Interest: It is the additional money besides the original money paid by the borrower to the money lender in lieu of the money used. Principal: The money borrowed (or the money lent) is called principal. Amount: The sum of the principal and the interest is called amount. Thus, amount = principal +interest. Rate: It is the interest paid on Rs 100 for a specified period. Time: It is the time for which the money is borrowed. Simple Interest: It is the interest calculated on the original money (principal) for any given time and rate. Formula: Simple Interest = (Principal x Rate x time)/100 Compound interestCompound interest (abbreviated C.I.) can be easily calculated by the following formula: C.I. = A -P = Remark. Interest may be converted into principal annually, semiannually, quarterly, monthly etc. The number of times interest is converted into principal in a year is called the frequency of conversion, and the period of time between two conversions is called the conversion period or interest period. Thus "rate of 5%" means a rate of 5% compounded annually; 12% compounded semi-annually means that each interest period of 6 months earns an interest of 6%. Thus the rate of interest per interest period is r = (annual rate of interest) / (frequency of conversion) and the number of interest periods is n = (given number of years) x (frequency of conversion). In solving problems on compound interest, remember the following: 1. A = P and C.I. =where A is the final amount, P is the principal, r is the rate of interest compounded yearly (or every interest period) and n is the number of years (or terms of the interest period). 2. When the interest rates for the successive fixed periods are r1 %, r2 %, r3 %, ..., then the final amount A is given by 3. S.I. (simple interest) and C.I. are equal for the first year (or the first term of the interest period) on the same sum and at the same rate. 4. C.I. of 2nd year (or the second term of the interest period) is more than the C.I. of Ist year (or the first term of the interest period), and C.I. of 2nd year -C.I. of Ist year = S.I. on the interest of the first year. 5. Equivalent, nominal and effective rates of interest
6. Present value or present worth of a sum of Rs P due n years hence at r% compound interest is P.V.= 7. Equal instalments (with compound interest) Note. If T = n years (or interest terms), then there will be n brackets. 8. Formulae for population If, however, there is annual decrease of r% per annum, the population after n years will be , and n years ago, the population was . DepreciationAll fixed assets such as machinery, building, furniture etc. gradually diminish in value as they get older and become worn out by constant use in business. Depreciation is the term used to describe this decrease in book value of an asset. The number of years a machine can be effectively used is called its life span. After that it is sold as waste or scrap. Illustrative ExamplesExampleFind the compound amount of Rs 1500 for 6 years 7 months, at 5·2% compounded semi annually. SolutionUsing formula, we could find the value of But in these kinds of problems, generally we use compound interest for full interest period and simple interest for fractional interest period. Here we find compound interest for 13 interest periods and simple interest for 1 month. Required compound amountA = 1500 =1500(1·026)13(1·0043) Taking logs, log A = log 1500 +13 log (1·026) +log (1·0043) = 3·1761 +13(0·0112) +0·0017) = 3·3234. A = antilog (3·3234) = 2144 Thus the required compound amount is Rs 2144. ExampleThe simple interest in 3 years and the compound interest in 2 years on a certain sum at the same rate are Rs 1200 and Rs 832 respectively. Find (i) the rate of interest. (ii) the principal. (iii) the difference between the C.I. and S.I. for 3 years. Solution
ExampleThe population of an industrial town is increasing by 5% every year. If the present population is 1 million, estimate the population five years hence. Also estimate the population three years ago. SolutionPresent population, P = 1 million, rate of increase = 5% per annum = 1000000 = 1000000 (1·05)5 = 1276280 Population three years ago = = 863838.ExampleAvichal Publishers buy a machine for Rs 20000. The rate of depreciation is 10%. Find the depreciated value of the machine after 3 years. Also find the amount of depreciation. What is the average rate of depreciation? SolutionOriginal value of machine = Rs 20000, Rate of depreciation, i = 10% Hence the book value after 3 years = 20000 = 20000 (0·9)³ = 20000 (0·729) = Rs 14580. Amount of depreciation in 3 years = Rs 20000 -Rs 14580 = Rs 5420 Average rate of depreciation in 3 years= (5420/20000) x (100/3) = 9·033% Exercise
Answers1. (i) Rs 784 (ii) Rs 6384 (iii) Rs 894 2. Rs 13240 |