Compound Interest and Depreciation
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 interest (abbreviated C.I.) can be easily calculated by the
A = P where A is the final amount, P is the principal, r is the rate of interest compounded yearly and n is the number of years.
C.I. = A -P =
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
Two annual rates of interest with different conversion periods are called equivalent if they yield the same compound amount at the end of the year. For example, consider an amount of Rs 10,000 invested at 4% interest compounded quarterly. So, the amount at the end of one year = 10000(1·01)4 = 10406. This is equivalent to interest of 4·06% compounded annually because 10000(1·0406) = 10406.
When interest is compounded more than once in a year, the given annual rate is called nominal rate or nominal annual rate. The rate of interest actually earned is called effective rate. In the above example, nominal rate is 4% while effective rate is 4·06%.
If nominal rate is r% compounded p times in year, then effective rate of interest is
6. Present value or present worth of a sum of Rs P due n
years hence at r% compound interest is
In particular, present value of sum of a Rs P due one year hence (i.e. n = 1) at r% (compound) interest is
7. Equal instalments (with compound interest)
Loan amount = , where
P = each equal instalment
R = rate of interest per annum (or per interest period)
T = time, say 3 years (or 3 interest terms).
Note. If T = n years (or interest terms), then there will be n brackets.
8. Formulae for population
If the present population of a town is P and its annual increase is r%, the population after n years will be P , and n years ago, the population was .
If, however, there is annual decrease of r% per annum, the population after n years will be , and n years ago, the population was .
All 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.
There are a number of methods of calculating depreciation. However, the most common method which is also approved by income tax authorities, is the Diminishing Balance Method. Here each years depreciation is calculated on the book value (i.e., depreciated value) of the asset at the beginning of the year rather than original cost. Note that as the book value decreases every year, the amount of depreciation also decreases every year. Therefore, this method is also called Reducing Instalment Method or "Written Down Value Method".
If the rate of depreciation is i% per year and the initial value of the asset is P, the depreciated value at the end of n years is and the amount of depreciation is . If n is large, log tables should be used for calculation.
The number of years a machine can be effectively used is called its life span. After that it is sold as waste or scrap.
Find the compound amount of Rs 1500 for 6 years 7 months, at 5·2% compounded semi annually.
Using 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 amount
A = 1500
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.
The 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.
- Let the principal be Rs P and rate of interest be R% p.a.
According to the first condition of the question,
(P x R x 3)/100 = 1200 => P x R = 40000 ... (1)
According to the second condition of the question,
=> 4[(100)² +R² +2. 100. R -(100)²] = 832 R
=> R² +200 R = 208 R => R² +200 R -208 R = 0
=> R² -8R = 0 => R(R -8) = 0
=> either R = 0 or R -8 = 0
=> either R = 0 or R = 8, but R cannot be zero.
Hence the rate of interest = 8% p.a.
- On using (1), we get P x 8 = 40000, so P = 5000
- Rate of compound interest = 8% p.a. and principal = Rs 5000
Amount due after 3 years = Rs 5000 x
= Rs 5000 x = Rs 6298·56
Hence C.I. for 3 years = A -P = Rs 6298·56 -Rs 5000 = Rs 1298·56
The difference between the C.I. and S.I. for 3 years = Rs 1298·56 -Rs 1200 = Rs 98·56
The 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.
Present population, P = 1 million, rate of increase = 5% per annum
Hence population after 5 years =
= 1000000 = 1000000 (1·05)5
Population three years ago = = 863838.
Avichal 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?
Original 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%
- A person invests Rs 5600 at 14% p.a. compound interest for 2 years. Calculate:
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year, correct to the nearest rupee.
- A man saves every year Rs 4000 and invests it at the end of the year at 10% per annum compound interest. Calculate the total amount of his savings at the end of the third year.
- The simple interest on a certain sum for 3 years is Rs 225 and the compound interest on the same sum at the same rate for 2 years is Rs 153. Find the rate of interest and the principal.
- A sum of money is lent out at compound interest for two years at 20% p.a., C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at the same rate percent per annum, C.I. being reckoned half-yearly, it would have fetched Rs 482 more by way of interest. Calculate the sum of money lent out.
- A man borrowed a certain sum of money and paid it back in 2 years in two equal instalments. If the rate of compound interest was 4 percent per annum and if he paid back Rs 676 annually, what sum did he borrow?
- A sum of Rs 32800 is borrowed to be paid back in 2 years by two equal annual instalments allowing 5% compound interest. Find the annual payment.
- A loan of Rs 4641 is to be paid back by 4 equal annual instalments. The interest is compounded annually at 10%. Find the value of each instalment.
- A man borrows Rs 5800 at 12% p.a. compound interest. He repays Rs 1800 at the end of every six months. Calculate the amount outstanding at the end of the third payment. Give your answer to the nearest Re.
- A man borrows Rs 5000 at 10% p.a. compounded annually. He repays Rs 1000 at the end of each of first three years. Find the amount which he has to pay at the end of the fourth year.
- Divide Rs 21866 in two parts such that the amount of one in 3 years is same as the amount of the second in 5 years. The rate of compound interest is 5% per annum.
- Two partners A and B together invest Rs 10000 at 12% per annum compounded annually. After 3 years, A gets the same amount as B gets after 5 years. Find their shares in the sum of Rs 10000.
- A debtor may discharge a debt by paying (a) Rs 80000 now, or (b) Rs 100000 five years from now. It money is worth 5% compounded semiannually to him, which alternative should he accept?
- At the birth of a daughter, a father wishes to invest sufficient amount to accumulate at 12% compounded semiannually to Rs 1 lac when the daughter is 16 years old. How much should he invest?
- In buying a house, X pays Rs 1 lac cash and agrees to pay Rs 75000 two years later. At 6% compounded semiannually, find the cash value of the home.
- The cost of a refrigerator is Rs 12000. If it depreciates at 10% per annum, find its value 3 years hence.
- The present value of a machine is Rs 160000. If its value depreciates 6% in the first year, 5% in the second and 4% in the third year, what will be its value after 3 years?
- I buy a mobike at Rs 20000 cash payment and three annual instalments of Rs 20000 each. If rate of interest is 15% compounded annually, what is the present worth of the mobike? If the rate of depreciation is 10%, what will be the resale value after 7 years?
- A person buys a land at Rs 30 lacs and a year later constructs a building on it at the cost of Rs 20 lacs. Assuming that land appreciates at 20% annually and building depreciates at 20% for first 2 years and at 10% thereafter, find the total value of property after 5 years from date of purchase of land.
- A loan of Rs 1 lac is to paid back in 5 equal annual instalments. The rate of interest charged is 20% annually. Find the amount of each instalment.
- The population of a town increased from 2 lacs to 8 lacs in last 10 years. If the same trend continues, in how many years will it become 1·6 million?
- Find the nominal rate compounded monthly equivalent to 6% compounded
semiannually. Also find the effective rate of interest.
[Hint. (1 +r/1200) 12 = (1·03) 2]
- The machinery of a certain factory is valued at Rs 18400 at the end of 1990. If it is supposed to depreciate each year at 8% of the value at the beginning of the year, calculate the value of the machine at the end of 1989 and 1991.
1. (i) Rs 784 (ii) Rs 6384
(iii) Rs 894
2. Rs 13240
3. 4% p.a.; Rs 1875 4. Rs 20000 5. Rs 1275
6. Rs 17640 7. Rs 1464·10 8. Rs 1177
9. Rs 3679·50 10. Rs 11466 and Rs 10400
11. Rs 5565, Rs 4435 [Hint. x (1·12)³ = (10000 - x) (1·12)5]
12. He should accept (b) 13. Rs 15496
14. Rs 166636·50 15. Rs 8748 16. Rs 137200
17. Rs 65664·50, Rs 47869·40 18. Rs 85 lacs approx.
19. Rs 33438 20. 5 years 21. 5·926%, 6·09%
22. Rs 20000; Rs 16928