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<body bgcolor="aqua"><center><h2>Compound Intrest</h2></center>
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<strong>Important Facts and Formulae:</strong>
<b>Compound Interest:</b>
Sometimes it so happens that the borrower and the lender
gree 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.
<b>Formulae:</b>
Let principal=p,Rate=R% per annum Time=nyears
1.When interest is compounded Annually,
Amount=P[1+(R/100)]n
2.When interest is compounded Halfyearly,
Amount=P[1+((R/2)100)]2n
3.When interest is compounded Quaterly,
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
<b>Simple Problems</b>
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
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<b>Complex Problems</b>
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
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