Some Questions On Compund Interest

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:
Method I:
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