The Leading Maths - 10

Arithmetic 50 Allied The Leading Mathematics-10 Compound Interest 51 Example-7 A woman deposits Rs. 7000 for three years if the rates are 9%, 9.5% and 10% for the 1st, 2nd and 3rd years respectively. (a) Which formula is used for finding the compound interest when the rates of interest are different per year ? (b) Find the compound interest in each year for the given information. (c) Verify that the sum of compound interest of three years taken in separate is the compound interest of three years taken at once. Solution : (a) The formula CI = P[(1 + r1) (1 + r2) (1 + r3) – 1] is used for finding the compound interest when the rates of interest are different per year. (b) For the 1st year, Principal (P1) = Rs. 7000, Time (T) = 1 year, Rate for the 1st year, r1 = 9%, Compound Interest (CI1) = ? Now, we have CI1 = P1[(1 + r1) – 1] = 7000[(1 + 9%) – 1] = 7000 × 0.09 = 630 For the 2nd year, Principal (P2) = Rs. 7000 + 630 = 7630, Time (T) = 1 year, Rate for the 2nd year, r2 = 9.5%, Compound Interest (CI2) = ? Now, we have CI2 = P2[(1 + r2) – 1] = 7630[(1 + 9.5%) – 1] = 7630 × 0.095 = 724.85 For 3rd year, Principal (P3) = Rs. 7630 + 724.85 = 8354.85, Time (T) = 1 year, Rate for the 3rd year, r3 = 10%, Compound Interest (CI3) = ? Now, we have CI3 = P3[(1 + r3) – 1] = 8354.85[(1 + 10%) – 1] = 8354.85 × 0.1 = 835.485 ∴ Total interest, CI = CI1 + CI1 + CI1 = 630 + 724.85 + 835.485 = 2190.335 (c) Here, Principal (P) = Rs. 7000, Time (T) = 3 years, Rates r1 = 9%, r2 = 9.5%, r3 = 10%, Compound Interest (CI) = ? Now, we have CI = P[(1 + r1) (1 + r2) (1 + r3) – 1] = 7000[(1 + 9%) (1 + 9.5%) (1 + 10%) – 1] = 7000[(1 + 0.09) (1 + 0.095) (1 + 0.1) – 1] = 7000[(1.09) (1.095) (1.1) – 1] = 7000[1.312805 – 1] = 7000 × 0.312805 = 2190.335 Hence, the sum of compound interest of three years taken in separate is the compound interest of three years taken at once.

Arithmetic 52 Allied The Leading Mathematics-10 Compound Interest 53 Example-8 A sum of money becomes 2.25 times of itself in 2 years compounded yearly. (a) Find the rate of compound interest. (b) At the same rate of compound interest, what will the amount of Rs. 125000 be for three years ? Solution : (a) Here, Rate (R) = ?, Time (T) = 2 years, Principal (P) = Rs. x (Suppose) Then, Amount (A) = Rs. 2.25x. ∴ CI = Rs. (2.25x – x) = Rs. 1.25x. We know, CA = P 1 + R 100 T or, 2.25x = x 1 + R 100 2 or, (1.5)2 = 1 + R 100 2 ∴ 1.5 = 1 + R 100 or, 0.5 = R 100 or, R = 50%. Example-9 If the compound amounts of a sum of money in 3 years and 4 years are Rs. 3240 and 3888 respectively, (a) find the rate of interest compounded yearly. (b) find the sum of money. Solution: (a) Case I; Compound amount in 3 years (CA1) = Rs. 3240 or, P(1 + R%)3 = Rs. 3240 .......... (i) Case II; Compound amount in 4 years (CA2) = Rs. 3888 or, P(1 + R%)4 = Rs. 3888 ......... (ii) By dividing eqn (ii) by the eqn (i), we get P(1 + R%)4 P(1 + R%)3 = 3888 3240 or, 1 + R% = 1.2 or, R 100 = 1.2 – 1 or, R = 0.2 × 100 = 20%. (b) Putting the value of R in the eqn (i), we get P(1 + 20%)3 = Rs. 3240 or, P 1 + 20 100 3 = Rs. 3240 or, P × 1.23 = Rs. 3240 (b) Principal (P) = Rs. 125000, Time (T) = 3 yrs, Interest rate, r = 50%, Compound amount (CA) = ? Now, we have CA = P(1 + r) = 125000(1 + 50%)T = 125000(1 + 0.5)3 = 125000 × 1.53 = 125000 × 3.375 = 421875 Hence, the amount of Rs. 125000 will be Rs. 421875 after 3 years. or, P × 1.728 = Rs. 3240 or, P = 3240 1.728 = Rs. 1875. Hence, the required rate of interest and the sum are 20% and Rs. 1875 respectively.

Arithmetic 52 Allied The Leading Mathematics-10 Compound Interest 53 "Alternatively" CA1 = Rs. 3240 and CA2 = Rs. 3888 ∴ Interest for 1 yr. on CA1=CA2 – CA1 or, SI = 3888 – 3240 or, PTR 100 = 648 or, 3240 × 1 × R 100 = 648 or, R = 648 × 100 3240 = 20%. Example-10 A man took a loan of Rs. 32,760 from a bank at 20% compounded annually and paid it back in three years in three equal annual installments. (a) Calculate the amount of each installment. (b) How much amount did he pay in total ? Find. Solution : (a) Let each installment be of Rs. x For 1st installment; A = P 1 + R 100 T or, x = P 1 + 20 100 1 or, P = Rs. 100x 120 Given, the total sum borrowed = Rs. 32760 i.e., 100x 120 + 100 × 100x 120 × 120 + 100 × 100 × 100x 120 × 120 × 120 = 32760 or, 100x 120 1 + 100 120 + 100 × 100 120 × 120 = 32760 or, 100x 120 14400 + 12000 + 10000 120 × 120 = 32760 or, x = 32760 × 120 × 120 × 120 100 × 36400 or, x = 15552 Hence, each installment is Rs. 15552. (b) The total paid amount by the man = Amount in 3 installments = 3 × Rs. 15552 = Rs. 46,656 Again, CA1 = P 1 + R 100 3 or, 3240 = P 1 + 20 100 3 or, 3240 = P × 1.728 or, P = 3240 1.728 = Rs. 1875. For 3rd installment; x = 1 + 20 100 2 or, P = 100 × 100 × 100x 120 × 120 × 120 For 2nd installment; x = P 1 + 20 100 2 or, P = Rs. 10000x 120 × 120

Arithmetic 54 Allied The Leading Mathematics-10 Compound Interest 55 PRACTICE 2.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the simple interest on the sum of P at the rate of R% in T years? (b) Write the formula to calculate the compound amount when the principal, rate and time are given. (c) What is the formula to compute the compound interest when the principal, rate and time are given? (d) Write the formula to calculate the compound interest when the principal, rate and time in years and months are given. Check Your Performance Answer the following questions for each problem. 2. Which values are inserted in the following blanks? Compute. Principal Rate of Compound interest Time (yrs) Compound Amount (a) Rs. 700 8% p.a. 3 (yearly) ------------ (b) Rs. 12000 6% p.a. 1 1 2 (half yearly) ------------ (c) Rs. 1,50,000 12% p.a. 2 (quarter yearly) ------------ (d) ------------ 13% p.a. 2 (yearly) Rs. 191535 (e) Rs. 200000 10% p.a. ----- (yearly) Rs. 292820 (f) Rs. 500000 ---------- 3 (yearly) Rs. 760437.50 3. (i) Sunita plans to deposit Rs. 21000 in any bank for 3 years. (a) Define compound interest. (b) How much amount will she get at 10% simple interest when she deposits it in a bank N ? Calculate it. (c) How much amount will she get at 91 2 % interest being compounded annually when she deposits it in another bank M ? Calculate it. (d) Which bank is better to deposit the sum and how much more amount will she get from the better bank ? (ii) Bishal thinks to burrow Rs. 15000 from Sovan for 2 years. (a) Write any one difference between simple interest and compound interest. (b) How much total amount would Bishal have to pay to Sovan at the rate of interest of 10% p.a. compounded annually after 2 years? Find. (c) How much total amount would Bishal have to pay to Sovan at the rate of simple interest of 11% p.a. after 2 years? Find. (d) In which process would Bishal pay more amount to Sovan and by how many more percentage ?

Arithmetic 54 Allied The Leading Mathematics-10 Compound Interest 55 4. (i) A girl plans to deposit Rs. 10000 in the fixed deposit account in any cooperative limited for 3 years 6 months. (a) How is compound interest yearly and half yearly different ? (b) A cooperative A pays compound interest compounded semi-annually at the rate of 16% p.a. (c) Another cooperative B pays compound interest compounded annually at the rate of 17% p.a. (d) In which bank will she deposit it to get more interest and by how much percent ? (ii) Rohit plans to deposit Rs. 150000 in the fixed deposit account in any bank. (a) A bank P pays compound interest compounded annually at the rate of 12.13% p.a. for 3 1 2 years. (b) A bank Q pays compound interest compounded semi-annually at the rate of 11.75% p.a. for the same time. (c) Which bank will be better to deposit the money ? Why ? 5. (i) A person borrowed Rs. 16000 from a bank at 12.5% per annum simple interest for 3 years and but he loans the whole amount to a shopkeeper at the same rate of compound interest for the same time. (a) Find the amount to be paid to the bank as simple interest. (b) Find the amount to be get to the shopkeeper as compound interest. (c) How much will he gain after 3 years ? (ii) Krishna borrowed Rs. 250000 from a bank at 8% per annum compound interest annually for 2 years and but the whole amount lent to his friend at the same rate of compound interest semi-annually for the same time. (a) Find the amount to be paid to the bank as compound interest annually. (b) Find the amount to be gotten to the shopkeeper as compound interest semiannually. (c) How much will he gain after 2 years? 6. (i) A bank provides compound interest and simple interest on Rs. 5120 for 3 years at 12.5% per annum in two ways. (a) What is the compound interest for the given information? (b) What is the simple interest for the given information? (c) Find the difference between compound interest and simple interest. (ii) The difference between the simple and compound interest on a sum of money at the rate of 10% p.a. is Rs. 240 in 2 years. (a) Let us suppose the required sum be x. Find the simple interest in term of x. (b) Find the compound interest in term of x. (c) Find the value of x. What is the sum ? Write. (iii) The simple interest on a sum of money in 2 years is Rs. 90 less than its compound interest. If the rate of interest in a year is 15%, find the sum.

Arithmetic 56 Allied The Leading Mathematics-10 Compound Interest 57 7. (i) A girl deposits Rs. 500 in a bank and gets Rs. 605 at 10% per annum compounded annually after some time. (a) Find the time when she gets Rs. 605 from the bank. (b) What is the simple interest on the same sum at the same rate for the same time ? (c) Which interest is more compound interest or simple interest and by how much percentage ? Find. (ii) A finance company provides the interest at 20 paisa per rupee per year. (a) What is the rate on compound interest ? (b) In what time will Rs. 1,500 yield Rs. 496.50 as compound interest at the same rate of compounded semi-annually ? (c) What is the compound interest on the same sum at the same rate for the same time compounded annually? Find it. (d) Which interest is more compound interest compounded semi-annually or compound interest annually and by how much percentage ? Find. 8. (i) A man gets Rs. 784 by depositing Rs. 625 in a bank for 2 years compounded annually. (a) What is the rate on compound interest ? (b) What is the compound interest on the same sum at the same rate for the same time compounded quarter annually? Find it. (ii) A girl deposits Rs. 2000 for 2 years at the rate of 10% per annum. (a) Find the compound interest for the given information. (b) At what rate of simple interest with the same sum will produce the same interest in the same period? (iii) The compound interest on a certain sum for 2 years at 10% per annum is Rs. 420. (a) What is the rate on compound interest ? (b) What would be the simple interest on the same sum at the rate for the same time ? 9. (i) Rs.16000 is deposited for 2 years at 15% per annum compounded quarter annually. (a) Which formula is used for finding compound amount as quarter yearly interest? (b) Find the compound amount for the given information. (c) Find the compound interest. (ii) Rs. 250000 is deposited for 1 3 4 years at 15% per annum compounded quarter annually. (a) Which formula is used for finding compound amount when the time is given in years and quarter year? (b) Find the compound amount for the given information. (c) Find the compound interest. 10. (i) A man borrowed Rs. 50000 for 2 years under the following terms, 8% per annum CI for the first year and 10% per annum CI for the second year. (a) Which formula is used for finding the compound interest when the rates of interest are different per year ?

Arithmetic 56 Allied The Leading Mathematics-10 Compound Interest 57 (b) Find the compound interest in each year for the given information. (c) Verify that the sum of compound interest of three years taken in separate is the compound interest of three years taken at once. (ii) Sarita deposits Rs. 15000 for 3 years at the rate of 4%, 5% and 6% per annum for the 1st, 2nd and 3rd year respectively. (a) Which formula is used for finding the compound interest when the rates of interest are different per year ? (b) Find the compound interest in each year for the given information. (c) Verify that the sum of compound interest of three years taken in separate is the compound interest of three years taken at once. 11. (i) A sum of money becomes 2 times of itself in 4 years compounded yearly. (a) Find the rate of compound interest. (b) At the same rate of compound interest, what will the amount of Rs. 300000 for three years ? (ii) A sum of money becomes 3 times of itself in 5 years compounded yearly. (a) Find the rate of compound interest. (b) At the same rate of compound interest, what will the amount of Rs. 750000 for 4 years only ? 12. (i) A certain sum amounts to Rs. 3380 in 2 years and to Rs. 3515.20 in 3 years at compound interest. (a) Find the rate of interest compounded yearly. (b) Find the sum of money. (ii) A certain sum of money at compound interest becomes Rs. 7396 in 2 years and Rs. 7950.70 in 3 years. (a) Find the rate of interest compounded yearly. (b) Find the sum of money. 13. (i) The compound interest on a certain sum for 2 years is Rs. 1025 and for 3 years is Rs. 1576.25. (a) Find the rate of interest compounded yearly. (b) Find the sum of money. (ii) The compound interest calculated yearly on a sum of 1 year and 2 years is Rs. 80 and Rs. 168 respectively. (a) Find the rate of interest compounded yearly. (b) Find the sum of money. 14. (i) A sum of money is paid back in two annual installments of Rs. 1764 each with adding 5% compound interest. (a) How much the total amount is to be paid? (b) What was the sum borrowed? (ii) A sum of Rs. 5040 is borrowed at 10% CI compounded yearly and is paid back in two years in two equal installments. (a) Calculate the amount of each installment. (b) How much amount was paid in total ? Find.

Arithmetic 58 Allied The Leading Mathematics-10 Compound Interest PB 15. (i) A woman borrowed a certain sum of money at 3% p.a. simple interest and invested it at 5% p.a. compounded interest. After three years she makes a profit of Rs. 1082. (a) What sum did she borrow ? (b) What percent of profit did she get? (ii) A man borrowed a sum of Rs. 500000 at 12% p.a. compounded interest. On the same day, he lent out this money to a woman at the same rate of compound interest half yearly. (a) How much money would she have to pay to the man in total ? Find. (b) What percent of profit did he get after 5 years? Find. 16. (i) A girl wants to deposit Rs. 50,000 in two banks such that they give equal amounts of compound interest after 2 years. If the bank A gives the interest at 10% p.a. and the bank B, 12% p.a., find the depositing amounts in both banks. (ii) Binod lend altogether Rs. 66000 to Pramod and Santosh for 2 years at 15% compounded yearly. If Santosh paid Rs. 1935 more than Pramod as the interest, find how much did Binod lend to each. 2. (a) Rs. 881.80 (b) Rs. 13,112.72 (c) Rs. 1,90,015.51 (d) Rs. 1,50,000 (e) 4 (f) 15% p.a. 3. (i) (b) Rs. 27,300 (c) Rs. 27,571.58 (d) Bank M, Rs. 271.58 (ii) (b) Rs. 18,150 (c) Rs. 18,300 (d) Simple interest, 0.826% 4. (i) (b) Rs. 7,138.24 (c) Rs. 7,377.50 (d) Bank B, 3.35% (ii) (a) Rs. 74,299.77 (b) Rs. 73,689.31 (c) Bank P 5. (i) (a) Rs. 22,020 (b) Rs. 22,781.25 (c) Rs. 781.25 (ii) (a) Rs. 2,91,600 (b) Rs. 2,92,464.64 (c) Rs. 864.64 6. (i) (a) Rs. 2,170 (b) Rs. 1,920 (c) Rs. 250 (ii) (a) x 5 (b) 21x 100 (c) Rs. 24,000 (iii) Rs. 4,000 7. (i) (a) 2 yrs. (b) Rs. 50 (c) CI, 110% (ii) (a) 20% (b) 1 1 2 yrs. (c) Rs. 480 (d) CIH, 3.44% 8. (i) (a) 12% (b) Rs. 166.73 (ii) (a) Rs. 420 (b) 10.5% (iii) (a) Rs. 2,000 (b) Rs. 400 9. (i) (b) Rs. 21,479.53 (c) Rs. 5,479.53 (ii) (b) Rs. 3,22,249.65 (c) Rs. 72,249.65 10. (i) (b) Rs. 94,000 (ii) (b) Rs. 2,362.80 11. (i)(a) 18.92% (b) Rs. 5,04,528.79 (ii)(a) 24.57% (b) Rs. 18,06,168.51 12. (i) (a) 4% (b) Rs. 3,125 (ii) (a) 10% (b) Rs. 10,000 13. (i) (a) 10% (b) Rs. 10,000 (ii) (a) 10% (b) Rs. 800 14. (i) (a) Rs. 3,528 (b) Rs. 3,200 16. (i) Rs. 27390.18, Rs. 22609.82 (ii) Rs. 30000, Rs. 36000 Answers Project Work Suppose you have some amount of money. In a bank, there are three types of accounts A, B and C. In account A, the simple interest rate is 12% p.a., in account B, the compound interest rate is 11% p.a. compounded annually and in account C, the compound interest rate is 10.75% p.a. compounded semi-annually. If you want to deposit the amount in the bank, in which account do you deposit and why?

PB Allied The Leading Mathematics-10 Compound Growth an Depreciation 59 Arithmetic 3.1 Compound Growth At the end of this topic, the students will be able to: ¾ solve the behaviour problems on compound growth. Learning Objectives I Introduction In our nature, population of human, the number of animals and plants, the number of virus and bacteria are increasing or decreasing. We cannot fully control them, but their increasing rates can be controlled. For example, in earlier, the human population was rapidly increasing in the world. Now, it is controlled by using various types of medicines and techniques. The population in some countries decreases due to war and epidemic. Similarly, Botanists reproduce many plants using different methods such as cutting, layering, grafting, budding, etc. These all are formed under the rule of powering or indices system like as compound interest. This is called compound growth. It is smoothly increasing the number of things at the fixed rate during the certain period. Activity 1 The population of a city in 2019 was 2,00,000. It has been reached 2,02,160 in 2020. The rate of growth of the population for the year 2019 was 202160 – 200000 200000 × 100% = ........ %. At the same rate the population of the city in 2021 has been reached ......................... . Here, the population growth is compound at the same rate of growth. Hence, the formula for the compound interest serves exactly in the same manner for the population/number growth. Therefore, if P is the population/number of certain place/things for a certain time T and R% is the rate of compound growth; then the total population/number PT after T years is, PT = P(1 + R%) T = P 1 + R 100 T , PT > P. And the grown population/increased number during these years is, PG = P(1 + R%) T – P = P 1 + R 100 T – 1 , for T = 1, PG = PTR 100 Population in Present Kathmandu Population in Previous Kathmandu CHAPTER 3 COMPOUND GROWTH AND DEPRECIATION Plant vegetative propagation

60 Allied The Leading Mathematics-10 Compound Growth an Depreciation 61 Arithmetic Remarks: (i) If the increased population by migration in is Min, the increased population by migration is Mout and the number of dead population/number is D then, PT = P 1 + R 100 T – Min – Mout – D. (ii) When the previous population/number and present population/number are given, the previous population is denoted by P and the present population will be PT. In this case, Population after T years (PT) = Present population = Previous Popn (P) × 1 + R 100 T . (iii) When the present population and future population are given, the present population is denoted by P and the future population will be PT. In this case, Population after T years (PT) = Future population = Present Popn (P) × 1 + R 100 T . (iv) When the population growth rates are different rates R1, R2, ..., RT for every year or any interval of time, (a) Population after T years (PT) = P 1 + R1 100 1 + R2 100 ....... 1 + RT 100 , for different rates in each year. (b) Population after T years (PT) = P 1 + R1 100 T1 1 + R2 100 T2 ....... 1 + RT 100 Tn , for different rates in any interval of time. (v) When the growth rate is negative i.e., the population is creasing, then Population after T years (PT) = P 1 – R 100 T , PT < P. (vi) If the growth rate changes (increasing and decreasing) from one conversion period to another conversion period (use '+' for increasing and '–' for decreasing population in the formula). Population after T years (PT) = P 1 ± R1 100 1 ± R2 100 ....... 1 ± RT 100 , PT > or < P. Activity 2 Test of the formula PT = P 1 + R 100 T for plant vegetation: Take a plant (rose) that is vegetating by cutting. Cut out three parts from it and grow another three new plants. Again, cut out three parts from each of these four plants. All old and new plants are continuously growing. So, the growing rate is 3:1 = 300%. How many plants are there now? Similarly, the old and growing new plants are cut down three parts each time and grow all plants. How many plants are there after 5 times cutting ? There is no cutting. It is only one plant. At first cutting, we get 4 plants with oldone and new 3. At second cutting of each plant with old and new, we get 12 next new plants and get 16 total plants. 1 + 3 = 4 1 1 1 1 1 + 3 + 4 × 3 = 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1st Cutting 2nd Cutting Old 1 plant Old 1 plant New 3 plants 3 new plants +

60 Allied The Leading Mathematics-10 Compound Growth an Depreciation 61 Arithmetic Cutting Times Process Total Number Using Formula T = 0 1 P0 = 1 P0 = P 1 + R 100 T = 1 × 1 + 300 100 0 = 1 T = 1 1 + 1 × 3 P1 = 4 P1 = 1 × 1 + 300 100 1 = 1 + 3 = 4 T = 2 1 + 3 + 4 × 3 P2 = 16 P2 = 1 × 1 + 300 100 2 = (1 + 3)2 = 16 T = 3 1 + 3 + 12 + 16 × 3 P3 = 64 P3 = 1 × 1 + 300 100 3 = (1 + 3)3 = 64 T = 4 1 + 3 + 12 + 48 + 64 × 3 P4 = 256 P4 = 1 × 1 + 300 100 4 = (1 + 3)4 = 256 T = 5 1 + 3 + 12 + 48 + 192 + 256 × 3 P5 = 1024 P5 = 1 × 1 + 300 100 5 = (1 + 3)5 = 1024 From the above table, when the old and all new plants are continuously cutting 5 times for vegetative propagation, there will be 1024 plants altogether. From the table, the number of plants each time of cutting forms the number patterns like as 1 + 3 + 12 + 48 + 192 + 768 + ...... . i.e., PT = 1 + Sum of Geometric Series with 1st term (a) = 3 and common ratio (r) = 1 + 300% = 4. For total number of plants, 1 + 4 + 16 + 64 + 256 + 1024 + ...... = ∑ n T = 1 (arn – 1) and PT = ST – ST–1 (T > 1), where ST = Sum of GS. Points to be Remembered 1. Compound growth is smoothly increasing the number of things at the fixed rate during a certain period. 2. The initial/original population/number of things between two number of things is called present population/number of things and another is total population/number of things. 3. If P is the population/number of certain place/things for a certain time T and R% is the rate of compound growth, then the total population/number PT after T years is, PT = P(1 + R%) T = P 1 + R 100 T , PT > P. 4. The growth population/increased number during T years is PG = P 1 + R 100 T – 1 . 5. When the population growth rates are different rates R1, R2, ..., RT, (a) Population after T years (PT) = P 1 + R1 100 1 + R2 100 ....... 1 + RT 100 , (b) Population after different T years (PT) = P 1 + R1 100 T1 1 + R2 100 T2 ....... 1 + RT 100 Tn , 6. When the growth rate is negative i.e., the population is decreasing, then Population after T years (PT) = P 1 – R 100 T , PT < P.

62 Allied The Leading Mathematics-10 Compound Growth an Depreciation 63 Arithmetic Example-1 The population of a town increases by 2% every year. If its present population is 2,16,000, (a) How much population will be increased in the first year ? (b) Find its population 2 years ago. (c) Find its population after 2 years. (d) How much percent did its population increase in 4 years ? Solution : (a) Here, present population of a town (P) = 2,16,000, Growth rate (R) = 20%, Increased Population in 1 year (PG) = ?, Time (T) = 1 yr. Now, we have The increased population in the first year = PTR 100 = 216000 × 1 × 20 100 = 43200 (b) Population after 2 years (PT) = ?, Time (T) = 2 yrs. Now, we have PT = P 1 + R 100 T = 2,16,000 1 + 20 100 2 = 2,16,000 × 1.44 = 3,11,040. Hence, the population of the town after 2 years is 3,11,040. (c) Population 2 years ago (P) = ?, Time (T) = 2 yrs. Now, we have PT = P 1 + R 100 T or, 2,16,000 = P 1 + 20 100 2 or, 2,16,000 = P × 1.44 or, 216000 1.44 = P or, 1,50,000 = P Hence, the population of the town 2 years ago is 1,50,000. I get the same answer by using the formula PG = P 1 + R 100 T – 1 . Population of Lalitpur some years ago Rapid population growth in Lalitpur

62 Allied The Leading Mathematics-10 Compound Growth an Depreciation 63 Arithmetic (d) Percentage of increased population in 4 years = 311040 – 150000 150000 × 100%. = 161040 150000 × 100% = 107.36%. Example-2 2 years ago, the price of one aana land was Rs. 6,25,000. Due to migration to city, its price decreases every year at the rate of 4% per annum. (a) What is the present price of one aana land ? Find. (b) 2 years hence, what is the price of the land at the same decreasing rate ? Compute. (c) How much price decreased during 4 years ? Solution: (a) Here, price of one aana land 2 years ago (P) = Rs. 6,25,000, Rate of decreased (R) = 4%,Time (T) = 2 yrs. Now, we have Present price = P 1 – R 100 T [ The case of negative growth] = 625000 1 – 4 100 2 = Rs. 5,76,000 Hence, the present price of one aana land is Rs. 5,76,000. (b) Here, price 2 years ago (Po) = Rs. 5,76,000, Rate of decreased (R) = 4%,Time (T) = 2 yrs. Now, we have Present price = P 1 – R 100 T [ The case of negative growth] = 5,76,000 1 – 4 100 2 = Rs. 530841 Hence, 2 years ago, its price was Rs. 5,30,841. (c) The decreased price during 4 years = 6,25,000 – 5,30,841 = Rs. 94,159. Example-3 2 years ago, the population of a town was 1,50,000. Now, it is 1,68,540. The growth rate each year was fixed. (a) What was the growth rate? Calculate. (b) If the growth rate is the same for the next 3 years, what will its population be ? Find. (c) What percent of the population increased during 5 years in total ? Find. Solution : (a) Here, Population of a town before 2 years (Po) = 1,50,000, Present Population of the town (PT) = 1,68,540, Duration of Time (T1) = 2 yrs, Rates of population growth (R) = ? Now, we know, PT = P 1 + R 100 T

64 Allied The Leading Mathematics-10 Compound Growth an Depreciation 65 Arithmetic or, 1,68,540 = 1,50,000 1 + R 100 2 or, 1,68,540 1,50,000 = 1 + R 100 2 or, 2809 2500 = 1 + R 100 2 or, 27 25 2 = 1 + R 100 2 ⸫ 27 25 = 1 + R 100 [ ⸫ By rule of indices] or, 27 25 – 1 = R 100 or, 27 – 25 25 = R 100 or, 2 × 100 25 = R or, 8% = R Hence, the population growth rate of the town is 8%. (b) If the growth rate is the same for the next 3 years, then growth rate (R) = 8%, time (T) = 3 yrs, present population (PT) = 168540 Now, Population after 3 years, PT = P 1 + R 100 T = 1,68,540 1 + 8 100 3 = 1,68,540 (1 + 0.08)3 = 1,68,540 × 1.083 = 212311.86 = 2,12,312 (c) The increased percent of the population during 5 years = 2,12,312 – 1,50,000 1,50,000 × 100% = 62,312 1,50,000 × 100% = 41.54% Example-4 The population of a town decreased 5% in 2017 and 10% in 2018. Its population in the beginning of 2017 was 32,000. (a) What was the population of the town in the beginning of 2018? Compute. (b) What was the population of the town in the beginning of 2019? Find. (c) How much percent of population decreased during 2 years in average ? Find. Solution : (a) Here, Population in the beginning of 2017 (P) = 32,000, Rates of decrease at 2017 (R1) = 5% and Time (T1) = 1 yr Now, we know,

64 Allied The Leading Mathematics-10 Compound Growth an Depreciation 65 Arithmetic Population in the beginning of 2018 (P1) = P 1 – R1 100 = 32,000 1 – 5 100 = 32,000 × 95 100 = 32,000 × 0.95 = 30,400. ∴ The population of the town in the beginning of 2018 was 30,400. (b) Here, Population in the beginning of 2019 (P1) = 30,400, Rates of decrease at 2018 (R2) = 10% and Time (T2) = 1 yr We know, Population in the beginning of 2019 (P2) = P1 1 – R2 100 = 32,000 1 – 10 100 = 32,000 × 90 100 = 30,400 × 0.90 = 27,360. ∴ The population of the town in the beginning of 2019 was 27,360. (c) Percent of population decreased during 2 years in average = 32,000 – 27,360 32,000 × 100% = 4640 32,000 × 100% = 14.5% "Alternatively" Population in the beginning of 2019 = P 1 – R1 100 1 – R2 100 = 32,000 1 – 5 100 1 – 10 100 = 32,000 × 95 100 × 90 100 = 32,000 × 0.95 × 0.9 = 27,360. PRACTICE 3.1 Read Think Understand Do Keeping Skill Sharp 1. (a) If the initial population P increases at the rate of R% p. a for T years, write the population after T years (PT) in terms P, T and R. (b) The initial population P decreases at the rate of R% per annum for T years, what is the population after T years (PT) in terms of P, T and R? (c) If the growth rate of the initial population P for first three years is R1 %, R2 % and R3% respectively then what is the population after three years ?

66 Allied The Leading Mathematics-10 Compound Growth an Depreciation 67 Arithmetic Check Your Performance Answer the following questions for each problem. 2. Which values are inserted in the following blanks? Compute. SN Place Population at present (PO) Growth rate (R) Time (T) Total population (PT) (a) (b) (c) (d) (e) A B C D E 5,00,000 2,50,000 --------- 25,000 2,50,000 2% – 2% 3% 4% ------- 3 yrs 3 yrs 2 yrs ------- 3 yrs --------- --------- 21218 27,040 3,14,928 3. (i) 3 year ago, the population of a place was 64,000 and the population growth rate of the village is 5%. (a) What is the present population of the place ? Calculate. (b) How much population will be reached after 1 year at the growth rate of 4% ? Find. (c) What percent did its population increase in 3 years ? Compute. (ii) John has 3200 shares of a company now. The number shares increases at the rate of 10% per annum. (a) What will be his number of shares after 2 years? Find. (b) Find his number of shares 2 years ago. (c) What percent did his shares increase in 4 years ? Compute. 4. (i) 2 years ago, the population of a village was 6,25,000. Due to epidemic disease to a village, its population decreases every year at the rate of 4% per annum. (a) What is the present population of the place ? Calculate. (b) 2 years hence, what will be the population of the village at the same decreasing rate ? Compute. (c) How much population decreased during 4 years ? Calculate. (ii) The present population of a city is 50,20,000. Due to COVID-19, its population decreases every year at the rate of 8% per year. (a) What is the population of the city before 2 years? Calculate. (b) 3 years hence, what will be the population of the city if its population will decrease at the same decreasing rate ? Compute. (c) What percent of the population decreased during 5 years ? Calculate. 5. (i) The cost of a plot of land was Rs. 1,00,000 last 2 years back. Now, it is Rs. 1,04,040. (a) What was the growth rate of the land? Calculate. (b) If the growth rate is the same for the next 2 years, what will be its price ? Find. (c) What percent of the price increased during 4 years in total ? Find.

66 Allied The Leading Mathematics-10 Compound Growth an Depreciation 67 Arithmetic (ii) The students enrolled in a university last 2 years back were 35,000. Now, there are 36,414. The admission rate each year was fixed. (a) What was the growth rate? Calculate. (b) If the growth rate is the same for the next 2 years, what will be its population ? Find. (c) What percent of the population increased during 4 years in total ? Find. 6. (i) The population of a town becomes 26,901 from 24,400 at the growth rate of 5% per annum after some time. (a) Which formula do you use to calculate the time ? (b) Find the required time for the above condition. (c) If the population growth rate had 4% per annum for the same time, what would be the population of the town? Calculate. (d) What percent is difference the final populations for the rates 5% and 4%? Find. (ii) The cost of gold ring is 40,000 and the price growth rate is 4% per year. (a) In how many years will the price of the ring will be Rs. 43,264? (b) If the price growth rate had 5% per annum for the same time, what would be the price of the ring? Calculate. (c) What percent is difference on the final price for the rates 4% and 5%? Find. 7. (i) The population of certain city in 2078 BS was 4,50,000. (a) Find its population in 2080 BS at the growth rate of 2% p.a. (b) What was its population in 2075 BS at the growth rate of 3% p.a. (c) Find the average growth rate for 5 years. (ii) The population of Maldives in the beginning of 2022 AD was 3,00,000 and the growth rate was 2.2%. (a) What will be the population in the beginning of 2024 AD? Find. (b) What was its population in the last of 2018 AD at the growth rate of 3% p.a.? (c) Find the average growth rate from the last of 2018 AD to the beginning of 2024 AD. 8. (i) The price of an ornament is Rs. 25,000. Its price grows at 4%, 5% and 8% during the first year, second year and third year respectively. (a) Find the its price after 3 years. (b) Verify your answer by finding the price of each year. (ii) The population of a municipality was 2,00,000 and the growth rate is estimated to be 2.1% p.a. in the last one year and then the population growth rate becomes 2.25% now. (a) Estimate the total population at the end of the next 3 years from now. (b) Verify your answer by finding the population of each year.

68 Allied The Leading Mathematics-10 Compound Growth an Depreciation 69 Arithmetic (iii) During one year, the population of a town increased by 2% but during the next year it diminished by 2%. If at the end of the second year the population was 6,25,000, what was the population at the beginning? 8. (i) The population of a city was 1,20,000 in the year 2078 and the population growth rate was 4.5%. 20,000 people migrated here from other places in the year 2079. (a) Find the population reached in the year 2079. (b) What will be the total population in the year 2081? (ii) In a certain city, the population in 2060 BS was 1,50,000. In 2061 BS; because of epidemic disease, 20,000 people died. The population growth rate was 3% p.a. (a) Find the population reached in the year 2061 BS. (b) What was the total population in 2063 BS? (iii) The population of a place increases every year by 2.5%. After two years, the total population of the place will be 24,895. 320 people migrate to the other places. (a) Find the population of the place before one year. (b) What was the population of the town before three years ? 10. (i) The population of a village was 15000 two years ago. 560 people migrated from the village, 45 people died and 145 people came as immigrant into this village in this period. If the rate of population growth is 4%, find the present population. (ii) Two years ago the population of a town was 120000. 1520 people migrated from the town. 120 people died and 2450 people came as immigrant into this town during this period. If the rate of population growth is 4%, find the population after three years. 2. (a) 5,30,604 (b) –2% (c) 20,000 (d) 2 yrs. (e) 8% 3. (i) (a) 74,088 (b) 77,051 (c) 20.39% (ii) (a) 3,520 (b) 2,644 (c) 33.13% 4. (i) (a) 5,76,000 (b) 5,30,841 (c) 15.67% (ii) (a) 59,31,001 (b) 39,09,013 (c) 34.09% 5. (i) (a) 2% (b) 1,08,243 (c) 8.243% (ii) (a) 2%, (b) 37885, (c) 8.24% 6. (i) (b) 2 yrs. (b) 26391 (d) 1.93% (ii) 2 yrs (b) 44100 (c) 1.93% 7. (i) (a) 768180 (b) 411813 (c) 13.69% (ii) (a) 313345 (b) 274542 (c) 14.13% 8. (i) (a) 29484 (ii) (a) 218295 (iii) 625250 9. (i) (a) 145400 (b) 158780 (ii) (a) 134500 (b) 142691 (iii) (a) 23289 (b) 22167 10. (i) 15744 (ii) 146909 Answers Project Work Collect the records of population of 2 simultaneous years of a Gaunpalika/Municipality/Submetropolitan city/ Metropolitan city. Find the rate of population growth and find the population of the place 3 years ago and estimate the population after 3 years.

68 Allied The Leading Mathematics-10 Compound Growth an Depreciation 69 Arithmetic 3.2 Compound Depreciation At the end of this topic, the students will be able to: ¾ solve the behaviour problems on compound depreciation. Learning Objectives I Introduction In our nature, the life of the mechanical capacity of the man-made things are slowly deceasing and their values or costs also decease according as their using periods. By using medicines, the number of bacteria and virus are decreased from the human body, other animals and plants. The reduced value of goods is known as depreciation. The value of the goods compounded reducing for fixed time is called compound depreciation. This is opposite to the compound growth. Activity 1 A second hand car was purchased last year for Rs. 12,20,000. This year, it was sold for Rs. 10,98,000. The price is less by Rs. ............ Therefore, in this year, the depreciated amount on the car = Rs. ......... In next year, the price of the car may go further down at the same rate. How much will be its cost in the next year ? How can it find ? At first, we find the depreciated rate. Depreciated rate = ................. The price of the car in the next year = ................... In this way, the value of machinery items like cars and other goods goes down year by year because it is already used, its machineries become less effective, it becomes old, damages, etc. If the depreciation is valuated at a constant rate, then depreciation is compounded annually. Activity 2 If a doctor proves that a dengue patient has 12,00,000 virus and he uses him/her medicines and the virus is depreciated by12% per day, then its number after 3 days will be, Initial number of virus = 1200000 Number of virus at the end of the 1st day = (100 – 12)% of 1200000 = ........ Number of virus at the end of the 2nd day = (100 – 12)% of ............. = ........ Number of virus at the end of the 3rd day = (100 – 12)% of ............. = ........ Did you notice here the depreciated number is subtracted at a constant rate every day from the revalued number. If the initial number/value of certain item is Vo, depreciates annually at the rate of R% p.a, then its value after T years is worked out as, Offset Printing Machine Corona Virus Dengue Virus

70 Allied The Leading Mathematics-10 Compound Growth an Depreciation 71 Arithmetic Depreciation at the end of the first year, D1 = VoR%. ∴ Value at the end of the first year, V1 = Vo – VoR% = Vo(1 – R%) Depreciation at the end of the 2nd year, D2 = Vo(1 – R%). R%. ∴ Value at the end of the 2nd year, V2 = Vo(1 – R%) – Vo(1 – R%) R%. = Vo(1 – R%) (1 – R%) = Vo(1 – R%) 2 Depreciation D1 at the end of the 3rd year, D3 = Vo(1 – R%) 2 . R% ∴ Value at the end of the 3rd year, V3 = Vo(1 – R%) 2 – Vo(1 – r)2 . R% = Vo(1 – R%) 2 (1 – R%) = Vo(1 – r)3 Proceeding in the same manner the depreciation after T years, DT = Vo(1 – R%) T – 1. R% And value after T years, VT = Vo(1 – R%) T = Vo 1 – R 100 T . i.e., Value after T years = Present value (Vo) 1 – R 100 T . Present value = Value T years ago (Vo) 1 – R 100 T . The depreciated value\population number during T years is DT = Vo 1 – 1 – R 100 T . Remarks: 1. When the population reduced rates are different as R1 , R2 , ..., RT, (a) Population after T years (PT) = P 1 – R1 100 1 – R2 100 ....... 1 – RT 100 (b) Population after different T years (PT) = P 1 – R1 100 T1 1 – R2 100 T2 ....... 1 – RT 100 Tn Points to be Remembered 1. The reduced value of goods is known as depreciation. 2. The value of the goods compounded reducing for fixed time is called compound depreciation. 3. If the initial number/value of certain item P and depreciates annually at the rate of R% p.a, then its value after T years is, PT = P(1 – R%) T = P 1 – R 100 T , PT < P. i.e., Value after T years = Present value (P) 1 – R 100 T . Present value = Value T years ago (P) 1 – R 100 T . 4. The depreciated population/decreased number during T years is PG = P 1 – 1 – R 100 T . 5. When the population growth rates are different rates R1, R2, ..., RT, (a) Population after T years (PT) = P 1 – R1 100 1 – R2 100 ....... 1 – RT 100 (b) Population after different T years (PT) = P 1 – R1 100 T1 1 – R2 100 T2 ....... 1 – RT 100 Tn

70 Allied The Leading Mathematics-10 Compound Growth an Depreciation 71 Arithmetic Example-1 The price of an electric scooter depreciates by 2% of its value at the beginning of each year. (a) Define compound depreciation. (b) Find its sales price after 2 years if its present cost is Rs. 140000. (c) Find the total depreciation in the transaction of the scooter. Solution : (a) If the reduced value of the goods is compounded for fixed time then it is called compound depreciation. (b) Here, Present price (P) = Rs. 140000, Rate of depreciation (R) = 2%, Time (T) = 2 years, Sale value after 2 years (V2) = ?, Depreciated amounts (D2) = ? Now, we know that price after T years (VT) = Present value 1 – R 100 T or, price after 2 years (V2) = Rs. 140000 1 – 2 100 2 = Rs. 140000 × 0.9604 = Rs. 134456. ∴ The selling price of the scooter after 2 years is Rs. 134456. (c) Total depreciation amount (D2) = Rs. (140000 – 134456) = Rs. 5544. Example-2 The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum and its value becomes Rs. 9680. (a) For how much was the refrigerator purchased? (b) Find the total depreciated amount in the 2 years for selling the second hand refrigerator. (c) How much percent depreciated in the refrigerator in total ? Solution : (a) Let the cost price of the refrigerator be Rs. V. Then, its present value (VT) = V × 1 – R 100 T or, 9680 = V × 88 100 × 88 100 or, V = Rs. 12500. Hence, the cost price of the refrigerator is Rs. 12500.

72 Allied The Leading Mathematics-10 Compound Growth an Depreciation 73 Arithmetic (b) The depreciated amount (DT) = VT – V = 12500 – 9680 = Rs. 2820. (c) The depreciated percent in the refrigerator in total = 2820 12500 × 100% = 22.56% Example-3 An industry was established with an initial investment of Rs. 10,00,00,000. At the end of 3 years the industry earned Rs. 1,50,00,000. The rate of depreciation in the assets was worked out at the rate of 3% p.a. compounded annually. The industry was sold after 3 years. (a) Find the value of the industry after three years. (b) How much earn did it with selling cost? (c) Work out the loss or profit percent ? Find Solution : (a) Here, Initial investment (P) = Rs. 100000000, Rate of depreciation (r) = 3% p.a. Time (T) = 3 yrs, Value after 3 years (VT) = ? Now, we have PT = P(1 – R%) T = Rs. 100000000 1 – 3 100 3 = Rs. 100000000 × (0.97)3 = Rs. 100000000. × 0.912673 = Rs. 9,12,67,300. The industry’s earning during these three years = Rs. 1,50,00,000. (b) Total revenues for the three years = Rs. 9,12,67,300 + Rs.1,50,00,000 = Rs. 10,62,67,300. (c) If the industry is sold, the profit after 3 years is Rs. 10,62,67,300 – Rs. 10,00,00,000 = Rs. 62,67,300. ⸫ The profit percent = 62,67,300 10,00,00,000 × 100% = 6.27%.

72 Allied The Leading Mathematics-10 Compound Growth an Depreciation 73 Arithmetic PRACTICE 3.2 Read Think Understand Do Keeping Skill Sharp 1. (a) If the initial value P of an article decreases at the rate of R% p. a. for T years, write the value of the article after T years (PT) in terms of P, T and R. (b) If the decreasing rate of the initial value P of an item for first three years is R1 %, R2 % and R3% respectively then what is the value of the item after three years ? (c) The machine costing Rs. a is depreciated at the rate of b% p.a. What will be its cost after c years ? Check Your Performance Answer the following questions for each problem. 2. (i) A motor-cycle was purchased for Rs. 185000. It is sold after three year at the depreciated rate of 5%. (a) What is compound depreciation ? Define. (b) What is the present value of the motor-cycle ? Find. (c) How much amount reduced on the value of the motor-cycle ? Find. (ii) 3 years ago, a house was bought for Rs. 2700000. The depreciated rate of the house is 8% p.a. (a) Which formula is used to find the value of a machine after T years that it depreciates at R% per annum ? (b) What is the present value of the house ? Find. (c) How much amount is reduced on the value of the house ? Find. 3. (i) The present value of an article is Rs. 15870 after compound depreciating of 8%. (a) Find the original cost of the article 2 years ago. (b) What was its reduced value in 2 years ? Find. (c) How much percent is reduced on its at all ? Find. (ii) The present cost of a TV is Rs. 19683 after reducing 10% compound depreciating rate. (a) Find the original cost of the TV before 3 years. (b) What was its reduced value in 3 years ? Find. (c) How much percent is reduced on its value? Find. 4. (i) A Sapha Tempo which was bought for Rs. 200000, was sold at Rs. 170000 after one year. (a) Which formula is used to find the rate of depreciation ? Write. (b) If the rate of depreciation is reduced by 1%, what is its selling price ? Find. (c) What percent is different in this transaction when the depreciation rate differs ?

74 Allied The Leading Mathematics-10 Compound Growth an Depreciation PB Arithmetic (ii) A press established in an initial investment of Rs. 50000000 was devaluated at the rate of 20% each year and was sold for Rs. 20480000. How long ago was the press established? (a) Which formula is used to find the used time of an article on compound depreciation? Write. (b) If the rate of depreciation is reduced by 3%, what is its selling price ? Find. (c) What percent is different in this transaction when the depreciation rate differs ? 5. (i) A machine bought for Rs. 1242540 is depreciated at 10% year for first 2 years and then the rate of depreciation will be 15% for 3 years. (a) What is the price of the machine after 2 years? (b) What is the price of the machine after 3 years? (c) Justify that your answer of the price of the machine after 2 years and 3 years is the price after 5 years directly? (ii) A motorcycle bought for Rs. 284000 is depreciated at 15% per year for first year and then the rate of depreciation will be 20% and 25% for 2 and 3 years respectively. (a) What is the price of the machine after 1, 2 and 3 years separately? (b) Justify that your answer of the price of the machine after 1 year, 2 years and 3 years is the price after 6 years directly? 6. (i) Jack bought a threshing machine and sold it to Rashila at 10% depreciation rate after using 2 years. Again, she sold it to Amar for Rs. 49766.40 at 20% depreciation rate after 3 years. (a) How much did Rashila sell for the threshing machine to Amar ? Find. (b) How much had Jack paid for it? Compute. (ii) The number of trees in a community forest was 45,00,000 in 2012. In 2014, it was only 36,45,000 by the causes of deforestation due to shortage of LP gas. If the deforestation was continued of the same rate. (a) Find the rate of compound deforestation. (b) Find the remaining trees in 2017. 2. (i)(b) Rs. 1,58,614.38 (c) Rs. 26,385.62 (ii)(b) Rs. 21,02,457.60 (c) Rs. 5,97,542.40 3. (i)(a) Rs. 18,750 (b) Rs. 2,880 (c) 15.36% (ii)(a) Rs. 27,000 (b) Rs. 7,317 (c) 27.10% 4. (i)(b) Rs. 1,72,000 (c) 1.18% (ii) (b) Rs. 2,48,93,568 (c) 21.55% 5. (i)(a) Rs. 10,06,457.40 (b) Rs. 6,16,090.65 (ii)(a) Rs. 2,41,400, Rs. 1,54,496, Rs. 65,178 6. (i) (a) Rs. 97,200 (b) Rs. 1,20,000 (ii) (a) 10% (b) 26,57,205 Answers

Arithmetic PB Allied The Leading Mathematics-10 Money Exchange 75 4.1 Currency and Exchange Rate At the end of this topic, the students will be able to: ¾ solve the behaviour problems related to currency and exchange rate. Learning Objectives I Introduction to Money and Currency Before money was invented, people exchanged for goods and services. It wasn't until about 5,000 years ago that the Mesopotamian people created the Shekel (ancient Mesopotamian coin made by silver), which is considered the first known form of currency. Gold and silver coins date back to around 650 to 600 BC. In Nepal, The earliest coin minted in today's territory of Nepal was in Shakya Mahajanapada, along the India-Nepal border at around 500 BC and King Mandev (464-505) used the coin Sri Mananka in Nepal valley and other states around in the 5th century. In the Lichhavi and Malla periods, Nepal exchanged money with Tibet. Nepal provided coins for Tibet and Tibet paid gold, silver and other materials to Nepal. The coins were very heavy in weight and were difficult in transportation from one place to another place. So, China first developed paper banknotes as receipts for value held on account "value received" during the Tang dynasty (618-907 AD), starting in the 7th century. European explorers like Marco Polo introduced the concept of paper notes in Europe during the 13th century. Napoleon Bonaparte (1769-1821) issued paper banknotes in the early 1800s. In October, 1945, the paper money of the amount Rs. 5, Rs. 10 and Rs. 100 was first introduced in Nepal, during the rule of King Tribhuvan. All different kinds of coins and paper notes are used for purchasing the goods from other places or counties. When the currency of one country is used for another country, what effects are on its values ? What is difference between money and currency ? Why do need we money exchange in our daily life ? Discuss. What are the buying rate and selling rate ? Discuss. Currency and money are the two words used in our everyday lives, often confused as being a similar things. But, they are not the same. Money is a broader term that refers to an intangible system of value that makes the medium of exchange of goods and services possible, now and in the future. Currency is simply one, tangible form of money. Now a days, all countries of the World use their own currency Shakya Janapada Coin (500 BC) Sri Mananka Coin of copper (576-605 AD) Electrum Shekel (600BC) Ancient Silver Shekel Banknote on the time of Napoleon Papernote on the time of Tribhuvan Banknote during Tang Dynasty CHAPTER 4 MONEY EXCHANGE

Arithmetic 76 Allied The Leading Mathematics-10 Money Exchange 77 or money as coins and banknotes. Their values are the same for the own country. But, the currencies of different counties do not have the same values. They have different values for other countries. For example, in the United States people use US dollars ($) and in the Great Britain they use Pound Sterling (£). In the US and the UK the Sterling Pound and the US dollar have different values. II Money Exchange and Exchange Rate The value of the currency of a country is flexible and changeable (devaluation and revaluation) due to the share market and other economic factors. When a government makes monetary policy to reduce a currency's value, then it is applied. As the different countries have different currencies with different values, there should be certain policy in the exchange rate. When the currency of a country converts into the currency of the other country, it is easy to transaction. An announced rate at which one currency will be exchanged for another currency and affects trade and the movement of money between countries, is known as exchange rate. The exchange rates are impacted by both the domestic currency value and the foreign currency value. There are two types of exchange rates: (i) Buying rate and (ii) Selling rate. As the selling rate is higher than the buying rate, the financial institutions make profit on the exchange of money. When the value of the currency goes to devaluation, the more amount needs to buy the foreign currency. When the value of the currency goes to revaluation, the less amount needs to buy the foreign currency. When the money exchanger takes commission on transaction of currency exchanging, the exchange rate in increased by the same commission rate. We use the Chain Rule by arrange two variables in two columns making the chain of variables by supposing unknown or required variable whose value to be found and put '=' in the middle and then multiply the variables of the first column and second column separately and then equal them. Now, find the value of the unknown term by solving the equation. In Nepal, the Rastra Bank announces the buying and selling rates of different currencies everyday as shown alongside. https://www.nrb.org.np/forex/ Money Exchange Rate for 23 January, 2023 (2079/10/09) Exchange Rates Fixed by Nepal Rastra Bank Currency Unit Buying/Rs. Selling/Rs. Indian Rupee 100 160.00 160.15 Open Market Exchange Rates (For the purpose of Nepal Rastra Bank) Currency Unit Buying/Rs. Selling/Rs. U.S. Dollar 1 129.50 130.10 European Euro 1 140.60 141.25 UK Pound Sterling 1 160.53 161.28 Swiss Franc 1 140.67 141.32 Australian Dollar 1 90.22 90.64 Canadian Dollar 1 96.78 97.23 Singapore Dollar 1 98.15 98.61 Japanese Yen 10 9.99 10.4 Chinese Yuan 1 19.09 19.18 Saudi Arabian Riyal 1 34.48 34.64 Qatari Riyal 1 35.35 35.52 Thai Baht 1 3.96 3.98 UAE Dirham 1 35.26 35.42 Malaysian Ringgit 1 30.22 30.36 South Korean Won 100 10.53 10.57 Swedish Kroner 1 12.58 12.64 Danish Kroner 1 18.90 18.98 Hong Kong Dollar 1 16.54 16.62 Kuwaity Dinar 1 424.24 426.21 Bahrain Dinar 1 343.50 345.09 Note: Under the present system the open market exchange rates quoted by different banks may differ. Table below gives the foreign exchange

Arithmetic 76 Allied The Leading Mathematics-10 Money Exchange 77 Points to be Remembered 1. A commodity accepted by general consent as a medium of exchanging goods and services, is called money. It has own price or value. 2. A form of money that circulates anonymously from person to person and country to country, thus facilitating trade, and it is the principal measure of wealth, is called Currency. 3. A rate at which one currency will be exchanged for another currency and affects trade and the movement of money between countries, is known as exchange rate. Exchange rates are impacted by both the domestic currency value and the foreign currency value. 4. A licensed business that allows customers to exchange one currency for another, is known as money exchange or currency exchange. Money exchanges make currency by charging a nominal fee and through the bid-ask spread in a currency. 5. A rate from which a bank buys the foreign currency from the customers or the customers sell the foreign currency to the bank, is called purchasing or buying rate. 6. A rate from which a bank or financial institute sells the foreign currency to the customers or the customers purchase the foreign currency from the financial institute, is called selling rate. 7. When the value of the currency goes down in the reference of other country's currency (specially dollar) due to bad economic condition, is called devaluation. When the currency of one country goes to devaluation, then the more amount needs to buy the foreign currency. 8. When the value of the currency goes up in the reference of other country's currency (specially dollar) due to good economic condition, is called revaluation. When the currency of one country goes to revaluation, then the less amount needs to buy the foreign currency. 9. When the money exchanger takes commission on transaction of currency exchanging, the exchange rate in increased by the same commission rate. 10. Arrange two variables in two columns making the chain of variables by supposing unknown or required variable whose value to be found and put '=' in the middle. Multiply the variables of the first column and second column separately and then equal them. Now, find the value of the unknown term by solving the equation. This method is called chain rule. Alert to Money Exchange 1. If our currency goes to devaluation by 10%, its exchange rate is increased by 10%, do not reduce by 10% on it. 2. If the currency of another country goes to a devaluation by 10%, its exchange rate is increased by 10%, do not increase by 10% on our currency. 3. If our currency is revaluated by 10%, its exchanging rate is decreased by 10%, , do not increase by 10% on it. 4. If the currency of another country goes to a revaluation by 10%, the exchange rate of the same country is decreased by 10%, do not decrease by 10% on our currency.

Arithmetic 78 Allied The Leading Mathematics-10 Money Exchange 79 Example-1 A Nepali citizen working in the UK brings an amount of £ 5000 to his home. He goes to the bank to exchange it into Nepali currency and the bank takes 2% commission on it. (Using buying rate, £ 1 = Rs. 133.62 and selling rate, £ 1 = Rs. 134.36) (a) What is exchange rate ? Define it. (b) How much Nepali currency does the bank take as commission ? Find it. (c) How much Nepali currency will s/he get after giving commission ? Find it. (d) What is the new exchange rate for her/him after commission ? Solution: (a) A rate at which one currency will be exchanged for another currency and affects trade and the movement of money between countries, is known as exchange rate. (b) Here, the person sales £ sterling in Nepali rupees. So, we use the buying rate for the conversion. i.e., £ 1 = Rs. 133.62 ∴ £ 5,000 = Rs. 133.62 × 5,000 = Rs. 6,68,100. Commission rate = 2% ∴ Commission amount= 2% of Rs. 668100 = 2 100 ×Rs.668100 = Rs. 13,362 Hence, the bank takes Rs. 13,362 as commission. (c) S/he gets Rs. 6,68,100 – Rs. 13,362 = Rs. 6,54,738 from £ 5000 in Nepal. (d) The new exchange rate after commission = Rs.6,54,738 £ 5000 = Rs.130.95/£. Example-2 A student studying in the UK has to arrange £ 5000 for the university fee. (Using buying rate, £ 1 = Rs. 133.62 and selling rate, £ 1 = Rs. 134.36) (a) Which exchange rate is used for converting Nepali currency into foreign currency ? (b) How much money (Nepali Rupees) does he need to pay for fee? Find. (c) If it is not needed for the fee after exchanging pound sterling because of a problem, he would sell it to the bank. How much Nepali currency does he get from it ? (d) How much percent does he gain or lose in the exchange transaction? Find. £ 5,000 "Alternatively" New exchange rate after com. £1 = Rs.133.62–2% of Rs.133.62 = 98% of Rs.133.62= Rs.130.9476 £ 5000 = Rs. 130.9476 × 5,000 = Rs. 6,54,738

Arithmetic 78 Allied The Leading Mathematics-10 Money Exchange 79 Solution: (a) The selling rate is used for converting Nepali currency into foreign currency. (b) Here, the student buys pound sterling (£) from Nepali currency. i.e., £ 1 = Rs. 134.36 (Selling rate) ∴ £ 5000 = Rs. 134.36 × 5000 = Rs. 671800. Hence, the student needs Rs. 671800 to pay for £ 5000 in Nepal. (c) For exchange pound sterling into Nepali currency, we use buying rate. i.e., £ 1 = Rs. 133.62 (buying rate) ∴ £ 5000 = Rs. 133.62 × 5000 = Rs. 668100. Hence, the student gets Rs. 668100 by selling £ 5000 in Nepal. (d) Here, his buying amount is more than the selling amount for exchange £ 5000. So, he gets loss. His loss percentage = 671800 – 668100 671800 × 100% = 3700 671800 × 100% = 0.55% Note: As the money is exchanged in different rates (the buying and the selling rates for Bank) the Nepali currency paid for the same amount differs in Example-1 and Example-2. Example-3 It is given that £ 50 = Rs. 6286 and Rs. 3900 = $ 50. A girl needs to exchange $ 10 into pound sterling (£). (a) How much US Dollar and pound sterling did she get from Re. 1 ? Find. (b) Find how much sterling pound she gets from $ 10. (c) Which currency is expensive the US dollar or pound sterling by what percent ? Solution: (a) Here, £ 50 = Rs. 6286 ⇒ Re. 1 = £ 50 6286 Rs. 3900 = $ 50 ⇒ Re. 1 = $ 50 3900 (b) Now, $ 50 3900 = £ 50 6286 or, $ 1 = £ 3900 6286 or, $ 10 = £ 3900 × 10 6286 = £ 6.20. (c) Here, $10 = £ 6.20. So, the pound sterling is expensive than the US dollar by, 10 – 6.20 10 × 100% = 3.80 10 × 100% = 38% "Alternative Method direct exchanging" Here, £ 50 = Rs. 6286 Rs. 3900 = $ 50 $ 10 = £ x ⸫ 50 × 3900 × 10 = 6286 × 50 × x or, 50 × 3900 × 10 6286 × 50 = x or, x = 6.20 Hence, £ 6.20 can be exchanged for US dollars 10.

Arithmetic 80 Allied The Leading Mathematics-10 Money Exchange 81 Example-4 A person buys US $ 5000 on 2079|09|20. Nepali rupees was devaluated by 12% next day. (a) What is the meaning of devaluation of Nepali currency? (b) How much Nepali currency is needed to buy $ 5000 at on 2079|09|20? Find. (c) What is the new rate after devaluation of Nepali currency by 12% ? Find. (d) What percentage of profit will he get if he sellsthe starling poundsthe next day? Solution: (a) The value of Nepali currency is decreasing and the value of foreign currency is increasing by the same rate. (b) A person buys US $ 5000 on 2079|09|20 that means the bank sells US $ to the person at selling rate for that day. US $ 1 = Rs. 109.57 ∴ US $ 5000 = 5000 × Rs. 109.57 = Rs. 547850. Hence, Rs. 547850 is needed to buy $ 5000 at on 2079|09|20. (c) The next day Nepali currency was devaluated by 12%. Then the US $ increases by 12% on the rate of foreign money exchange. For this, the US $ increases on buying rate on next day when the person buys US $. Hence, US $ 1 = Rs. 108.97 + 12% of Rs. 108.97 = Rs. 122.0464. (d) Now, US $ 5000 = 5000 × Rs. 122.0464 = Rs. 610232. Now, the profit made in this exchange = Rs. 610232 – Rs. 547850 = Rs. 62382. ∴ Percentage of profit (P%) = Rs. 62382 Rs. 547850 × 100% = 11.39%. Example-5 The selling rate of Qatar Riyal 1 was Rs. 29.93 yesterday. Today, Nepali currency is devaluated by 15%. The cost of mobile set costs Riyal 665 in Qatar. A employee wants to send the mobile set to Nepal through an agent. Nepal government charges a custom duty of 60%, takes 40% commission in Nepal by the agent and VAT is 13% on all. (a) How much amount is to be paid for custom duty? Find. (b) How much amount is to be taken for commission? Find. (c) How much amount is to be paid for VAT? Find. (d) What is the cost of the mobile set in Nepal ? Calculate. Solution: (a) Here, Qatar Riyal 1 = Rs. 29.93 (yesterday) Today, Nepali currency is devaluated by 15%. Then the new rate as Qatar Riyal 1 = Rs. 29.93 + 15% of Rs. 29.93 = Rs. 34.4195 Now, Cost of a Mobile set = Qatar Riyal 665 = Rs. 34.4195 × 665 = Rs. 22888.9675 Custom duty = 60% of Rs. 22888.9675 = Rs. 13733.3805

Arithmetic 80 Allied The Leading Mathematics-10 Money Exchange 81 (b) Cost of a Mobile set with custom duty = Rs. 22888.9675 + Rs. 13733.3805 = Rs. 36622.348 Commission on Mobile set = 40% of Rs. 36622.348 = Rs. 16648.9392 ∴ Cost of a Mobile set with custom duty and commission = Rs. 36622.348 + Rs. 16648.9392 = Rs. 51271.2872 (c) VAT amount = 13% of Rs. 51271.2872 = Rs. 6665.2673 (d) Cost of a Mobile set in Nepali currency = Rs. 51271.2872 + Rs. 6665.2673 = Rs. 57936.55454 = Rs. 57936.55 PRACTICE 4.1 Read Think Understand Do Keeping Skill Sharp Use the exchange rate given above for the date 2079/10/09 to solve the following problems: 1. How much Nepali currency is paid to buy the following currencies by the bank ? (a) £ 5562 (b) $ 3546 (c) Euro 5248 (d) Indian rupees 15,228 (e) Japanese yen 250368 (f) Canadian dollars $ 14,256 (g) Australian dollars $ 607 (h) Singapore dollars $ 3,568. 2. (a) If $1 = Rs. 130.85, how much Nepalese rupees are exchanged from $ 5? (b) How much pounds will be equal to Rs. 36410.25 if 1 pound (£) = Rs. 157.62 ? (c) How much Nepali currency will be needed to buy Canadian dollar 4500 if 1 Canadian dollar = Rs. 96.90? (d) Using the rate of Rs. 129.85 per US dollar, what is the US dollar for Rs. 3246.25 ? (e) If IC Rs. 2500 is equal to NC Rs. 4009.75, write down the rate of IC Re. 1. (f) How much Euro can be exchanged with Rs. 56000? (1 Euro = Rs. 139.46) 3. (i) Indian Rs. 100 worths Rs. 160 for buying purpose. A Nepali worker has Rs. 2,00,765 (Indian)? (a) What is currency ? Define it. (b) What amount of money will the worker get in Nepal ? Find.

Arithmetic 82 Allied The Leading Mathematics-10 Money Exchange 83 (ii) Sohan has to arrange Indian rupees 5,00,500 for the treatment of health. (IC Rs. 100 = NC Rs. 160.15) (a) What is exchange rate ? Define it. (b) How much Nepali rupees does he need for this purpose? 4. (i) According to the notice of Nepal Rastra Bank, $ 1 = Rs. 115.76. If a commercial bank takes 2% commission on it, how much Nepali currency does $ 1 get ? (ii) According to the notice of Nepal Rastra Bank, Chinese Yuan 1 = Rs. 16.68. If the Nepali currency is devaluated by 10%, how much Nepali currency does 1 Chinese Yuan get ? Check Your Performance Answer the following questions for each problem. 5. (i) A Nepali citizen working in Malaysia brings an amount of Malaysian Ringgit (RM) 75000 to his home. He goes to the bank to exchange it into Nepali currency and the bank takes 2% commission on it. [Using buying rate, RM 1 = Rs. 29.86 and selling rate, RM 1 = Rs. 30.20] (ii) A Nepali citizen working in Dubai brings an amount of Emirati Dirham (AED) 75000 to his home. He goes to the bank to exchange it into Nepali currency and the bank takes 2% commission on it. [Using buying rate, AED 1 = Rs. 36.03 and selling rate, AED 1 = Rs. 36.52] (a) What is money exchange ? Define it. (b) How much Nepali currency does the bank take as commission ? Find it. (c) How much Nepali currency will he get after giving commission ? Find it. (d) What is the new exchange rate for him after commission ? 6. (i) A Nepali goes to London and back from London from time to time. He is exchanging the pound sterling each time. The bank takes 2% commission each time also. [Buying rate £ 1 = Rs. 133.62, selling rate £ 1 = Rs. 134.36] (a) Find out how much Nepali rupees can be exchanged for £ 15,500 when coming from London. (b) When he wants to go back to London, how many Nepali rupees is required to get $15,500? (c) How much percent does he gain or lose in the exchange transaction? Find. (ii) A business person comes from USA and goes to USA from time to time. He is exchanging the US dollar each time. The bank takes 3% commission each time also. [Buying rate $ 1 = Rs. 132.17, selling rate $ 1 = 132.97] (a) Find out how much Nepali rupees can be exchanged for £ 18,200 when coming from London. (b) When he wants to go back to London, how many Nepali rupees is required to get $18,200? (c) How much percent does he gain or lose in the exchange transaction? Find.

Arithmetic 82 Allied The Leading Mathematics-10 Money Exchange 83 (iii) A student studying in Australia has to arrange AUD $ 24200 for the university fee. (Using buying rate, AUD $ 1 = Rs. 87.32 and selling rate, AUD $ 1 = Rs. 88.03) (a) Which exchange rate is used for converting Nepali currency into foreign currency? (b) How much money (Nepali Rupees) does he need to pay for fee? Find. (c) If it is not needed for the fee after exchanging Australian dollar because of a problem, he would sell it to the bank. How much Nepali currency does he get from it? (d) How much percent does he gain or lose in the exchange transaction? Find. 7. (i) Given that $ 25(US) = Rs. 1596.26 and Rs. 4036.20 = $ 62 (CA). (a) How much US Dollar and pound sterling can we get from Re. 1 ? Find. (b) Find how much US dollars can we get for $ 82 (CA). (c) Which currency is expensive the US dollar or Canadian dollar by how much percent ? (ii) It is given that ¥ 100 Japanese = Rs. 55.90 and Rs. 4397 (Nep) = $ 100 (Singapore). (a) How much Japanese Yens and Singapore Dollars can we get from Re. 1? Find. (b) Find how much Singapore Dollars she gets for Japanese ¥ 5,00,500 (CA). (c) Which currency is expensive the Japanese Yen or Singapore dollar by how much percent? (iii) (a) How much will be the profit percentage if an article is bought for $ 1 and sold for £ 1? [$ 1 = Rs. 112.56 and £ 1 = Rs. 168.34] (b) How much will be the profit or loss percentage if an article bought for £ 1 is sold for $ 1? [$ 1 = Rs. 112.56 and £ 1 = Rs. 168.34] 8. (i) A man buys US $ 15000 in one day. The next day Nepali currency is devaluated by 20%. (Before devaluation buying rate, 1 US $ = Rs. 132.53, selling rate, US $ 1 = Rs. 133.17) (a) What is the meaning of devaluation of Nepali currency? (b) How much Nepali currency is needed to buy US $ 15000 on the first day? Find. (c) What is the new rate after devaluation of Nepali currency by 20% ? Find. (d) What percentage of profit will he get if he sells the US dollars the next day? (ii) Sunaina purchased UK £12000 in the usual selling rate. The next day Nepali currency was revaluated by 6%. (Before devaluation buying rate, 1 UK £ = Rs. 172.43, selling rate, 1 UK £ 173.48) (a) What is the meaning of revaluation of Nepali currency? (b) How much Nepali currency is needed to buy UK £12000 on the first day? Find. (c) What is the new rate after revaluation of Nepali currency by 6% ? Find. (d) What percentage of profit or loss will she get if she sells the UK £ the next day?

Arithmetic 84 Allied The Leading Mathematics-10 Money Exchange 85 9. (i) How many US $ can be bought with the amount of Nepali currency that can buy £ 200 after allowing 10% devaluation in Nepali currency and 4% bank commission ? (Before devaluation selling rate, 1 US $ = Rs. 109.45, buying rate £ 1 = Rs. 150.26) (ii) How many pound sterling (£) can be bought with the amount of Nepali currency that can buy $ 1200 after allowing 20% devaluation in Nepali currency and 3% bank commission? (Before devaluation, buying rate 1 US $ = Rs. 109.45, selling rate £1 = Rs. 150.26) 10. (i) A washing machine is purchased for Dinar 85 in the Kuwait paying 1.5% transportation cost, 110% custom was paid in the custom office and sold for a profit of 150%. (Dinar 1 = Rs. 355.70) (a) How much amount is to be paid for transportation in Nepali currency? Find. (b) How much amount is to be paid for custom duty? Find. (c) How much amount would gain on selling the machine? Find. (d) What is the selling price of the machine in Nepal ? Calculate. (ii) 3 Refrigerators bought for Japanese Yen ¥ 25,000 each paying 2% transportation charge and 115% customs was paid in the customs office. The refrigerators were sold at a profit of 50%. (¥ 10 = Rs. 9.22) (a) How much amount is to be paid for transportation in Nepali currency? Find. (b) How much amount is to be paid for custom duty? Find. (c) How much amount would gain on selling the machine? Find. (d) What is the selling price of the machine in Nepal ? Calculate. 11. (i) A businessman paid 5% inter-port tax in Kolkata port and 150% custom duty at Kathmandu airport to bring some domestic goods costing $ 24835.50. He takes 25% commission on it and 13% VAT in total. ($ 1 = Rs. 115.52) (a) How much amounts are to be paid for inter-port tax and custom duty? Find. (b) How much amount is to be paid for commission? Find. (c) How much amount is to be paid for VAT? Find. (d) Find the cost of the goods when arrived at Kathmandu. (ii) A woman bought 2500 pieces of Nepali caps for Rs. 250 each in Nepal and exported in Germany including 15% export tax and 10% transpiration charge. She made a profit of 150%. (Euro 1 = Rs. 132.57) (a) How much amounts are to be paid for export tax? Find. (b) How much amount is to be paid for transportation charge? Find. (c) How much amount does she get profit? Find. (d) How much amount is the selling price of each cap in Germany? Find.

Arithmetic 84 Allied The Leading Mathematics-10 Money Exchange 85 2. (a) Rs. 654.25 (b) £ 231 (c) Rs. 436050 (d) $25 (e) Rs. 1.6039 (f) EUR 401.55 3. (i) Rs. 321224 (ii) Rs. 801550.75 4. (i) Rs. 113.45 (ii) Rs. 18.35 5. (i) (b) Rs. 44790 (c) 2194710 (d) Rs. 29.26 (ii) (b) Rs. 54045 (c) Rs. 2648205 (d) Rs. 35.31 6. (i) (a) Rs. 2029687.80 (b) Rs. 2124251.60 (c) 1.96% (ii) (a) Rs. 2333329.18 (b) Rs. 2492655.62 (c) loss, 2.91% (iii) (b) Rs. 2130326 (c) Rs. 2113144 (d) loss, 0.81% 7. (i) (a) $0.0157, £ 0.0154 (b) $83.60 (c) 1.96% (ii) Yen 1.7889, Singapore $ 0.0227 (b) $ 6362.96 (Singapore) (c) 98.73% (iii) (a) 49.75% (b) loss, 33.14% 8. (i) (b) Rs. 1997550 (c) Rs. 159.04 (d) 19.43% (ii) (b) Rs. 2081760 (c) 162.08 (d) 6.57% 9. (i) $240 (ii) £ 707.19 10. (i) Rs. 453.52 (b) Rs. 33257.95 (c) Rs. 95918.96 (d) Rs. 159864.93 (ii) Rs. 1383 (b) Rs. 79522.50 (c) Rs. 51977.75 (d) Rs. 155933.25 11. (i) (a) Rs. 4303495.44 (b) Rs. 1828985.56 (c) Rs. 1188840.62 (d) Rs. 10333768.40 (ii) (a) Rs. 93750 (b) Rs. 62500 (c) Rs. 1171875 (d) Euro 5.89 Answers Project Work Cut down the foreign exchange rate of currency from the daily newspaper. Study the buying rate and selling rate of Nepal Rastra Bank or other commercial banks. Collect the record of buying and selling the foreign currency in any bank by two customers. When these currencies are exchanged into Nepali currencies by using the foreign exchange rate how much does Nepali currency need or get for it. If Nepali currency devaluated by 10% and exchanged these currencies which bought or purchased by the customers, who much percent is the loss or gain to the customers ?

Arithmetic 86 Allied The Leading Mathematics-10 Money Exchange 87 CONFIDENCE LEVEL TEST - II Unit II : Arithmetic Class: 10, The Leading Maths Time: 45 mins. FM: 23 Attempt all questions. 1. Ramesh plansto deposit Rs. 200000 for 3 yearsin a bank. Bank has offered two schemes as given below. Scheme A Rate of Interest (R) = 12% p.a. Interest compounded annually Scheme B Rate of Interest (R) = 11% p.a. Interest compounded semi-annually (a) Write the formula for finding the compound interest annually. [1] (b) How much interest will he get for three years when he deposits in the scheme A? [1] (c) If Ram withdrew Rs. 100000 at the end of the first year, how much interest will he get when he deposited in the scheme B? Calculate it. [2] (d) Which is better scheme A or B? Why? [1] 2. A sum of money amounts to Rs. 220000 at a certain rate of yearly compound interest in 1 year and Rs. 266200 in 3 years. (a) What is compound interest ? Define it. [1] (b) Find the rate of interest and the sum of money. [3] 3. Ram bought a house. The price of the house is Rs. 5000000 and its cost decreases at 10% every year. (a) After how many years will the cost of house be Rs. 3645000? Find it. [2] (b) If he deposited Rs 5000000 in a bank at the rate of 10% p.a. compounded annually instead of buying this house, how much profit will he get after paying Rs. 1400000 for rent and household expenses in 3 years? Calculate it. [2] 4. The number of trees in a community forest was 45000 in 2017 AD. In 2019, only remained 36450 due to deforestation due to a shortage of LP gas. If deforestation continued at the same rate up to 2021 but the rate of deforestation in 2022 was only half of the previous rate. (a) Find the rate of compound depreciation up to 2021 and 2022 respectively. [2] (b) How many trees were cut down in 2022 only? Find it. [2] 5. A Nepali Babu Ramesh working in Malaysia brings an amount of Malaysian Ringgit MYR 75000 to his home. He goes to the bank to exchange it into Nepali currency and the bank takes 2% commission on it. [RM 1 = Rs. 30.10] (a) What is Nepali currency in Nepal after giving commission? Write it. [1] (b) Ramesh deposits the two-thirds of the amount in a bank at the rate of 12% compounded annually, how much amount does he get after 3 years? [2] (c) Ramesh has bought a second hand car from the remaining amount. What will be the price of the car at the decreasing rate of 15% compounded annually after 3 years? [1] (d) At the end of 3 years, how much amount does he have with the valuation of the car and cash in the bank? [1] Best of Luck

Arithmetic 86 Allied The Leading Mathematics-10 Money Exchange 87 Additional Practice – II Compound Interest 1. (a) The compound interest and compound on a sum P in T years at R% p.a. are CI and CA respectively. Write the relation among P, T, R and CI. (compounded annually). (b) The compound interest and compound on a sum P in T years at R% p.a. are CI and CA respectively. Write the relation between P, T, R and CI for (compounded semi-annually) 2. Find the amount and the compound interest in each of the following compounded annually: Principal Rate Time (yrs) (a) Rs. 5000 8.5% p.a. 3 (b) Rs. 15000 8% p.a. 2 1 2 3. What sum of money will amount to Rs. 7123.20 in one and a half year at 12% per annum compounded yearly? 4. A person deposits a sum of Rs. 8000 for some period in a finance company which pays an interest of 5% compounded annually. After maturity he gets Rs. 9261. Find the time for which he deposited the money. 5. A invested Rs. 25000 for 3 years at the rate of 12% p.a. simple interest. B invested the same amount for the same time at the rate of 10% p.a. interest compounded annually. Who received more interest at the end of 3 years and by how much? 6. Anil invested Rs. 150000 in a small scale industry. He receives 5% compound interest compounded annually. How much money does Bishal have to invest to get the same interest in 3 years at the rate of 6% p.a. simple interest? 7. The difference between the compound interest and the simple interest on a sum of money at the rate of 8% p.a. is Rs. 40 in 3 years. Find the amount of money invested. 8. Find the amount and compound interest on Rs. 5000 at the rate of 10% p.a. for 2 years 3 months. 9. A sum of money amounts to Rs. 60500 at a certain rate of yearly compound interest in 2 years and Rs. 73205 in 4 years. Find the rate of interest and the sum. 10. The compound amount on a certain sum for a year is Rs. 396900 and for 3 years is Rs. 497871.36. Calculate the sum and the rate of interest. 11. A sum of money lent out on CI doubles in 3 years. In how many years, would the sum of money (lent out on the same rate of interest) become eight times? 12. Hari borrows a sum of money and agrees to pay off by paying Rs. 3150 at the end of the first year and Rs. 4410 at the end of the second year. If the rate of compound interest is 5% p.a., how much does he borrow?

Arithmetic 88 Allied The Leading Mathematics-10 Money Exchange 89 Population Growth 13. If the initial population P increase at the rate of R% p. a for T years what is the population migrated in after T years ? 14. The present population of a village is 8500. If the rate of population growth was 2% per annum, what is the total population after one year? 15. The population of a town decreases 5% per year. If its present population is 1,20,000, find its population after 3 years. 16. The population of the country Syria decreases 2.05% per year due to domestic war. If its present population is 2,20,44,000, find its population after 3 years. 17. The population in a village in 2060 was 5000; and the growth rate was 2%. In 2061, because of economic crisis 1000 people migrated to another village. What was the population in the village in 2063? 18. The population of a village is 53,045 in 2074. What was the population growth rate if the population in 2072 was 50,000. 19. In Germany the population in 2014 was 8,26,52,256 with population growth rate 1.2%. In 2015 because of political situation in Syria, 2,50,000 refugees migrated in Germany from Syria. If the population growth rate is 1.3% p.a., find the total population of Germany in 2017. 20. If the population of a village becomes 1331 1000 times in three years, find the population growth rate. 21. Three years ago the number of students in a university was 120000. 52 students are restigated from the university and 82 students are entered from the other university. If the increasing rate of students is 4.5% every year, find the number of present students in this period. If the increasing rate of students will be 5.5% for next two years, what will be the number of students two years hence? 22. The population of a village was 30000 in the beginning of the year 2072. 136 people died and 175 children born within the last of the year 2073. Find the population till 2073. Again, 1270 people migrated from the village, 95 people died, 75 children born and 245 people came as immigrant into this village in the last of the year 2074. If the rate of population growth is 2%, how many people will there be in the village at the last of the year 2074? Compound Depreciation 23. An article costing Rs. x is depreciated at the rate of b% p.a. What will be its cost after 3 years? 24. The cost of a machine is 850000. If the rate of depreciation is 2% per annum, what is the cost of the machine after one year? 25. A machine costing Rs. 3500 is depreciated in a year by 10%. What is the amount of depreciation?

Arithmetic 88 Allied The Leading Mathematics-10 Money Exchange 89 26. A piece of land was bought for Rs. 2200000 before 3 years. Now it is sold for Rs. 1886225 only. At what rate was the value depreciated? 27. A car bought for Rs. 5542000 is depreciated at 10% year for the first 2 years and with different rate afterward. It was sold for Rs. 2298378.24. What was the later rate of depreciation if the machine was sold after 5 years? 28. A domestic industry was established with an initial investment of Rs. 1350000. At the end of 2 years the industry earned Rs. 340000 and the rate of compounded depreciation is 10% p.a. If the industry was sold after 2 years, work out the loss or profit. 29. A house bought for Rs. 12500000 is depreciated at 3%, 4% and 6% for 2, 3, and 4 years respectively. What is the price of the house after 2 years, 3 years and 4 years respectively? 30. A car is sold for some amount by a businessman to a man after using 2 years at 10% rate of depreciation. The man again sold it to a woman at 20% rate of depreciation after using 3 years and she also sold it for Rs. 1219276.80 at 30% rate of depreciation after using 2 years. How much was the businessman paid for the car? Money Exchange Use the exchange rate given above for the date 2073/09/20 to solve the following problems: 31. The following amounts have to be changed by overseas going students from the bank. How much the Nepali currency is needed to purchase these amounts ? (a) $ 3567 (b) £ 4289 (c) $ 10,920 (Canadian) (d) $ 42617 (Australia) (e) Euro 2985 32. What is the amount of the following currencies that could be purchased for Rs. 125678.00? (a) US $ (b) UK £ (c) $ (Ca) (d) $ (Aus) 33. Sarmila works in Japan. He brought home for Japanese Yen ¥ 50,00,000. Find the Nepali equivalence for the exchange of this amount. (Use today's currency exchange rate) 34. If Euro 60 = Rs. 5557.88 and Rs. 4423.20 = $ 80 (Australian), find the amount of Euro that can be purchased for $ 240 (Australian) 35. If 1 London pound (£) = Rs. 155.78 and 1 US dollar ($) = Rs. 109.75, how much dollars exchanged can be for 1 London pound (£)? 36. An American tourist can purchase air ticket in Chicago for Kathmandu at $ 950 and Rs. 105600 in Kathmandu for Chicago. From where he gets cheaper and by how much? [$ 1 = Rs. 112.53] 37. Bishal brought US $ for Rs. 590215. The next day US $ was devaluated by 5%. If he exchanges Nepalese rupees next day, what is his loss percentage ? (Before devaluation, buying rate, 1 US $ = Rs. 112.53, selling rate, 1 US $ = Rs. 113.78)

Arithmetic 90 Allied The Leading Mathematics-10 Money Exchange PB 38. How much will be profit or loss percent if an article bought for $ 6 is sold for £ 6? [$ 1 = Rs. 102.56 and £ 1 = Rs. 138.34] 39. The selling rate of $ 1 was Rs. 112. 45 yesterday. Today, Nepali currency is devaluated by 15%. What will be the cost of a TV in Nepali currency that costs $ 150 if the custom duty is 120% and VAT is 13% once the custom duty is added ? 40. The selling rate of UK £ 1 was Rs. 172. 75 yesterday. Today, Nepali currency is devaluated by 10%. What will be the cost of a mobile set in Nepali currency that costs UK £ 70 if the custom duty is 50%, 13% VAT is added once including the custom duty and 25% commission on it? 2. (a) Rs. 6386.45, Rs. 1386.45 (b) Rs. 18195.84, Rs. 3195.84 3. Rs. 6000 4. 3 years 5. A, Rs. 725 6. Rs. 131354.17 7. Rs. 2029.21 8. Rs. 6201.25, Rs. 1201.25 9. 10%, Rs. 50000 10. 12%, Rs. 354375 11. 9 years 12. Rs. 7000 14. 8670 15. 1,02,885 16. 20715897 17. 4266 18. 3% 19. 86089508 20. 10% 21. 152452 22. 30832 24. Rs. 867000 25. Rs. 350 26. 5% 27. 20% 28. Profit, Rs. 83500 29. Rs. 8124162.38 40. Rs. 6000000 34. Euro 143.25 35. $ 1.42 36. Kathmandu, Rs. 1303.50 37. 55.80% 38. Profit, 34.89% 39. Rs. 48222.50 40. Rs. 28183.08 Answers

Mensuration PB Allied The Leading Mathematics-10 Area and Volume 91 Analyse knowledge, skill and concept related to surface area and volume of solid objects in behavior problem solving and use them. Marks Weightage : 13 Estimated Working Hours : Competency Learning Outcomes 1. To find the surface area and volume of pyramid. 2. To solve the problems related to area and volume of combined solid. 3. To solve the problems related to estimation by using the properties of different solid object or geometric shapes in constructing field. 5. Area and Volume 5.1 Right Pyramid 5.2 Right Cone 5.3 Combined Solid 5.4 Cost Estimation Chapters / Lessons MENSURATION III UNIT 28 (Th. + Pr.) Nyatapola Pagoda Temple Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of Items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks III Mensuration 28 2 2 2 3 2 5 2 3 8 3 13 HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School.

Mensuration 92 Allied The Leading Mathematics-10 Area and Volume 93 R E V I S I T I N G M E N S U R A T I O N WARM-UP Mensuration Mensuration isthe study ofthemeasurement of geometricmeasurements ormagnitudessuch aslength, area and volume.What type of geometric figure is called triangle and quadrilateral?Tell their types and properties. What is the area and perimeter of a plane figure ? Tell and write the formulae of the perimeter and area of the plane figures studied in the class 8 and 9 mentioned in the table below. SN Name Definition Properties Figure Perimeter Area 1. Rectangle Equiangular quadrilateral Opposite sides and diagonals are equal, diagonals bisect and every angle is 90o . b l P = ... A = ... 2. Square Rectangle with equal adjacent sides All sides and diagonals are equal, diagonals bisect perpendicularly and every angle is 90o . l P = ... A = ... 3. Right Triangle Triangle with right angle Sum of two acute angles is 90o , sum of squares of legs is equal to square of hypotenuse (Pythagoras' Theoremp2 + b2 = h2 ). b p h P = ... A = ... 4. Isosceles Right Triangle Right triangle with equal legs Two legs are equal, acute/base angles are equal to 45o . p p h P = ... A = ... 5. Triangle Closed plane figure of three line segments Sum of length of any two sides is longer than third side, sum of angles is 180o . b a h c P = ... A = ... 6. Isosceles Triangle Triangle with equal any two sides Base angles are equal, median perpendicularly bisects the base and vertical angle. b a a P = ... A = ... 7. Equilateral Triangle Triangle with equal sides All sides angle are equal and every angle is equal to 60o . a a a P = ... A = ... 8. Quradrilateral Clased plane figure of four line segments Sum of angles is 160 degrees a b h1 c h2 e d P = ... A = ... 9. Trapezium One pair of opposite sides are paraallel Base sides are parallel a b1 b2 c h P = ... A = ...

Mensuration 92 Allied The Leading Mathematics-10 Area and Volume 93 R E V I S I T I N G M E N S U R A T I O N 10. Parallelogram Opposite sides are parallel Opposite sides and angles are equal, diagonas are biseced a a b b h P = ... A = ... 11. Rhombus Equal adjacent sides in oparallelogram Diagonals are perpendicular, all sides are equal a a a a d1 d2 P = ... A = ... 12. Kite Opposite adjacent sides are equal Diagonals are perpendicular a a b b d2 d1 P = ... A = ... 8. Scalene Triangle Triangle with unequal sides Opposite side of the greatest/ smallest angle is longest/ shortest and vice versa. b c a P = ... A = ... 9. Circle Curved line equidistant from a fixed point (centre) Radii are equal, the length of diameter is twice of radius, the ratio of circumference and diameter is π. r P = ... A = ... 10. SemiCircle Half of circle divided by diagonal Each diameter bisectthe circle, the ratio of half circumference and diameter is π 2. r P = ... A = ... 11. Sector A part of circle divided by any two radii Area of sector depends up on the arc length, the central angle is positive. r q P = ... A = ... Surface Area and Volume of Prism, Cylinder, Cone and Sphere, Hemi-sphere Tell and write the volumes of the solids mention in the table below. SN Name Definition Properties Figure RLA/LSA TSA Volume 1. Prism A geometric solid object whose polygonal crosssection is identical and parallel to its bases. The faces of prism are rectangles or parallelograms, called rectangular or lateral faces. A h LSA = .......... TSA = .......... V = .......... 2. Right Prism A prism whose bases are perpendicular to its bases. All faces are rectangles. A h RSA = .......... TSA = .......... V = ..........

Mensuration 94 Allied The Leading Mathematics-10 Area and Volume 95 3. Triangular Prism A prism with triangular bases It has three rectangular faces. h A RSA = .......... TSA = .......... V = .......... 4. Square based Prism A prism with square bases It has 4 rectangular faces. a h RSA = .......... TSA = .......... V = .......... 5. Rectangular Prism (Cuboid) A prism with rectangular bases It has 4 rectangular faces. a b h RSA = .......... TSA = .......... V = .......... 6. Cube A prism with identical square bases and faces It has 6 square surfaces. a a a RSA = .......... TSA = .......... V = .......... 7. Polygonal Prism A prism with polygonal bases It hasthe same number of faces and the number of sides of the base. h A RSA = .......... TSA = .......... V = .......... 8. Cylinder A prism with circular bases It has one curved lateral or round face. h r CSA = .......... TSA = .......... V = .......... 9. HemiCylinder Diagonally half cylinder along its height It has one half round face and one rectangular face. h r CSA = .......... TSA = .......... V = .......... 10. Quarter Cylinder One-fourth cylinder along its height from two radii It has one-fourth round face and two rectangular faces. r h CSA = .......... TSA = .......... V = .......... 11. Sphere A perfectly round solid object with every point on its surface equidistant from its centre It has only one round face, no edge, no vertex and no flat face and height = diameter d = 2r r SA = .......... V = .......... 12. Hemisphere One half of sphere along great circle or diameter It has half round face and one flat circle. r CSA = .......... TSA = .......... V = ..........

Mensuration 94 Allied The Leading Mathematics-10 Area and Volume 95 5.1 Right Pyramid At the end of this topic, the students will be able to: ¾ find the CSA, TSA and Volume of right pyramid and solve related problems. Learning Objectives I Introduction Lets discuss the following activity on Geoboard. Activity 1 Make a triangle by using rubber band or thread. Tie three vertical nails with other rubber bands or threads and take up these three rubber bands or threads as shown in the figure. What do you see ? How many vertices, edges and plane faces does it have ? This is exactly looked like "The Great Pyramid of Giza (Khufu), Egypt". This is another solid object different from the prisms and spheres that have 5 faces, 5 vertices and 8 edges. This is satisfied for the Euler's formula F + V - E = 2. This is a three-dimensional geometric object; having length, breadth and height. In this object, all the laterals faces meet at a point, called apex. This types of geometric object is called Pyramid. Pyramid is a monumental structure. The form was used in Ancient Egypt for funerary structures: celebrated examples are the pyramids at Giza, near Cairo, Egypt (2551 - 2472 BC). Other types of pyramid include the stepped form found in Meso-American Pre-Columbian architecture in the period of Mayan Civilization, built by Imhotep for King Zoser (2630 - 2611 BC). Measurements of different parts and relation between different solids have been derived by Greek mathematician, physicist, engineer, astronomer and inventor Archimedes of Syracuse (287- 212 BC). In Nepal, the pyramid types of temple, Pashupati Temple, was build around the first century BC, called Pagoda Temple. The idea and measurement of pyramids are used tent or shelter, top of pillar, roof of house, needle of fence, temples, etc. This pyramid is formed by a prism. When all the points containing in a base, say PQRS, of the prism combine to a single point, it forms a pyramid as shown in the figure. Great Pyramid of Giza, Egypt Nyatapola Pagoda Temple V A P Q S R B C D CHAPTER 5 AREA AND VOLUME

Mensuration 96 Allied The Leading Mathematics-10 Area and Volume 97 II Parts of Pyramid A pyramid is a solid with convex polygonal base and other faces are triangles that meet at one common point outside the polygon with straight lines. The polygonal face is called the base of the pyramid. All other triangular faces are called its lateral faces. The external point at which the triangles meet, is called the vertex or apex of the pyramid. The lines joining the external point to the vertices of the base (or the intersections of the lateral faces) are called its edges or slanting edges. If the base is a regular polygon and the perpendicular from the vertex passes through the centre of the base, the pyramid is called a regular or right pyramid. Otherwise, it is oblique pyramid as shown in the adjoining figures. A pyramid with regular base is called the regular right pyramid. In the above figures, the Nyatapola temple and Great Pyramid of Giza are the regular right pyramids. In particular, if the base of a pyramid is a triangle, it is a triangular pyramid. It is also called a tetrahedron. A rectangular pyramid is a pyramid with rectangular base. If the base of a pyramid is square, it is called square pyramid and so on. In any pyramid, the length of the perpendicular from the vertex to a side of the base is called the slanting height of the corresponding face of the pyramid; and the length of the perpendicular from the vertex to the base is called its height. III Surface Area of Square based Pyramid The total area of the triangles is called the lateral surface area (LSA) of the pyramid and the sum of the base area and lateral area is the total surface area (TSA) of the pyramid. (a) Base Area (A) The base area of a square base pyramid is the area of the base of the pyramid. Its base is square. Therefore, Area of base square (A) = Area of square = a2 . ∴ Area of base square (A) = a2 , where a = Length of base side B B C D C O Vertex Vertex Lateral faces Lateral faces Slanting edges Slanting edges Base Base Triangular pyramid Hexagonal pyramid Rectangular pyramid Square pyramid O A B C D F E O A B D C O A A Oblique Pyramid Right Pyramid Ab Ab h Square base LSA TSA a a a

Mensuration 96 Allied The Leading Mathematics-10 Area and Volume 97 (b) Surface Area (SA) Activity 2 To visualize the surface area, we draw the nets of the square based pyramids. The lateral faces of the square based pyramid are congruent isosceles triangles. If a denotes the length of the sides of the base square and l, the slant height of each slant isosceles triangle then, Area of one slant isosceles triangle = 1 2 al Therefore, lateral surface area of square based pyramid (LSA) = Area of 4 slant isosceles triangle = 4 × 1 2 al = 2al In other hand, LSA = 4 × 1 2 al = 1 2 4a × l = 1 2 Pl ∴ LSA of Square based Pyramid = 2al or 1 2 Pl, where a = Length of side of base square, l = Slant height of lateral face, P = Perimeter of base square Total surface area of a pyramid (TSA) = Lateral Surface Area (LSA) + Area of Base Square (A) = 2al + A or 1 2 Pl + A ∴ TSA of Square based Pyramid = 2al + a2 or 1 2 Pl + a2 . For more, (i) If the base of the pyramid is an equilateral triangle with side a then, LSA = 3 2 al and TSA = 3 2 al + 3 4 a2 . (ii) If the base of the pyramid is a rectangle with sides a and b, slant heights l1 and l2 then, LSA = 2 × 1 2 al1 + 2 × 1 2 bl2 = al1 + bl2 and TSA = al1 + bl2 + ab. IV Volume of Square based Pyramid Activity 3 Make six congruent pyramids with base side 2x and height x. Combine them to each other as shown in the figure, we obtain a cube of side 2x. Hence, we can say that the volume of each pyramid with base side 2x and height x is the one-sixth of the volume of the cube with side 2x. = = = 6 pyramids 6 pyramids 1 cube a a a a a a a a l l l l l l a l l2 l1 a b

Mensuration 98 Allied The Leading Mathematics-10 Area and Volume 99 Let's formulate the volume of the pyramid: Take a cube of side 2x. Draw its all space diagonals. The diagonals intersect at a point. All the vertices of the cube are joined with the intersecting point of the space diagonals. Observe in the given figure. How many equal pyramids do you find?You can see 6 equal pyramids. Forinstance OABCD is a pyramid with bottom face ofthe cube asits base. Let the volume of a pyramid be V. It is obvious that the total volume of 6 pyramids is equal to the volume of the cube. i.e., 6V = (2x)3 V = 1 6 (2x)2 .2x = 1 3 (2x)2 . x [where x is the height of pyramid OABCD.] ∴ Volume of Square based pyramid (V) = 1 3 Base Area(A) × Height of pyramid (h) = 1 3 a2 h Activity 4 "Alternatively" From the adjoining activities, we can easily say that the volume of each pyramid is one-third of the volume of the cube. i.e., Vol. of a pyramid (V) = 1 3 Vol. of Cube = 1 3 l 3 = 1 3 l 2 × l = 1 3 Base Area (A) × Height (h). ∴ Volume of Square based pyramid (V) = 1 3 Base Area (A) × Height of pyramid (h) = 1 3 a2 h Activity 5 "Alternatively" Make a cuboid/cube and pyramid with the same base and same height. Fill the pyramid with any materials and pour into the cuboid/cube at three times, the cuboid/cube is exactly fulfilled. Then we can say that the volume of pyramid is one-third of the volume of cuboid/cube with the same base and same height. i.e., Vol. of Squared based pyramid (V) = 1 3 Vol. of Cuboid = 1 3 l × b × h = 1 3 A × h = 1 3 Base Area (A) × Height of Pyramid (h) = 1 3 a2 h. H G E F A B x D O C 2x 2x 2x O A B D C x 2x 2x = = 3 pyramids 3 pyramids 1 cube 3 pyramids 1 Cube

Mensuration 98 Allied The Leading Mathematics-10 Area and Volume 99 For more, (i) If the base of the pyramid is an equilateral triangle with side a, then V = 1 3 A × h = 1 3 3 4 a2 h = 3 12 a2 h. (ii) If the base of the pyramid is a rectangle with sides a and b, slant heights l1 and l2, then V = 1 3 A × h = 1 3 abh. V Relation among Side, Height, Slant height and Edge of Square based Pyramid In the adjoining figure, OABCD is a squared base pyramid with base sides 'a', height 'h', slant height 'l' and edge side 'e'. Draw the diagonals and a median of the squareABCD, which are intersected, isthe center of the square ABCD. Therefore, AQ = BQ = PQ = a 2 . Also, draw OQ⊥AB. Now, (a) Relation between l, h and a In rt.∠d ∆POQ, OQ = OP2 + PQ2 = h2 + a2 4 = 4h2 + a2 4 = 1 2 4h2 + a2 ∴ Slant height (l) = 1 2 4h2 + a2 , Height (h) = 1 2 4l 2 – a2 and Base side (a) = 2 l 2 – h2 . (b) Relation between e, l and a In rt.∠d ∆AOQ, OQ = OA2 – AQ2 = e2 – a2 4 = 4e2 – a2 4 = 1 2 4e2 – a2 Slant height (l) = 1 2 4e2 – a2 , Edge side(e) = 1 2 4l 2 + a2 and Base side (a) = 2 e2 – l 2 . (c) Relation between e, d and h In rt. ∠d ∆OPA, AP = 1 2 AC = 1 2 d, d = diagonal and, OA = OP2 + AP2 = h2 + d 2 2 = h2 + d2 4 = 4h2 + d2 4 = 1 2 4h2 + d2 ∴ Edge (e) = 1 2 4h2 + d2 , Height (h) = 1 2 4e2 – d2 , diagonal (d) = 2 e2 – h2 (d) Relation between e, h and a, In rt.∠d ∆ABC, AC = AB2 + BC2 = a2 + a2 = 2a2 = 2 a. ∴ AP = 1 2 AC = 1 2 2 a = a 2 In rt.∠d ∆AOP, OP = OA2 – AP2 = e2 – a2 2 = 2e2 – a2 2 = 1 2 2e2 – a2 ∴ Height (h) = 1 2 2e2 – a2 , Edge side (e) = 1 2 2h2 + a2 and Base side (a) = 2(e2 – h2 ) . From the above relations, we derive the relations between h, l and e as given below: Height (h) = 2l 2 – e2 , slant height (l) = 1 2 h2 + e2 and edge side (e) = 2l 2 – h2 . a a a h h a b a 2 a 2 a 2 A B D C a a a e h O Q P

Mensuration 100 Allied The Leading Mathematics-10 Area and Volume 101 Points to be Remembered 1. A monumental structure with a convex polygonal base and sloping sides or faces that meet at a point at the top is called pyramid. 2. The sloping sides or triangular faces of a pyramid are called lateral surfaces. 3. A common meeting point of the lateral surfaces of a pyramid is called apex or vertex. 4. The polygonal face of a pyramid is called the base. The length of the base side is denoted by a. 5. The lines joining the external point to the vertices of the base (or the intersections of the lateral faces) are called its edges or slanting edges. 6. The length of the perpendicular from the vertex to a side of the base is called the slanting height (l). 7. The length of the perpendicular from the vertex to the base is called its height (h). 8. If the base is a regular polygon and the perpendicular from the vertex passes through the centre of the base, the pyramid is called a right pyramid. Otherwise, it is oblique pyramid as shown in the adjoining figures. 9. A pyramid with regular base is called the regular right pyramid. 10. A pyramid with triangle base is called triangular pyramid. It is also called a tetrahedron. 11. A pyramid with rectangle base is called rectangular pyramid. 12. A pyramid with square base is called square pyramid. 13. Area of base of a squared pyramid (A) = a2 , where a = Length of base side 14. Lateral surface Area of Square based Pyramid (LSA) = 2al or 1 2 Pl, where a = Length of side of base square, l = Slant height of lateral face, P = Perimeter of base square 15. Total surface Area of Square based Pyramid (TSA) = 2al + a2 or 1 2 Pl + a2 . 16. Volume of Square based pyramid (V) = 1 3 Square Base Area (A)×Height (h) = 1 3 a2 h B B C D C O Vertex Lateral faces Slanting edges Base Triangular pyramid Square pyramid O A A Oblique Pyramid Right Pyramid Ab Ab h a a a a l l l l 1 cube