Compound Interest

Concept:

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. Formulae: 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

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

Medium 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

Complex Problems

1.A certain sum amounts to Rs.7350 in 2 yrs and to Rs.8575 in 3 yrs.Find the sum and rate%? sol: S.I. on Rs.7350 for 1yr =Rs.(8575-7350) =Rs.1225 S.I. on Rs.7350 for 2yrs=Rs.2*1225 =Rs.2450 PTR/100=2450 =>P*2*R/100=2450 Since S.I. on Rs.7350 for 1yr =Rs.(8575-7350) =Rs.1225 Rate R=100*1225/(7350*10 =16 2/3% Since it is C.I ,Let sum be Rs.X Then X[1+(R/100)]2=7350 =>X[1+(50/100)]2=7350 =>X=7350*(36/49) Sum=Rs.5400 2.If the difference between C.I compounded halfyearly and S.I on a sum at 10% per annum for one yr is Rs.25 the sum is sol: p[1+((R/2)/100)]2n-PTR/100=25 P[1+((10/2)/100)]2n-P*1*10/100=25 =>P=Rs.400 3.A man borrowed Rs.800 at 10 % per annum S.I and immediately lent the whole sum at 10% per annum C.I What does he gain at the end of 2yrs? sol:

C.I.=Rs.[800[1+(10/100)2]-800]=Rs.168 S.I=Rs.[800*10*2/100]=Rs.160 Gain=C.I-S.I=Rs(168-160) =Rs.8 4.On what sum of money will be S.I for 3 yrs at 8% per annum be half of C.I on Rs.400 for 2 yrs at 10% per annum? sol: C.I on Rs.400 for 2yrs at 10%=Rs.[400*[1+(10/100)]2-400] =Rs.84 Required S.I =1/2*84=42/- New S.I=Rs.42,Time=3yrs Rate=8% Sum=Rs.[100*42/(3*8)] =Rs.175 5.A sum of money placed at C.I doubles itself in 5yrs .It will amount to 8 times itself in------------- sol: p[1+(R/100)]5=2P =>[1+(R/100)]5=2 To become 8 times =>8P p[1+(R/100)]5=2^3P =[1+(R/100)]^(5*3) =[1+(R/100)]^15 n=15years