 2 or, 24,200 = P0 × 121 100 or, P0 = 24,200×100 121 or, P0 = 20,000 Alternative method
Oasis School Mathematics-10 53 or, 24200 = P0 × 121 100 or, P0 = 24 200 100 121 , × or, P0 = 200 × 100 or, P0 = 20,000 ∴ Population at the beginning = 20,000 example 5 two years ago the population of a town was 40,000. the population of the town at present is 44,100. Find the population growth rate. Solution: Here, Time (T) = 2 years Population before 2 years (P) = 40,000 Present population (PT) = 44,100 Growth rate (R) = ? We have, PT = P R T 0 1 100 + 44,100 = 40,000 P R 1 100 2 + or, 44100 40000 = P R 1 100 2 + or, 441 400 = P R 1 100 2 + or, 21 20 2 = P R 1 100 2 + or, = P R 1 100 2 + or, P R 1 100 + = 21 20 or, P R 1 100 + = 21 20 – 1 or, P R 1 100 + = 1 20 or, R = 5% ∴ Population growth rate = 5% 21 20
54 Oasis School Mathematics-10 example 6 In how many years will the population of a town be 26,901 from 24,400 at a growth rate of 5% p.a.? Solution: Here, P0 = 24,400, PT = 26,901 R = 5% p.a. T = ? We have, PT = P R T 0 1 100 − 26,901 = 24,400 1 5 100 + T or, 26901 24400 = 105 100 T or, 441 400 = 21 20 T or, 21 20 2 = 21 20 T ∴ Time (T) = 2 years example 7 Population of a city is 10,00,000. (a) the birth and death rate are 2%and 1% respectively; Find the net growth rate of the population if 3% people migrated in and 1% population migrated out from the city. (b) Find the population of the city in the next year. (c) In the next year, birth rate and death rate of the population is 3% and 2% respectively. During this time 15,000 people migrated here from different places and 12,500 people migrated out from in city. Find the population of the city at the end of second year. (d) How much population is added on the city in the interval of 2 years? Solution: (a) Net growth rate of population = Birth rate – Death rate + Rate of migrant in – Rate of migrant out = 2% – 1% + 3%– 1% = 3%
Oasis School Mathematics-10 55 (b) Here, P0 = 10,00,000 T = 1 year, R = 3% We have, PT = P0 (1 + R 100) T = 10,00,000(1 + 3 100) 1 = 10,00,000 × 1.03 = 10, 30,000. (c) For next year, P0 = 10,30,000, T = 1 year, Min = 15000, Mout = 12,500 Growth rate (R) = 3% – 2% = 1% We have, PT = P0 (1 + R 100) T + Min – Mout = 10,30,000(1 + 1 100) 1 + 15,000 – 12,500 = 10,30,000 × 1.01 + 15,000 – 12,500 = 10,42,800. (d) total population increased in 2 years = 40, 42,800 – 10,00,000 = 42,800 exercise 3.2 1. (a) If the initial population (Po ) is increasing at the rate of R% per annum for T years, write the relation of population after T years (PT) with Po , R and T. (b) Write the relation among the initial population (Po ), population after T years (PT) and time T years if the population is decreasing at the rate of R% per annum. (c) If the population (Po ) of a certain village is increasing at the rate of R% per annum and D be the number of dead people in an interval of T years, write the relation among PT , Po , R, T and D. (d) If Po be the original population, PT the population after T years R the population growth rate, Min the number of people who are migrated in and Mout the number of people who migrated out, write the relation among Po , PT , T, R, Min and Mout. (e) If R1 be the birth rate and R2 the death rate of the population, then what is the population growth rate. 2. Find the present population in the following cases. (a) Population before 2 years = 3,600, growth rate = 5% p.a. (b) Population before 3 years = 2,50,000, growth rate = 2% p.a.
56 Oasis School Mathematics-10 3. Find the population after the following period of time. (a) Present population = 5,000, growth rate = 10% p.a., time = 2 years. (b) Present population = 1,20,000, growth rate = 5% p.a., time = 3 years. 4. Find the population before the following period of time. (a) Present population = 5,408, growth rate = 4% p.a., time = 2 years. (b) Present population = 33,075, growth rate = 5% p.a., time = 2 years. (c) Present population = 10,648, growth rate = 10% p.a., time = 3 years. 5. (a) The population of a village is 20,400. If the yearly growth rate of the population is 2%, what will be the population after one year? (b) The present population of a village is 30,000. If it is increased at the rate of 10% p.a., what will be the population after 2 years? (c) 2 years ago, the population of a town was 1,20,000. If the rate of growth of population is 3% p. a., what is the present population of the town? (d) One year ago, the population of Kusunde village was 10,000. If the population growth rate is 2%, what will be the population of the village after 1 year? 6. (a) The present population of a town is 2,74,000. If the annual birth and death rate are 8% and 3% respectively, what will be the population of the town after 2 years? (b) The present population of Birtamod Bazaar is 45,000. If the annual birth rate is 1%, and 1% people migrated from different places, find the population of Birtamod after 2 years. 7. (a) The present population of a country is 2,58,519. What was its population one year ago if it increased by 2% annually? (b) The population of a village is 17,640. If the population growth rate is 5% p.a., find its population 2 years ago. (c) The present population of a town is 33,800. If the birth rate is 6% p.a. and death rate is 2% p.a., what was the population of the town 2 years ago? 8. (a) The population of a village 2 years ago was 4,000. If the annual growth rate of population in the last 2 years was 5% and 10% respectively, find the present population. (b) The population of a town 3 years ago was 45,000. In the last 3 years the annual growth rate of the population was 4%, 5% and 10% respectively, find the present population of the town. 9. (a) The present population of a village is 2,205. 2 years ago, the population was 2000. Find the annual growth rate. (b) One year ago the population of a town was 10,000. After one year the population will be 12,100. Find the growth rate of the population. (c) The population of a village was 10,000 one year ago. The population at present is 9,780. Find the decreasing rate of the population.
Oasis School Mathematics-10 57 10. (a) In how many years does the population of a village increases from 2500 to 2525 at the rate of 1% per annum? (b) The present population of a town is 50,000. If the annual growth rate is 4%, after how many years will the population be 54,080? (c) The present population of a town is 66,550. If the annual growth rate is 10%, how many years ago was the population 50,000? 11. (a) The number of curd clotting bacteria increases at the rate of 5% per hour. At 5 P.M., if the number of bacteria is 2 × 108 , find its number at 7 P.M. of the day. (b) The number of bacteria in an infected person increases every day at the rate of 10%, if the number of bacteria in a day is 3.2 × 107 , find its number after 2 days. 12. (a) The population of a town increases every year by 10%. At the end of two years the total population of the town was 30,000. If 5,800 people were added by migration, what was the population of the town at the beginning? (b) Population of Chhaimale village increases every year by 5%. At the end of 2 years, the total population of the village is 3,835. If 134 people migrated to other places during this interval, what was the population of the village at the beginning? 13. Population of a rular municipality in the beginning of 2077 B.S is 12,000. If the population growth rate is 5%, (a) Find the population of the municipality at the end of 2077 B.S. (b) If 1400 people migrated in from different villages in the beginning of 2078 B.S., find the population at the end of 2079 B.S. (c) What percent of population in increased in 3 years? 14. In the beginning of 2076 B.S, the population of Butwal was around 1,00,000. If the growth rate of population is 2%, (a) Find the population of the town at the end of 2076 B.S. if 8000 people were migrated here from different places with in this period. (b) Find the population of the town at the end of 2078 B.S. (c) What is the total increment of the population in 3 years. 15. Three year ago, the population of Barpak village was 5,000. The population growth rate was 1% (a) What was the population before 2 years? (b) What is the present population? (c) If the population increases in the same way, what will be the population after 2 years? 16. 10000 workers were employed to construct a bridge in four years. At the end of the first year 10% workers left. At the end of second year, 5% of the workers at that time left.
58 Oasis School Mathematics-10 However to complete the project in time, the number of workers was increased by 10% at the end of the third year. (a) How many workers left the work at the end of the first year? (b) How many workers were there at the end of the second year? (c) How many workers were working during the fouth year? 17. Nutan's monthly income of Baishakh is Rs. 1,50,000 and her household expense is Rs. 1,00,000. In Jestha income decreased by 10% and expenses increased by 10%. In Ashar income decreased by 15% and expense increased by 15%. (a) What are her income and expenses in Jestha? (b) Can she manage the expenses in Asar? If so how it not why? Answers 1. Consult your teacher 2. (a) 3,969 (b) 26,5,302 3. (a) 6,050 (b) 1,32,300 4. (a) 5,000 (b) 30,000 (c) 8,000 5. (a) 20,808 (b) 36,300 (c) 1,27,308 (d) 10,404 6. (a) 302085 (b) 46,818 7. (a) 25,3,450 (b) 16,000 (c) 31,250 8. (a) 4,620 (b) 54,054 9. (a) R = 5% (b) 10% (c) 2.2% 10. (a) 1 year (b) 2 years (c) 3 years 11. (a) 2.205 × 108 (b) 3.872 × 107 12. (a) 20,000 (b) 3600 13. (a) 12,600 b) 15435 c) 28.63% 14. (b) (a) 1,10,000 (b) 1,14,444 (c) 14444 15. (a) 5050 (b) 5151 (around) (c) 5255 16. (a) 1000 (b) 8550 (c) 9405 17. (a) 1,35,000 (b) 1,10,000 (c) No. Project Work 1. Visit the office of your ward or municipality. Take information of the population of your village according to the latest census. Assuming different growth rates like 1%, 1.5%, 2%, etc., estimate the present population. 2. Take the record of number of students in your school. Calculate the number of increased / decreased students form the record of number of students in the last year. Using this fact project the number of students after 2 years, 3 years, and 4 years. 3.3 Compound Depreciation Let's discuss these questions in your class. A mobile set is bought for Rs. 15,000. What will be the price of the mobile after 2 years? • More than Rs. 15,000 • Rs. 15,000 • Less than Rs. 15,000 A man bought a motorcycle for Rs. 2,00,000.
Oasis School Mathematics-10 59 What will be its value after 2 years? • More than Rs. 2,00,000 • Rs. 2,00,000 • Less than Rs. 2,00,000 The value of fixed assets like machines, equipment decreases gradually after being used for a certain period of time. The decline in the original value of fixed assets is called depreciation. There are many causes of depreciation like wear and tear, efflux of time, obsolescence, fall in market price, etc. The depreciation of fixed assets in unit time is the rate of depreciation. The residual value of the fixed assets to which it is reduced with time is known as scrap value. Depreciation may be simple or compound. In simple depreciation, the reduction in the value of the fixed asset is the same in each year. But in compound depreciation, reduction in the value of the assets is not the same. It is calculated from the depreciated value of the asset at the end of the previous year. If, V0 = Original value of the asset R = Yearly rate of depreciation T = Time period VT = Value after T years then VT = V R T 0 1 100 − Amount of depreciation = V0 – VT If the rate of depreciation is different in different years, then, VT = V0 (1– R1 100) (1– R2 100) (1– R3 100)... Where, R1 = rate of depreciation in first year R2 = rate of depreciation in second year R3 = rate of depreciation in third year, and so on Worked Out examples example 1 The original value of a machine is Rs. 60,000. Find its value after 2 years if the annual rate of depreciation is 4%. Solution: Here, Original value (V0 ) = Rs. 60,000 Rate of depreciation (R) = 4% p.a. Time period (T) = Rs. 2 years Value after T years (VT) = ?
60 Oasis School Mathematics-10 We have, VT = V R T 0 1 100 − = 60 000 1 4 100 2 , – = 60 000 96 100 2 , = 60 000 24 25 2 , = Rs. 55,296 ∴ Value of the machine after 2 years = Rs. 55,296 example 2 the present value of a machine is Rs. 17496. If the annual rate of depreciation is 10%, what was the value of the machine 3 years before? Solution: Here, Original value (Vo ) = ? Rate of depreciation (R) = 10% Time period (T) = 3 years Value after T years (VT) = Rs. 17496 We have, VT = V R T 0 1 100 − 17,496 = V0 3 1 10 100 – or, 17,496 = V0 3 90 100 or, 17,496 = V0 3 9 10 or, 17,496 = 729 1000 Vo or, V0 = Rs. 24,000 ∴ Original value of the machine = Rs. 24,000 example 3 the original value of a television set was Rs. 16,000. Find the annual rate of depreciation if its value after 3 years is Rs. 13,718. Solution: Here,
Oasis School Mathematics-10 61 Original value (V0 ) = Rs. 16,000 Time (T) = 3 years Annual rate (R) = ? Value after T years (VT) = Rs. 13,718 We have, VT = V R T 0 1 100 − or, 13718 = 16000 1 100 3 – R or, 13718 16000 16000 = 1 100 3 – R or, 6859 8000 16000 = 1 100 3 – R or, 19 20 3 16000 = 1 100 3 – R or, 19 20 16000= 1 – 1 100 3 – R or, 16000 1 100 3 – R = 1 – 19 20 or, 16000 1 100 3 – R = 1 20 ∴ R = 5% Hence, the rate of depreciation = 5% example 4 A motorcycle is bought for Rs. 1,80,000. If the rate of compound depreciation is 8% p.a., when will its value be Rs. 1,52,352 ? Solution: Here, Original value (Vo ) = Rs. 1,80,000 Rate (R) = 8% p.a. Value after T years (VT) = Rs. 1,52,352 Time (T) = ? We have, VT = V R T 0 1 100 − or, 1,52,352 = 1 80 000 1 8 100 , , − T
62 Oasis School Mathematics-10 or, 1 52 352 1 80 000 , , , , = 92 100 T or, 4761 5625 = 23 25 T or, 529 625 = 23 25 T or, 23 25 2 = 23 25 T ∴ T = 2 years example 5 A man bought a taxi for Rs. 4,50,000. Using the taxi, he earned Rs. 50,000 in 2 years. If he sold the taxi after 2 years at the rate of 10% p.a. compound depreciation, calculate his gain or loss in 2 years. Solution: Here, Original value (V0 ) = Rs. 4,50,000 Time (T) = 2 years Rate (R) = 10% p.a. Value after T years (VT) = ? We have, VT = V R T 0 1 100 − or, PT = 4 50 000 1 10 100 2 , , – = 4 50 000 90 100 2 , , = 4 50 000 9 10 9 10 , , × × = Rs. 3,64,500 Amount earned by using taxi = Rs. 50,000 Now, value after depreciation + amount earned by taxi = Rs.(3,64,500 + 50,000) = Rs. 4,14,500 Since the original value is greater than this amount, loss = Rs. 4,50,000 – Rs. 4,14,500 = Rs. 35,500
Oasis School Mathematics-10 63 example 6 Plot of land is bought at Rs. 8,00,000 and an apartment worth Rs. 27,00,000 is made on it. If the value of land increases 20% every year and the value of of the apartment decreases 20% every year. (a) Find the price of the land after T years. (b) Find the price of the apartment after T years. (c) What is the value of T if the value of land and apartment are same after T years. Solution:(a) For the land V0 = 8,00,000, R = 20% Its value after T years VT = VO (1 + R 100) T = 8,00,000 (1 + 20 100) T = 8,00,000 (1.2)T (b) For apartment, V0 = 27,00,000, R = 20% We have, VT = V0 (1 – R 100) T = 27,00,000(1 – 20 100) T = 27,00,000 (0.8)T (c) Now, 8,00,000 (1.2)T = 27,00,000 (0.8)T (1.2)T (0.8)T = 27,00,000 8,00,000 ( 1.2 0.8 ) T = 27 8 ( 1.2 0.8 ) T = 27 8 ( 3 2 ) T = ( 3 2 ) 3 ∴ T = 3 years Hence, after 3 years the value of the land and apartment remains same. exercise 3.3 1. (a) If Vo be the original value of an asset, VT the value of the asset after T years, R the rate of depreciation, write the relation among Vo , VT , R and T. (b) If Vo be the value of a piece of land, R be the growth rate of the value, T the time, then write the relation of the value after T years VT with Vo , R and T. (c) If Vo be the original value and VT the value of the asset after T years. Then, what is the amount of depreciation? (d) If R1 and R2 be the rate of depreciation for two consecutive years and Vo the original value, write the relation of VT with Vo , R1 and R2 . 2. (a) A machine costing Rs. 10,000 is depreciated 5% every year, what will be its value next year?
64 Oasis School Mathematics-10 (b) A mobile set costing Rs. 15,000 is depreciated 10% every year, what will be its value next year? (c) The value of a piece of land is Rs. 15,00,000. Its value is increasing at the rate of 20%, what will be its value next year? 3. Find the value afterthe given period oftime with given rate in each ofthe following cases. (a) Original value = Rs. 15,000, annual rate of depreciation = 8%, time = 2 years. (b) Original value = Rs. 36,000, annual rate of depreciation = 10%, time = 3 years. 4. Find the original value in each of the following cases. (a) Value after 2 years = Rs. 10,000, annual rate of depreciation = 10% (b) Value after 3 years = Rs. 1,20,000, annual rate of depreciation = 5% 5. (a) Present value of a machine is Rs. 2,50,000. Its value depreciated by 12% per annum, then (i) what will be its value next year? (ii) what was its value one year before? (b) A man bought a second hand computer at Rs. 24,000 in 2060 B.S. If the annual rate of depreciation is 20%, (i) what was its value in 2058 B.S.? (ii) what was its value in 2062 B.S.? (c) The present price of a scooter is Rs. 95,000. If it is depreciated at 6% per year, what will be the price of the scooter after 2 years? (d) A television set costing Rs. 11,600 is depreciated at the rate of 15% per annum. What will be the cost of the television set after 2 years ? 6. (a) A man bought 250 shares of Rs. 100 each. The value of the shares got devaluated by 5% every year due to political instability. Find his profit or loss at the end of 2 years. (b) Sangita bought 150 secondary shares of a bank at the rate of Rs. 500 per share. The value of share depreciated by 3% for 2 years. Find her profit or loss. 7. (a) A company bought 5 ropanis of land at Rs. 2,80,00,000. Due to the political instability, its value got devaluated by 5% every year. Find the price of the land after 2 years. (b) A man sold a piece of land at Rs. 19,44,000 after the devaluation of the price of land by 10% every year. Find at what price he had bought the land 2 years before. 8. (a) The number of WBC in the infected blood was 14 × 109 per litre. After using antibiotics its number decreased by 10% every day. Find the number of WBC per liter after 3 days. (b) After using medicine the number of bacteria decreased by 5% per hour. If its number was 2.5×105 per ml at 11:00 A.M., what was its number at 9:00 A.M.? 9. (a) A machine costing Rs. 50,000 is sold at Rs. 40,500 after 2 years. Find the annual rate of depreciation. (b) A man bought a motorcycle at Rs. 1,50,000. After 2 years, he sold it at Rs. 1,16,160. Find the annual rate of depreciation. 10. (a) In how many years is the value of an article depreciated from Rs. 18000 to Rs. 14580 at the rate of 10% p.a.?
Oasis School Mathematics-10 65 Answers 1. Consult your teacher 2. (a) Rs. 9,500 (b) Rs. 13,500 (c) Rs. 18,00,000 3. (a) Rs. 12,696 (b) Rs. 26,244 4. (a) Rs. 12345.68 (b) Rs. 1,39,962.09 5. (a) (i) Rs. 2,20,000, (ii) Rs. 2,84,090.90 (b) (i) Rs. 37,500, (ii) Rs.15,360, (c) Rs. 83,942, d. Rs. 8,381 6. (a) Loss Rs. 2,437.50 (b) Loss Rs. 4,432.50 7. (a) Rs. 2,52,70,000 (b) Rs. 24,00,000 8. (a) 10.206 × 109 (b) 2.77 × 105 9. (a) 10% (b) 12% 10. (a) 2 years (b) 1 year (c) 2 years 11. (a) Rs. 1,65,000 profit, (b) Profit 3,58,768.50 12. (a) Rs. 14,52,000 (b) Rs. 36,10,000 (d) decrease by 2.65% 13. (a) Rs. 10, 58, 400 (b) Rs. 18, 89, 280 (c) decrease by 2.07% Project Work 1. Visit different showrooms and garages. Ask the resale value of the vehicles, their original value, time period and calculate the rate of depreciation. 2. Collect the original price list of different assets like bus, car, mobile, computer available in your home or school. Give the projection of their price after 1 year, 2 yeas, 3 years, 4 years using different rate of depreciation. (b) In how many years is the value of an article depreciated from Rs. 2,00,000 to Rs. 1,70,000 at the rate 15% p.a.? (c) A radio costing Rs. 1200 depreciates 10% every year. After how many years will its value be reduced by Rs. 228? 11. (a) A microbus was bought for Rs. 15,00,000. It was used for 2 years and the net profit during that time was Rs. 4,50,000. After 2 years it was sold at the rate of compound depreciation of 10% p.a. Find the profit or loss. (b) Bhai Raja bought a tractor for Rs. 8,50,000. He earned Rs. 4,80,000 from the tractor. If he sold the tractor after 3 years at compound depreciation of 5%, calculate his profit or loss. 12. A man bought a land at Rs. 12,00,000. He made a house at Rs.40,00,000. (a) What will be the price of land after 2 years if its price increase at the rate of 10% p.a.? (b) What will be the price of house after 2 years if its price decreases at the rate of 5% every year? (c) By what percent, the total price of land and house decreases or increases in 2 years? 13. The price of land is Rs. 9,60,000. Its value increases every year by 5% per annum. The price of the house on it is Rs. 20,50,000. It is depreciated at the rate of 4% per year. (a) What will be the price of the land after 2 years? (b) What will be the price of the house after 2 years? (c) By how much percent the price of house and land together increased or decreased in 2 years?
66 Oasis School Mathematics-10 Unit 4 Money exchange 4.1 Warm-up Activities • Kalpana's father works in Europe. He has send 5000 Euro. How to covert 5000 Euro into Nepali currency? She went to money exchange office, the exchange rate of 1 Euro = Rs. 132.50. • As if wants to visit USA. He needs 7000 US dollars. If 1 US dollar = Rs. 126.30. How much Nepali currency does he need? If bank takes the commission of 2%, how much commission does he have to pay to the bank? 4.2 Money exchange • Gaurav's father works in the USA. He sent 10,000 U.S. Dollars to Nepal. How to convert 10,000 U.S. Dollars into Nepali currency ? • Dawa has to go to Europe. He needs some Euros. How do you convert Nepali currency into Euros? • Sundar works in Qatar. He sent 5000 Qatari Riyal to his family. How much Nepali Rupees does his wife receive ? There are many such examples in our daily life related to money. Every country has its own currency. It has its own value. The value of different currencies is different, like Rs. 100 I.C. is equal to Rs. 160 N.C. In our country, Nepal Rastra Bank determines the conversion rate of different currencies into Nepali currency. The given table shows the rate of exchange for Poush 13, 2073.
Oasis School Mathematics-10 67 exchange rates of by Nepal Rastra Bank Currency Unit Rate Ba ying/Rs. Rate Selling / Rs. Indian Rupee 100 160.00 160.15 U.S. Dollar 1 131.34 131.94 European Euro 1 138.66 139.29 UK Pound Sterling 1 160.89 161.63 Swiss Franc 1 140.62 141.26 Australian Dollar 1 88.79 89.20 Canadian Dollar 1 96.31 96.75 Japanese Yen 10 9.65 9.69 Chinese Yuan 1 18.93 19.03 Saudi Arabian Riyal 1 34.93 36.09 Qatari Riyal 1 35.89 36.06 Thai Bhat 1 3.77 3.79 UAE Dirham 1 35.76 35.12 Malaysian Ringgit 1 29.82 29.95 South Korean Won 100 10.08 10.12 Swedish Kroner 1 12.71 12.77 Bahrain Dinar 1 348.40 149.99 Kuwaity Dinnar 1 428.25 430.21 Note • Under the present system, the open market exchange rates quoted by different banks may differ. • Use buying rate while converting other currencies into Nepali currency and use selling rate while converting Nepali currency into other currencies. What happens if Nepali currency is devaluated on the comparison of foreign currency? • If Nepali currency is devaluated in the comparison of other currency, Nepali currency becomes weak. We need to pay more Nepali currency ot buy foreign currency. Let's be clear with an example, If 1 US dollar = Rs. 125. Nepali currency is devaluated by 2% on the comparison of US
68 Oasis School Mathematics-10 dollar, what is the new rate of exchange? Given, 1 US dollar = Rs. 125. Since, Nepali currency is devaluated, we have to pay more Nepali currency, So, New rate of exchange, 1 US dollar = Rs. 125 + 2% of Rs. 125 = Rs. 125 + 2 100 ×125 = Rs. 125 + Rs. 2.50 = Rs. 127.50 What happens if Nepali currency is revaluated? Revaluation of Nepali currency means increasing the strength of Nepali currency. Lets be clear with an example. If £ 1 = Rs. 145, what is the new rate of exchange if the Nepali currency is revaluated by 2%? Given rate of exchange, £ 1 = Rs. 145 =Rs. 145 – 2% of Rs. 145 [needs to pay less money as Nepali currency become strong] = Rs. 145 – 2 100 × Rs. 145 = Rs. 145 – Rs. 2.90 = Rs. 142.10. How to calculate the commission to be paid while converting Nepali currency into foreign currency? Let's be clear with an example. A man need 5000 Euro for his Europe tour, if bank takes 1% commission, how much Nepali currency does he need if 1 Euro = Rs. 135.50 Here, 1 Euro = Rs. 135.50 500 Euro = Rs. 5000 × 135.50 = Rs. 677500 Commission to be paid to the bank = 2% of 6, 77, 500 = 2 100 × 6, 77, 500 = 2 × 6.775 = Rs. 13, 550. Let's take one more example, A man has Rs. 7,00,000. He has to convert it into US dollar. How much US dollar does he get if $ 1= Rs. 126 and 2% commission is to be paid to the bank. Here, commission = 2% of 7,00,000 = 2 100 × 7,00,000
Oasis School Mathematics-10 69 = Rs. 14,000. Remaining sum after deducting commission = Rs. 7,00,000 – Rs. 14,000 = Rs. 6,86,000. Now, converting this amount into dollar. = $ 6,86,000 125 = $ 5444.44 Worked Out examples example 1 Using the above rate of exchange, convert $ 125 into Nepali rupees. (use buying rate) Solution: Using the buying rate , $1 = Rs. 131.34 ∴ $125 = Rs.125 × 313.34 = Rs. 16417.50 example 2 Using the above table, convert the following currencies into Nepali rupees. (use buying rate) a. 1500 Qatari Riyal b. £ 250 c. 3,250 Euro d. 8,500 South Korean Won e. 50,000 Japanese Yen. Solution : a. 1 Qatari Riyal = Rs. 35.89 Then, 1500 Qatari Riyal = Rs. 1,500 × 35.89 = Rs. 53,835 c. 1 Euro = Rs. 138.66 then, 250 Euros = Rs. 138.66 × 250 = Rs. 34,665 e. 10 Japanese Yen = Rs. 9.65 b. 1 Pound Sterling = Rs. 160.89 then, 250 Pound Sterling = Rs. 250 × 160.89 = Rs. 40,222.50 d. 100 South Korean won= Rs. 10.08 1 South Korean won = Rs. 10.08 100 8500 South Korean won = Rs. 10.08 100 × 8,500 = Rs. 8568
70 Oasis School Mathematics-10 1 Japanese Yen = Rs. 9.65 10 50,000 Japanese Yen = Rs. 9.65 10 × 50,000 = Rs. 48,250 example 3 Using the buying rate, convert 50 U.S. Dollars into Japanese yen. Solution: First, let's convert 50 U.S. Dollars into Nepali rupees. From the above table, 1 U.S. Dollar = Rs. 131.34 then, 50 U.S. Dollars = Rs. 50 × 131.34 = Rs. 6,567. Again, convert Nepali Rupees into Japanese Yen Now, Rs. 9.65 = 10 Japanese Yen Re 1 = 10 9.65 Japanese Yen Rs. 5409 = 10 9.65 × 5409 Japanese Yen = 5605.52 Japanese Yen example 4 Using the above rate, convert 1 euro into South Korean won. (Use buying rate) Solution : From the above table, 1 Euro = Rs. 138.66 Let's convert Rs. 138.66 into South Korean Won. Rs. 10.08 = 100 South Korean Won Re 1 = 100 10.08 South Korean Won. Rs. 138.66 = 100 10.08 × 138.66 South Korean Won Rs. 138.66 = south Korean Won ∴ 1 Euro = 1375.59 South Korean Won. example 5 A man needed 6,000 U.S. Dollars. While converting Nepali currency into Dollars, the bank took a commission of 1%, find how much Nepali currency does he need? Solution : Let's convert 6,000 U.S. Dollars into Nepali Rupees. Here, 1 U.S. Dollar = Rs. 131.34 Let, $ 50 = x Yen 10 Yen = Rs. 9.65 Rs. 131.34 = $ 1 Using chain rule, 50 × 10 × 131.34 = x × 9.65 or, x = 50×10×108.18 9.65 = 5605.52 Yen ∴ $ 50 = 5605.52. Yen Alternative method
Oasis School Mathematics-10 71 then, 6,000 U.S. Dollars = Rs. 131.34 × 6000 = Rs. 7,88,040 Now, commission = 1% of Rs. 7,88,040 = 1 100 × Rs. 7,00,040 = Rs. 7880.40 ∴Total money required = Rs. 7,88,040 + Rs. 7,880.40 = Rs. 7,65,920.40. example 6 Himanka converted Rs. 6,00,000 into Pound Sterling. After one week, Nepali rupees got devalued by 5%, find whether he has made a profit or loss. (Given 1 Pound Sterling = Rs. 160.89) Solution : 1 Pound Sterling = Rs. 160.89 i.e., Rs. 160.89 = 1 Pound Sterling Re. 1 = 1 160.89 Pound Sterling Rs. 6,00,000 = 1 160.89 × 6,00,000 Pound Sterling = 3729.25 Pound Sterling. After the devaluation of the Nepali rupees, 1 Pound Sterling = Rs. 160.89 + 5% of Rs. 160.89 = Rs. (160.89 + 5 100 × 160.89) = Rs. 168.93 Now, 1 Pound Sterling = Rs. 168.93 4,514.33 Pound Sterling = Rs. 168.93 × 3729.25 = Rs. 6,29,982.20 His profit = Rs. 6,29,982.20 – Rs. 6,00,000 = Rs. 29,982.20 example 7 A machine is bought from the Indian market at Rs. 64,000 I.C. Find its cost in the Nepali market if 50% customs duty and 13% VAt are added. Solution: Cost of the machine = Rs. 64,000 I.C. Since, Rs. 100 I.C. = Rs. 160 N.C. Re 1 I.C. = Rs. 160 100 N.C.
72 Oasis School Mathematics-10 Rs. 64,000 I.C. = Rs. 160 100 × 64,000 NC = Rs. 102,400 NC Now, customs duty = 50% of Rs. 1,02,400 = Rs.50 100 × 1,02,400 = Rs. 51,200 Cost of the machine in N.C. = Rs. 1,02,400 + Rs. 51,200 = Rs. 1,53,600 VAT = 13% of Rs. 1,53,600 = Rs.13 100 × 1,53,600 = Rs. 19,968 ∴ S.P. with VAT = Rs. (1,53,600+19,968) = Rs. 1,73,568. example 8 Kumar bought 500 Nepali thanka scrolls at Rs. 1500 per piece. At what rate can he sell it in the Chinese market to make a profit of 80%, if he paid 5% export tax? [Given, 1 Chinese Yuan = Rs. 18.93) Solution: Cost of a piece of Thanka = Rs. 1,500 Export tax = 5% of Rs. 1500 = Rs.5 100 × 1500 = Rs. 75 ∴ Total cost including export tax = Rs. 1500 + Rs. 75 = Rs. 1575 Profit % = 80% Selling price per piece = Rs. 1575 + 80% of Rs. 1575 = Rs. 1575 + 80 100 × 1575 = Rs.1575 + Rs. 1260 = Rs. 2835 Let's convert Rs. 2835 into Chinese Yuan From the above table, 1 Chinese Yuan = Rs. 18.93 i.e. Rs. 18.93 = 1 Chinese Yuan Re. 1 = 1 18.93 × Chinese Yuan Rs. 2835 = 1 18.93 × 2835 Chinese Yuan = 149.76 Chinese Yuan. ∴ He can sell it for 149.76 Chinese yuan per Thanka.
Oasis School Mathematics-10 73 example 9 A man converted Rs. 6,70,000 to euro for his europe tour at the rate of 1 euro = Rs. 134. During his stay in europe he spent 4500 euro. Before his arrival in Nepal Nepali currency is devaluated by 2%, find how much Nepali rupees is left with him. Solution: Given, 1 Euro = Rs. 134 Re 1 = 1 134 Euro Rs. 6,50,000 = 1 134 × 6,70,000 Euro = 5000 Euro. Since, he spent 4500 Euro on his Europe tour Money left with him = 5000 Euro – 4500 Euro = 500 Euro Again, Nepali currency is devaluated by 2%, New rate of exchange 1 Euro = Rs. (134 + 2% of 134) = Rs. (134 + 2 100 × 134) = Rs. 136.68 Now, converting 500 Euro into Rupees, 500 Euro = Rs. 500 × 136.68 = Rs. 68,340 exercise 4.1 1. (a) If 1 U.S. Dollar = Rs. 131.34, convert 150 U.S. Dollars into rupees. (b) If 1 Pound Sterling = Rs. 160.89, convert 250 Pound Sterling into rupees. (c) If 1 Euro = Rs. 138.66, convert 425 Euro into rupees. (d) If 1 Qatari riyal = Rs. 35.89, convert 500 Qatari riyal into rupees. (e) If Rs. 100 I.C. = Rs. 160 N.C., convert Rs. 10,600 I.C. into N.C. rupees. 2. (a) Using the table of exchange rate and applying the buying rate convert Rs. 60,000 into i. U.S. Dollars ii. Euro iii. Indian Rupees iv. Pound Sterling v. Chinese Yuan vi. Thai Baht (b) Using the above table of exchange rate and applying the selling rate, convert Rs. 25,000 into i. Japanese Yen ii. Kuwaity Dinar iii. Canadian Dollar iv. UAE Dirham v. Swiss Franc 3. (a) Using the table given alongside convert: i. 6500 Thai baht into U.S. Dollar. ii. 3000 sterling pound into Japanese yen iii. 450 Australian Dollar into Indian rupees iv. 1200 Chinese yuan into Euro (Use buying rate only) Currency Unit Buying/Rs. Selling/Rs. Indian Rupees 100 160.00 160.15 U.S. Dollar 1 131.34 134.94 European Euro 1 138.66 139.29 UK Pound Sterling 1 160.89 161.63 Australian Dollar 1 88.79 81.20 Japanese Yen 10 9.65 9.69 Chinese Yuan 1 18.93 19.02 Thai Baht 1 3.77 3.79
74 Oasis School Mathematics-10 4. (a) According to exchange rate of Nepal Rastra Bank, the buying and selling rates of $ 1 are Rs. 131.34 and Rs. 131.94 respectively. If a man bought $ 6,000 and sold it, find his profit. (b) The buying rate and selling rate of £1 are Rs. 160.89 and Rs. 161.63 respectively. If £8,000 is bought and sold by money exchange centre, what is the profit in rupees. (c) According to the exchange rate of Nepal Rastra Bank, the buying rate and selling rate of 1 Euro is equal to Rs. 138.66 and Rs. 139.29, find (i) how much Euros can be exchanged with Rs. 40,000? (ii) how much Nepali rupees can be exchanged with 6,000 Euros? (d) Selling rate of 1 Australian Dollar is equal to Rs. 89.20. A bank bought 6,000 Australian Dollar and made a profit of Rs. 4,200. Find the buying rate of Australian Dollar. 5. (a) If 1 US Dollar = Rs. 132 and £1 = Rs. 162.50, convert 500 US Dollar into Pound sterling. (b) If Rs. 100 NC = Rs. 160 IC and 1 Chinese Yuan = Rs. 18.60, NC, find the exchange rate of Chinrse Yuan and Indian Rupees. 6. (a) 1 American dollar = Rs. 131.50. If Nepali currency is devaluated by 5%, find the new rae of exchange. (b) 1 Euro = Rs. 138.50. If Nepali currency is devaluated by 2%, find new rate of exchange. Using new rate of exchange convert 200 Euro into Nepali currency. (c) Given that £1 = Rs. 160. If Nepali currency is devaluated 2%, find (i) new rate of exchange. Using new rate convert £50 into Nepali currency and convert Rs. 60,000 into sterling Pound. 7. (a) Abdul works in a gas station in the US. He earns $ 25.6 per hour. i. Convert his income per hour into rupees. ii. If he works 9 hours a day, find his daily income in rupees. (Given $1 = Rs. 131.34) (b) Sonam works in a departmental store in Malaysiya. He makes 30 ringgit per hour. If 1 Malaysian ringgit = Rs. 29.82, i. find his hourly income in rupees. ii. find his daily income in rupees if he works 10 hours per day. iii. find his monthly income. (c) Aadhya works in a bank in the UK. She gets £ 45 per hour. Find i. her income per hour in rupees. ii. her daily income in rupees if she works 7 hours per day. iii. her weekly income in rupees if she works 5 days a week. [Use £1 = Rs.160.85] 8. (a) Panna Kaji needs $6000 for his tour. How much money does he need if $ 1 = Rs. 131.34 and the bank takes the commission of 2%? (b) Shaily is planning to visit Europe and she needs 3500 Euros. How much money does she need if 1 Euro = Rs. 138.50 and the bank takes the commission of 2%? (c) A man has to send £1500 to the UK for his sons study. How much money does he need if the bank takes a commission of 1.5% and the exchange rate is £ 1 = Rs. 161.80? (d) A man has to convert Rs. 5,00,000 into US dollar. How much dollar can be exchanged with the amount it bank takes the commission of 2%? 9. (a) A man converted Rs.7,34,500 into Euros for his business. After one week, the Nepali currency is devaluated by 2%. If the exchange rate before the devaluation was 1 Euro = Rs. 138.50, find -
Oasis School Mathematics-10 75 i. new exchange rate ii. his profit or loss. (b) A man converted Rs. 5,00,000 into Euro using the rate 1 Euro = Rs. 138.50. Immediately after that Nepali currency is devaluated by 2% and he again converted Euro into Nepali currency. Find (i) new rate of exchange (ii) his profit or loss by the devaluation of Nepali currency. 10. (a) A laptop is bought in the Indian market at Rs. 15,000 I.C. i. find its price in N.C. if 50% custom is added. ii. find its selling price if 13% VAT is added while selling. (b) A machine is bought in China at 2700 Chinese Yuan- i. find its price in the Nepali market if 80% customs duty is added. ii. find its selling price if it is sold at a profit of 20%. iii. how much does a customer have to pay if 13% VAT is levied? [Given 1 Chinese yuan = Rs. 18.60] (c) A merchant bought 15 tolas of gold at 350 U.S. Dollars per tola. What should be its cost in the Nepali market after paying 20% custom and adding 13% VAT. [Given 1 U.S. Dollar = Rs. 130.50] (d) Senmikha bought a photocopy machine from Japan at 2,50,000 Japanese Yen. It is imported to Nepal after paying 30% transportation charge and 120% customs duty. If he wants to make a profit of 80% on his total expenditure, find the selling price of the machine in Nepali market including 13% VAT. Given that 10 Japanese Yen = Rs. 9.65. 11. (a) The aeroplane fare from Kathmandu to Bangkok is Rs. 25,000 and from Bangkok to Kathmandu is 6,000 Thai Baht. Which one is cheaper? [Given 1 Baht = Rs. 3.77] (b) In the Nepali market, the cost of ghee is Rs. 800 per kg. In the Indian market, the cost is Rs. 500 per kg. Find which market is cheaper one and by what percent? [Rs. 100 IC = Rs.160 NC] 12. (a) Santosh bought 15 Nepali handicrafts at the rate of Rs. 1,800 per piece. He paid 10% export tax and sold them for 20 Euros per piece in Europe. Find his profit or loss. [1 Euro = Rs. 138.66] (b) Lakpa bought 600 Thanka at the rate of Rs. 800 per piece. He paid 15% export tax and sold them at £15 per piece in London, find how much profit he made. Also find the profit percentage. [Given, £1 = Rs. 160.89] (c) Ram Bharosh bought 120 carved windows at the rate of Rs. 2,500 per piece. After paying 10% export tax, at what rate should he sell it in Japan to make a profit of 50%. [Given 10 Yen = Rs. 9.65] 13. (a) A man converted Rs. 9,45,000 into Euro for his Europe tour at the rate of 1 Euro = Rs. 138.66. Later on his tour was cancelled because of the effect of Covid-19. At the same time Nepali currency is devaluated by 5%, find his profit or loss. (b) A man converted Rs. 5,00,000 to Euro for his Europe tour at the rate of 1 Euro = Rs. 138.66. He spent 500 Euro on his tour and returned back to Nepal. Find much Nepali currency is left with him?
76 Oasis School Mathematics-10 (c) A businessman converted Rs. 8,80,000 into American dollar at the rate of 1 dollar = Rs. 131.34. Altogether he spend 5000 dollar. During that time Nepali currency is devaluated by 2%, find how much Nepali currency is left with him. 14. Zenith works 5 days in a week in Australia and works 8 hours in a day. He earns 15 Australian dollars per hour. (a) How many hour does Zenith work in a week? (b) How much Nepali rupees does he earn in a week? (c) Keeping with him 1400 Australian dollar from his monthly income, he sent rest money to his home in Kathmandu. How much Nepali rupees will be obtained? [1 Aus. dollar = Rs. 90.40] 15. Srijana bought some Australian dollar fro Rs. 1,50,000. After 4 days the Nepali Rupees devaluted by 5%. On the same day she had to exchange the dollar to Nepali rupees. (a) How much Australian dollars did she buy? (b) Find the new exchange rate after 4 days. (c) How much profit or loss did she get? [1 Aus. dollar = Rs. 82.83] 16. You are going to educational tour to America. (a) Can you take Nepali rupee to America and spend this directly? (b) Who declares the currency exchange rate in our country Nepali? (c) If he needs $ 750 for tour, how much Nepali rupee does he need to save? [$ 1 = Rs. 106.80] 17. A project was constructed with total amount of Chinese Yuan 10,00,000. Afterthe contract, Nepali rupee was devaulated by 10%. [ 1 Yuan = Rs. 19.20] (a) What was the contract amount in Nepali rupee? (b) Find the new exchange rate offer devaluation? (c) How much should a contractor add to complete the project? 18. the exchange rate of money between various countries are the followings Rs. 134.60 = £ 1.50 = 1.52 marks, 4.8 francs = $ 1 and 1.66 marks = $ 1. (a) Find the rate of exchange between France and Kathmandu? (b) If direct rate of exchange is $ 1 franc = Rs. 28.50 find the profit if 100 franc are bought directly and exchange indirectly? 19. A shopkeeper has to pay 5% bonus for selling goods. He paid Rs.17500 bonus and then added 13% value added tax (VAt). (a) How much should he/she mark the goods for ? (b) What is value added tax (VAT)? In which fiscal year VAT rate as 13% was implemented? (c) Find the price of the goods with 13% VAT.
Oasis School Mathematics-10 77 Project Work 1. Collect the names of 10 people from your family or locality who are working in foreign country and from the source of the family ask their monthly income. Go through the exchange rate in the daily papers, convert their income into Nepali currency and present that in your class. 2. See the price of gold, silver and oil in the international market (in U.S. Dollar) and find out the rate of custom duty, VAT, etc. Convert their price into Nepali currency, including all taxes. Present your report in class. 3. Visit money exchange office nearby your home or take the exchange rate from daily newspaper. Convert Rs. 5,00,000 into US dollar, Euro, Sterling pound, etc. Show in the table how much foreign currency can we get if 2% commission is to be given. 4. From the different sources, find the exchange rate of currency of different SAARC countries with US dollar. Using this fact, find the exchange rate of Nepali currency with the e currency of SAARC countries. Tabulate it and present it in classroom. 1. (a) Rs. 19,701 (b) Rs. 40,222.50 (c) Rs. 58,930.50 (d) Rs. 17,945 (e) Rs. 16,960 2. (a) (i) 456.83 US Dollar (ii) 432.71 Euro (iii) Rs. 37,500 (iv) £372.92 (v) 3169.576 Chinese Yuan (vi) 15915.12 Thai Baht (b) (i) 25906.73 Japanese Yen (ii) 58.11 Kuwait Dinnar (iii) 250.39 Canadian Dollar (iv) 695.99 UAE Dirham (v) 176.97 Swiss Franc 3. (a) (i) 187.57 US. Dollar (ii) 5,00,176.17 Japanese Yen (iii) Rs. 24,972.19, (iv) 163.83 Euro 4. (a) Rs.3600 (b) Rs. 5920 (c) (i) 287.17 Euro, (ii) Rs. 8,31,960 (d) Rs. 88.50 5. (a) 408.41 (b) Rs. 9.75 6. (a) Rs. 115.50 (b) Rs. 120.36, Rs. 24072 (c) Rs. 142.80, Rs. 7140, 420.17 pound 7. (a) (i) Rs. 3362.30 (ii) Rs. 30260.74, (b) (i) Rs.894.60 (ii) Rs. 8946 (iii) Rs. 268380 (c) (i) Rs. 7238.25 (ii) Rs. 50,667.75 (iii) Rs. 253338.75 8. (a) Rs. 8,03,800.80 (b) Rs. 4,94,445 (c) Rs. 2,46,340.50 9. (a) (i) 1 Euro = Rs. 141.27 (ii) Profit = Rs. 14,690, (b) (i) Rs. 141.27 (ii) Profit Rs. 10,000 10. (a) (i) Rs. 36,000 (ii) Rs. 40,680 (b) (i) Rs. 75,816, (ii) Rs. 1,08,475.20 (iii) Rs. 1,22,576.98 (c) Rs. 61935.70 per tola (d) Rs. 1226756.25 11. (a) From Bangkok to Kathmandu (b) Same price 12. (a) Profit = Rs. 11,898 (b) Profit = Rs. 8,96,010, 162.32% (c) 4274.61 Yen per piece 13. (a) Profit = Rs. 47, 229.55, (b) Rs. 4,30,670 (c) Rs. 2,27,771.44 14. (a) 40 hours (b) Rs. 54,240 (c) Rs. 90,400 15. (a) 1810.94 Aus. dollars (b) 1 Aus. dollars = Rs. 86.97 (c) Profit Rs. 7497.50 16. (a) No. (c) Rs. 80,100 17. (a) Rs. 19200000 (b) 1 Yuan = Rs. 21.12 (c) Rs. 1920000 18. (a) 1 Franc = Rs. 30.62 (b) Profit Rs.212 19. (a) Rs. 3,50,000 (b) Consult your teacher. (c) Rs. 3,95,500 Answers
78 Oasis School Mathematics-10 Miscellaneous exercise Compound Interest 1. A sum of Rs. 5,120 in kept for 3 years at 10% per annum in the bank. Calculate S.I. and C.I. (a) Which formula isued to calculate C.I and S.I. (b) Calculate C.I. and S.I. separately (c) Which one is greater C.I. or S.I., why? Also find their difference. [(b)Rs. 1536, 1694.72 (c) Rs. 158.72] 2. A certain sum invested at 4% per annum compounded semi-annually amounts to Rs. 7,803 at the end of one year. Find the sum. (a) Write the value of R, C.A. and T. and write the formula to calculate P in this condition. (b) Calculate the value of P using this relation. (c) What is the value of C.I.? [(b) Rs. 7500 (c) Rs. 303] 3. The compound interest of a certain sum of money for 2 years at the rate of 5% p.a. is Rs. 410. Find the principal. (a) Write the value of T, R and C.I. and write their relation with P. (b) Calculate the value of P, using this relation. (c) What is the value of compound amount? [Ans: (b) Rs. 400 (c)Rs. 4410] 4. Rs. 16,000 invested at 10% p.a. compounded semi-annually amounts to Rs. 18,522. Find the time period of the investment. (a) Write the value of P, C.A., C.I. and R. (b) Which formula is used to calculate T in this condition? (c) Calculate the value of T. [Ans: (c) 3 years] 5. A man invested Rs. 60,000 at the rate of 1 paisa per rupee per month. In how many years will he get an interest of Rs. 7,200? (a) What is the rate percent per annum? (b) Write the value of P, C.I and C.A.? (c) Calculate the value of T using the relation P, C.I., R and T. [Ans: (c) 1 year] 6. A sum of Rs. 1,25,000 is invested in a bank, for 3 years. If the interest of Rs. 91,000 is generated, then (a) Write the value of P, C.I and T. (b) Write their relation with R. (c) Calculate the value of R. [Ans: (c) 20%] 7. A sum of money is invested for 2 years at the rate of 8% per annum. (a) Which scheme gives more interest by simple interest or compound interest? (b) If their difference is Rs. 360, calculate the value of P. [Ans: Rs. 56.250]
Oasis School Mathematics-10 79 8. A certain sum of money is invested for 2 years at the rate of 15% p.a. (a) Write the formula to calculate the S.I. and compound interest. (b) Using the above information, calculate the value of P if their difference is Rs. 90. (c) Calculate the C.I in this case. [(b) Rs. 4,000, (c) 9360] 9. The compound interest on a sum of money in 2 years at 8% per annum is Rs. 360 more than the simple interest. (a) Find the sum (b) Find the compound interest if it is compounded half yearly. (c) By how much C.I compounded half yearly is more than S.I? [(a) Rs. 56, 250, (b) Rs.9554.54 (c) Rs.554.59] 10. According to the system of yearly compound interest, a sum of money amounts in 2 years and 3 years for Rs. 12,100 and Rs. 13,310 respectively. (a) Find the rate of interest (b) Find the sum (c) Find the sum compounded half yearly in 2 years. [Ans: (a) 10%, (b) Rs. 10,000 (c) Rs. 1025] 11. A sum of money amounts to Rs. 19,360 in 2 years and Rs. 23,425.60 in 4 years at the rate of compound interest compounded annually. (a) Find the rate of interest (b) Find the sum (c) How much it would be compounded in 3 years? [Ans: (a) 10%, (b) Rs. 16,000 (c) Rs.4608.16] 12. 'A' invested Rs. 25000 for 3 years at the rate of 12% p.a. simple interest and 'B' invested the same amount for the same time at the rate of 10% annual compound interest. Calculate (a) interest received by 'A'. (b) interest received by 'B'. (c) without altering the time and the rate of interest, how much more or less money should 'A' invest for equal interest? [Ans: Rs. 9,000, Rs. 8,275, Rs. 2,013.89 less] 13. (a) What will be the simple interest on Rs. 20,000 for 2.1 years at the rate of 10% per annum ? (b) At what time does the same sum produce the same interest at the same rate of compound interest compounded annually ? [Ans: Rs. 4,200, 2 yrs.] 14. A man borrowed Rs. 30,000 for 2 years at the rate of 10% p.a. compounded annually. (a) How much money does he have to pay at the end of 2 years to clear the loan? (b) If he paid only half of the principal at the end of second year, how much he hare to pay to clear the loan after next 2years? (c) If the rate of interest is changed to 10% p.a. compounded annually to 10% p.a. compounded semi annually for last 2 years, how much more money does he have to pay? [Ans: (a) Rs. 36,300 (b) Rs. 25773 (c) Rs. 117.28 more]
80 Oasis School Mathematics-10 15. A man took a loan for 2 years. The interest in 1 year is Rs. 2,100 and the interest is 2 years 4641 respectively. (a) If the interest is compounded semi annually, find the rate of interest. (b) Find how much loan has he taken? (c) What would be the interest in 2 years if it is compounded yearly at the same rate? [Ans: (a) 20% p.a (b) Rs. 10,000 (c) Rs. 241 less] 16. A man took a loan of Rs. 5,00,000 at the rate of 10% p.a. compound interest compounded yearly. If he has to pay the loan in 3 equal instalment in 3 year, (a) What is the amount of loan at the end of first year? (b) If the yearly instalment is Rs. x, what amount of loan will be left at the end of second year, find it in terms of x. (c) Find the value of x. [Ans: (a) Rs. 5,50,000, (b) Rs. 6,05,000 – 0.1x (c) Rs.2,01,057.40] 17. A man took a loan at the rate of 10% p.a. compounded yearly. He paid the loan in three equal instalment of Rs. 50,500 per year. (a) If he has taken the loan of Rs. x, find how much loan is left to pay after paying the first instalment. (b) Find how much loan is left to pay after paying second instalment (c) Find the amount of loan that he has taken. [Ans: (a) (1.1x – 50,500) (b) Rs. (1.12x – 1,06,050) (c) Rs. 1,25,586] Growth and Compound Depreciation 1. The birth rate of the population of town is 8% every year and death rate is 1% per year. The present population of the town is 338000. (a) What is the population of growth rate? (b) What was the population before 3 years? (c) If 338000 was the population of the beginning of 2068 B.S., find the population borned within that year. Ans: [(a) 5%, (b) 291978, (c) 20280] 2. In the beginning of 2065 B.S, the population of a town was 100000 and the rate of population growth is 2% every year. In the beginning of 2066 B.S., 8000 people migrated there from different places. (a) What is population? (b) What was the population of the town in the beginning 2066 B.S.? (c) What was the population at the end of 2067 B.S? Ans: [(b) 1,20,000, (c) 1,44,444]
Oasis School Mathematics-10 81 3. The population of a city increases each year by 4% of what if had been at the beginning of the each year. The population in the beginning of 2075 B.S. was 6760000. (a) What was the population in the beginning of 2077 B.S.? (b) What was the population in the beginning of 2073 B.S.? (c) If 2073 B.S. was the base year, what was the population growth rate from 2073 B.S. to 2073 B.S.? Ans: [(a) 7311616, (b) 6250000] 4. Suman bought a taxi of Rs. 800000. He collected Rs. 300000 from the taxi fare but spent at Rs. 5,000 to his repair in 3 years. At the end of 3 years he sold it at a price which is depreciated at the rate of 10% p.a. (a) What was the cost of the taxi at the end of 3 years? (b) How much did he earn in 3 years? (c) What was his profit or loss in 3 years? Ans: [(a) 583200, (b) 250000, (c) 33,200] 5. Due the political instability of nation, a company's share price depreciated at the rate of 12% p.a. for 3 years. The present value of the shares is Rs. 85,184. (a) What was the value of shares before 3 years? (b) How many shares of Rs. 100 were sold 3 years ago? (c) What is cost of 1 kitta of share now? Ans: [(a) 1,25000, (b) 12,50, (c) 68.14] 6. 24000 blood donors were registered with a charitable hospital. The number of donors increased at the rate of 5% every 6 month. (a) In which system, the number of donors has increased? (b) Find the time period at the end of which the total number of blood donors be come 27783. (c) What was the flat rate of growth in 1 1 2 years? Ans: [(a) half yearly compound growth, (b) 1 1 2 years, (c) 55.76%] 7. The cost of a TV was quoted Rs. 17,000 at the beginning at the year 2013, the price was increased by 5% in 2014. Because of decreased in the demand, the cost was reduced by 4% in the beginning of 2015. (a) Write the relation between demand and price of the goods when other factors are constant. (b) What is the cost of the TV in 2015? Ans: [(b) 17,136] 8. The population of a village increased every year by 5%. At these of two years, the total population was 10000. If 1025 were migrate to other places. (a) Which model of population growth is in this context?
82 Oasis School Mathematics-10 (b) The present population P increased every year by R%. At the end of T years x people were migrated to other places. What will be the population of that place at the end of T years? (c) What was the population of the village in the beginning? Ans: [(a) PT = P (1+ R 100)T–x] 9. The cost of a refrigerator is Rs. 20000. It's value depreciates at the rate of 10% per annum. (a) What is the nature of depreciation? Simple or compound? (b) What will be its value in 3 years? (c) Find the depreciation in its value of 3 years? Ans: [(b) 14,580, (c) 5,420] Money exchange 1. Given, $ 1 = Rs. 130 and 1 Euro = Rs. 135 (a) Find how much Euros can be bought for $ 800. (b) If the Nepali rupees is devaluated by 2%, find how many Euro can be bought for $800 after giving 3% commission to the bank. (c) Find the rate of exchange between Euro and US dollar. [Ans: (a)Rs. 1,04,000, (b) 754.96 Euro, (c) 1 Euro = 1.04 US dollar] 2. Some electronic goods costing $ 25,000 were imported from America through India. A 2% shipment charge was paid in India. These goods were imported to Nepal from India by paying 150% customs duty. If $1 = Rs. 125, then: (a) Find the cost of goods in Nepali market. (b) What is the selling price to make a profit of 10%? (c) What is the net selling price including 13% VAT? [Ans:(a) 78,75,000, (b) 86,62,500, (c) 97,88,625] 3. Ramesh exchanged Rs. 2,70,000 into Euro for his Europe tour at the rate of 1 Euro = Rs. 135. (a) How much Euro can be bought from that amount? (b) If he spent 1500 Euro in his tour how much Euro is left with him at the end of his tour? (c) During his arrival in Nepal, Nepali currency is devaluated by 2% on the comparison of Euro, find how much Nepali currency is left with him? Ans: [(a) 2000 Euro, (b) 500 Euro, (c) Rs. 68.840] 4. 5,000 Nepali caps were bought in Nepal at the rate of Rs. 400 per cap. (a) It is exported to UK after paying 20% export tax. How much tax is to be paid altogether?
Oasis School Mathematics-10 83 (b) If it is sold at UK at the profit of 150%, at what rate should it be sold if Pound sterling 1 = Rs. 160. (c) Find the profit percent if it is sold at the rate of Pound sterling 8. Ans: [(a) Rs. 4,00,000, (b) Pound 7.5, (c) 166.67%] 5. A machine is bought from the Indian market at Rs. 80,000 (IC). (a) Find the price of the machine in NC if Rs. 100 (IC) = Rs. 160 (NC) (b) Find the price in NC if 80% custom is added on the boarder. (c) If person spent Rs. 8,000 (NC) in transportation and wants to make the profit of 50% on his total expense, find at what price did he sell in Nepali market. (d) What is the total amount paid by the customer if 13% VAT is composed on it. Ans: [(a) Rs. 1, 28,000, (b) Rs.2,30,400, (c) 3,57,600, (d) 4,04,088] 6. A man exchanged some Nepali currency into US dollar at the rate of $1= Rs. 130. After some days, Nepali currency devaluated by 5%, and he converted dollar into Nepali rupees. (a) If he made the profit of Rs. 55,000, find how much Nepali currency does he have at first? (b) What would be the profit if Nepali currency is devaluated at 2% only? (c) What is conclusion from the result of (a) and (b)? Ans: [(a)Rs. 11,00,000, (b) Rs. 22,000 ] 7. Sapana bought 500 Nepali Thanka at the rate of Rs. 200 per piece. (a) What is the net C.P. of a Thanka set if he paid 10% export tax? (b) She sold it Europe at the rate 4 Euro per piece, what is his profit percent ? [Given that: 1 Euro = Rs. 138] (c) At what price per piece should he sell it to make the profit of 80%? Ans: [(a) Rs. 1,10,000, (b)150.9%, (c) 2.87 Euro]
84 Oasis School Mathematics-10 Attempt all the questions. 1. A man kept Rs. 20,000 in a bank at the rate of 10% p.a. compound interest compounded annually. (a) Find the C.I. received by him at the end of first year. (1) (b) If he withdrew Rs. 5,000 from the bank at the end of first year, find how much money is left at the end of second year. (2) (c) If he again deposited Rs. 8,000 in the bank. Find the compound interest received by him in third year, provided that interest is compounded half yearly in the last year. (2) 2. Some electrical goods costing Rs. 80,000 (IC) is bought from the Indian market. (a) Find its cost in NC if Rs. 100 (IC) = Rs. 160 (NC) (1) (b) If 120% custom is imposed, what is the net cost price of the items. (1) (c) If a man paid Rs. 5,000 transportation charge and made the profit of 50^ on his total expenditure, find the selling price and net selling price if 13% VAT is imposed.(3) 3. A man bought a land costing Rs. 10,00,000 and made a home with the budget of Rs. 25,000,000. (a) What is his total expenditure in the land and home? (1) (b) If value of land increases 10% every year what will be its value after 2 years. (2) (c) If the value of house decreases at 5% every year, what will be its value after 2 years? (2) (d) What percent of increment or decrement in the total value is there after 2 years? (2) 4. A shopkeeper bought a machine from the Chinese market at Rs. 5,000 Chinese Yuan. (a) What is its cost price in Nepali rupees if 1Chinese Yuan = Rs. 16.50. (1) (b) If 20% of the cost price is spent on transportation charge, how much be spent for the machine. (1) (c) If 80% customs is imposed, what is the net C.P of the machine including transportation charge. (2) (d) If he wants to make the profit of 50% on his total expenditure, what should be the market price? (1) (e) If 20% discount is allowed and 13% VAT is imposed, at what price he has to sell it to be the customer? (2) Full Marks – 24
Oasis School Mathematics-10 85 MeNSURAtION Contents • Pyramid • Combined Solids • Cost Estimation expected Learning Outcomes At the end of this unit, students will be able to develop the following competencies: • To calculate the surface area and volume of different types of (circular based, square based) Pyramid • To calculate the surface area and volume of given combined solids • To estimate the cost of construction of different solid object or different Geometrical bodies. Materials Required • Model of different types of Pyramid, model of different combined solids. Estimated Teaching Hours 28
86 Oasis School Mathematics-10 Look at the given figures. • It has circle on its base. It is a circular based Pyramid. It has a triangular base, it is a triangular-based pyramid. It has a rectangular base. So it is a rectangular-based pyramid. Vertex Height Base Height Vertex Edge Base Unit 5 Pyramid 5.1 Warm-up Activities Observe the given pictures and discuss the answer of following questions. • What is the name of given pictures? • What is the shape of its surface? • How many such faces are there? • What is the shape of its base? Vertex Edge Height Base
Oasis School Mathematics-10 87 It has a square base. So it is called a square-based pyramid. If a pyramid has a pentagon as a base, then it is called a pentagonal-based pyramid. If a pyramid has a hexagonal base, it is a hexagonal-based pyramid. Draw out the conclusion from the figures given above. • What is shape of the base of the pyramid. • What is the possible shape of its lateral surface? • What is the common part of different plane of its lateral surface? • What is the relation of height with the base? • The base of the pyramid may be circular triangular, square, rectangular, polygonal, etc. • Every lateral surface of a pyramid is triangular. • Each triangular shape of the lateral surface has a common vertex. • The height of a pyramid is perpendicular to the base. Activity Draw a circle and draw a sector AOB as shown in the figure. Cut the sector AOB with the help of scissors. Fold this sector such that OA just coincides with OB. The structure so formed is cone. B O A O A B O Vertex Edge Height Base base Vertex Edge Height Base vertex height height Slant base Edge
88 Oasis School Mathematics-10 Surface area of circular based pyramid Take a hollow cone and cut it along its slant height as shown in the figure. You will get a sector as shown in the figure. The arc length of the sector is equal to 2πr as it is equal to the circumference of the base of the original cone. Its radius is equal to l. Now, length of this shape = 2πr 2 = πr Breadth = l Area of this shape = πr × l = πrl Which represents the CSA of the cone. ∴ Curved surface area of the cone (CSA) = πrl. Again, Total surface area = Curved surface area + Area of base = πrl + πr2 = πr (l + r) Hence, TSA of a cone = πr (r + l). Volume of circular based pyramid (cone) Take a hollow cone and a hollow cylindrical jar having the same radius and height. Fill the cone with water and pour it into the cylindrical jar. You can see that three cones full of water just fill the cylinder. Then, Volume of cylinder = 3 × Volume of cone Volume of cone = 1 3 Volume of cylinder = 1 3 (πr2 h) = 1 3 πr2 h Hence, volume of cone = 1 3 πr2 h = 1 3 (area of base) × height = 1 3 A×h r " 1 2 3 4 5 6 7 8 l 2πr [Circumference of a cone is equal to the arc length of the sector] 1 2 3 4 5 6 7 8 l πr πr
Oasis School Mathematics-10 89 Remember ! • CSA = πrl • TSA = πr (r + l) • Volume (V) = 1 3 πr²h • Volume (V) = 1 3 area of base × height • TSA = CSA + Area of base • Relation among r, l and h is l2 = h2 + r2 r h l For circular based Pyramid, Worked Out examples example 1 the area of the circular base of a cone is 100 cm2 and its height is 18 cm. Find the volume of the given circular based pyramid.. Solution: Here, In the given circular based pyramid, Area of the circular base (A) = 100 cm2 Height (h) = 18 cm We have, Volume (v) = 1 3 area of circular base × height = 1 3 × 100 × 18 cm3 = 600 cm3 example 2 the radius of the circular base of a cone is 7 cm and slant height is 20 cm. Find its curved surface area and total surface area. Solution: Here, Radius of the circular base (r) = 7 cm Slant height (l) = 20 cm We have, the curved surface area (CSA) = π r l. = 22 7 × 7 × 20 cm2 = 440 cm2 Again, and total surface area (TSA) = π r (l + r) = 22 7 × 7 (20 + 7) = 22 × 27 cm2 = 594 cm2
90 Oasis School Mathematics-10 example 3 Find the volume of the given circular based pyramid. Solution: Here, Radius of circular base (r) = 8 cm Slant height (l) = 10 cm so, height of cone (h) = l 2 – r2 = 100 – 64 = 36 = 6 cm We have, volume (V) = 1 3 π r2 h = 1 3 × 22 7 × 82 × 6 cm3 = 402.28 cm3 example 4 Curved surface area and total surface area of a cone are 1430 7 cm² and 1980 7 cm² respectively. Find the slant height of the cone. Solution: Given, curved surface area = 1430 7 cm2 Total surface area = 1980 7 cm2 We have, TSA = CSA + area of base or, area of base = 1980 7 – 1430 7 or, πr² = 550 7 or, 22 7 r² = 550 7 or, r² = 550 22 or, r2 = 25 or, r = 5 cm Again, we have, CSA = πrl 1430 7 = 22 7 × 5 × l or, 22 × 5 l = 1430 or, l = 1430 22 × 5 or, l = 13 cm Slant height = 13 cm P O h 10 cm 8 cm
Oasis School Mathematics-10 91 example 5 Find the total surface area of the given cone, whose semi-vertical angle is 30° and the radius is 7 cm. Solution: In the given cone Radius (r) = 7 cm ∠BOC = 30° In right-angled triangle BOC, sin 30° = BC OB 1 2 = r l or, l = 2r or, l = 2×7 cm = 14 cm We have, TSA of a cone = πr (r + l) = 22 7 × 7 (7 + 14) = 22 × 21 cm² = 462 cm² exercise 5.2 1. (a) If 'A' be the area of the base of the cone and 'h' be its height, write the formula to calculate the volume of the cone. (b) Write the formula of the volume of a cone in terms of 'r' and 'h'. (c) Write the relation among radius (r), height (h) and slant height (l) of a cone. (d) Write the formula for the curved surface area of a cone. (e) Write the formula for the total surface area of a cone in terms of CSA of cone and area of base. (f) What is the formula of the TSA of a cone in terms of 'r' and 'l'? 2. Find the curved surface area and total surface area of a cone having, (a) radius of the base = 7 cm, slant height = 25 cm . (b) radius of the base = 10 cm, slant height = 15 cm. 3. Find the volume of the cone having (a) radius of the base = 7 cm, height = 24 cm. (b) radius of the base = 5 cm, height = 12 cm. (c) area of the base = 300 cm², height = 10 cm. 4. (a) If the slant height and radius of a cone are 10 cm and 6 cm respectively, find its curved surface area. (b) The diameter of the circular base of a cone is 20 cm and its slant height 21 cm. Find its curved surface area and total surface area. O A B 30° C 7 cm
92 Oasis School Mathematics-10 (c) The circumference of the base of a cone is 22 cm, and the sum of its radius and slant height is 16 cm. Find its total surface area. (d) If the circumference of the base is 44 cm and the slant height is 15 cm, find the total surface area of the cone. 5. (a) The radius of the circular base of a cone is 14 cm and its height is 10 cm. Find its volume. (b) If the area of the base of the cone is 250 cm² and the height is 5 cm, find its volume. (c) If the perimeter of the base of the cone is 88 cm and the vertical height is 10 cm, find the volume of the cone. 6. Find the curved surface area and total surface area of the following solids. (a) (b) (c) (d) 7. Find the volume of the given solids. (a) (b) (c) (d) 8. (a) A cone whose radius of base is 7 cm, has a volume of 1,232 cm³. Find the height of the cone. (b) If the volume of the given cone is 23,100 cm³ and its height is 50 cm, what is the radius of the base? (c) A cone 21 cm high has a volume of 1,078 cm3 . Calculate the radius of its base. (d) The volume of the cone is 324 π cm³ and the radius of the base is 9 cm. Find the curved surface area of the cone. 9. (a) The total surface area of the cone whose radius is 10 cm is 594 cm2 . Find the slant height of the cone. (b) The curved surface area of a cone is 8,880 cm². If the slant of the height is 100 cm, find the height of the cone. (c) The total surface area and the diameter of the base of the cone are 300π cm² and 24 cm respectively. Find the slant height of the cone. (d) Total surface area of the cone is 2,816 cm². If the radius is 14 cm, find (i) slant height of the cone (ii) height of the cone (iii) volume of the cone (e) Curved surface area of a cone is 550 cm². If its diameter is 14 cm, find its volume. 15 cm 7 cm 12 cm 12 cm 5 cm 25 cm 16 cm 6 cm 12 cm 5 cm 24 cm 25 cm B O 3 cm A 5 cm B O 14 cm 25 cm A
Oasis School Mathematics-10 93 10. (a) In the given cone, its radius and height are equal. If the volume of the cone is 9702 cm³, find the height of the cone. (b) The radius and the height of the cone are in the ratio 3:4. If the volume of the cone is 96π cm3 , find the slant height. (c) The ratio of height and slant height of a cone is 24:25. If its volume is 1,232 cm3 , find its lateral surface area and total surface area. (d) The vertical height of a cone is 3 times its radius. If its volume is 27π cm3 , find its radius, height and slant height. 11. (a) The total surface area and the curved surface area of a cone are 704 cm² and 550 cm² respectively. Find the slant height of the cone. (b) The total surface area and the curved surface area of a cone are 2,816 sq. cm and 2,200 sq. cm respectively. Find the radius and slant height of the cone. 12. (a) If the given figure of the solid cone has a base diameter of 10 cm and slanting height 13 cm, find: (i) area of the curved surface. (ii) total surface area. (iii) volume of the cone. [π = 22 7 ] (b) If the given figure of the solid cone has a base radius of 7 cm and slanting height 25 cm, find: (i) area of the curved surface. (ii) total surface area. (iii) volume of the cone. [π = 22 7 ] (c) In the given figure, the semi-vertical angle of the cone is 30°. If its diameter is 14 cm, find (i) its CSA. (ii) its TSA. (iii) its volume. (d) In the given figure, the semi-vertical angle of the cone is 30°. If its height is 20 cm, find (i) its CSA. (ii) its TSA. (iii) its volume. 13. (a) If the water filled in the cone is poured into the cylinder, to what height will the surface of the water reach? 10 cm 13 cm 7 cm 25 cm 30° 14 cm 20cm 30° 14 cm 24 cm 14 cm
94 Oasis School Mathematics-10 (b) A cylinder of height 40 cm and diameter 14 cm is melted to form a right circular cone of height 30 cm. Find the radius of the cone. 14. A conical shaped tent is made as shown in the figure. Its height is 24ft and diameter is 14 ft. (a) Find the CSA of the tent. (b) If you have 660 ft² cloths to cover the tent, how much more or less cloths is required for the tent? (c) Keeping the radius same, what should the slant height of the tent if 660 ft² cloths is used to cover the tent. 15. A cylindrical tent is made whose radius is 7m, with the cloths having area 1034 cm². The top of the tent is covered by the cloths. (a) Find the height of the tent. (b) With the same cloths if you are going to make the conical tent of same radius, what will be its height. Answers 1. Consult your teacher 2. (a) 550 cm2 , 704 cm2 (b) 471.42 cm2 785.71 cm2 3. (a) 1,232 cm3 (b) 314.28 cm3 (c) 1,000 cm3 4. (a) 188.57 cm2 (b) 660 cm2 , 974.28 cm2 (c) 176 cm2 (d) 484 cm2 5. (a) 2053.33 cm3 (b) 416.67 cm3 (c) 2,053.33cm3 6. (a) 330 cm2 484cm2 (b) 471.43 cm2 , 584.57 cm2 (c) 204.29 cm2 , 282.86 cm2 (d) 251.42 cm2 , 452.57 cm2 7. (a) 314.28cm3 (b) 1,232 cm3 (c) 37.71cm3 (d) 1,232cm3 8. (a) 24cm (b) 21cm (c) 7cm (d) 424.29cm2 9. (a) 8.9 cm (b) 96cm (c) 13 cm (d) (i) 50 cm (ii) 48 cm (iii) 9,856cm3 (e) 1,232 cm3 10. (a) 21cm (b) 10cm (c) 550cm2 , 704cm2 , (d) r = 3cm, h=9 cm, l = 9.49cm 11. (a) 25cm (b) 14 cm, 50cm 12. (a) (i) 204.28 cm2 (ii) 282.86 cm2 (iii) 314.28 cm3 (b) (i) 550 cm2 (ii) 704 cm2 (iii) 1,232 cm2 (c) (i) 308 cm2 (ii) 462cm2 (iii) 622.38 cm3 (d) (i) 761.9cm2 (ii) 1,180.94cm2 (iii) 2,793.65cm3 13. (a) 4 cm (b) 14cm 14. (a) 550 ft² (b) 110 f² less (c) 30ft. 15. (a) 20m (b) 46.48m Project Work Draw the net of circular based pyramid. Make a circular based pyramid. Measure its radius, height and slant height then find its base area, curved surface area, total surface area and volume. Present the result in your 24ft 14ft
Oasis School Mathematics-10 95 5.3 Square Based Pyramid Observe the given solid objects and answer the question given below. • How many faces does it have? • How many square face is there? • How many triangular faces are there? • Are the area of all triangular faces equal. Prepare the solid object like this and find its different feature. Volume of a pyramid Prepare a net of a pyramid as shown in the figure. Fold and paste them to form a pyramid. Make three pyramids of the same size. Paste 3 Pyramids, then a cube is formed. Now, 3 × volume of the pyramid = Volume of a cube Volume of the pyramid = 1 3 volume of the cube = 1 3 area of base × height = 1 3 a2 h Hence, the volume of the pyramid = 1 3 area of base × height = 1 3 a2 h Lateral surface and total surface area of square-based pyramid The given figure is a square-based pyramid, where ABCD is its base. OM is its slant height. ON is its height. Let, AB = BC = CD = AD = a OM = l and ON = h Area of ∆OBC = 1 2 BC × OM = 1 2 × a × l. Lateral surface area (LSA) = 4 × ∆OBC = 4 × 1 2 al. = 2al Total surface area (TSA) = LSA + area of base = 2al + a2 ∴ TSA of square-based pyramid = 2al + a2 Where a = length of a side of the base, O A D B C a a 2 M D C N A B O h l e BD = d DN = NB = d 2
96 Oasis School Mathematics-10 l = slant height of pyramid (height of the triangle) In the given figure, ABM is a right angled triangle. So, OB2 = OM2 + BM2 e2 = l2 + a2 4 Similarly, in right angled triangle ONB, OB2 = ON2 + NB2 e2 = h2 + d2 4 Remember ! • Area of all triangular faces of a square-based pyramid is equal. • Area of opposite triangular faces is equal in a rectangular pyramid. In a square-based pyramid, ∆ OEF is a right-angled triangle. • l 2 = h2 + a2 4 • e2 = l 2 + a2 4 • Area of ∆OBC = 1 2 BC × OF = 1 2 al • EF = 1 2 DC • Area of base = a2 = 1 2 d2 if diagonal is given. • LSA = 4 ∆OBC = 2al • TSA = 4 ∆ + area of base = 2al + a2 • Volume = 1 3 area of ABCD × OE = 1 3 area of base × height = 1 3 a2 h • Volume = 1 6 d2 h, if diagonal ‘d’ is given. e Base O E A B D C F Worked Out examples example: 1 Find the total surface area and volume of the given pyramid. Solution: In the given square-based pyramid, • If diagonal ‘d’ of the base is given, area of base = 1 2 d2 . Note • If 'e' be the length of the edge of the pyramid then, e2 = l 2 + ( a 2 ) 2 , e2 = l 2 + a2 4 • If 'd' be the diagonal of the base, 'e' the length of the edge and h the vertical height of the pyramid then, e2 = h2 + ( d 2 )2 = h2 + d2 4 . 16 cm 6 cm O A B C D P Q