WEBVTT
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Welcome to the Lecture 6 on Time value of
money-Concepts and Calculation. The present
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lecture is devoted to Continuous Compounding.
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In last few lecture we have discussed simple
and compound interest and in these lectures
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consider the form of interest in which periodic
payments are charged at discrete intervals
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of time with discrete amount of interest accumulating
as the end of each interest period. In practice
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the basic time interval for interest accumulation
is usually taken as 1 year. However shorter
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periods like 6 months, 1 month, 1 day, 1 hour
or 1 second can also be used. In the extreme
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case when the time interval approaches to
0. So, that the interest is compounded continuous
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is called Continuous Compounding.
Continuous compounding is useful because,
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it gives straight forward way to compare interest
rates between different types of compounding.
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Second, it illustrates that compounding more
often does not really give you as much of
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a boost as you might imagine. And third, sometimes
it really is useful to invest money for very
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short amount of time.
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Now, development of equation for continuous
compounding the basic equations which we know
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that S is equal to P 1 plus r divided m to
the power m N which is use for discreet compounding
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represents the basic equation for which continuous
interest relationships can be derived. The
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symbol r represents the nominal interest rate
with m interest period per year. If the interest
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is compounded continuously m approaches infinity
and the above equation can be rewritten has
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S equal to P limit m tends to infinity in
bracket 1 plus r by m whole to the power m
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N is equal to P limit m tends to 0 in the
bracket 1 plus r by m whole to the power m
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by r into r into N.
The fundamental definition for the base of
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the natural system of logarithms that is e
is equal to 2.71828 is limit m tends to infinity
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1 plus r by m to the power m by r is equal
to e. Thus with continuous interest compounding
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at nominal interest rate of r the amount S
N for initial principal P 0 compounded for
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m years is equal to S equal to P e to the
power rN. The term e to the power rN is known
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as the continuous single payment compound
amount factor.
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Now, let us see at what type of problems can
be created for this continuous compound interest.
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The equation is S is equal to P e to the power
rN. The above equation which contains four
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variables that is S P r and N can be solve
to find out the value of single unknown variable
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only when other three variables are known.
Thus 4 type of problem can be generated out
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of this equation. The problem matrix of the
above equation is shown below.
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Now if P r and N are given we can find out
the value of S, where S is equal to P into
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e to the power rN. In this case we have to
find out the value of S that is future value
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when present value P interest rate r and number
of compounding years N r known. This type
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of problem will be called problem type A.
The second type of problem will be given S
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r and N, we have to calculate the value of
p. So, the formula we should be used for this
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purpose is P is equal to S divided by e to
the power rN. Find the present value P provided
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S r and N are known. Such type of problem
will be called problem type B.
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The third type problem S N and P will be given
and r as to be found out. For this problem
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r is equal to 1 by N ln S by P. Find the interest
rate r where S P and N are known. Such type
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of problem is called problem type C.
The forth and the last type of problem is
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S P and r are given N has to be found out.
An N is equal to ln S by P whole divided by
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r. In this case number of periods N has to
be found out when S P and r are known. This
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type of problem will be called problem type
D.
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Let us start with different problems and it
is solution.
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Now problem number A or the problem type A
for example 1; if the nominal interest rate
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is 8, percent how much is Rupees 10000 towards
in 15 years in a continuous compounded account?
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There are four answers, the solution is P
is given as 10000, nominal interest rate is
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given as 8 percent, and N is given has 15
years. So, S as to be found out S is P into
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e to the power rN.
So, S is equal to 10000 into e to the power
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0.08 into 15. This comes out to be 10000 into
3.3201169 comes out to be Rupees 33201.17.
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That means if we invest 10000 we can nominal
interest rate of 8 percent for 15 years it
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will provide you Rupees 33201.17 if the compounding
is continuous.
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Again problem type A; example 2. Consider
nominal annual interest rate of 15 percent
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per year; determine the total amount which
will be accumulated after 5 years for an initial
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investment of 1000 if continuous compounding
is employed. Solution; we know what is the
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value of P that is 1000, we know the value
of interest rate which is nominal interest
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rate it is 15 percent, we know the value of
N which is 5 years. So, we have to find out
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S or FV and FV is equal to P into e to the
power rN is equal to 1000 into e to the power
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0.15 the value of r into 5 which is the value
of N.
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And this gives us Rupees 2117. That is if
1000 Rupees invested today for 5 year with
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a nominal annual interest rate of 15 percent
and the compounding is continuous it will
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convert into Rupees 2117.
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Problem type B example 3, considering nominal
annual interest rate of 16 percent per, year
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and the total amount which will be accumulated
after 6 year as 20896.6 for this find initial
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investment if continuous compounding is employed.
Solution S is given or FV is given has 20896.6,
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r is given has 16 percent, N is 6. So, S equal
to P into e to the power rN or P is equal
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to S divided e to the power rN.
So, this is equal to 20896.6 into 0.3828929
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and this value 0.3828929 is equal to 1 by
e to the power rN. So, the multiplication
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gives Rupees 8001.16. That means, if I invest
8001.16 for 6 year with nominal annual interest
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rate of 16 percent it will give me 20896.6
Rupees.
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Now problem C, this is given in example 4.
Find the nominal annual interest rate per
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year if the total amount accumulated after
10 years as 54446.83 for this we find the
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initial investment of Rupees 9000 if continuous
compounding is employed. So, if we analyze
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this problem we find that S is given 54446.83,
P is given as Rupees 9000, N is given as 10
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years, but what is the value of r if value
of r is not there. So, we have to find it
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out so again we use the same equation S is
equal to P into e to the power rN or P is
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equal to S divided by e to the power rN or
we can write down S by P is equal to e to
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the power rN.
And if we take log both the side becomes ln
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S by P is equal to rN or r is equal to 1 by
N into ln S by P. If we substitute the value
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of the N S and P N into this, so this is equal
to 0.1 into ln 6.049647 and this is equal
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to 0.174444 and that is approximately 18 percent.
That means, 9000 will be converted into 54446.83
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Rupees in 10 years if nominal interest rate
r is equal to 18 percent.
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Now problem type D for example 5; considering
the nominal annual interest rate per year
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to be 15 percent and if the total amount accumulated
after given number of S is Rupees 19676.8.
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Find initial investment of 8000; find the
number of years if continuous compounding
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is employed. Solution; given S is equal to
Rupees 19676.8, P is 8000, r is equal to 15
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percent and what is the value of N.
So, again we use the same equation S is equal
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to P into e to the power rN or P is equal
to S by e to the power rN or we can write
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down e to the power rN is equal to S by P
if we take log both the sides ln S by P is
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equal to rN or N is equal to 1 by r into ln
S by P. This is equal to 6.6666 into ln 2.4596
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and we will multiply these two it comes out
to be 5.99999 which is equivalent to 6 years.
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That means, the Rupees 8000 invested with
nominal interest rate of 15 percent we will
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convert into Rupees 19676.8 in 6 years if
continuous compounding is employed.
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Now, again problem D this is example 6. Find
out time period in years required to double
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an investment within continuous compounding
if nominal interest rate is 10 percent per
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year.
Solution; here S by P is equal to 2 and we
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know that for continuous compounding S is
equal to P e to the power rN or S by P e to
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the power rN or e to the power rN is equal
to 2, or e to the power 0.1 N when we put
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the value of r as 0.1 because it is 10 percent
then it becomes e to the power 0.1 N is equal
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to 2 and if we take log of both the sides
then N is equal to ln 2 divided by 0.1 or
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N is equal to 0.693147 which is the natural
logarithmic value of two divided by 0.1 is
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equal to 6.93 years.
So, a value will double in 6.93 years if nominal
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interest rate is 10 percent and continuous
compounding is used.
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Let us take a mix problem for this we considering
example 7. For the case of nominal annual
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interest rate of 20 percent per year determine
the total amount to which Rupees 1 of initial
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principal would accumulate after 1 year with
continuous compounding and the effective annual
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interest rate. What are the things which are
given P is equal to Rupees 1, r is equal to
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20 percent, N is equal to 1, and S is what,
that means we have to find out S when the
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value of P r and N are known.
So, again we use the formula for continuous
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compounding S is equal to P into e to the
power rN or S equal to 1 e to the power 0.2
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into one because r is 0.2 and N is equal to
1. So, S is equal to Rupees 1.2214. Now I
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effective is equal to e to the power r minus
1 and this comes out to be 0.2214. That means,
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if I invest Rupees 1 I can get Rupees 1.2214
in 1 year if nominal interest rate is 20 percent
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and continuous compounding is used. And that
is why the i effective is 22.14 percent and
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in fraction it is 0.2214.
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Now problem type D, example 8. How long will
it take Rupees 5000 to triple if it is invested
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at 6 percent compounded continuously? So,
we have 4 options and let us take the solution.
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What have given P is equal to Rupees 5000,
S is equal to Rupees 15000 because it is going
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to triple 5000 into 3; 15000, r is 6 percent
and what is the value of N.
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So, we use again the formula for continuous
compounding S is equal to P e to the power
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rN or e to the power rN equal to S by P is
equal to 3, taking log both sides, rN is equal
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to log 3 is equal to 1.09861229. So, 0.06
which is the value of r into N is equal to
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1.09861229, so N is equal to 18.31. So, your
answer d is correct.
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So, it tells that a 5000 Rupees input can
be triple to 15000 Rupees if nominal interest
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rate is 6 percent in 18.31 years if continuously
compounded.
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Again a mix problem, example 9; the difference
between simple and compound interest compounded
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continuously on a certain some of money for
2 years at 8 percent per annum is Rupees 135.1.
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Find the amount of the money invested. Let
the sum of money is x. So, continuous compound
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interest for 2 years will be x into e to the
power rN minus x which comes out to be 0.173510871x.
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Now simple interest for 2 years is x into
0.08 into to 2 which is 0.16x.
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So, continuous interest compounding interest
minus simple interest is equal to 0.1735101
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x minus 0.16x and this is equal to 0.0135101x
is equal to 135.1, so x is equal to 9999.93.
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Now again problem type mixed, example 10.
The difference between compound interest and
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simple interest on an amount of Rupees 20000
for 2 years is Rupees 270.22. What is the
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rate of interest per annum? Given that P is
equal to Rupees 20000, N is equal to 2 and
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what is the value of r. So, we have to find
out r, when P and N are given. Let the rate
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of interest per annum is r. So, continuous
compound interest C.C.I for 2 years is 2000
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into e to the power rN minus 2000 which comes
are to be 20000 in brackets e to the power
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rN minus 1.
Simple interest S.I for 2 years is 20000 into
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i into 2, so it is 20000 2i. So, C.C.I minus
S.I that is continuous compound interest manners
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simple interest is equal 20000 in bracket
e to the power rN minus 1 into 20000 into
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2 into i is called to 270.22. By trial and
error if we put r is equal to 0.06 the left
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hand side becomes 149.937, if we take r equal
to 0.07 the left hand side which is equal
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to 20000 into e to the power rN minus 1 minus
2 r becomes205.476 and you we take r is equal
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to 0.08 it is 270.2174. So, right hand side
is 270.22 and at the value of r becomes 0.08
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it is 270.2174 and hence the correct value
of r is 0.08 that is 8 percent.
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Thank you.