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
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