What will be the compound interest earned on an amount of Rs 12000 for 2 years at the rate of 10% pa compounded annually?

Solution : Principle`=` Rs .`12000`<br>Time`(n) = 2`year<br>Rate`= 20%`<br>Now , `A= P(1+R/100)^n`<br>`A= 12000 ( 1+20/100)^2`<br>`A= 12000 (1+1/5)^2`<br>`A= 12000 (6/5)^2`<br>`A= 12000xx 36/25`<br>`A=17280`

Compound Interest: The future value (FV) of an investment of present value (PV) dollars earning interest at an annual rate of r compounded m times per year for a period of t years is:

FV = PV(1 + r/m)mtor

FV = PV(1 + i)n

where i = r/m is the interest per compounding period and n = mt is the number of compounding periods.

One may solve for the present value PV to obtain:

PV = FV/(1 + r/m)mt

Numerical Example: For 4-year investment of $20,000 earning 8.5% per year, with interest re-invested each month, the future value is

FV = PV(1 + r/m)mt   = 20,000(1 + 0.085/12)(12)(4)   = $28,065.30

Notice that the interest earned is $28,065.30 - $20,000 = $8,065.30 -- considerably more than the corresponding simple interest.

Effective Interest Rate: If money is invested at an annual rate r, compounded m times per year, the effective interest rate is:

reff = (1 + r/m)m - 1.

This is the interest rate that would give the same yield if compounded only once per year. In this context r is also called the nominal rate, and is often denoted as rnom.

Numerical Example: A CD paying 9.8% compounded monthly has a nominal rate of rnom = 0.098, and an effective rate of:

r eff =(1 + rnom /m)m   =   (1 + 0.098/12)12 - 1   =  0.1025.

Thus, we get an effective interest rate of 10.25%, since the compounding makes the CD paying 9.8% compounded monthly really pay 10.25% interest over the course of the year.

Mortgage Payments Components: Let where P = principal, r = interest rate per period, n = number of periods, k = number of payments, R = monthly payment, and D = debt balance after K payments, then

R = P � r / [1 - (1 + r)-n]

D = P � (1 + r)k - R � [(1 + r)k - 1)/r]

Accelerating Mortgage Payments Components: Suppose one decides to pay more than the monthly payment, the question is how many months will it take until the mortgage is paid off? The answer is, the rounded-up, where:

n = log[x / (x � P � r)] / log (1 + r)

Future Value (FV) of an Annuity Components: Ler where R = payment, r = rate of interest, and n = number of payments, then

FV = [ R(1 + r)n - 1 ] / r

Future Value for an Increasing Annuity: It is an increasing annuity is an investment that is earning interest, and into which regular payments of a fixed amount are made. Suppose one makes a payment of R at the end of each compounding period into an investment with a present value of PV, paying interest at an annual rate of r compounded m times per year, then the future value after t years will be

FV = PV(1 + i)n + [ R ( (1 + i)n - 1 ) ] / i where i = r/m is the interest paid each period and n = m � t is the total number of periods.

Numerical Example: You deposit $100 per month into an account that now contains $5,000 and earns 5% interest per year compounded monthly. After 10 years, the amount of money in the account is:

FV = PV(1 + i)n + [ R(1 + i)n - 1 ] / i =
5,000(1+0.05/12)120 + [100(1+0.05/12)120 - 1 ] / (0.05/12) = $23,763.28

Value of a Bond:

V is the sum of the value of the dividends and the final payment.

You may like to perform some sensitivity analysis for the "what-if" scenarios by entering different numerical value(s), to make your "good" strategic decision.

Replace the existing numerical example, with your own case-information, and then click one the Calculate.

The compound interest accrued on Rs.12000 in two years is Rs.1996.80. What will be the simple interest accrued at the same rate of interest for the third year on the total amount at the end of two years?

1. Rs. 1256.63
2. Rs. 8500.25
3. Rs. 9845.5
4. Rs. 1119.74

Answer (Detailed Solution Below)

Option 4 : Rs. 1119.74

Free

10 Questions 10 Marks 6 Mins

Given:

The sum = 12000

The time = 2 years

The compound interest for two years = 1996.80

Formula used:

\({CI}\ =\ {P\ × \ [({1\ +\ {R\over100}})^T\ -\ 1]}\)

\({SI}\ =\ {P\ ×\ T\ ×\ R\over 100}\)   Where, CI = The compound interest, P = The principle, R = The rate of the interest, SI = The simple interest, and T = The time

Calculation:

Let us assume the rate of interest be R

⇒ According to the question

⇒ \({1996.8}\ =\ {12000\ × \ [({1\ +\ {R\over100}})^2\ -\ 1]}\)

⇒ \({1996.8}\ =\ {12000\ × \ [({100\ +\ R\over100})^2\ -\ 1]}\)

⇒ \({1996.8}\ =\ {12000\over10000} × \ [({100\ +\ R})^2\ -\ 10000]\)

⇒ \({1996.8}\ =\ {1.2} × \ [({100\ +\ R})^2\ -\ 10000]\)

⇒ \({1996.8}\ =\ {1.2} × \ [({100^2\ +\ R^2\ +\ 200R})\ -\ 10000]\)

⇒ \({1996.8}\ =\ {1.2} × \ [{10000\ +\ R^2\ +\ 200R}\ -\ 10000]\)

⇒ \({1996.8}\ =\ {1.2} × \ [{\ R^2\ +\ 200R}]\)

⇒ 1996.8 = 1.2R2 + 240R

⇒ 19968 = 12R2 + 2400R

⇒ 1664 = R2 + 200R

⇒ By solving the quadratic equation

⇒ R = 8, and -208

⇒ The value of R = 8

⇒ The total amount after two years = 12000 + 1996.8 = 13996.8

⇒ The simple interest for third year on 13996.8 at 8% per annum

⇒ According  to the question

⇒ \({SI}\ =\ {13996.8\ ×\ 1\ ×\ 8\over 100}\)

⇒ SI = 139.968 × 8 = 1119.744

⇒ The simple interest for the third year = 1119.744

∴ The required result will be 1119.74.

Last updated on Sep 26, 2022

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