business mathematics

Compound Interest:
Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time ,say yearly or half-yearly or quarterly to settle the previous account.
In such cases ,the amount after the first unit of time becomes the principal for the 2nd unit ,the amount after second unit becomes the principal for the 3rd unit and
so on. After a specified period ,the difference between the amount and the money borrowed is called Compound Interest for that period.

Formula:

Let principal=p, Rate=R% per annum, Time=n years

1.When interest is compounded Annually,
Amount=P[1+(R/100)]n
2.When interest is compounded Half yearly,
Amount=P[1+((R/2)100)]2n
3.When interest is compounded Quarterly,
Amount=P[1+((R/4)100)]4n
4.When interest is compounded Annually,but time in fractions
say 3 2/5 yrs Amount=P[1+(R/100)]3[1+((2R/5)/100)]
5.When rates are different for different years R1%,R2%,R3%
for 1st ,2nd ,3rd yrs respectively
Amount=P[1+(R1/100)][1+(R2/100)][1+(R3/100)]
6.Present Worth of Rs.X due n years hence is given by
Present Worth=X/[1+(R/100)]n

Interest in Maths
Interest:
Interest is the cost of borrowing money. An interest rate is the cost stated as a percent of the amount borrowed per period of time, usually one year.


Simple Interest:
Simple interest is calculated on the original principal only. Accumulated interest from prior periods is not used in calculations for the following periods. Simple interest is normally used for a single period of less than a year, such as 30 or 60 days.

Simple Interest = p * i * n

Example: You borrow $10,000 for 3 years at 5% simple annual interest.
interest = p * i * n = 10,000 * .05 * 3 = 1,500


Compound Interest:
Compound interest is calculated each period on the original principal and all interest accumulated during past periods. Although the interest may be stated as a yearly rate, the compounding periods can be yearly, semiannually, quarterly, or even continuously.You can think of compound interest as a series of back-to-back simple interest contracts. The interest earned in each period is added to the principal of the previous period to become the principal for the next period.


For simple interest problems click here

For compound interest problems click here

Compound Interest problems

Compound Interest problems


Simple Problems:

1.Find CI on Rs.6250 at 16% per annum for 2yrs ,compounded
annually.

Sol: Rate R=16,n=2,Principle=Rs.6250

Method1:
Amount=P[1+(R/100)]n
=6250[1+(16/100)]2
=Rs.8410
C.I=Amount-P
=8410-6250
=Rs.2160
Method2:

Iyear------------------6250+1000
\\Interest for 1st yr on 6250
II yr---------------6250+1000+160
\\Interest for I1yr on 1000
C.I.=1000+1000+160
=Rs.2160

2.Find C.I on Rs.16000 at 20% per annum for 9 months
compounded quaterly

Sol:

MethodI:
R=20%
12months------------------------20%
=> 3 months------------------------5%
For 9 months,there are '3' 3months
--------16000+800
--------16000+800+40
--------16000+800+40+10+2
=>Rs.2522

MethodII: Amount=P[1+(R/100)]n
=16000[1+(5/100)]3
=Rs.18522
C.I=18522-16000
=Rs.2522



Complex Problems

1.The difference between C.I and S.I. on a certain sum
at 10% per annum for 2 yrs is Rs.631.find the sum

Sol:

MethodI:
NOTE:

a) For 2 yrs -------->sum=(1002D/R2)
b) For 3 yrs -------->sum=(1003D/R2(300+R))
Sum=1002*631/102
=Rs.63100
MethodII:

Let the sum be Rs.X,Then
C.I.=X[1+(10/100)]2-X
S.I=(X*10*2)/100
=X/5
C.I-S.I.=21X/100-X/5
=X/100
X/100=631
X=Rs.63100

2.If C.I on a certain sum for 2 yrs at 12% per annum is
Rs.1590. What would be S.I?

sol:
C.I.=Amount-Principle
Let P be X
C.I=X[1+(12/100)]2-X
=>784X/625-X=1590
=>X=Rs.6250
S.I=(6250*12*2)/100=Rs.1500

3.A sum of money amounts to Rs.6690 after 3 yrs and to
Rs.10035b after 6 yrs on C.I .find the sum

sol:
For 3 yrs,
Amount=P[1+(R/100)]3=6690-----------------------(1)
For 6 yrs,
Amount=P[1+(R/100)]6=10035----------------------(2)
(1)/(2)------------[1+(R/100)]3=10035/6690
=3/2
[1+(R/100)]3=3/2-----------------(3)
substitue (3) in (1)
p*(3/2)=6690
=>p=Rs.4460
sum=Rs.4460
4.A sum of money doubles itself at C.I in 15yrs.In how many
yrs will it become 8 times?

sol: Compound Interest for 15yrs p[1+(R/100)]15
p[1+(R/100)]15=2P
=>p[1+(R/100)]n=8P
=>[1+(R/100)]n=8
=>[1+(R/100)]n=23
=>[1+(R/100)]n=[1+(R/100)]15*3
since [1+(R/100)] =2
n=45yrs

5.The amount of Rs.7500 at C.I at 4% per annum for 2yrs is

sol:
Iyear------------------7500+300(300------Interest on 7500)
IIyear ----------------7500+300+12(12------------4% interest
on 300)
Amount=7500+300+300+12
=Rs.8112

6.The difference between C.I and S.I on a sum of money for
2 yrs at 121/2% per annum is Rs.150.the sum is

sol:
Sum=1002D/R2=( 1002*150) /(25/2)2=Rs.9600

7.If the S.I on sum of money at 15% per annum for 3yrs is Rs.1200,
the C.I on the same sum for the same period at same rate is------

sol:
S.I=1200
P*T*R/100=1200
P*3*5/100=1200
=>P=Rs.8000
C.I for Rs.8000 at 5% for 3 yrs is-------------8000+400
-----8000+400+20
-------------8000+400+20+20+1
C.I =400+400+20+400+20+20+1
=Rs.1261