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
- Amount = Principal + Interest
- Simple Interest = (Principal x Rate x Time)/100
- A = P
Here A = amount, P = principal, r = rate percent yearly (or every fixed period) and n is the number of years (or terms of the fixed period). - C.I. = P
, where C.I. = compound interest - If the interest rates for the successive fixed periods are r1%, r2%, r3% ..., then A (amount) is given by
A=P
... - S.I. (simple interest) and C.I. (compound interest) are equal for the first year (or the first term of the fixed period) on the same sum and at the same rate.
- C.I. of 2nd year (or the second term of the fixed period) is more than the C.I. of 1st year or the first term of the fixed period), and C.I. of 2nd Year -C.I. of 1st year = S.I. on the interest of the first year.
- Equal installments (with compound interest):
Loan amount = P
where P = each equal installment
R = rate of interest per annum (or per specified period)
T = time, say 3 years (or 3 specified terms).
Note. If T = n years (or specified terms), then there will be n brackets.
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).
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).
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
A =
A =
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 interestTwo 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
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
P.V. =
In particular, present value of sum of a Rs P due one year hence (i.e. n = 1) at r% (compound) interest is
P.V.=
P.V. =
In particular, present value of sum of a Rs P due one year hence (i.e. n = 1) at r% (compound) interest is
P.V.=
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.
, whereP = 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 populationIf 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 
.
, 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 .Depreciation
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.
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.
Example
Find the compound amount of Rs 1500 for 6 years 7 months, at 5·2% compounded semi annually.
Solution
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
=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.
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
=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.
Example
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.
Solution
- 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
Example
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.
Solution
Present population, P = 1 million, rate of increase = 5% per annum
Hence population after 5 years =
= 1000000
= 1000000 (1·05)5
= 1276280
Population three years ago =
= 863838.
Hence population after 5 years =
= 1000000
= 1000000 (1·05)5= 1276280
Population three years ago =
= 863838.Example
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?
Solution
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%
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%
