WEBVTT
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Welcome to the lecture series on Time value
of money-Concepts and Calculation. This lecture
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is devoted to Compounding and that too compounding
annually. Time value of money says that, the
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worth of a unit of money is going to be changed
with time.
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To quantify this change two tools are used;
the first tool is compounding methods use
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to know the future value of present money
and discounting method use to compute present
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value of future money. Both the tools are
very important to know the worth of money
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at different point of time. We start with
the present value and using compounding reach
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to the future value, and we start with the
future value and reach to present value using
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discounting method has shown in the figure.
Let us see the comparison between compounding
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and discounting methods based on different
basis.
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Now if we see the basis is meaning then compounding
is the method used to determine the future
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value of present investment is known as Compounding.
And what is discounting the method used to
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determine the present value of future cash
flows is known as Discounting. Now if we say
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the basis is concept then for compounding
if some money is invested today what will
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be the amount one gets at a future date. And
the same concept for discounting is what should
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be the amount we need to invest today to get
a specific amount in future.
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Now, the basis is use of then compounding
and the compounding who is compound interest
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rate and in the discounting we use discount
rate. What it computes or notes? The compounding
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gives us present value to future value, and
discounting gives future value to present
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value. That means, for compounding known is
present value and we calculate future value
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based on it. And in discounting known is future
value and we calculated present value based
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on the discount rate.
What is the factor which is used for compounding?
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For compounding future value factor or compounding
factor is used, and for discounting present
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value factor or discounting factor is used.
And if you see the formula for compounding
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FV is equal to PV in brackets 1 plus r to
the power N. And for discounting PV is equal
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to FV divided by in brackets 1 plus r to the
power N.
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Now, let us see the definition of compounding.
Compounding refers to the process of earning
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interest on both the principle amount, as
well as accrued interest by reinvesting the
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entire amount in order to generate more interest.
That means, compounding of the principle as
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well as the interest is done to find out the
final sum.
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Compounding is the method used in finding
out the future value of the present investment.
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The future value can be computed by applying
the compound interest formula which we have
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given in the last slide. As per the Benjamin
Franklin, ‘money makes money, and the money
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that money makes more money’.
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Now the compound interest can be divided into
three parts; the first is annually compounding,
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the second is discrete annually compounding,
and the third is continuously compounding.
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In these three methods what is changing is
the time period.
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If the time period is 1 year then it is annually
compounding. If the time period is less than
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1 year say every month then it is discrete
annually compounding. And if the time period
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is very small then we call it a continuously
compounding.
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In this lecture we will deal with annually
compounding only. Now let us see the derivation
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of formula for annually compounding. The compound
amount due after any discrete number of interest
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periods in this present case a year can be
determined as follows. Let us take the period
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1 that means, 1 year. The principle at the
start of the period is a P then interest earned
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during that period is P into i, if i is the
interest rate per year then the compounded
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amount S that is which includes the principle
as well as interest earned is P plus Pi which
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is equal to P into brackets 1 plus i.
Now for the second year the principle is P
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into 1 plus i, because compound interest the
interest is also charged on principle as well
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as interest earned. So, at the start of second
year the principle is P in brackets 1 plus
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i. Now the interest earned in the second year
will be P in brackets 1 plus i into i. This
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is the interest which will be earned in second
year. So, the compound amount will be now
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P in brackets 1 plus i plus P in brackets
1 plus i into i. And if I take P 1 plus i
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common then it becomes P 1 plus i into 1 plus
i which becomes P into 1 plus i whole square.
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So, end of the second year the compound amount
will be P 1 plus i in brackets whole square.
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At the start of the third year the P 1 plus
i whole square will now work as the principle,
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and the interest at the end of the third year
will be P 1 plus i square into i. So, the
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compounded amount will be P 1 plus i whole
square plus P into 1 plus i whole square into
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i, and if I take common then it becomes P
1 plus i whole cube. Similarly, if I can write
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extend this logic to the nth period then the
start of the nth period the principle will
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be P 1 plus i to the power N minus 1, because
in the third period start of the third period
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the principle is P 1 plus i whole square which
I can write down as P 1 plus i to the power
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3 minus 1.
So, on this same logic for nth period I can
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write down the principle will be P 1 plus
i to the power N minus 1. Then interest earned
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on this principle will be P 1 plus i to the
power N minus 1 into i. And the compounded
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amount will be P 1 plus i to the power N minus
1 plus P 1 plus i to the power N minus 1 into
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i. And if I solved this it becomes equal to
P 1 plus i to the power n.
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Now, this gives us the formula for annually
compounding.
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Therefore, if the total amount of principles
plus compounded interest after N interest
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period is designated as S, then S is equal
to P 1 plus i to the power N. Now for all
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compounding problems which is compounding
annually we will use this formula which is
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S is equal to P 1 plus i to the power N. The
term 1 plus i to the power N is commonly referred
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to as the compound value interest factor CVIF.
So, I can write down CVIF is equal to 1 plus
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i to the power N.
Now, let us see how many type of problems
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can be created on annual compound interest.
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So, I have written my equation S is equal
to P 1 plus i to the power N. Now this equation
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can solve only one unknown, because I have
only one equation and using one equation I
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can solve only one unknown. The above question
which contains four variable; S, P, i, N,
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can be solved to find out the value of a single
unknown variable only when other three variables
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are known.
That means, in these equations if I feed three
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known I can find out the fourth unknown variable.
Thus four type of can be generated out of
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these equations. So, what I will see now that
four type of problems can be created using
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this equations. The problem matrix for these
equations is shown in the next slide. This
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is a problem matrix.
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So, the first type of problems which I call
problem type A in which P will be given, i
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will be given, and N will be given, and what
has to be found out the S has to be found
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out. So, the equation which will be used for
this purpose is S is equal to P 1 plus i to
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the power N. So, the problem is find future
value S when present value P interest rate
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i and number of period N are known. So, when
I find such types of problem I call it a problem
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A.
The second type of problem will be in which
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S will be given, i will be given, and N will
be given, and we have to find out P. So, the
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equation used in this problem will be P is
equal to S divided by 1 plus i to the power
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N. And the problem is to find the present
value P when future value S interest rate
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i and number of periods known. Such type of
problems will be called problem type B.
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Now, the third type of problem which I will
call it problem type C. In this S will be
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supplied, N will be supplied, and P will be
supplied, and one has to calculate the value
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of i. And the equation will be i is equal
to in brackets S divided by P to the power
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1 by N minus 1. In this discussion find the
interest rate i when future value S, present
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value P and the no of periods N is known.
And will call this type of problems problem
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type C.
Now in the last problem type which will called
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D; S is given, P is given, i is given, and
we have to find out N. So, equation will be
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N is equal to ln S by P divided by ln 1 plus
i in brackets. So, here one has to find the
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number of periods when future value S, present
value P, and interest rate i are known.
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Now let us take problems based on this matrix.
So, first we take problem type A.
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Example 1; a person deposits Rupees 10,000
in his bank accounts to earn compound interest
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for a period of 3 years when annual interest
rate is 8 percent. What will be the accrued
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interest at the end of third year part a.
And part b, what is the future FV or S of
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the deposit at the end of the third year.
So, if you analyze the problem matrix then
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we find that it demands accrued interest and
final value enhance false in the category
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of problem type A.
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Now let us see the solution. In the A part
P is the principle amount because it is invested
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at time equal to 0 in the time line, i equal
to interest rate per year, N is number of
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years. So in this problem P is equal to Rupees
10,000, i is equal to 8 percent, and N is
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equal to 3. And what is demand at here, that
interest earned after period N. Now this problem
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can be solved in many a ways, but I am solving
it using the first principle.
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So, my 1st year interest will be P into i
divided by 100. It should be noted always
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that i is always given in percentage like
in this case 8 percent, but when it will be
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used in the formula it will be used as a ratio.
So, i will be converted into 8 divided by
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a 100, so 0.08. So the 1st years interest
will be P into i divided by 100, and the second
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years interest will be P 1 plus i divided
by 100 in brackets to the power 1 into i divided
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by 100. And for the third year’s interest
will be P 1 plus i divided 100 to the power
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2 into i by 100.
This has been taken from the derivation of
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the formula for annually compound interest.
So, when we put values this is 10,000 into
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0.08 plus 10,000 into 1.08 to the power 1
into 0.08 plus 10,000 into 1.08 to the power
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2 into 0.08; because 1 plus i divided by 100
gives 1.08. So, when we solved it we find
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that interest is 2597.12. Now if we calculate
the future value, future value is principle
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amount P plus accrued interest in 3 years.
So, this is 10,000 plus 2597.12 is equal to
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Rupees 12597.12.
Now the same can be directly calculated by
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the using the formula which we have learned,
S is equal to P into 1 plus i to the power
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N. So, if we use that formula then it is 10,000
into 1.08 to the power 3 is equal to 12597.12.
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And the part a from this data can be back
calculated that means, the interest earned
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in 3 years will be equal to Rupees 12597.12
minus 10,000 which will come out to be 2597.12.
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Let us take another problem, and the problem
is also type A. It is example 2; the question
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is will you prefer to get 5000 today or the
same amount after 5 years. The annual interest
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rate is 8 percent. This is the time value
problem, clearly tells that if somebody offers
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you 5000 Rupees today or offers you the same
5000 Rupees after 5 years and gives you alternative
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that you can select one of this, which one
you should select. Let us say analyze this
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problem. After 5 years the principle amount
of Rupees 5000 will grow to if compound interest
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compounding annually is considered then future
value of this 5000 will be 5000 in the brackets
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1 plus 8 divided by 100 to the power 5.
This 5000 in 5 years will convert into 7346.64
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Rupees. As the future value of this 5000 after
5 years is far more than the present value
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of 5000 we should prefer to get 5000 today
than the same after 5 years. After getting
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the money today we should deposit this in
the bank and get Rupees 7346.64 after 5 years
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which will be around 2346.64 Rupees more than
the 5000 Rupees. Note, future value FV is
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the value at some future time of the principle
amount evaluated at a given interest rate
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Another problem we take, the problem type
is A.
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For principle amount of 1000 deposited in
a bank what will be final value of money at
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the end of 1st year, 10th year and 15th year,
if annual interest rate is 10 percent and
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simple interest is charged. Annual interest
rate 10 percent and compound interest compounding
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annually is considered. This problem gives
or permits us to find out if I invest my money
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based on simple interest or invest the same
money based on compound interest what will
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be the difference.
Let us see the solution a part; that means
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simple interest. Simple interest at the end
of the 1st year will be 1000 into 10 divided
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by 100 into 1 which is 1 year will be 100
Rupees. So, the final value will be 1000 plus
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100 is equal to Rupees 1100. That means if
I use the simple interest of 10 percent then
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my final value will be 1100. That means, from
the principle of 1000 two it will grow to
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1100 after 1 year. At the same way the simple
interest at the end of 10th year is equal
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to 1000 10 by 100 into 10 which comes out
to be 1000. So, the final value is 1000 plus
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the interest 1000 comes out to be 2000. That
means, if I invest 1000 today at the rate
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of 10 percent and simple interest is used
then at the end of 10th year I will get a
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value of Rupees 2000 only.
In the similar way if I calculate what I will
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get at the end of 15th year then the interest
will be 1000 into 10 by 100 into 50 will be
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5000. And my final value will be 1000 my principle
value and my interest 5000 which will be 6000.
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So, what we conclude here that after 1 year
we get 1100, after 10 years we get 2000 and
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after 15th year we get 6000. So, the money
will grow like this, but if we see the same
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how it is growing when compound interest is
charged will find a large change.
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Now, in the part b we are using the compound
interest. So, final value at the 1st year
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when compound interest is charged is 1000
in the brackets 1 plus 10 divided by 100 to
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the power 1 is 1100. Final value at the 10th
year is 1000 into 1 plus 10 by 100 to the
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power 10 comes out to be 2593.72. So, what
we see here the final value is the same for
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the compound interest as well as the simple
interest when the it is 1st year. In the second
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year the difference is less only a 593.72
Rupees.
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But if we see the 15th year then the final
value is 1000 into 1 plus 10 by 100 to the
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power 50 which comes out to be 1,17,390.85,
whereas for the simple interest it was far
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less. So, what we conclude that we get more
benefit if the money is invested for a longer
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in duration of time using compound interest.
Now, let us take another problem which is
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problem type B.
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Presume that you are going to receive an amount
of Rupees 50,000 6 years from now. How much
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loan should you take from a bank today such
that the money received after 5 years from
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now will be exactly enough to pay off the
loan plus interest at that time. Assume the
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bank charges and annual interest rate of 10
percent.
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Now, if you analyze this problem that our
future value is 50,000, number of years is
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6 i is 10 percent, and the demand what should
be my principle amount. P is the present value
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of the money which will be borrowed from the
bank as loan. So, P is equal to future value
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divided 1 plus i by 100 to the power 6, so
this is 50,000 divided by 1 plus 0.1 to the
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power 6 is comes out to the 28223.69. So,
if we takes loan of 28223 today then after
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6 years it will grow to 50,000 Rupees for
an interest rate of 10 percent.
29:43.410 --> 29:52.980
Now, another problem type B. To get Rupees
20,000 15 years hence how much should be invested
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now at 10 percent annually interest rate.
It tells that how much money I should invest
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today at 10 percent annual interest rate so
that after 15 years I get 20,000 Rupees. The
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solution of this problem is the final value
is equal to principle which is present value
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into 1 plus i divided by 100 to the power
N.
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For the present problem final value is given
as 20,000, N is 15 years, i is 10 percent
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and it ask what is the present value or the
principle amount. So, we will use the same
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equation present value is equal to final value
divided by 1 plus i divided 100 to the power
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N. I would say here that always i is expressed
in terms of percentage, but when i is used
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in the equation it is used as a fraction and
that is why 10 is divided by 100. So, it comes
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out to be 20,000 divided by 1 plus 0.1 to
the power 15 which gives us Rupees 4787.84.
31:15.850 --> 31:30.090
That means, if I invest 4787.84 today at an
annual interest rate of 10 percent then after
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15 years I will get 20,000 Rupees.
31:33.560 --> 31:47.960
Again problem type B, what principle amount
will grow to 5000 after 10 years if the annual
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interest rate is 10 percent and compound interest
is charged. This is also a problem where principle
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is amount is amount unknown, final amount
is known, number of years for which the money
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is invested is known, and interest rate known.
So, it is a problem type B and for problem
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type B our formula is final value is equal
to present value into 1 plus 10 by 100 to
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the power 10 or P is equal to final value
divided by 1 plus 10 divided by 100 to the
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power 10.
So, from this equation we can very well calculate
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the value of P which come at to be 1927.718.
That means, if today I spend 1927.718 Rupees
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at 10 percent compound interest annually then
it will grow to 5000 Rupees after 10 years.
33:05.330 --> 33:12.120
Now, take a different problem which is problem
type C. This is the example 7. I assume the
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total cost of a college education will be
60,000 Rupees, when your child enters college
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in 12 years. You have Rupees 10,000 to invest
today. What annual rate of interest must you
33:36.650 --> 33:46.750
earn on your investment to cover the cost
of your child’s education? You have 10,000
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in your hand you want 60,000 Rupees after
12 years what should be the interest rate
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which will grow this 10,000 Rupees to 60,000
Rupees after 12 years. So, our equation is
34:03.280 --> 34:14.740
final value is 60,000, present value is 10,000,
interest rate is unknown and time is 12 years.
34:14.740 --> 34:25.060
So, I can write down that final value is 60,000
is equal to present value 10,000 into 1 plus
34:25.060 --> 34:38.190
0.01 into i to the power 12. Here I have used
i divided by 100 and that 1 by 100 is giving
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this factor 0.01. That means, I am considering
here i in percentage so whatever result will
34:46.799 --> 34:55.869
come out by solving this equations i it will
be in percentage and not in fraction. Or I
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can write down 1 plus 0.1 into i is equal
to 60,000 divided by 10,000 to the power 1
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by 12 this is equal to 6 to the power 1 by
12 which comes out to be 1.1610366. So, 1
35:16.660 --> 35:28.780
plus 0.01 i is equal to 1.1610366.
So, from here I can find out i which will
35:28.780 --> 35:42.979
be 1.1610366 minus 1 divided by 0.01 which
comes to be 16.104 percent. That means, if
35:42.979 --> 35:55.890
the interest rate is 16.104 percent then only
the Rupees 1000 which we you will invest today
35:55.890 --> 36:06.599
will grow to 60,000 Rupees in 12 years. And
if you get interest rate below this then it
36:06.599 --> 36:15.640
will not grow to 60,000 Rupees. So, the person
we have to search a bank which will give him
36:15.640 --> 36:19.520
16.104 percent interest rate.
36:19.520 --> 36:30.349
Take another problem of problem type C. This
is example; at what annual interest rate at
36:30.349 --> 36:40.999
the end of 5 years an amount of Rupees 1000
will be doubled, if the money is compounded
36:40.999 --> 36:49.869
annually. Again we know the time its 5 years,
P this is principle is 1000 this will double
36:49.869 --> 36:57.099
that means, FV which is final value is 2000,
so we need here what is the annual interest
36:57.099 --> 37:04.589
rate. Again the same equation you can use
final value is equal to present value into
37:04.589 --> 37:15.609
1 plus i divided by 100 to the power N.
In this case the final value is equal to 2000
37:15.609 --> 37:28.140
which is equal to 1000 into 1 plus i divided
by 100 to the power 5. We can write down 1
37:28.140 --> 37:36.239
plus 0.01 into i to the power 5 is equal to
2. Now this 0.01 factor which is multiplied
37:36.239 --> 37:45.119
by i is basically comes from 1 by 100, because
I have used i divided by 100 so I can write
37:45.119 --> 37:56.079
this as 0.01 into i. So, 1 plus 0.01 into
i to the power 5 is 2. Again I will emphasize
37:56.079 --> 38:04.029
that here I am using i as percentage directly
so whatever result I will get after solving
38:04.029 --> 38:16.749
the i will be in percentage not has a fraction.
So, if you proceed 1 plus 0.01 i is equal
38:16.749 --> 38:33.780
to 2 to the power 1 by 5 which come out to
be 1.14869 or i is equal to 1.14869 minus
38:33.780 --> 38:42.309
1 and this is in brackets divided by 0.01
which comes out to be 14.87 percent. That
38:42.309 --> 38:53.500
means if 1000 it will be invested for 5 years
it will be doubled in 5 years provided the
38:53.500 --> 38:57.049
annual interest rate is 14.87 percent.
38:57.049 --> 39:07.330
Let us take another problem, the problem type
is D. The example is 9; how long export it
39:07.330 --> 39:17.130
demands approximate years, it will take to
double Rupees 1000 at an annual interest rate
39:17.130 --> 39:28.380
of 10.41 percent if the compound interest
is paid. If you analyze these examples then
39:28.380 --> 39:35.460
it gives interest rate 10.41 it gives the
value of P that is principle or present value
39:35.460 --> 39:43.289
to be 1000, it gives the final value because
the money has to be doubled. So, the final
39:43.289 --> 39:55.079
value is 2000 which is 1000 into 2. And it
requires the years that is N that for how
39:55.079 --> 40:06.299
many years this 1000 should be invested at
10.41 percent interest rate so that it doubles
40:06.299 --> 40:12.369
Let us see the solution we consider that after
10 years the money will be doubled. So, the
40:12.369 --> 40:21.200
final value is 2000 is equal to 1000 1 plus
i divided by 100 to the power N. So, is becomes
40:21.200 --> 40:30.369
1000 into 1.1041 to the power [noise], and
we have to find out this value of N. So, this
40:30.369 --> 40:40.059
becomes 1 plus 1 to the power N is equal to
2 or I should say that 1.1041 to the power
40:40.059 --> 40:52.269
N is equal to 2. So, if I take natural logarithm
of both the sides then N into ln 1.1041 is
40:52.269 --> 41:00.750
equal to ln 2 or N will be equal to ln 2 divided
by ln 1.1041.
41:00.750 --> 41:12.140
If I take the natural logarithmic of 2 it
is 0.6931478 and if I take the logarithmic
41:12.140 --> 41:25.359
of 1.1041 it is 0.09903. So when we divide
it, it becomes 6.999 years which is approximately
41:25.359 --> 41:38.519
7 years. That means, if I invest Rupees 1000
today at an interest rate of 10.41 per year
41:38.519 --> 41:43.180
then it will takes 7 years to double this
money.
41:43.180 --> 41:49.579
Another problem we take which is problem type
D. This is example 10; the compound interest
41:49.579 --> 42:05.069
on Rupees 40,000 at 8 percent per annum for
a certain period is Rupees 58773.12. Find
42:05.069 --> 42:13.369
the period in years. Here the compound interest
is given which is 8 percent per annum, P is
42:13.369 --> 42:24.300
given which is 40,000 and FV which is the
final value is given 58773.12, and what is
42:24.300 --> 42:30.650
unknown is the N which is number of years.
Let us take the solution. Again the similar
42:30.650 --> 42:41.059
way let N be the number of years which will
grow the 40,000 Rupees to 58773.12 at the
42:41.059 --> 42:48.619
interest rate of 8 percent per annum. So,
final value is 58773.12 which is equal to
42:48.619 --> 42:57.249
40,000 into 1 plus i divided by 100 to the
power N which comes out to be 40,000 into
42:57.249 --> 43:08.999
1.08 to the power N. So, the 1.08 to the power
N is equal to 58773.12 divided by 40,000 which
43:08.999 --> 43:16.460
comes out to be 1.469321.
Now the find out the value of N what we have
43:16.460 --> 43:21.529
to do, we have to take the log of both the
sides. So, if I take the log of both the sides
43:21.529 --> 43:36.509
then it converts into N into log 1.08 is equal
to ln 1.469321. Or N is equal to ln 1.08 divided
43:36.509 --> 43:47.289
by ln 1.469321. Now if I take this natural
logarithm of these values 1.08 I find it to
43:47.289 --> 44:00.089
be 0.3848. Similarly, if I take the natural
logarithm of 1.469321 it is 0.076961. So,
44:00.089 --> 44:13.829
once I divide 0.3848 with 0.076961 it comes
out 4.9994 which is approximately equal to
44:13.829 --> 44:28.069
5 years. That means if 40,000 Rupees is invested
for 5 years at the rate of 8 percent per annum
44:28.069 --> 44:41.509
then it will convert into Rupees 58773.12.
Now, let us take some problems which are mixed
44:41.509 --> 44:49.640
in nature. With problem type A, problem type
B, problem type C, problem type D we have
44:49.640 --> 44:55.589
covered the matrix which is possible matrix.
And after we have covered the possible matrix
44:55.589 --> 45:04.049
then we see some problems which are of mix
type. For that I have taken several problems
45:04.049 --> 45:06.910
and an example 11 is one such problems.
45:06.910 --> 45:17.869
Example 11 says there is 90 percent increase
in an amount in 9 years at simple interest.
45:17.869 --> 45:27.619
What will be the compound interest of Rupees
20,000 after 5 years at the same rate? So,
45:27.619 --> 45:34.119
is a mixture of simple interest and compound
interest problems and the from the first statement
45:34.119 --> 45:45.859
I have to find out what is the interest rate
and what is the simple interest rate per annum.
45:45.859 --> 45:52.009
And in the second parts of the questions I
have to use that interest rate to compute
45:52.009 --> 46:00.430
the compound interest of 20,000 Rupees for
5 years. So, let us see the solution.
46:00.430 --> 46:08.989
Let the rate of interest per annum is i, let
the principle amount be P. So, 90 percent
46:08.989 --> 46:17.180
increase in amount of P is 0.9 P. So, interest
received after 9 years is equal to when we
46:17.180 --> 46:28.839
use it as simple interest P into i divided
by 100 into 9 is equal to 0.9 P, because P
46:28.839 --> 46:38.809
into 1 i divided by 100 into 9 is the interest
earned by the principle amount P in 9 years
46:38.809 --> 46:47.059
using simple interest. So, i comes out to
be 90 divided by 9 as 10 percent. Once i is
46:47.059 --> 46:54.880
known, that is i of the simple interest is
known same i interest rate has to be applied
46:54.880 --> 47:01.670
for the compound interest so the problem becomes
simpler, now we know the value of i. Compound
47:01.670 --> 47:09.460
interest after 5 years for Rupees 20,000 will
be now 20,000 into 1 plus 10 divided by 100
47:09.460 --> 47:19.979
to the power 5 minus 20,000 which will be
equal to 32210.2 minus 20,000 which come out
47:19.979 --> 47:27.750
to be 12210.2.
Lets taken an another problem, and this problem
47:27.750 --> 47:31.650
is mixed type.
47:31.650 --> 47:39.170
Example 12; the difference between compound
interest and simple interest on an amount
47:39.170 --> 47:49.510
of Rupees 20,000 for 2 years is Rupees 162.
What is the rate of interest per annum? The
47:49.510 --> 47:57.789
solution is let the rate of the interest per
annum is i, so the compound interest abbreviated
47:57.789 --> 48:10.730
as C I for 2 years will be 20,000 into 1 plus
i divided by 100 square minus 20,000. 20,000
48:10.730 --> 48:17.820
into 1 plus i divided by 100 square is the
FV value that is final value. And 20,000 is
48:17.820 --> 48:22.789
the principle which is the present value so
final value minus present value is the interest.
48:22.789 --> 48:32.019
So, this comes out to be 20,000 in brackets
1 plus 0.1 into i whole square minus 1.
48:32.019 --> 48:43.809
This 0.01 factor has come from i divided by
100. So, i divided by 100 can be written as
48:43.809 --> 48:52.529
i into 1 by 100 and 1 by 100 will convert
into 0.01, so i divided by 100 is 0.01 into
48:52.529 --> 49:02.749
i. It should be noted that i here is percent
is not a fraction, so when we solve for i
49:02.749 --> 49:11.819
whatever we get the results will be in percentage.
So, the compound interest now is 20,000 into
49:11.819 --> 49:19.299
in the brackets 1 plus 0.1 into i whole square
minus 1. Similarly, the simple interest for
49:19.299 --> 49:28.930
2 years is 20,000 i divided by 100 into 2
which is 200 into 2 i.
49:28.930 --> 49:38.009
Now, the problem gives that compound interest
minus simple interest is 162. So, C I minus
49:38.009 --> 49:47.109
S I is equal to 20,000 in the brackets 1 plus
0.1 i whole square minus 1 minus 20,000 into
49:47.109 --> 50:02.529
0.02 i is equal to 162. Now if we solve this
then we find that 20,000 into 0.001 i square
50:02.529 --> 50:11.299
is equal to 162 because while solving you
will find that 1 plus minus 1 cancels 0.02
50:11.299 --> 50:19.440
i also cancels with minus 0.02 i and hence
left out is the 20,000 into 0.001 i square
50:19.440 --> 50:30.739
is equal to 162. And hence 2.0 i square is
equals to 162 and i square is equal to 81
50:30.739 --> 50:37.890
and i is equal to 9 percent. So, if i is taken
to be 9 percent then the condition which is
50:37.890 --> 50:45.319
given in the example two will be satisfied.
Again we take another example which is a problem
50:45.319 --> 50:49.789
type mixed it is example 13.
50:49.789 --> 50:55.200
The difference between simple and compound
interest compounded annually on a certain
50:55.200 --> 51:05.819
sum of money for 2 years at 8 percent annum
is Rupees 6.4. Find the amount of money invested.
51:05.819 --> 51:14.999
Now, here amount of money is required. So,
if you see the solution we presume that let
51:14.999 --> 51:23.150
the some of the amount is x then compound
interest C I for 2 is equal to x into 1 plus
51:23.150 --> 51:32.119
8 divided by 100 whole square minus x which
comes out to be 0.1664 x. Similarly, if we
51:32.119 --> 51:39.990
calculated the simple interest for 2 years
it is x into 0.08 into 2 which comes out to
51:39.990 --> 51:52.450
be 0.16 x. So, compound interest minus simple
interest is equal 0.1664 x minus 0.16 x which
51:52.450 --> 52:06.380
is equal to 0.0064 x is equal to 6.4. And
hence x is equal to 1000.
52:06.380 --> 52:19.089
Thank you.