Compound Interest and Growth Comparisons

Lesson 13 of 1825 minPDF notesFree

What you will learn

Compound interest is a repeated-percentage model. At the end of each stated compounding period, the interest is added to the amount. That new amount becomes the base for the next period. The important question is therefore: “What amount earns interest in this period?”

You will build this model year by year, match the rate to the period, compare it fairly with simple interest, and use the same multipliers for hypothetical growth, depreciation and simple reverse questions. Read this lesson in three parts. Pause at each checkpoint before continuing. You do not need logarithms.

These are original learning exercises, not official past questions or predictions of exam frequency. All monetary situations are mathematical models with no additional deposits, withdrawals, repayments, fees or taxes. A stated model does not describe every real financial arrangement.

Before you start

Try these without looking ahead:

What is 10% of 4,500?
What multiplier represents an 8% increase?
How many six-month intervals are in 18 months?
What is the positive square root of 1.21?

Check: 450; 1.08; 3 intervals; 1.1. If percentage multipliers or exact decimals are difficult, revisit Percentages and Successive Changes or Fractions Decimals and Exact Calculation. If the last answer is unfamiliar, revisit Squares Square Roots and Sensible Estimates. The existing Simple Interest lesson explains the fixed-original-principal model.

Part 1 What changes each period

Start with a period ledger

The original principal is P. An amount at the end of a period includes the principal and all interest accumulated so far. Under compound interest, that amount becomes the next period’s opening principal. Under simple interest, the interest base remains the original P.

Example 1 Annual compound interest beside simple interest

Start with ₹4,000 at 10% per year for three full years, compounded annually. The rate is 10% for each one-year period.

Compound-interest ledger:

Year 1: opening base ₹4,000; interest = 4000 × 10/100 = ₹400; closing amount = ₹4,400
Year 2: opening base ₹4,400; interest = 4400 × 10/100 = ₹440; closing amount = ₹4,840
Year 3: opening base ₹4,840; interest = 4840 × 10/100 = ₹484; closing amount = ₹5,324

The second year’s interest is ₹440. Total interest after two years is ₹840. These are different quantities. After three years, total CI = 400 + 440 + 484 = ₹1,324. Check: final amount minus original principal = 5324 − 4000 = ₹1,324.

For the matching simple-interest calculation, use the same original principal ₹4,000, annual rate 10% and three-year duration. The interest is ₹400 each year. The amounts are ₹4,400, ₹4,800 and ₹5,200. Three-year SI = ₹1,200. Therefore annual CI exceeds SI by ₹124 after three years. The first-year amounts agree; the later interest bases do not.

Derive the formula from one step

Let i be the rate per compounding period written as a fraction or decimal. For example, 10% per period means i = 10/100 = 0.10.

One period changes any opening amount B to B + Bi = B(1+i). Thus:

After one period: A₁ = P(1+i)
After two periods: A₂ = P(1+i)(1+i) = P(1+i)^2
After N full periods at the same period rate: A = P(1+i)^N
Total compound interest: CI = A − P

The power counts repeated full periods. It is not automatically the number of calendar years. Work with the period rate and period count together. For zero periods, A=P; for a zero rate, the amount remains P after every period.

Why the two-year difference works

Use exactly two years, annual compounding, the same original principal P and the same constant annual percentage rate R in both SI and CI. Put i = R/100.

Both models earn Pi in year one. In year two, SI again earns Pi. CI earns interest on P+Pi, so its second-year interest is (P+Pi)i = Pi + Pi². The extra interest is therefore Pi²:

CI − SI = P(R/100)^2, for these two annual periods only.

In Example 1, the two-year difference is 4000 × (0.10)^2 = ₹40, agreeing with 4840 − 4800. After three years the difference was ₹124, so the two-year result cannot simply be reused.

More generally, compare the same P over the same N full periods with the same nonnegative rate i per period. SI = PNi. If i>0 and P>0, SI and CI agree for zero or one period, and CI is larger for N≥2. At i=0 they always agree. Do not claim that CI is always larger when the two questions have different rates, durations or compounding conditions.

Example 2 When the annual rate changes

₹6,000 is compounded annually at 5% in the first year and 8% in the second. Use the stated rate for each year:

End of year 1: 6000 × 1.05 = ₹6,300
End of year 2: 6300 × 1.08 = ₹6,804
Total CI = 6804 − 6000 = ₹804

Equivalently, A = 6000 × 1.05 × 1.08. The two rates act on different bases. The overall increase is 804/6000 × 100 = 13.4%, not 5%+8%=13%.

If the question instead specified simple interest on the unchanged original principal at those yearly rates, SI would be 6000 × (0.05+0.08) = ₹780. The CI result is ₹24 larger. Do not change the model just because the rate changes.

Pause and check the base

At 20% per year, compounded annually, ₹2,500 becomes ₹3,000 after one year. Before calculating, name the base for year two. Then find the second-year interest and compare the two-year amount with simple interest at the same rate.

Check: the second-year base is ₹3,000, so that year’s interest is ₹600 and the CI amount is ₹3,600. SI earns ₹500 each year and gives ₹3,500. The difference is ₹100. Explain why using ₹2,500 again would switch to the simple-interest model.

Part 2 Match the rate and interval

Name the convention before dividing a rate

A rate quoted per year and a compounding interval are separate pieces of information. In the subannual examples below, the annual rate is explicitly nominal: the question defines it to be divided equally among the stated compounding periods in one year. This is the convention used for those exercises, not a rule to impose on every unexplained annual quote.

Write these three things before calculating:

Length of one compounding period
Percentage rate for that one period
Number of full periods in the stated duration

Let R be the nominal annual percentage number, m the number of equal compounding periods per year, and T the duration in years. Then, under the stated convention:

Annual: m=1; percentage per period j=R; period count N=T
Half-yearly: m=2; j=R/2; N=2 T
Quarterly: m=4; j=R/4; N=4 T

These substitutions here require N to be a whole number. The fractional rate used in the formula is i=j/100, so A=P(1+i)^N. For example, R=12 means 12%, while i=0.06 means 6% per period. Do not substitute 12 or 6 as if either were already a decimal fraction.

If the rate is already stated as 3% per quarter, use i=0.03 for each quarter. There is no annual rate to divide by four again.

Example 3 Half-yearly compounding over eighteen months

Find the amount and CI on ₹12,500 for 18 months at a nominal annual rate of 12%, compounded half-yearly. Here “nominal annual” means divide 12% equally between the two half-years.

Setup: one period = 6 months; j=12/2=6% per period; N=18/6=3 full periods. Hence i=0.06.

Timeline:

Start: ₹12,500
After 6 months: 12500 × 1.06 = ₹13,250
After 12 months: 13250 × 1.06 = ₹14,045
After 18 months: 14045 × 1.06 = ₹14,887.70

A = 12500 × (1.06)^3 = ₹14,887.70. CI = 14887.70 − 12500 = ₹2,387.70. Check: the interests ₹750, ₹795 and ₹842.70 sum to ₹2,387.70.

Using N=1.5 with the half-year rate would count too few periods. Using 12% for each half-year would apply the whole nominal annual quote twice per year.

Example 4 Quarterly compounding over nine months

Find the amount and CI on ₹10,000 at a nominal annual rate of 8%, compounded quarterly for 9 months. Divide the annual quote equally among four quarters.

Setup: one period = 3 months; j=8/4=2%; N=9/3=3. Thus: A = 10000 × (1.02)^3 = ₹10,612.08 CI = 10612.08 − 10000 = ₹612.08

Check the amounts: ₹10,200, ₹10,404 and ₹10,612.08. Matching SI for 9 months at 8% per year is ₹600, so CI is ₹12.08 larger.

If the question had directly stated “2% per quarter, compounded quarterly”, j would already be 2%. The result would be identical; dividing 2 by four again would be wrong.

Example 5 The annual quote and the one-year increase

Compare ₹7,500 for one year at the same nominal annual 12% under annual and half-yearly compounding. The half-year rate is explicitly 12/2=6%.

Annual amount = 7500 × 1.12 = ₹8,400. Half-yearly amount = 7500 × (1.06)^2 = ₹8,427. Difference = 8427 − 8400 = ₹27.

For half-yearly compounding, the one-year percentage increase is (8427−7500)/7500 × 100 = 12.36%. The stated nominal annual quote remains 12%. This difference comes from the first half-year’s interest earning interest in the second half-year. We are comparing two specified mathematical models, not recommending a product.

An incomplete final period needs a rule

Annual compounding alone does not specify what happens after 18 months, because only one full annual period has ended. Here is a counterexample using the same positive principal ₹1,000 and annual rate 10%:

Explicit rule A: after the first annual compounding, calculate proportional simple interest on the new amount for the remaining six months. The first-year amount is ₹1,100; extra interest is 1100 × 0.10 × 1/2 = ₹55; final amount is ₹1,155
Explicit rule B: no further interest is credited until the next complete annual period ends. At 18 months the credited amount is ₹1,100

The answers differ because the added rules differ. Neither is an automatic default. A fractional-power growth rule would also be an additional assumption; “annual compounding” by itself does not select it. If the question leaves the incomplete-period rule unstated, identify the missing information instead of inventing a rule. This issue does not arise in Example 3, where 18 months contains exactly three stated half-year periods.

Pause and check the period

For 9 months at a nominal annual 16%, compounded quarterly, state j and N without calculating an amount. Then do the same for a quote of 4% per quarter over 9 months.

Check: the nominal quote gives j=16/4=4% and N=9/3=3. The per-quarter quote directly gives j=4% and N=3. Both use i=0.04 three times. Explain why the second quote must not be divided by four again.

Part 3 Transfer the multiplier and work backwards

Growth and depreciation are stated models

When a quantity grows by g% of its current value every stated period, each step multiplies by 1+g/100. When a value decreases by d% of its current value, each step multiplies by 1−d/100. These factors describe the question’s assumption; they are not a prediction that real populations, machines or prices will follow an exact constant rate.

Example 6 Repeated growth

A hypothetical count starts at 10,000 and grows by exactly 5% of its current count at each of two annual model steps.

After the first step: 10000 × 1.05 = 10,500. After the second: 10500 × 1.05 = 11,025. So final count = 10000 × (1.05)^2 = 11,025, and total increase = 1,025.

Check: the second increase is 5% of 10,500, or 525. Adding two increases of 500 would keep the old base and give 11,000 instead.

Example 7 Repeated depreciation

In a hypothetical model, a machine worth ₹32,000 loses 25% of its current value at each year-end for two years. It retains 75%, so the factor is 0.75.

After year 1: 32000 × 0.75 = ₹24,000. After year 2: 24000 × 0.75 = ₹18,000. Total depreciation = 32000 − 18000 = ₹14,000.

The decreases are ₹8,000 and ₹6,000, not ₹8,000 twice. A 25% decrease followed by a 25% increase would not undo the loss: the factors would be 0.75×1.25, not 1.

For ordinary nonnegative-value depreciation, use 0≤d≤100. If d=100 (a 100% decrease), the value becomes zero and stays zero under further proportional decreases. A final zero then cannot reveal a unique starting value. Reverse depreciation therefore requires d<100 and enough information about the periods and rates. For changing stated rates, multiply each period’s own factor rather than averaging the rates.

Example 8 Recover the original principal

An amount is ₹10,890 after two full years at 10% compounded annually. Find the original principal.

Forward model: 10890 = P × (1.10)^2. Reverse both multipliers: P = 10890/(1.10)^2 = 10890/1.21 = ₹9,000.

Check forward: 9000 → 9900 → 10890. Subtracting 21% of the final ₹10,890 would use the wrong base; the 21% increase was measured against the unknown original principal.

Example 9 Recover a simple annual rate

₹1,600 becomes ₹1,936 after two annual compounding periods at a constant rate that is zero or positive. Let i be its fractional annual rate.

(1+i)^2 = 1936/1600 = 1.21 = (1.1)^2. Because the growth factor 1+i is positive, take the positive square root: 1+i=1.1. Thus i=0.1 and the annual rate is 100×0.1=10%.

Check: 1600 × 1.1 × 1.1 = 1936. The overall two-year increase is 21%; dividing it by two would give 10.5%, which ignores the changing base. The value 110% describes the amount factor, not the interest rate.

Example 10 Recover the number of whole periods

₹5,120 becomes ₹10,000 at 25% per year, compounded annually. The duration is a whole number of years.

The yearly factor is 1.25=5/4. The amount ratio is 10000/5120=125/64=(5/4)^3. Therefore N=3 annual periods, or 3 years.

Check with a short ledger: 5120 → 6400 → 8000 → 10000. This uses a small exact power, not a simple-interest doubling-time rule. At a positive period rate and positive principal, the amounts increase each period, so only one whole-period count can give that amount.

If the period rate is zero, A=P for every whole N. Then A=P does not determine a unique time, and A≠P has no solution under that zero-rate model. We will not solve arbitrary reverse-time questions with logarithms here.

Pause and check a reverse calculation

An annual-CI amount is ₹5,808 after two years at 10%. Is the original principal found by subtracting 21% of ₹5,808 or by dividing by 1.21?

Check: divide by 1.21 to get ₹4,800; then 4800 × 1.21 = 5808. Also explain why a value reduced by 100% cannot be uniquely reconstructed from the resulting zero.

Final method check

Before choosing an option:

Name the model: original-base SI, compounding, stated growth or stated depreciation
Separate principal, final amount, total interest and interest for one particular period
Write the percentage per period and count the full periods
Divide an annual quote only under an explicit nominal-rate convention; never divide an already per-period quote again
State any missing incomplete-period rule instead of guessing
Keep exact arithmetic through the calculation, then follow any stated final rounding instruction
For a reverse problem, put the answer back into the forward model

Independent practice

Try all 16 questions before reading the explained answers. This is free, untimed learning practice: +1 for a correct answer, 0 for an incorrect or unattempted answer. There is no pass cutoff and no negative marking. It is not a full RRB CBT mock.

1. ₹3,000 earns compound interest at 10% per year, compounded annually. What is the principal used to calculate only the second year’s interest?

A. ₹3,000 B. ₹3,300 C. ₹300 D. ₹3,630

2. Find the amount on ₹6,400 after two full years at 12.5% per year, compounded annually.

A. ₹8,000 B. ₹1,700 C. ₹7,200 D. ₹8,100

3. Find the total compound interest on ₹7,500 for three full years at 4% per year, compounded annually.

A. ₹936.48 B. ₹900 C. ₹612 D. ₹8,436.48

4. On the same ₹18,000 at a fixed 5% per year for exactly two years, find CI minus SI. For CI, compounding is annual.

A. ₹900 B. ₹1,800 C. ₹45 D. ₹137.25

5. ₹8,000 is compounded annually at 5% in the first year and 10% in the second year. Find the final amount.

A. ₹9,200 B. ₹9,240 C. ₹9,245 D. ₹1,240

6. A nominal annual rate of 8% is divided equally between two half-years. Interest is compounded half-yearly for 18 months. Which pair gives the rate per compounding period and the number of full periods?

A. 4% per half-year; 3 periods B. 8% per half-year; 3 periods C. 4% per half-year; 1.5 periods D. 4% per half-year; 6 periods

7. Find CI on ₹15,000 for one year at a nominal annual 8%, compounded half-yearly. The nominal annual rate is divided equally between two half-years.

A. ₹1,200 B. ₹600 C. ₹16,224 D. ₹1,224

8. Find the amount on ₹20,000 for six months at a nominal annual 12%, compounded quarterly. Divide the nominal annual rate equally among four quarters.

A. ₹20,600 B. ₹21,200 C. ₹21,218 D. ₹1,218

9. The rate is explicitly 3% per quarter, compounded quarterly. Which calculation correctly gives the amount on ₹10,000 after six months?

A. ₹10,000 × (1+0.03/4)^2 B. ₹10,000 × (1.03)^2 = ₹10,609 C. ₹10,000 × (1+0.03×2) = ₹10,600 D. ₹10,000 × 1.03 = ₹10,300

10. For ₹10,000 over one year at the same nominal annual 10%, how much does the half-yearly-compounded amount exceed the annually-compounded amount? For half-years, divide the nominal annual rate equally by 2.

A. ₹0 B. ₹500 C. ₹1,025 D. ₹25

11. A hypothetical count starts at 8,000 and grows by exactly 10% of its current count at each of two annual model steps. Find the final count.

A. 9,680 B. 9,600 C. 1,680 D. 8,800

12. A hypothetical value of ₹48,000 falls by 25% of its current value at each year-end for two years. Find the remaining value.

A. ₹24,000 B. ₹75,000 C. ₹27,000 D. ₹21,000

13. An amount is ₹8,470 after two full years at 10% per year, compounded annually. Find the original principal.

A. ₹6,691.30 B. ₹7,000 C. ₹7,700 D. ₹8,470

14. ₹3,600 becomes ₹5,184 after two annual compounding periods at a constant rate that is zero or positive. Find the annual percentage rate.

A. 20% B. 44% C. 22% D. 120%

15. ₹6,400 becomes ₹10,000 at 25% per year, compounded annually. The duration is a whole number of years. Find it.

A. 1 year B. 2.25 years C. 4 years D. 2 years

16. A question supplies a positive principal and a positive annual rate, says interest is compounded annually, and asks for the amount after 18 months. It gives no rule for the final six months. Which statement is justified?

A. Automatically divide the annual rate by 2 and use three half-years. B. Automatically raise the annual multiplier to the power 1.5. C. The treatment of the final incomplete period must be specified for a unique amount. D. Automatically use simple interest on the original principal for all 18 months.

Explained answers

1. Answer B

After year 1, amount = 3000 + 3000×0.10 = ₹3,300. This closing amount is the opening principal for year 2. Check: second-year interest would be 3300×0.10=₹330, not ₹300.

Check every option:

A: ₹3,000 keeps the original SI base.
B: ₹3,300 includes the first-year interest and is the required base.
C: ₹300 is only first-year interest.
D: ₹3,630 is the amount after year 2, not its opening base.

2. Answer D

12.5%=1/8, so each annual factor is 9/8. A=6400×(9/8)^2=₹8,100. Check the ledger: 6400→7200→8100. Total CI is ₹1,700, but amount is asked.

Check every option:

A: ₹8,000 is the two-year SI amount.
B: ₹1,700 is CI alone.
C: ₹7,200 stops after one year.
D: ₹8,100 applies the factor for both years.

3. Answer A

The annual factor is 1.04. Amounts are ₹7,800, ₹8,112 and ₹8,436.48. Hence CI=8436.48−7500=₹936.48. Check: 300+312+324.48=936.48.

Check every option:

A: ₹936.48 is the sum of all three period interests.
B: ₹900 uses the original principal for all three years, giving SI.
C: ₹612 omits the third year.
D: ₹8,436.48 is amount, not interest.

4. Answer C

For two matched annual periods, extra CI is interest on the first year’s interest: 18000×0.05×0.05=₹45. Check: SI=₹1,800; CI=18000×1.05^2−18000=₹1,845; difference=₹45.

Check every option:

A: ₹900 is one year’s interest.
B: ₹1,800 is the full two-year SI.
C: ₹45 is the requested difference.
D: ₹137.25 is the three-year CI−SI difference for these values; the question gives two years.

5. Answer B

A=8000×1.05×1.10=₹9,240. The second rate applies to ₹8,400. Check: interest ₹400+₹840=₹1,240, so amount is ₹9,240.

Check every option:

A: ₹9,200 adds 5% and 10% on the original base.
B: ₹9,240 uses the stated rates in order.
C: ₹9,245 uses the average 7.5% twice, which is not the stated sequence.
D: ₹1,240 is CI, not final amount.

6. Answer A

Under the stated nominal convention, j=8/2=4% per half-year. One period lasts six months, so N=18/6=3. Check: 3×6=18 months. Use i=0.04 three times.

Check every option:

A: Both rate and count match half-years.
B: The annual rate has not been divided by 2.
C: 1.5 is time in years, not the half-year period count.
D: Six half-years would last 36 months, not 18.

7. Answer D

Period rate=4%; N=2. A=15000×1.04^2=₹16,224; CI=₹1,224. Check: first-half interest ₹600; second-half interest 15600×0.04=₹624; total ₹1,224.

Check every option:

A: ₹1,200 is the annual-compounding result over one year, or the matching SI.
B: ₹600 is only the first half-year’s interest.
C: ₹16,224 is amount.
D: ₹1,224 includes both half-year interests.

8. Answer C

j=12/4=3% per quarter and N=6/3=2. A=20000×1.03^2=₹21,218. Check: 20000→20600→21218; interest totals ₹600+₹618=₹1,218.

Check every option:

A: ₹20,600 stops after one quarter.
B: ₹21,200 adds two lots of 3% of the original principal.
C: ₹21,218 compounds for both quarters.
D: ₹1,218 is interest only.

9. Answer B

The quote already gives the rate for one quarter. Use i=0.03, not 0.03/4. Six months contains two quarters, so A=10000×1.03^2=₹10,609. Check: 10000→10300→10609.

Check every option:

A: Divides an already per-quarter rate by 4 again.
B: Uses 3% for each of the two quarters.
C: Uses the original base twice, giving SI-like addition.
D: Counts only one quarter.

10. Answer D

Annual amount=10000×1.10=₹11,000. Half-yearly amount=10000×1.05^2=₹11,025. Difference=₹25. Check: the first-half interest ₹500 earns an extra 5%, or ₹25, in the second half.

Check every option:

A: Same nominal quote does not force the same final amount.
B: ₹500 is first-half interest.
C: ₹1,025 is all half-yearly CI, not the difference.
D: ₹25 is the extra interest on first-half interest.

11. Answer A

Use the growth factor 1.10 twice: 8000×1.10^2=9,680. Check: first increase 800; second increase 880; total increase 1,680, giving 9,680.

Check every option:

A: 9,680 updates the base after the first step.
B: 9,600 adds 20% of the original count.
C: 1,680 is the increase, not the final count.
D: 8,800 stops after the first step.

12. Answer C

Each year 75% remains, so factor=0.75. Final value=48000×0.75^2=₹27,000. Check: 48000→36000→27000. Total depreciation is ₹21,000.

Check every option:

A: ₹24,000 subtracts 25% of the original value twice.
B: ₹75,000 uses two 25% increases.
C: ₹27,000 uses two current-value decreases.
D: ₹21,000 is the loss in value, not what remains.

13. Answer B

A=P×1.10^2, so P=8470/1.21=₹7,000. Check forward: 7000→7700→8470. Reverse multiplication by division, not by subtracting 21% of the final amount.

Check every option:

A: ₹6,691.30 subtracts 21% of the final amount.
B: ₹7,000 reverses both annual factors.
C: ₹7,700 reverses only the last factor.
D: ₹8,470 is already the final amount.

14. Answer A

(1+i)^2=5184/3600=1.44. The positive square root gives 1+i=1.2, so i=0.2 and the annual rate is 20%. Check: 3600×1.2×1.2=5184.

Check every option:

A: 20% produces the required annual factor 1.2.
B: 44% is the whole two-year increase.
C: 22% halves the total increase as if the base stayed fixed.
D: 120% expresses the amount factor, not the interest rate.

15. Answer D

Annual factor=1.25=5/4. The ratio 10000/6400=25/16=(5/4)^2, so N=2 years. Check: 6400→8000→10000. With positive rate these amounts increase, so the whole-year answer is unique.

Check every option:

A: One year gives ₹8,000, not ₹10,000.
B: 2.25 years comes from treating the ₹3,600 increase as SI at ₹1,600 per year.
C: Four years is the SI doubling time at 25%, which solves a different problem.
D: Two annual factors give ₹10,000.

16. Answer C

Only one full annual period ends in 18 months. Annual compounding does not state what happens in the remaining six months. For example, on ₹1,000 at 10%, an explicitly allowed simple stub on the ₹1,100 year-end amount gives ₹1,155, while an explicit rule of no further credit before the next full year gives ₹1,100. Different added rules give different results; neither may be silently assumed.

Check every option:

A: Changes the compounding frequency without permission.
B: Adds an unstated fractional-power rule.
C: Correctly identifies the missing convention.
D: Changes the given compound-interest model to simple interest.

Sources for further reading

The textbooks support the mathematical models. The RRB list names Simple and Compound Interest and is illustrative, not exhaustive; it does not prescribe these lessons or question counts. The explanations and exercises here are original. The explicit nominal-rate and incomplete-period statements are part of each model, not current financial advice. No recruitment dates, eligibility or logistics are taught here.

Analogy

Imagine a 100 mm line enlarged by 10% of its current length at each pass through a mathematical resizing rule. After one pass it is 110 mm; after the next it is 121 mm, because 10% is now taken from 110. By contrast, adding a fixed 10 mm each pass gives 110 mm then 120 mm. Compound interest resembles the repeated multiplier; fixed-rate simple interest resembles the fixed addition calculated from the original base. The comparison explains the arithmetic only. The actual interest question must still specify its rate, period and model.

Quick reference

Compound interest quick reference

Use positive principal P>0 and a nonnegative whole-period count N. P is the original principal; A is final amount; total CI=A−P. Interest for one period is not the same as total CI
Each period changes its opening base B to B(1+i), where i is that period’s fractional rate. A fixed period rate over N full periods gives A=P(1+i)^N
A rate of j% per period means i=j/100. For example, j=6 gives i=0.06
Under an explicitly stated nominal annual convention only: m equal periods per year give j=R/m and N=mT, with R the annual percentage number and T in years. Here N must be a whole number. Annual m=1; half-yearly m=2; quarterly m=4
If the rate is already quoted per half-year or quarter, use it directly for that period. Do not divide it by m again
For changing stated period rates, multiply the separate factors: A=P(1+i₁)(1+i₂)…; do not add or average the rates
Match P, period rate and N when comparing SI with CI. SI=PNi. For P>0 and i>0, the models agree at N=0 or 1, and CI>SI for whole N≥2. At i=0 they agree for all N
Exactly two annual periods, same P and constant annual percentage rate R: CI−SI=P(R/100)^2. Do not use this two-year rule for another duration or compounding frequency
Growth by g% of current value uses factor 1+g/100. Depreciation by d% uses 1−d/100, with 0≤d≤100 for ordinary nonnegative values. Reverse depreciation requires d<100
Reverse principal: divide A by the complete multiplier. For fixed i, P=A/(1+i)^N. Never subtract the total growth percentage from the final amount to undo it
For two equal-rate periods, positive factor 1+i=√(A/P). Then rate%=100 i. Use the stated valid domain; do not confuse the factor with the interest rate
For a simple whole-period reverse-time question, recognise a small exact power or build a short ledger, then check forward. At zero rate, A=P gives no unique time; A≠P is impossible under that model
An incomplete final period needs an explicit rule. Do not silently assume fractional powers, simple interest for the remainder or a different compounding frequency
Keep exact arithmetic until the final rounding instruction. Distinguish money, percentage, interval and period count. These formulas assume the stated model with no additional cashflows, fees or taxes

Notes for this lesson

Compound Interest and Growth Comparisons: Bilingual Lesson and Practice33 pages

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