Fractions in Disguise · Lesson 9 of 12
Growth and Compounding
“Derive simple and compound interest relationships and apply repeated percentage growth.”
• Explain principal, interest, amount and rate per year. • Compare interest paid out each year with interest added back each year. • Derive and apply formulas for growth with and without compounding. • Use a decimal rate consistently and match the rate period to the number of periods. • Apply repeated percentage growth to populations and other quantities.
Imagine placing ₹6000 in a model savings arrangement that pays 10% interest each year. At the end of the first year, the interest is ₹600. Now consider two choices. You could take that ₹600 away and leave the original ₹6000 in the arrangement. Or you could leave the interest there as well, so that the next year begins with ₹6600. The first year is identical in both choices, but the amounts used to calculate later interest are different.
This difference creates compounding. When earlier interest joins the principal, later interest is earned on a larger amount. Nothing mysterious is added: you simply apply the same percentage to the current balance rather than repeatedly to the original deposit.
We will use clear mathematical models with a fixed rate, equal time periods, no fees and no extra deposits or withdrawals. Real financial arrangements may use other conditions, so the calculations depend on the conditions stated in each problem.
The starting sum on which interest is calculated. Without compounding, the original principal stays the calculation base. With compounding, each period begins with the current balance.
The extra money earned on a deposit or charged on a loan under a stated arrangement.
A yearly rate. The abbreviation p.a. means per annum, or per year. A rate of 10% p.a. represents 10% over one year under the specified interest model.
The principal together with the interest included in the stated calculation. When interest is paid out separately, distinguish the balance still deposited from the total money received.
For ₹6000 at 10% over one year, the interest is 0.10 × ₹6000 = ₹600. If that interest is added to the deposit, the balance becomes ₹6600. This is 110% of the original, so multiplying by 1.10 finds the final balance directly.
There are two different multipliers here. Multiplying by 0.10 finds the interest alone. Multiplying by 1.10 finds principal plus interest. Confusing them would turn a final balance of ₹6600 into an answer of only ₹600.
Problem
Find the interest and final amount on ₹6000 at a stated 10% yearly rate for one year.
- 1.Convert the rate to a decimal: 10% = 0.10.
- 2.Interest = 6000 × 0.10 = ₹600.
- 3.Amount = principal + interest = ₹6000 + ₹600 = ₹6600.
- 4.Alternatively, amount = 6000 × 1.10 = ₹6600. The larger multiplier includes the original deposit.
No Compounding
Suppose the ₹600 interest is paid out at the end of each year and is not added back. The deposit used for the next year’s calculation remains ₹6000. The same 10% of the same ₹6000 produces ₹600 every year.
After three years, three interest payments total 3 × ₹600 = ₹1800. When the original principal is returned, the total received is ₹6000 + ₹1800 = ₹7800. The balance left in the deposit during the arrangement is still ₹6000; ₹7800 describes all payments received together.
Because the same amount of interest is earned in each equal period, total interest grows by equal additions. This is the simple-interest or non-compounding model.
Interest calculated on the same original principal for each period. With a fixed rate, each equal period earns or costs the same interest amount.
| Year | Principal used | Interest paid out | Principal left |
|---|---|---|---|
| First | ₹6000 | ₹600 | ₹6000 |
| Second | ₹6000 | ₹600 | ₹6000 |
| Third | ₹6000 | ₹600 | ₹6000 |
Let P be the original principal and r the rate written as a decimal per period. Interest for one period is Pr. For t equal periods, total interest is Pr repeated t times, or Prt. Adding the original principal gives P + Prt = P(1 + rt).
Here r = 0.10 for 10%, r = 0.06 for 6%, and r = 0.025 for 2.5%. If you prefer to keep a percentage number R such as 10, replace r by R/100. Both conventions work, but you must not mix them.
Problem
Find total interest and total received on ₹50,000 at 7% p.a. for three years without compounding.
- 1.P = ₹50,000, r = 0.07 per year and t = 3 years.
- 2.Yearly interest = 50,000 × 0.07 = ₹3500.
- 3.Total interest = 3500 × 3 = ₹10,500.
- 4.Total received = ₹50,000 + ₹10,500 = ₹60,500.
- 5.Check: the three-year interest is 21% of the original, and 21% of ₹50,000 is ₹10,500.
With Compounding
Now leave the first ₹600 interest in the original ₹6000 deposit. The second year begins with ₹6600. Ten per cent of ₹6600 is ₹660, so the second year ends with ₹7260. The interest has increased because its base has increased.
The third year begins with ₹7260. Its interest is ₹726, giving a final balance of ₹7986. At each year end, the current balance is multiplied by 1.10. The repeated multiplication is the defining pattern of compounding.
The rate did not rise: it stayed at 10%. The interest amounts rose from ₹600 to ₹660 to ₹726 because the quantities to which 10% applied became larger. That distinction is essential.
The process of adding each period’s interest or percentage growth to the current amount, so that the next period’s calculation uses the updated amount.
| Year | Starting balance | Interest added | Ending balance |
|---|---|---|---|
| First | ₹6000 | ₹600 | ₹6600 |
| Second | ₹6600 | ₹660 | ₹7260 |
| Third | ₹7260 | ₹726 | ₹7986 |
Problem
Find the amount and interest on ₹6000 at 10% p.a., compounded annually for three years.
- 1.After the first year: 6000 × 1.10 = ₹6600.
- 2.After the second year: 6600 × 1.10 = ₹7260.
- 3.After the third year: 7260 × 1.10 = ₹7986.
- 4.Total interest = ₹7986 − ₹6000 = ₹1986.
- 5.Without compounding the total was ₹7800. Compounding gives ₹7986 − ₹7800 = ₹186 more under these conditions.
To express the final totals as percentages of the initial deposit, compare each with ₹6000. Without compounding, 7800/6000 = 1.30, so the total received is 130% of the deposit and the gain is 30%. With compounding, 7986/6000 = 1.331, so the final amount is 133.1% of the deposit and the gain is 33.1%. The additional 3.1% of the original deposit equals ₹186.
The first period ends at P(1 + r). The second ends at P(1 + r)(1 + r). Continuing for t periods gives t copies of the same multiplier, written as P(1 + r)ᵗ. The exponent counts how many times the growth factor is applied.
The formula gives the final amount, not the interest alone. Subtract the original principal to find compound interest. For positive rates and more than one whole period, compounding earns more than simple interest at the same rate, principal and duration.
Problem
Compare ₹20,000 for two years at 10% p.a., with and without annual compounding.
- 1.Without compounding: interest = 20,000 × 0.10 × 2 = ₹4000. Total = ₹24,000.
- 2.With compounding: first-year balance = 20,000 × 1.10 = ₹22,000.
- 3.Second-year balance = 22,000 × 1.10 = ₹24,200, giving ₹4200 interest.
- 4.The extra ₹200 is 10% interest earned in the second year on the first year’s ₹2000 interest.
Problem
Compare ₹20,000 for four years at 5% p.a., with and without annual compounding.
- 1.Without compounding: I = 20,000 × 0.05 × 4 = ₹4000. Total = ₹24,000.
- 2.With compounding, successive balances are ₹21,000, ₹22,050, ₹23,152.50 and ₹24,310.125.
- 3.Round the final balance to paise: ₹24,310.13. Compound interest is approximately ₹4310.13.
- 4.The non-compounding total equals the previous two-year, 10% example because 0.05 × 4 = 0.10 × 2. Their compound totals differ because the growth factors and numbers of repetitions differ.
The same mathematics describes more than money. A population growing by 3% each year is multiplied by 1.03 each year in a constant-growth model. A laboratory count growing by 2.5% each hour is multiplied by 1.025 each hour.
Match the rate period to the exponent. A rate per hour with two hours uses t = 2. An annual rate with three years also uses t = 3, but the periods mean years. We are not changing the time unit without adjusting the model.
Problem
An initial count of 5,06,000 bacteria grows by 2.5% per hour. Find the model count after two hours.
- 1.The hourly multiplier is 1 + 0.025 = 1.025.
- 2.After one hour: 506,000 × 1.025 = 518,650.
- 3.After two hours: 518,650 × 1.025 = 531,616.25.
- 4.The model predicts approximately 531,616 bacteria after rounding to a whole count. Intermediate values are mathematical estimates, not fractional physical bacteria.
Problem
Giridhar borrows ₹12,500 at 12% p.a. without compounding for three years. Raghava borrows the same amount at 10% p.a., compounded annually for three years. Who pays more interest?
- 1.Giridhar’s interest = 12,500 × 0.12 × 3 = ₹4500.
- 2.Raghava’s amount = 12,500 × 1.10³ = 12,500 × 1.331 = ₹16,637.50.
- 3.Raghava’s interest = ₹16,637.50 − ₹12,500 = ₹4137.50.
- 4.Giridhar pays ₹4500 − ₹4137.50 = ₹362.50 more interest.
- 5.Compounding alone does not determine which arrangement costs more when rates differ. Calculate using all the stated conditions.
How long would ₹1000 take to double at 10% per year? Without compounding, ₹100 is added each year, so ten years are needed to add another ₹1000. With annual compounding, the balance after seven years is about ₹1948.72 and after eight years about ₹2143.59. It first exceeds double at the eighth year-end.
This shows the difference between linear growth through equal additions and exponential growth through equal multiplication factors. You can investigate the pattern with a table without needing any advanced equation-solving method. State that you are comparing balances at complete year-ends in the annual model.
Begin with 100 units. For five periods, build one column by adding 10 units each period and another by multiplying the previous value by 1.1. The additive column ends at 150; the compound column ends at 161.051. Explain why the difference is zero after the first period and then increases. Keep full intermediate values before rounding the final result.
Quiz
In A = P(1 + r)ᵗ, what value of r represents 6% per period?
Why does compound interest grow each year at a fixed positive rate?
What is the amount on ₹1000 after two years at 10% compounded annually?
Which expression gives simple interest on P for four years at 6% p.a.?
A count grows at 2.5% per hour for two hours. Which multiplier applies to the initial count?
Practice Problems
- Find simple interest and total amount on ₹8000 at 5% p.a. for three years. Solution: I = 8000 × 0.05 × 3 = ₹1200. A = ₹8000 + ₹1200 = ₹9200.
- Find compound amount and interest on ₹5000 at 10% p.a. for two years with annual compounding. Solution: After one year: ₹5500. After two years: 5500 × 1.1 = ₹6050. Interest = ₹6050 − ₹5000 = ₹1050.
- Compare interest on ₹50,000 at 7% p.a. for three years with and without annual compounding. Solution: Simple interest = ₹10,500. Compound amount = 50,000 × 1.07³ = ₹61,252.15, so compound interest = ₹11,252.15. Compounding gives ₹752.15 more interest.
- A model city population is initially 1.5 crore and grows 3% each year. Find its expected population after three years. Solution: Use the yearly factor 1.03 three times: 1.5 × 1.03³ = 1.6390905 crore. This equals 16,390,905 people in the model. Do not use 1.09, because each yearly increase has a different base.
- Compare ₹1000 growing at 10% annually with and without compounding. At which complete year-end does each first reach or exceed ₹2000? Solution: Without compounding, yearly interest is ₹100; ten years gives ₹2000. With compounding, 1000 × 1.1⁷ ≈ ₹1948.72 and 1000 × 1.1⁸ ≈ ₹2143.59. Thus the first qualifying year-end is the eighth. This comparison concerns complete annual periods.
Key Takeaways
• Simple interest repeatedly uses the original principal. • Compounding repeatedly uses an updated balance that includes earlier interest. • With decimal rate r, simple interest is Prt and total amount is P(1 + rt). • The compound amount after t equal periods is P(1 + r)ᵗ. • Subtract the original principal to obtain compound interest. • Match rate periods to the exponent and keep the rate convention consistent. • Equal additions and repeated multiplication create different growth patterns.