8.5 Growth and Compounding

Banks pay interest on money deposited, usually quoted “p.a.” (per annum, per year). The sum deposited is the principal p, and the rate r is the interest earned per year.

Two Kinds of Growth - Without compounding (simple): each year’s interest is paid out and pocketed, so the principal never grows. After t years, A = p + p\,r\,t = p(1 + rt). - With compounding: each year’s interest is folded back in, so the principal swells. After t years, A = p\,(1+r)^{t}.

Worked Example

Deposit ₹25,000 at 10% p.a. for 2 years. Compare the two schemes.

Here p = 25000, r = 0.10, t = 2.

  • Simple: 25000\,(1 + 0.10\times 2) = 25000 \times 1.2 = ₹30{,}000, a gain of ₹5000 (20%).
  • Compound: 25000\,(1.1)^2 = 25000 \times 1.21 = ₹30{,}250, a gain of ₹5250 (21%).

Compounding earns more, ₹250 extra, because you collect interest on the interest.

Worked Example: interest on interest, year by year

A ₹8000 deposit grows at 12% p.a., compounded annually, for 2 years. Track it.

  • After Year 1: 8000 \times 1.12 = ₹8960.
  • After Year 2: the new principal ₹8960 grows again: 8960 \times 1.12 = ₹10{,}035.20.

The formula agrees in one line: 8000\,(1.12)^2 = 8000 \times 1.2544 = ₹10{,}035.20. Under simple interest the same deposit would have reached only 8000(1 + 0.12\times 2) = ₹9920, so compounding has handed over an extra ₹115.20, and that gap widens fast as the years pile up.

Try This: a compounding explorer Set a principal, rate and number of years, then watch simple growth race against compound growth.

Decline (depreciation)

Things also bleed value over time. A fall of r each period multiplies the value by (1-r) each time.

Worked Example

A ₹32,000 laptop depreciates 8% in a year. Find its value.

It holds onto 92\% of its worth: 0.92 \times 32000 = ₹29{,}440.

A small town of 1600 residents shrinks 12% per decade. Population after 3 decades? 1600 \times (0.88)^3 = 1600 \times 0.681472 = 1090.36 \approx \mathbf{1090\ \text{people}}.

Figure it Out: Growth and Compounding

Practice

  1. A bank offers 10% p.a. Compare ₹25,000 for 2 years, with and without annual compounding.
  2. Ishaan invests p for 4 years at 6% p.a. without compounding. Which expression(s) give the total?   (i) p\times 6\times 4   (iii) p\times\tfrac{0.6}{100}\times 4   (iv) p\times\tfrac{0.06}{100}\times 4   (vi) p\times 1.06\times 4   (vii) p + (p\times 0.06\times 4).
  3. A post office offers 8% p.a. on ₹60,000 for 3 years. How much interest without compounding, and how much more with it?
  4. A city’s population climbs about 4% a year. If it is 2 crore now, estimate it after 3 years.
  5. Bacteria grow 3% per hour. Starting at 4,80,000, how many after 2 hours?
  1. Without: 25000(1 + 0.1\times 2) = ₹30{,}000.   With: 25000(1.1)^2 = ₹30{,}250. Compounding gives ₹250 more.
  2. The valid expression is (vii) p + (p\times 0.06\times 4), which equals p(1 + 0.24) = 1.24p. (Each of (i), (iii), (iv) and (vi) is wrong: 6 and 1.06 are mis-scaled, and \tfrac{0.06}{100} divides by 100 a second time.)
  3. Simple interest = 60000\times 0.08\times 3 = ₹14{,}400. Compound amount = 60000(1.08)^3 \approx ₹75{,}582.72, i.e. interest \approx ₹15{,}582.72, about ₹1182.72 more.
  4. 2\text{ cr}\times (1.04)^3 \approx 2\text{ cr}\times 1.1249 \approx \mathbf{2.25\ \text{crore}} (about 2,24,97,280 people).
  5. 480000\times (1.03)^2 = 480000\times 1.0609 \approx \mathbf{5{,}09{,}232} bacteria.