Formula - Simple & Compound Interest


INTEREST

Interest is the fixed amount paid on borrowed money.
The sum lent is called the Principal.
The sum of the principal and interest is called the Amount.

Interest is of two kinds :

(i) Simple interest

(ii) Compound interest

(i) Simple Interest :

When interest is calculated on the original principal for any length of time, it is called simple interest.

Simple interest = \(\frac{Principal \times Time \times Rate}{100}\)

i.e. \(S.I. = \frac{P \times R \times T}{100}\)

Amount = Principal + Interest

i.e. A = P + I = \(P + \frac{PRT}{100} = P\left ( 1 + \frac{RT}{100} \right )\)

Principal (P) = \(\frac{100 \times S.I.}{R \times T}\)

Rate (R) = \(\frac{100 \times S.I.}{P \times T}\)

Time (T) = \(\frac{100 \times S.I.}{P \times R}\)

If rate of simple interest differs from year to year, then

S.I. = \(P \times \frac{\left ( R_{1} + R_{2} + R_{3} + ... \right )}{100}\)

Example 1

1. Find the interest to be paid on a loan of  6000 at 5% per year for 5 years

Sol. P =  6000, R = 5% and T = 5 years

S.I. = \(\frac{P \times R \times T}{100} = \frac{6000 \times 5 \times 5}{100}\) = 1500

Example 2

2. Find the amount to be paid back on a loan of 18,000 at 5.5% per annum for 3 years

Sol. P = 18000, R =5.5%, T= 3 years

S.I. = \(\frac{P \times R \times T}{100} = \frac{18000 \times 5.5 \times 3}{100}\) = 2970

Amount = P + I = 18000 + 2970 = 20970

Example 3

3. In how many years will a sum of money triple itself, at 25% per annum simple interest.

Sol. Let the sum of money be P.So, A = 3P

and S.I. = A – P = 3P – P = 2P

R = 25%

∴ T = \(\frac{100 \times S.I.}{P \times R} = \frac{100 \times 2P}{P \times 25}\) = 8 years

Example 4

4. What rate per cent per annum will produce 250 as simple interest on 6000 in 2.5 years

Sol. P = 6000; Time (T) = 2.5 years; S.I.= 250

∴ Rate = \(\frac{100 \times S.I.}{P \times T} = \frac{100 \times 250}{6000 \times 2.5} = \frac{10}{6} = \frac{5}{3} = 1\tfrac{2}{3}\%\)

Example 5

5. To buy furniture for a new apartment, Sylvia Chang borrowed 5000 at 11 % simple interest for 11 months.How much interest will she pay?

Sol. From the formula, I = Prt, with P = 5000, r =.11, and t = 11/12(in years). The total interest she will pay is

I = 5000 (.11) (11/12) = 504.17 or  504.17

 (ii)  Compound Interest:  

Money is said to be lent at compound interest when at the end of a year or other fixed period, the interest that has become due is not paid to the lender, but is added to the sum lent, and the amount thus obtained becomes the principal in the next year or period.The process is repeated until the amount for the last period has been found. Hence,When the interest charged after a certain specified time period is added to form new principal for the next time period,the interest is said to be compounded and the total interest accrued is compound interest.

C.I. = \(P\left [ \left ( 1 + \frac{r}{100} \right )^{n} - 1\right ];\)

Amount (A) = \(P\left ( 1 + \frac{r}{100} \right )^{n}\)

If rate of compound interest differs from year to year, then

Amount = \(P\left ( 1 + \frac{r_{1}}{100} \right )\left ( 1 + \frac{r_{2}}{100} \right )\left ( 1 + \frac{r_{3}}{100} \right )......\)

Example 6

6. Find the compound interest on 70000 for 4 years at the rate of 14% per annum compounded annually.

Sol. P = 70000, n = 4, r = 14%

A = \(P\left [ \left ( 1 + \frac{r}{100} \right )^{n} \right ] = 70000\left [ \left ( 1 + \frac{14}{100} \right )^{4} \right ] = 118227.20\)

C.I. = A – P = 118227.20 – 70000 = 48227.20

Example 7

7. If  60000 amounts to  68694 in 2 years then find the rate of interest.

Sol. Given : A = 68694

P = 60000

n = 2 years

r =?

\(P\left [ \left ( 1 + \frac{r}{100} \right )^{n} \right ]\)

\(68694 = 60000\left [ \left ( 1 + \frac{r}{100} \right )^{2} \right ]\)

=> \(\frac{68694}{60000} = \left [ \left ( 1 + \frac{r}{100} \right )^{2} \right ] \)

=> \(\frac{11449}{10000} = \left [ \left ( 1 + \frac{r}{100} \right )^{2} \right]\)

=> \(1 + \frac{r}{100} = \sqrt{\frac{11449}{10000}} = \sqrt{1.1449}\)

=> \(1 + \frac{r}{100} = 1.07\)

=> r100 = 1.07 – 1 = 0.07

∴ r = 0.07 × 100 = 7%

Example 8

8. In how many years, the sum of 10000 will become 10920.25 if the rate of compound interest is 4.5% per annum?

Sol. A = 10920.25

P = 10000

Rate of interest = 4.5%

Time (n) = ?

\(P\left [ \left ( 1 + \frac{r}{100} \right )^{n} \right ]\)

\(10920.25 = 10000\left [ \left ( 1 + \frac{4.5}{100} \right )^{n} \right ]\)

\(\frac{10920.25}{10000} = \left ( 1 + \frac{0.9}{20} \right )^{n} = \left ( \frac{20.9}{20} \right )^{n}\)

=> \(\frac{436.81}{400} = \left ( \frac{20.9}{20} \right )^{n}\)

=> \(\left ( \frac{20.9}{20} \right )^{2} = \left ( \frac{20.9}{20} \right )^{n}\)

Hence 10000 will become 10920.25 in 2 years at 4.5%.

Example 9

9. Suppose 1000 is deposited for 6 years in an account paying 8.31% per year compounded annually.

(a) Find the compound amount.

In the formula above,P = 1000, i = .0831, and n =6.

The compound amount is

A = P (1 + i)n

A = 1000 (1.0831)6

A = 1614.40.

(b) Find the amount of interest earned.

Subtract the initial deposit from the compound amount.

Amount of interest = 1614.40 – 1000 = 614.40.

Compound interest – when interest is compounded annually but time is in fraction

If time = tpq years, then

\(A = P\left ( 1 + \frac{r}{100} \right )^{t}\left ( 1 + \frac{\frac{p}{q}r}{100} \right )\)
Example 10

10. Find the compound interest on 8000 at 15% per annum for 2 years 4 months, compound annually.

Sol. Time = 2 years 4 months = 2412 years = 213 years

Amount = \( \left [ 8000\left \{ \left ( 1 + \frac{15}{100} \right ) \right \}^{2}\left ( 1 + \frac{\frac{1}{3} \times 15}{100} \right ) \right ]\)

= \( \left ( 8000 \times \frac{23}{20} \times \frac{23}{20}\times \frac{21}{20}\right ) = \; 11109\)

∴ C.I. = (11109 –8000) = 3109.

Compound interest – when interest is calculated half-yearly
Since r is calculated half-yearly therefore the rate per cent will become half and the time period will become twice, i.e.,

Rate per cent when interest is paid half-yearly = r2%

and time = 2 × time given in years

Hence,

\(A = P\left ( 1 + \frac{r}{2 \times 100} \right )^{2n}\)
Example 11

11. What will be the compound interest on 4000 in 4 years at 8 per cent annum. If the interest is calculated half-yearly.

Sol. Given : P = 4000, r = 8%, n = 4 years

Since interest is calculated half-yearly, therefore,

r = 82% = 4% and n = 4 × 2 = 8 half years

∴ A = \(4000\left(1 + \frac{4}{100} \right)^{8} = 4000 \times \left ( \frac{26}{25} \right )^{8}\)

= 4000 × 1.3685 = 5474.2762

Amount = 5474.28

∴ Interest = Amount – Principal

= 5474.28 – 4000 = 1474.28

Compound interest – when interest is calculated quarterly

Since 1 year has 4 quarters, therefore rate of interest will become 14th of the rate of interest per annum, and the time period will be 4 times the time given in years

Hence, for quarterly interest

\(A = P\left ( 1 + \frac{\frac{r}{4}}{100} \right )^{4 \times n} = P\left ( 1 + \frac{r}{400} \right )^{4n}\)
Example 12

12. Find the compound interest on 25625 for 12 months at 16% per annum, compounded quarterly.

Sol. Principal (P) = 25625

Rate (r) = 16% = 164% = 4%

Time = 12 months = 4 quarters

A = \(25625\left ( 1 + \frac{4}{100} \right )^{4} = 25625\left (\frac{26}{25} \right )^{4}\)

C.I. = A – P = 29977.62 – 25625 = 4352.62

Difference between Compound Interest and Simple Interest

When T = 2

(i) C.I. – S.I. = \(P\left ( \frac{R}{100} \right )^{2}\)

(ii) C.I. – S.I. = \(\frac{R \times S.I.}{2 \times 100}\)

When T = 3

(i) C.I. – S.I. = \(\frac{PR^{2}}{10^{4}}\left ( \frac{300 + R}{100} \right )\)

(ii) C.I. – S.I. = \(\frac{S.I.}{3}\left [ \left ( \frac{R}{100} \right )^{2} + 3\left ( \frac{R}{100} \right ) \right ]\)

Example 13

13. The difference between compound interest and simple interest on a certain amount of money at 5% per annum for 2 years is 15. Find the sum :

(a) 4500
(b) 7500
(c) 5000
(d) 6000

Sol.(d) Let the sum be 100.

Therefore, SI = \(\frac{100 \times 5 \times 2}{100} = 10\)

and CI = \(100\left ( 1 + \frac{5}{100} \right )^{2} ? 100\)

\(= 100 \times \frac{21 \times 21}{20 \times 20} ? 100 = \; \; \frac{41}{4}\)

Difference of CI and SI = 414 – 10 = 14

If the difference is 14 , the sum = 100

=> If the difference is 15, the sum

= 400 × 15 = 6000

Example 14

14. The difference between the simple interest and the compound interest compounded annually at the rate of 12% per annum on 5000 for two years will be :

(a) 47.50
(b) 63
(c) 45
(d) 72

Sol.(d) Required difference

= \(\left [ 5000\left ( 1 + \frac{12}{100} \right )^{2} ? 5000\right ] ? \frac{5000 \times 12 \times 2}{100}\)

= \(5000\left ( \frac{28}{25} \times \frac{28}{25} ? 1\right ) ? 1200\)

= \(5000\left ( \frac{784 ? 625}{625}\right ) ? 1200 = \; 72\)

EFFECTIVE RATE

If 1 is deposited at 4% compounded quarterly,a calculator can be used to find that at the end of one year, the compound amount is 1.0406, an increase of 4.06% over the original 1.The actual increase of 4.06% in the money is some what higher than the stated increase of 4%. To differentiate between these two numbers, 4% is called the nominal or stated rate of interest, while 4.06% is called the effective rate. To avoid confusion between stated rates and effective rates, we shall continue to use r for the stated rate and we will use rc for the effective rate.

Example 15

15. Find the effective rate corresponding to a stated rate of 6% compounded semiannually.

Sol. A calculator shows that 100 at 6% compounded semiannually will grow to

A = \(100\left ( 1 + \frac{0.06}{2} \right )^{2} = 100(1.03)^{2} = \$106.09\)

A = \(100(1.03)^{2} = \$106.09\)

Thus, the actual amount of compound interest is 106.09 – 100 = 6.09. Now if you earn 6.09 interest on 100 in 1 year with annual compounding, your rate is 6.09/100 = .0609 = 6.09%.

Thus, the effective rate is re = 6.09%.

In the preceding example we found the effective rate by dividing compound interest for 1 year by the original principal.The same thing can be done with any principal P and rate r compounded m times per year.

\(Effective \; rate = \frac{compound \; interest}{principal}\)

re = \(\frac{compound \; amount \; ? \; principal}{principal}\)

= \(\frac{P\left ( 1 + \frac{r}{m}\right )^{m} ? P }{P} = \frac{P\left [ \left ( 1 + \frac{r}{m} \right )^{m} ? 1 \right ]}{P}\)

= re = \(\left ( 1 + \frac{r}{m}\right )^{m} ? 1\)

Example 16

16. A bank pays interest of 4.9% compounded monthly. Find the effective rate.

Sol. Use the formula given above with r = .049 and m = 12.

The effective rate is re = \(\left ( 1 + \frac{.049}{12} \right )^{12} ? 1\)

= 1.050115575 – 1 ≈ .0501 or 5.01%

Present worth of P due n years hence

Present worth = \(\frac{P}{\left ( 1 + \frac{r}{100} \right )^{n}}\)

Equal annual instalment to pay the borrowed amount

Let the value of each instalment = x

Rate = r% and time = n years

Then, Borrowed Amount

\(= \frac{x}{\left ( 1 + \frac{r}{100} \right )} + \frac{x}{\left ( 1 + \frac{r}{100} \right )^{2}} + ?+ \frac{x}{\left ( 1 + \frac{r}{100} \right )^{n}}\)
Example 17

17. Subash purchased a refrigerator on the terms that he is required to pay 1,500 cash down payment followed by 1,020 at the end of first year, 1,003 at the end of second year and 990 at the end of third year. Interest is charged at the rate of 10% per annum. Calculate the cash price :

(a) 3,000
(b) 2,000
(c) 4,000
(d) 5,000

Sol.(c) Cash down payment = 1500

Let x becomes 1020 at the end of first year.

Then, 1020 = \(x\left(1 + \frac{10}{100} \right)\)

or x = \(\frac{1020 \times 100}{110}\) = 927.27

Similarly, \(1003 = y\left(1 + \frac{10}{100} \right )^{2}\)

or y = \(\frac{1003 \times 20 \times 20}{22 \times 22}\) = 828.92

and z = \(\frac{990 \times 20 \times 20 \times 20}{22 \times 22 \times 22}\) = 743.80

Hence, CP = 1500 + 927.27 + 828.92 + 743.80

= 3999.99 or 4000.

Example 18

18. The difference between the interest received from two different banks on 500 for 2 yrs is 2.5. Find the difference between their rates.

Sol. I1 = \(\frac{500 \times 2 \times r_{1}}{100} = 10r_{1}\)

I2 = \(\frac{500 \times 2 \times r_{2}}{100} = 10r_{2}\)

I1 – I2 = 10r1 – 10r2 = 2.5

Or, I1 – I2 = 2.510 = 0.25%

Examination method :

When t1 = t2,

(r1 – r2) = \(\frac{I_{d} \times 100}{sum \times t} = \frac{2.5 \times 100}{500 \times 2} = 0.25\%\)

Example 19

19. At what rate percent compound interest does a sum of money becomes nine – fold in 2 years?

Sol. Let the sum be x and the of compound interest be r% per annum; then

9x = \(x\left ( 1 + \frac{r}{100} \right )^{2}\)

or, 9 = \(x\left ( 1 + \frac{r}{100} \right )^{2}\)

or, 3 = 1 + r100; or, r100 = 2

∴ r = 200%

Examination method :

The general formula of compound interest can be changed to the following form :

If a certain sum becomes ‘m’ times in ‘t’ years, the rate of compound interest r is equal to 100 [(m)1t – 1]

In this case ,r = 100 [(9)1t – 1]

= 100 (3 – 1) = 200%

Example 20

20. The simple interest on a certain sum of money at 4% per annum for 4 years is 80 more than the interest on the same sum for 3 years at 5% per annum. Find the sum.

Sol. Let the sum be x, then at 4% rate for 4 years the simple

Interest = \(\frac{x \times 4 \times 4}{100} = \; \frac{4x}{25}\)

At 5% rate for 3 yrs the simple interest

= \(\frac{x \times 5 \times 3}{100} = \; \frac{3x}{20}\)

Now, we have, \(\frac{4x}{25} = \frac{3x}{20} = 80\)

or \(\frac{16x ? 15x}{100} = 80\)

∴ x = 8000

Examination Method :

For this type of question

Sum = \(\frac{Difference \times 100}{\left [ r_{2}t_{1} ? r_{2}t_{2} \right ]} = \frac{80 \times 100}{4 \times 4 ? 3 \times 5} = 8000\)

Example 21

21. Some amount out of 7000 was lent at 6 % per annum and the remaining at 4 % per annum. If the total simple interest from both the fractions in 5 years was 1600,find the sum lent at 6 % per annum.

Sol. Suppose x was lent at 6% per annum.

Thus, \(\frac{x \times 6 \times 5}{100} + \frac{\left(7000 ? x \right )\times 4 \times 5}{100} = 1600\)

or, \(\frac{3x}{10} + \frac{7000 ? x}{5} = 1600\)

or, \(\frac{3x + 14,000 ? 2x}{100} = 16000\)

∴ x = 16000 – 14000 = 2000

By Method of Alligation :
Overall rate of interest

= \(\frac{1600 \times 100}{5 \times 7000} = \frac{32}{7}\%\)

∴ ratio of two amounts = 2 : 5

Image 1

∴ amount lent at 6 % = 70002 × 2 = 2000

Example 22

22. As n amount of money grows upto  4840 in 2 years and upto  5324 in 3 years on compound interest. Find the rate percent

Sol. We have,

P + CI of 3 yrs = 5324…….(1)

P + CI of 2 yrs =  4840……..(2)

Subtracting (2) from (1), we get

CI of 3rd year = 5324 – 4840 =  484.

Thus, the CI calculated in the third year which is  484 is

basically the amount of interest on the amount generated after 2 years which is  4840.

∴ r = \(\frac{484 \times 100}{4840 \times 1} = 10\%\)

Examination method :

\(\frac{Difference \; of \; amount \; after \; n \; years \; and \; \left ( n + 1 \right ) \; years \times 100}{Amount \; after \; 2 \; years}\)

In this, n = 2.

∴ rate = \(\frac{Difference \; of \; amount \; after \; 2 \; years \; and \; 3 \; years \times 100}{Amount \; after \; 2 \; years}\)

= \(\frac{\left ( 5324 ? 4840 \right )}{4840} \times 100 = \frac{484 \times 100}{4840} = 10\%\)

Example 23

23. A certain amount of money at compound interest grows upto 51168 in 15 yrs and upto 51701 in 16 years. Find the rate per cent per annum.

Sol. Rate = \(\frac{\left ( 51701 ? 51168 \right ) \times 100}{51168} = \frac{533 \times 100}{51168}\)

= \(\frac{100}{96} = \frac{25}{24} = 1\tfrac{25}{24}\%\)

Example 24

24. Find the compound interest on  18,750 in 2 years the rate of interest being 4% for the first year and 8% for the second year.

Sol. After first year the amount

= \(18750\left ( 1 + \frac{4}{100} \right ) = 18750\left ( \frac{104}{100} \right )\)

After 2nd year the amount = \(18750\left ( \frac{104}{100} \right )\left( \frac{108}{100} \right )\)

= \(18750\left ( \frac{26}{25} \right )\left( \frac{27}{25} \right ) = 21060\)

∴ CI = 21060 – 18,750 = 2310.

A computer gives the following results for various values of n.
Interest in Compound n (1 + 1n)n
Annually 1 (1 + 11)1 = 2
Semiannually 2 (1 + 12)2 = 2.25
quarterly 4 (1 + 14)4 ≈ 2.4414
Monthly 12 (1 + 112)12 ≈ 2.6130
Daily 365 (1 + 1365)365 ≈ 2.71457
Hourly 8760 (1 + 18760)8760 ≈ 2.718127
Every Minute 525,600 ≈ 2.7182792 (1 + 1525,600)525,600
Every Second 31,536,00092 (1 + 131,536,000)31,536,000 ≈ 2.7182818