What time will it take to obtain double of the interest on Rs 2000 for 3 years at the rate of 15% per annum?

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.

How to calculate simple interest?

Principal: The money which we deposit in or the lower from the bank or the money learned called the principal.

Rate of interest: The interest paid on $ 100 for one year is called the rate per cent per year or rate per cent per annum.

Time: The period of time for which the money is lent or invested.

Interest: Additional money paid by the borrowed to the lender for using the money is called interest.

Simple Interest: If the interest is calculated uniformly on the original principal throughout the lone period, it is called simple interest.

Amount: The total money paid back to the lender is called the amount.

Calculate Simple Interest

Formula to calculate Simple Interest?

If P denotes the principal ($), R denotes the rate (percentage p.a.) and T denotes time (years), then:-

S.I = (P × R × T)/100

R = (S.I × 100)/(P × T)

P = (S.I × 100)/(R × T)

T = (S.I × 100)/(P × R)

If the denotes the amount, then A = P + S.I

Note:

● When we calculated the time period between two dates, we do not could the day on which money is deposited but we count the day on which money is retuned.

● Time is always taken according to the per cent rat.

● For converting time in days into years, divide th number of days by 365 (for ordering or lap year.)

● For converting time in month into years, divide th number of month by 12 (for ordering or lap year.)

Examples to find or calculate simple interest when principal, rate and time are known

Calculate Simple Interest

Find the simple interest on:

(a) $ 900 for 3 years 4 months at 5% per annum. Find the amount also.

Solution:

P = $ 900,

R = 5% p.a.

T = 3 years 4 months = 40/12 years = 10/3 years

Therefore, S.I = (P × R × T)/100 = (900 × 5 × 10)/(100 × 3) = $ 150

Amount = P + S.I = $ 900 + $ 150 = $ 1050

(b) $ 1000 for 6 months at 4% per annum. Find the amount also.

Solution:

P = $ 1000,

R = 4% p.a.

T = 6 months = 6/12 years

S.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20

Therefore, A = P + I = $( 1000 + 20) = $ 1020

(c) $ 5000 for 146 days at 15¹/₂% per annum.

Solution:

P = $ 5000, R = 151/2% p.a. T = 146 days

S.I = ( 5000 × 31 × 146)/(100 × 2 × 365)

= $ 10 × 31 = $ 310

(d) $ 1200 from 9ᵗʰ April to 21ˢᵗ June at 10% per annum.

Solution:

P = $ 1200, R = 10% p.a. T = 9th April to 21st June

= 73 days [April = 21, May = 31, Jun = 21, 73 days]

= 73/365 years

S.I = (1200 × 10 × 73)/(100 × 365) = $ 24

Examples to find or calculate Time when Principal, S.I and Rate are known

Calculate Simple Interest

1. In how much time dose $ 500 invested at the rate of 8% p.a. simple interest amounts to $ 580.

Solution:

Here P = $ 500, R = 8% p.a A = $ 580

Therefore S.I = A - P = $ (580 - 500) = $ 80

Therefore T = (100 × S.I)/(P × R) = (100 × 80)/(500 × 3) = 2 years

2. In how many years will a sum of $ 400 yield an interest of $ 132 at 11% per annum?

Solution:

P = $ 400, R = 11% S.I = $ 132

T = (100 × S.I)/(P × R) = (132 × 100)/(400 × 11) = 3 years

Calculate Simple Interest

3. In how many years will a sum double itself at 8 % per annum?

Solution:

Let Principal = P, then, Amount = 2P

So , S.I. = A - P = 2P – P = P

T = (100 × S.I)/(P × R) = ( 100 × P)/(P × 8) = 25/2 = 121/2 years

Calculate Simple Interest

4. In how many years will simple interest on certain sum of money at 6 1/4% Per annum be 5/8 of itself?

Solution:

Let P = $ x, then S.I = $ 5/8 x

Rate = 6 1/4% = 25/4 %

Therefore T = ( 100 × S.I)/(P × R) = ( 100 × 5/8)/(x × 25/4) x = ( 100 × 5 × x × 4)/(x × 8 × 25)

Examples to find or calculate Rate per cent when Principal, S.I. and Time are known

1. Find at what rate of interest per annum will $ 600 amount to $ 708 in 3 years.

Solution:

P= $ 600 , A = $ 708 Time = 3 years

Therefore S.I. = $ 708 - $ 600 = $ Rs. 108

Now, R = ( 100 × S.I)/(P × R) = (100 × 108)/(600 × 3) = 6% p.a.

Calculate Simple Interest

2. Simple interest on a certain sum is 36/25 of the sum. Find the Rate per cent and time if they are both numerically equal.

Solution:

Let the Principal be $ X Then S.I. = 36/25 x

R = ? T = ?

Let Rate = R % per annum, then Time = R years.

So S.I. = (P × R × T)/100 → 36/25 x = (x × R × T)/100

--- ( 36 × 10 × x)/(25 × x) = R2

----- R2 = 36 × 4 ----- R = √(36 × 4) = 6 × 2

Therefore Rate = 12 % p.a. and T = 12 years

Calculate Simple Interest

3. At what rate per cent per annum will $ 6000 produce $ 300 as S.I. in 1 years?

Solution:

P= $ 600, T = 1 year S.I. = $ 300

Therefore R = ( S.I × 100)/(P × R) = ( 300 × 100)/(6000 × 1) = 5% p.a

4. At what rate per cent per annum will a sum triple itself in 12 years ?

Solution:

Let the sum be $ P, then Amount = $ 3P

S.I. = $ 3P – P = $ 2P, Time = 12 years

Now, R =( S.I × 100)/(P × R) = (100 × 2P)/(P × 12) = 50/3 = 16.6 %

Examples to find or calculate Principal when Rate, Time and S.I. are known

Calculate Simple Interest

1. What sum will yield $ 144 as S.I. in 21/2 years at 16% per annum?

Solution:

Let P = $ x, S.I. = $ 144

Time = 21/2 years or 5/2 years, Rate = 16%

So, P = ( 100 × S.I)/(P × R) = ( 100 × 144)/(16 × 5/2) = ( 100 × 144 × 2)/(16 × 5) = $ 360

2. A some amount to $ 2040 in 21/2 years at , P = ?

Solution:

Let the principal = $ x

S.I. = $ (x × 11 × 5/2 × 1/100) = $ 11x/40

Amount = P + S.I. = x/1 + 11x/40 = (40x × 11x)/40 = 51x/40

But 51x/40 = 2040

51x = 2040 × 40 ---- x = (2040 × 40)/51 = $ 1600

Calculate Simple Interest

3. A certain sum amounts to $ 6500in 2 years and to $ 8750 in 5 years at S.I. Find the sum and rate per cent per annum.

Solution:

S.I. for 3 years = Amount after 5 years – Amount after 2 years

= $ 8750 – $ 6500 = 2250
S.I. for 1 years = Rs. 2250/3 = $ 750

Therefore S.I. for 2 years = $ 500× 2 = $ 1500

So, sum = Amount after 2 years – S.I.for 2 years

= $ 6500- 1500 = $ 5000

Now, P = Rs.5000, S.I. = $ 1500, Time = 3 years

R = ( 100 × S.I)/(P × T) = (100 × 1500)/(5000 × 2) = 15%

Therefore The sum is $ 5000 and the rate of interest is 15%

4. Divide $ 6500 in to two parts , such that if one part is lent out at 9% per annum and other at 10% per annum, the total yearly income is $ 605

Solution:

Let the first part be $ x. Second part = $ (6500 - x )

Now S.I. on $ X at 9% per annum for 1 year = $ (x × 9 × 1)/100 = 9x/100

S.I. on $ (6500 – x ) at 10% per annum

1 year = $ ((6500-x) × 10 × 1)/100 = $ ((6500 - x))/10

Total S.I = $ (9x/100+ (6500 - x)/10) = ((9x + 6500 - 10x)/100)

= $ ( 65000 - x)/100

But given that total S.I.= $ 605

So, (6500 - x)/100 =605 -----65000 - x = 60500

----- 65000 – 60500 = x ---- x = $ 4500

Now, second part = 6500 – x = 6500 – 4500 = $ 2000

Hence, first part = $ 2000 and second part = $ 4500

5. When the rate of interest in a bank is increased from 9% to 10% per annum; A person deposits $ 500 more into his account. If the annual interest now Received by him is $ 150more then before, find his original deposit.

Solution:

Let the original deposits be $ x

Then, S.I. on $ x for 1 year at (10 - 9 )% = 1% per annum + S.I. on $ 500

For I year at 10% per annum = $ 15

----- ( x × 1 × 1)/100 + ( 500 × 10 × 1)/100 = 150

----- x/(100 ) + 50 = 150 ---- x/(100 ) + 150 – 50 ----- x/(100 ) + 100

----- x = 100 × 100 = $ 10,000

Therefore, the original deposit is $ 10,000.

Calculate Simple Interest

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