Wednesday, January 29, 2014
COMPOUND INTEREST
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 first unit of time becomes the principal for the second unit,the amount after second unit becomes the
principal for the third unit and so on.
After a specified period, the difference between the amount
and the money borrowed
is called the Compound Interest (abbreviated as C.I.) for that period.
IMPORTANT FACTS AND FORMULAE
Let Principal = P, Rate = R% per annum, Time = n years.
I. When interest is compound Annually:
Amount
= P(1+R/100)n
II. When interest is compounded Half-yearly:
Amount = P[1+(R/2)/100]2n
III. When interest is compounded Quarterly:
Amount = P[ 1+(R/4)/100]4n
IV. When interest is compounded
AnnuaI1y but time is in fraction, say 3(2/5) years.
Amount = P(1+R/100)3 x (1+(2R/5)/100)
V.
When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd
and 3rd year
respectively.
Then, Amount = P(1+R1/100)(1+R2/100)(1+R3/100)
VI.
Present worth of Rs.x due n years hence is given by :
Present Worth = x/(1+(R/100))n
SOLVED EXAMPLES
Ex.1. Find compound interest on Rs. 7500 at 4% per annum
for 2 years, compounded annually.
Sol.
Amount = Rs
[7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112.
therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.
Ex. 2. Find compound interest on Rs.
8000 at 15% per annum for 2 years 4 months,
compounded annually.
Sol. Time = 2 years 4
months = 2(4/12) years = 2(1/3) years.
Amount =
Rs'. [8000 X (1+(15/100))2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs.
(11109 - 8000) = Rs. 3109.
Ex. 3. Find the compound
interest on Rs. 10,000 in 2 years at 4% per annum, the
interest being
compounded half-yearly. (S.S.C.
2000)
Sol.
Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.
Amount =
Rs
[10000 * (1+(2/100))4] = Rs(10000 * (51/50) * (51/50) * (51/50) *
(51/50))
= Rs. 10824.32.
:. C.I. = Rs. (10824.32 -
10000) = Rs. 824.32.
Ex.
4. Find the compound interest on Rs. 16,000 at 20% per annum
for 9 months,
compounded quarterly.
Sol. Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3]
= Rs. 18522.
CJ. = Rs. (18522 - 16000) = Rs. 2522.
Ex. 5. If the simple interest on a sum of money at
5% per annum for 3 years is Rs. 1200, find the compound
interest on the same sum for the same period at the same rate.
Sol.
Clearly, Rate = 5% p.a., Time = 3 years,
S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 - = Rs.
9261.
.. C.I. = Rs. (9261 - 8000) = Rs. 1261.
Ex. 6. In what time will
Rs. 1000 become Rs. 1331 at 10% per annum compounded annually? (S.S.C. 2004)
Sol.
Principal
= Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let
the time be n years. Then,
[ 1000 (1+
(10/100))n ] = 1331 or
(11/10)n = (1331/1000) =
(11/10)3
n = 3 years.
Ex. 7. If
Rs. 600 amounts to Rs. 683.20 in two years compounded annually,
find the
rate of interest
per annum.
Sol. Principal = Rs. 500;
Amount = Rs. 583.20; Time = 2 years.
Let the rate be R%
per annum.. 'Then,
[ 500 (1+(R/100)2 ] = 583.20 or [ 1+ (R/100)]2 = 5832/5000
= 11664/10000
[ 1+ (R/100)]2 = (108/100)2 or 1 +
(R/100) = 108/100 or R = 8
So, rate = 8% p.a.
Ex. 8. If the
compound interest on a certain sum at 16 (2/3)% to 3 years is Rs.1270,
find the simple interest on the same sum at the same rate
and f or the same period.
Sol.
Let the sum be Rs. x. Then,
C.I. = [
x * (1 + (( 50/(3*100))3 - x ] = ((343x / 216)
- x) = 127x / 216
127x /216 = 1270 or x = (1270 * 216) / 127 =
2160.
Thus, the sum is Rs. 2160
S.I. = Rs ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.
Ex. 9. The difference between the compound interest and simple
interest on a
certain
sum at 10% per annum for 2 years is Rs. 631. Find
the sum.
Sol. Let the sum be Rs. x.
Then,
C.I. = x ( 1
+ ( 10 /100 ))2 -
x =
21x / 100 ,
S.I. = ((
x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) =
x / 100
( x / 100 ) =
632 x
= 63100.
Hence,
the sum is Rs.63,100.
Ex. 10. The difference between the compound
interest and the simple interest accrued on an amount of
Rs. 18,000 in 2 years was Rs. 405. What was the rate of
interest p.c.p.a. ? (Bank
P.O. 2003)
Sol. Let the rate
be R% p.a. then,
[ 18000 ( 1 + ( R
/ 100 )2 ) - 18000 ] - ((18000 * R * 2) / 100 ) = 405
18000 [ ( 100 + (R / 100 )2 / 10000)
- 1 - (2R / 100 ) ]
= 405
18000[(
(100 + R ) 2 - 10000 - 200R) / 10000 ] = 405
9R2 / 5 =
405 R2 =((405
* 5 ) / 9) = 225
R = 15.
Rate = 15%.
Ex. 11. Divide
Rs. 1301 between A and B, so that the amount of A after 7 years
is equal to the amount of B after 9 years, the interest being
compounded at 4% per annum.
Sol. Let the two parts be Rs. x and Rs. (1301 - x).
x(1+4/100)7 =(1301-x)(1+4/100)9
x/(1301-x)=(1+4/100)2=(26/25*26/25)
625x=676(1301-x)
1301x=676*1301
x=676.
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625.
Ex.12. a certain sum amounts to rs.7350 in 2 years and to
rs.8575 in 3 years.find the sum and rate percent.
S.I on rs.7350 for 1 year=rs.(8575-7350)=rs.1225.
Rate=(100*1225/7350*1)%=16 2/3%
Let the sum be rs.x.then,
X(1+50/3*100)2=7350
X*7/6*7/6=7350
X=(7350*36/49)=5400.
Sum=rs.5400.
Ex.13.a sum of money amounts to rs.6690 after 3 years and
to rs.10,035 after 6 years on compound interest.find the sum.
Sol. Let the sum be rs.P.then
P(1+R/100)3=6690…(i) and P(1+R/100)6=10035…(ii)
On dividing,we get (1+R/100)3=10025/6690=3/2.
Substituting this value in (i),we get:
P*3/2=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.
Ex.14. a sum of money doubles itself at compound interest
in 15 years.in how many years will it beco,e eight times?
P(1+R/100)15=2P
(1+R/100)15=2P/P=2
LET P(1+R/100)n=8P
(1+R/100)n=8=23={(1+R/100)15}3[USING
(I)]
(1+R/100)N=(1+R/100)45
n=45.
Thus,the required time=45 years.
Ex.15.What annual payment will
discharge a debt of Rs.7620 due in 3years at
16 2/3% per annum interest?
Sol. Let each installment
beRs.x.
Then,(P.W. of Rs.x due 1 year hence)+(P>W
of Rs.x due 2 years hence)+(P.W of Rs. X due 3
years hence)=7620.
\
x/(1+(50/3*100))+ x/(1+(50/3*100))2 + x/(1+(50/3*100))3=7620
Û(6x/7)+(936x/49)+(216x/343)=7620.
Û294x+252x+216x=7620*343.
Û
x=(7620*343/762)=3430.
\Amount
of each installment=Rs.3430.
