Simple and Compound Interest

Simple interest

Simple interest charges the same amount every period, always calculated on the original principal $P$ :

SI = \dfrac{P \times N \times R}{100}

where $N$ is the number of years and $R$ the annual rate as a percentage.

Worked example: the simple interest on ₹25,000 borrowed for 3 years at 9% per annum is $\dfrac{25000 \times 3 \times 9}{100} = 6750$ , so the total repaid is $25000 + 6750 = 31{,}750$ .

Compound interest: interest on interest

Compound interest adds each period's interest to the principal before working out the next period's, so later interest is earned on more than the original sum. With the same ₹25,000 at 9%, compounded yearly:

I_1 = 25000 \times \dfrac{9}{100} = 2250 \;\Rightarrow\; 27{,}250, \qquad I_2 = 27250 \times \dfrac{9}{100} = 2452.5 \;\Rightarrow\; 29{,}702.5

I_3 = 29702.5 \times \dfrac{9}{100} = 2673.225

The total interest over 3 years is $2250 + 2452.5 + 2673.225 = 7375.725$ , more than the ₹6,750 of simple interest at the same rate. The formula gives the same result directly, with $n$ compounding periods a year over $t$ years:

CI = P\left(1+\dfrac{r}{100n}\right)^{nt} - P = 25000(1.09)^3 - 25000 = 7375.725

Solved example: interest for part of a year

Find the simple interest on ₹60,000 for 6 months at 5% per annum.

Solution: Time must be in years, so $N = \tfrac{6}{12} = 0.5$ : $SI = \dfrac{60000 \times 0.5 \times 5}{100} = 1500$ .

Solved example: finding the principal

The simple interest on a sum for 3 years at 7% is ₹900 more than the simple interest on the same sum for 2 years at 6%. Find the sum.

\dfrac{3 \times 7}{100}x - \dfrac{2 \times 6}{100}x = 900 \;\Rightarrow\; \dfrac{9x}{100} = 900 \;\Rightarrow\; x = 10{,}000

Check: $2100 - 1200 = 900$ .

Solved example: splitting an investment

₹50,000 is split between two investments paying 11% and 14% a year. Together they return 12.8% of the total. How much went into each?

Solution: Let $x$ go into the 11% investment. The total return is $0.128 \times 50000 = 6400$ :

0.11x + 0.14(50000 - x) = 6400 \;\Rightarrow\; 7000 - 0.03x = 6400 \;\Rightarrow\; x = 20{,}000

So ₹20,000 earns 11% and ₹30,000 earns 14%. Check: $2200 + 4200 = 6400$ .

Solved example: compound interest over five years

The compound interest on ₹25,000 for 5 years at 10% per annum, compounded yearly, is $25000(1.1)^5 - 25000 = 25000 \times 1.61051 - 25000 = 15{,}262.75$ , against $12{,}500$ of simple interest.

Solved example: a different rate each year

₹50,000 grows at 8% in the first year and 9% in the second, compounded yearly. Each year's factor multiplies:

A = 50000 \times 1.08 \times 1.09 = 58{,}860

Solved example: finding the time

In how many years will ₹6,250 grow to ₹7,290 at 8% per annum, compounded yearly?

\dfrac{7290}{6250} = 1.1664 = (1.08)^2 \;\Rightarrow\; t = 2 \text{ years}

Solved example: quarterly compounding

A loan of ₹3,90,625 is charged 16% a year, compounded quarterly. What is owed after 1 year?

Solution: Each quarter earns $16 \div 4 = 4\%$ , and there are 4 quarters:

A = 390625 \times (1.04)^4 = 390625 \times 1.16985856 = 4{,}56{,}976

The result is exact: $390625 = 5^8$ , and it cancels the denominator of $(1.04)^4 = \tfrac{26^4}{25^4}$ .

Solved example: a fractional number of years

Find the amount and compound interest on ₹30,000 for $3\tfrac{3}{4}$ years at 8%, compounded yearly.

Solution: A common textbook convention compounds for the 3 whole years and then adds simple interest for the remaining $\tfrac{3}{4}$ year, at $\tfrac{3}{4} \times 8 = 6\%$ :

A = 30000 \times (1.08)^3 \times 1.06 \approx 40{,}058.84, \qquad CI = A - P \approx 10{,}058.84

Note the difference between the two: ₹40,058.84 is the amount, and the interest is what's left after subtracting the principal.

Solved example: population growth and decline

Growth at a steady rate works like compound interest. A town of 1,25,000 growing at 2% a year reaches $125000 \times (1.02)^3 = 1{,}32{,}651$ after 3 years. A town of 1,20,000 that grows 6% and then shrinks 5% ends at $120000 \times 1.06 \times 0.95 = 1{,}20{,}840$ : the changes multiply, they don't simply add to +1%.