Compound interest

In com­pound inter­est, the inter­est of each year is added to the prin­ci­pal, and the next year's inter­est is worked out on that big­ger amount. So the money grows faster than with sim­ple inter­est. The same idea describes any­thing that grows or shrinks by a fixed per­cent­age each period: a town's pop­u­la­tion, a fac­to­ry's out­put, or bac­te­ria in a dish. In this les­son you will learn the com­pound inter­est for­mula and use it for all of these.

The for­mula#

Let the prin­ci­pal be $P$ and the rate be $R\%$ per annum. If the inter­est is com­pounded every year, the amount after $n$ years is

$$\displaystyle A = P\left(1 + \frac{R}{100}\right)^n$$

Why it works.

Amount after 1 year $\displaystyle = P + \frac{R}{100}P = P\left(1 + \frac{R}{100}\right)$.

In the sec­ond year, this whole amount earns inter­est:

$$\displaystyle \begin{aligned} \text{amount after 2 years} &= P\left(1 + \frac{R}{100}\right) + \frac{R}{100} \times P\left(1 + \frac{R}{100}\right) \\ &= P\left(1 + \frac{R}{100}\right)\left(1 + \frac{R}{100}\right) \\ &= P\left(1 + \frac{R}{100}\right)^2 \end{aligned}$$

Each year mul­ti­plies the amount by $\displaystyle \left(1 + \frac{R}{100}\right)$ once more. After $n$ years, it has been mul­ti­plied $n$ times.

The com­pound inter­est is the amount minus the prin­ci­pal:

$$\text{C.I.} = A - P$$

Exam­ple 1#

Find the com­pound inter­est on ₹12,600 for 2 years at 10% per annum.

$$\displaystyle A = 12600\left(1 + \frac{10}{100}\right)^2 = 12600 \times \frac{11}{10} \times \frac{11}{10} = 15246$$

$$\text{C.I.} = 15246 - 12600 = 2646$$

This is the same ₹2646 we found year by year in the last les­son. The sim­ple inter­est was only ₹2520.

Inter­est added more than once a year#

Some­times inter­est is added every half year, or every three months. Let the inter­est be added $k$ times a year. Then in each part of the year the rate is $\displaystyle \frac{R}{k}\%$, and in $n$ years it is added $nk$ times:

$$\displaystyle A = P\left(1 + \frac{R}{100k}\right)^{nk}$$

1. Com­pounded yearly: take $k = 1$.
2. Com­pounded half yearly: take $k = 2$.
3. Com­pounded quar­terly (every three months): take $k = 4$.

Exam­ple 2#

Find the amount on ₹10,000 for 1 year at 10% per annum, com­pounded half yearly.

Here $k = 2$ and $n = 1$, so the rate for each half year is 5%, and it is added twice.

$$\displaystyle A = 10000\left(1 + \frac{10}{200}\right)^2 = 10000 \times \frac{21}{20} \times \frac{21}{20} = 11025$$

The amount is ₹11,025. With yearly com­pound­ing it would be only ₹11,000.

Growth and decrease#

The same for­mula works for any quan­tity that changes by a fixed per­cent­age each year. For a decrease, use $\displaystyle \left(1 - \frac{R}{100}\right)$.

Exam­ple 3: A falling pop­u­la­tion#

The pop­u­la­tion of a town 2 years ago was 62,500. Because peo­ple moved to cities, it decreased by 4% each year. Find the present pop­u­la­tion.

$$\displaystyle \begin{aligned} A &= 62500\left(1 - \frac{4}{100}\right)^2 \\ &= 62500\left(1 - \frac{1}{25}\right)^2 \\ &= 62500 \times \frac{24}{25} \times \frac{24}{25} \\ &= 57600 \end{aligned}$$

The present pop­u­la­tion is 57,600. (Here $25 \times 25 = 625$, and $62500 \div 625 = 100$, so the answer is $100 \times 24 \times 24$.)

Exam­ple 4: Going back in time#

The present pop­u­la­tion of a city is 92,61,000. It has been increas­ing at 5% per annum. Find its pop­u­la­tion 3 years ago.

Let the pop­u­la­tion 3 years ago be $P$. Then

$$\displaystyle \begin{aligned} P\left(1 + \frac{5}{100}\right)^3 &= 9261000 \\ P\left(\frac{21}{20}\right)^3 &= 9261000 \\ P &= 9261000 \times \frac{20}{21} \times \frac{20}{21} \times \frac{20}{21} \\ &= 8000000 \end{aligned}$$

The pop­u­la­tion 3 years ago was 80,00,000. (Note that $21 \times 21 \times 21 = 9261$.)

Exam­ple 5: Find­ing the rate#

The pro­duc­tion of scoot­ers in a fac­tory rose from 40,000 to 48,400 in 2 years. Find the rate of growth per annum.

$$\displaystyle \begin{aligned} 48400 &= 40000\left(1 + \frac{R}{100}\right)^2 \\ \left(1 + \frac{R}{100}\right)^2 &= \frac{48400}{40000} = \frac{121}{100} = \left(\frac{11}{10}\right)^2 \\ 1 + \frac{R}{100} &= \frac{11}{10} \\ \frac{R}{100} &= \frac{1}{10} \\ R &= 10 \end{aligned}$$

The rate of growth is 10% per annum.

Dif­fer­ent rates in dif­fer­ent years#

If the rate changes from year to year, mul­ti­ply by a dif­fer­ent fac­tor for each year:

$$\displaystyle A = P\left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)$$

For a decrease in some year, that fac­tor has a minus sign.

Exam­ple 6: Bac­te­ria#

Bac­te­ria in a dish grow by 10% in the first hour, decrease by 10% in the sec­ond hour, and grow by 10% again in the third hour. At the start there are 10,000 bac­te­ria. How many are there after 3 hours?

$$\displaystyle \begin{aligned} & 10000\left(1 + \frac{10}{100}\right)\left(1 - \frac{10}{100}\right)\left(1 + \frac{10}{100}\right) \\ &= 10000 \times \frac{110}{100} \times \frac{90}{100} \times \frac{110}{100} \\ &= 10890 \end{aligned}$$

There are 10,890 bac­te­ria. A rise of 10% and a fall of 10% do not can­cel out.

Exam­ple 7: Three dif­fer­ent growth rates#

The pop­u­la­tion of a town was 1,60,000 three years ago. It grew by 3%, 2.5% and 5% in the last three years. Find the present pop­u­la­tion.

$$\displaystyle \begin{aligned} & 160000\left(1 + \frac{3}{100}\right)\left(1 + \frac{2.5}{100}\right)\left(1 + \frac{5}{100}\right) \\ &= 160000 \times \frac{103}{100} \times \frac{102.5}{100} \times \frac{105}{100} \\ &= 177366 \end{aligned}$$

The present pop­u­la­tion is 1,77,366.

Exam­ple 8: Profit and loss over three years#

Aashish opened a book­shop with ₹32,000. In the first year he made a loss of 5%. In the sec­ond year he made a profit of 10%, and in the third year the profit rose to $\displaystyle 12\frac{1}{2}\%$. Find his net profit over the three years.

Here $R_1 = -5$, $R_2 = 10$ and $\displaystyle R_3 = 12\frac{1}{2} = \frac{25}{2}$.

$$\displaystyle \begin{aligned} A &= 32000\left(1 - \frac{5}{100}\right)\left(1 + \frac{10}{100}\right)\left(1 + \frac{25}{200}\right) \\ &= 32000 \times \frac{95}{100} \times \frac{110}{100} \times \frac{225}{200} \\ &= 37620 \end{aligned}$$

Net profit $= 37620 - 32000 = 5620$ rupees.

Prac­tice#

1. Find the com­pound inter­est on ₹5000 for 2 years at 10% per annum.
2. Find the amount on ₹8000 for 3 years at 5% per annum, com­pounded yearly.
3. Find the com­pound inter­est on ₹16,000 for 1 year at 10% per annum, com­pounded half yearly.
4. A vil­lage has 25,000 peo­ple and grows by 4% a year. Find its pop­u­la­tion after 2 years.
5. A car worth ₹5,00,000 loses 10% of its value each year. Find its value after 2 years.
6. A town has 17,640 peo­ple now. It has grown by 5% a year. Find its pop­u­la­tion 2 years ago.
7. ₹2500 becomes ₹3025 in 2 years at com­pound inter­est. Find the rate.
8. Find the dif­fer­ence between the com­pound inter­est and the sim­ple inter­est on ₹6000 for 2 years at 10% per annum.
9. A busi­ness worth ₹40,000 grows by 10% in the first year and falls by 5% in the sec­ond year. Find its value after 2 years.
10. Find the amount on ₹10,000 for 6 months at 8% per annum, com­pounded quar­terly.

Answers#