Simple and Compound Interest

Interest is the price of money over time, and almost every numerical reasoning section in a judiciary or CLAT-PG paper tests whether you can compute it quickly and correctly. Simple interest grows in a straight line; compound interest grows on a curve, because each period's interest is itself put to work. The examiner rarely asks you to plug into a formula blindly. Instead the question buries the variable you must isolate, mixes compounding frequencies, or hides a difference between the two interest regimes that collapses to a one-line shortcut. This chapter builds the full toolkit, from first-principles derivations to the speed formulae that turn a four-step calculation into a single multiplication, with worked examples in the style of UPPSC, Delhi Judicial Service and CLAT-PG aptitude papers.

What Interest Measures and Why It Splits Into Two Families

Interest is the charge a borrower pays a lender for the use of a sum of money, or equivalently the return a depositor earns, expressed as a rate over a unit of time. Three quantities anchor every problem: the principal (P), the original sum lent or invested; the rate (R), almost always quoted per annum unless stated otherwise; and the time (T or n), the duration the money is outstanding. The fourth quantity, the interest itself, is what the formula produces, and the amount (A) is simply principal plus interest.

The split between the two families turns on one question: does interest earned in earlier periods itself earn interest? If it does not, the interest each period is computed only on the original principal, and we call it simple interest. If it does, the interest is added back to the principal at the end of each period and the next period's interest is computed on the larger base, which we call compound interest. This single difference produces a straight line in the first case and an exponential curve in the second, and the gap between the two lines is itself a favourite examiner target. Because the two regimes share the same inputs, a great deal of exam difficulty comes from converting cleanly between them, a skill that rewards the same fluency with fractions and ratios you built in percentage, ratio and proportion.

The Simple Interest Formula and Its Rearrangements

Simple interest is computed on the principal alone for the whole term. The governing relation is SI = (P × R × T) / 100, where R is the rate percent per annum and T is the time in years. The amount is A = P + SI = P(1 + RT/100). Because the formula is a product of three terms over a constant, you can solve for any one unknown by rearranging: P = 100·SI/(R·T), R = 100·SI/(P·T), and T = 100·SI/(P·R). Examiners exploit this symmetry by giving you the interest and two of the three inputs and asking for the third.

A worked example: a sum of Rs 12,000 is lent at 8% per annum simple interest for 3 years. Then SI = (12000 × 8 × 3)/100 = Rs 2,880, and the amount repayable is Rs 14,880. Reverse the question: if Rs 2,880 interest accrued on Rs 12,000 over 3 years, the rate is R = (100 × 2880)/(12000 × 3) = 8%. The most common trap is a time period quoted in months or days; always convert to years (for example 9 months is 9/12 = 3/4 year) before substituting. Note also that under simple interest the money grows by the same rupee amount every year, so the interest for any single year equals SI/T, a fact that lets you interpolate the principal from two amounts at different times. This proportional reasoning is the same machinery you use in profit, loss and discount.

Recovering Principal and Rate From Two Amounts

A recurring simple-interest format gives the amount at two different times and asks for the principal or the rate. Because simple interest adds a constant slice each year, the difference between the two amounts is pure interest for the intervening years. Suppose a sum amounts to Rs 9,800 in 5 years and to Rs 12,005 in 8 years under simple interest. The extra Rs 2,205 accrued over the 3 intervening years, so one year's interest is Rs 735, and over 5 years the interest is Rs 3,675. Subtracting that from the 5-year amount gives the principal: 9800 − 3675 = Rs 6,125. The rate then follows from R = (100 × 735)/6125 = 12% per annum.

This subtraction method is faster and safer than building two simultaneous equations, and it generalises: whenever a problem provides two amounts at two times under simple interest, the per-year interest is the amount-difference divided by the time-difference. The same idea underlies questions where a principal "doubles" or "trebles". If a sum doubles in T years at simple interest, the interest earned equals the principal, so P = (P × R × T)/100, giving R × T = 100; to treble, the interest is twice the principal, so R × T = 200. These compact relations let you answer "in how many years will a sum become n times itself" in a single line, a frequent CLAT-PG and state-judiciary item.

The doubling and trebling identities also work in reverse to recover a missing rate or term. If a sum doubles in 8 years under simple interest, then R × 8 = 100, so R = 12.5% per annum; and to become five times itself the interest must be four times the principal, requiring R × T = 400, hence 12.5 × T = 400 and T = 32 years. Because the relation is purely linear, the time to reach successive multiples grows in equal steps: the same sum reaches double in 8 years, triple in 16, quadruple in 24, and so on, adding 8 years for each further multiple of the principal. This arithmetic-progression pattern is the simple-interest counterpart to the geometric power law of compounding, and keeping the two pictures side by side is the surest defence against the examiner who swaps one regime for the other mid-question.

The Compound Interest Formula From First Principles

Compound interest folds each period's interest back into the principal. Start with principal P. After one year the amount is P(1 + R/100). Treat that as the new principal: after the second year it becomes P(1 + R/100)², and after n years, by induction, A = P(1 + R/100)ⁿ. The compound interest itself is the amount minus the principal: CI = P[(1 + R/100)ⁿ − 1]. This is the exponential growth that distinguishes compounding from the linear simple-interest line.

A worked example mirroring the verified sources: Rs 5,000 at 9% per annum compounded annually for 2 years gives A = 5000(1.09)² = 5000 × 1.1881 = Rs 5,940.50, so CI = Rs 940.50. The matching simple interest is (5000 × 9 × 2)/100 = Rs 900, so compounding adds Rs 40.50 over the term. For mental arithmetic at small n, expand the bracket rather than reaching for a calculator: for 2 years the multiplier (1 + R/100)² = 1 + 2(R/100) + (R/100)². The first two terms reproduce simple interest; the last term, P(R/100)², is exactly the excess of CI over SI, a result we sharpen into a shortcut below. Recognising powers of common multipliers (1.1² = 1.21, 1.2² = 1.44, 1.05² = 1.1025) saves precious seconds, and that pattern-recognition is the same arithmetic fluency developed in number system and arithmetic foundations.

It helps to read the year-by-year structure explicitly. On the Rs 5,000 example, the first year's interest is 5000 × 0.09 = Rs 450, lifting the base to Rs 5,450. The second year's interest is then 5450 × 0.09 = Rs 490.50, computed on the enlarged base rather than the original. The extra Rs 40.50 above the flat Rs 450 is interest on the first year's interest, 450 × 0.09, which is precisely P(R/100)². Seeing the mechanism this way demystifies why the difference formula takes the shape it does, and it gives you an independent check: at any point the running amount equals the previous amount times (1 + R/100), so you can rebuild a compound schedule line by line if a formula slips your mind. Examiners who supply a partial schedule, such as the amount after one year, are inviting exactly this stepwise reconstruction.

The CI Minus SI Shortcut for Two and Three Years

The single most lucrative compound-interest shortcut is the gap between CI and SI at the same P, R and n. Expanding the compound multiplier and subtracting the linear simple-interest growth, the algebra collapses to clean closed forms. For two years, CI − SI = P(R/100)². For three years, CI − SI = P(R/100)²(3 + R/100), which many candidates remember in the equivalent form P[3R²/100² + R³/100³].

These let you answer a whole class of questions instantly. If the difference between CI and SI on a sum for 2 years at 10% is Rs 60, then P(10/100)² = 60, so P/100 = 60, giving P = Rs 6,000. Reverse it just as easily: the 2-year CI−SI gap on Rs 8,000 at 5% is 8000(0.05)² = 8000 × 0.0025 = Rs 20. For three years at 10% on Rs 5,000, the gap is 5000 × (0.1)² × (3 + 0.1) = 5000 × 0.01 × 3.1 = Rs 155. A useful sanity check: the 2-year gap is always one year's interest on the first year's interest, which is why P(R/100)² appears, the rate applied twice to the principal. Memorising these two forms converts a multi-step computation into one multiplication and is heavily tested in Delhi Judicial Service and bank-pattern aptitude sections.

Half-Yearly, Quarterly and Other Compounding Frequencies

The annual formula assumes interest is added once a year. When compounding is more frequent, the rate per period shrinks and the number of periods grows in proportion. The general adjustment is: divide the annual rate by the number of compoundings per year and multiply the number of years by that same count. For half-yearly compounding the amount is A = P(1 + R/(2×100))²ⁿ; for quarterly compounding it is A = P(1 + R/(4×100))⁴ⁿ; for monthly, divide by 12 and multiply the exponent by 12.

Worked example: Rs 10,000 at 10% per annum compounded half-yearly for 1 year. The half-year rate is 5% and there are 2 periods, so A = 10000(1.05)² = 10000 × 1.1025 = Rs 11,025, against Rs 11,000 under annual compounding. The extra Rs 25 is the reward for compounding twice. A subtle examiner trick combines a non-integer year with sub-annual compounding, for instance "compounded half-yearly for 1.5 years", which means 3 half-year periods. Another classic mixes regimes within a single term: a sum compounded annually for whole years and then charged simple interest on the fractional remainder, for which A = P(1 + R/100)ⁿ(1 + fR/100), where f is the fractional year. Always read whether the quoted rate is "per annum" or already "per period"; misreading this is the single biggest source of error here.

Doubling, Trebling and Growth Multiples Under Compounding

Compound-interest growth-multiple questions exploit the power law. If a sum becomes n times itself in t years, then n = (1 + R/100)^t. The handy consequence is that if a sum doubles in t years, it becomes 2² = 4 times in 2t years, 2³ = 8 times in 3t years, and in general 2^k times in kt years, because the multiplier simply raises to successive powers. So a sum that doubles in 5 years under annual compounding becomes 8-fold in 15 years and 16-fold in 20 years, with no rate computation required.

This power-law reasoning answers questions that look forbidding. "A sum becomes 3 times in 4 years at compound interest; in how many years will it become 27 times?" Since 27 = 3³, the answer is 3 × 4 = 12 years. Compare this with the simple-interest version covered earlier, where doubling obeyed the linear relation R×T = 100 and trebling needed R×T = 200; under compounding the relationship is multiplicative rather than additive, so the two regimes give very different answers for the same data. Examiners frequently pair the two in a single question to test whether you can keep the regimes distinct. The exponential structure here is the same growth pattern you meet when chaining successive percentage changes in percentage, ratio and proportion.

Finding the Rate or Time Hidden Inside a Compound Problem

When the unknown is the rate, you usually have an amount-to-principal ratio that resolves to a recognisable power. If Rs 4,000 grows to Rs 4,410 in 2 years compounded annually, then (1 + R/100)² = 4410/4000 = 1.1025 = (1.05)², so R = 5%. The trick is to recognise the right-hand side as a perfect square, cube or familiar power; the verified multipliers (1.05² = 1.1025, 1.1² = 1.21, 1.2² = 1.44, 1.1³ = 1.331) cover most exam cases. If the ratio is not a clean power, the figures are usually chosen so a successive-difference method works: split the growth period into single years and read the rate off the first year's increase.

A powerful linking relation connects consecutive years' compound interest. The interest for any year equals the previous year's interest multiplied by (1 + R/100), because each year's interest is computed on a base that grew by that factor. So if the CI for the second year is Rs 660 and for the third year is Rs 726, the rate satisfies (1 + R/100) = 726/660 = 1.1, giving R = 10%, and the principal follows from the second-year interest. This ratio-of-successive-interests method is far quicker than setting up the full amount equation and is a signature CLAT-PG and judiciary shortcut. For the hub of related quantitative chapters, see the aptitude and reasoning hub.

Equal Installments Under Simple and Compound Interest

Installment problems ask what equal periodic payment fully repays a debt. The governing idea is that the present values of the installments, each discounted back to the loan date, must sum to the loan. Under compound interest, if a loan A is repaid in n equal annual installments at rate R, each installment x satisfies A = x/(1+R/100) + x/(1+R/100)² + … + x/(1+R/100)ⁿ. Equivalently, the verified closed form gives the installment as x = A·(R/100) / [1 − (1 + R/100)⁻ⁿ], and for the special two-installment case you can solve directly.

A clean worked example: a sum of Rs 1,820 is to be repaid in two equal annual installments at 20% per annum compound interest. Let each installment be x. The first installment, paid after one year, has present value x/1.2; the second, after two years, has present value x/1.2² = x/1.44. Setting x/1.2 + x/1.44 = 1820 and solving gives x = Rs 1,200. Under simple interest the discounting is linear rather than geometric: each installment carries simple interest from the loan date to its payment date, so the equation uses factors like (1 + RT/100) on each payment. Read the question carefully to see whether "installment" means equal principal plus interest or a fully amortising equal payment; the phrasing decides which model applies. Installment items reward methodical present-value setup, the same disciplined structuring you practise in time and work.

Reading the Gap: When CI and SI Coincide and When They Diverge

For the first year, simple and compound interest are always equal, because compounding has not yet had a chance to act on accumulated interest; the divergence begins only from the second year. This is why every CI−SI shortcut starts at n = 2. The gap then widens each year at an accelerating pace, and for a fixed rate the ratio of CI to SI grows without bound as n increases. Conversely, for very small rates or very short terms the two are nearly indistinguishable, which is why short-tenor problems are often safely approximated by simple interest.

Examiners test this boundary directly. A question may state that the difference between CI and SI for the first year is zero (a deliberate trap to confirm you know they coincide) and then ask for the second-year gap, which is P(R/100)². Another format gives the SI for 2 years and asks for the CI, exploiting CI = SI + P(R/100)²: if 2-year SI is Rs 400 at 10%, then P = 400×100/(10×2) = Rs 2,000, the gap is 2000(0.1)² = Rs 20, and CI = Rs 420. Holding the relationships between the regimes in view, rather than memorising isolated formulae, is what lets you convert between them under time pressure, and it draws on the same comparative rate reasoning used in time, distance and speed.

Common Traps, Unit Errors and Reading Pitfalls

Most lost marks here are not conceptual but clerical. First, time units: rates are per annum, but terms arrive in months, quarters or days, and you must convert before substituting (8 months is 2/3 year, 73 days is 73/365 = 1/5 year). Second, compounding frequency: "10% per annum compounded half-yearly" does not mean 5% per half-year applied once; it means 5% applied twice, which is why the effective annual yield (10.25%) exceeds the nominal rate. Third, which quantity is asked: the problem may want the interest, the amount, or the difference between two scenarios, and these differ by a principal.

Fourth, mixing regimes: a single question may charge simple interest on one loan and compound on another and ask for the combined or differential figure; keep the two computations on separate lines. Fifth, the effective-rate trap in comparing offers: a scheme at 8% compounded quarterly beats one at 8.1% compounded annually only after you compute each effective annual rate, so never compare nominal rates directly when compounding frequencies differ. Sixth, rounding too early; carry the multiplier to four decimal places until the final step, because rounding 1.1025 to 1.1 can shift a rupee answer by a wide margin. A final cross-check that catches many slips: compute SI as a baseline and confirm that your CI exceeds it by a plausible amount, and that the excess matches the P(R/100)² estimate for short terms.

Exam Strategy: Choosing the Fastest Route Under Time Pressure

In a timed paper, the order in which you reach for tools matters as much as knowing them. For any 2-year or 3-year CI−SI question, go straight to P(R/100)² or P(R/100)²(3 + R/100) rather than computing both interests in full. For growth-multiple questions, use the power law n = (1+R/100)^t and look for the answer as a power of the given multiple, avoiding logarithms entirely since exam data is always engineered to factor cleanly. For "find the rate" questions, set the amount-to-principal ratio equal to a recognisable square or cube. For successive-year interest data, divide consecutive interests to read off (1 + R/100) in one step.

Treat the formula choice as a triage: identify whether the problem is single-regime or mixed, whether compounding is annual or sub-annual, and whether the unknown is interest, principal, rate or time, before writing anything. Verify the units in the first ten seconds, because a months-to-years slip propagates through every later step. Finally, sanity-check magnitudes against simple interest as a floor for CI and a ceiling for any discounted installment. These habits, combined with the closed forms in this chapter, turn the interest section into reliable, fast marks. To round out the quantitative syllabus and the chart-and-table questions that often embed interest data, work through data interpretation alongside this chapter.

Frequently asked questions

What is the core difference between simple and compound interest?

Simple interest is computed only on the original principal for the whole term, so it grows by the same rupee amount each year in a straight line. Compound interest adds each period's interest back to the principal, so the next period's interest is computed on a larger base, producing exponential growth. The formulae are SI = PRT/100 and A = P(1 + R/100)ⁿ respectively.

What is the quickest formula for the difference between CI and SI?

For two years at the same principal P and rate R, the difference CI − SI equals P(R/100)². For three years it equals P(R/100)²(3 + R/100). These closed forms turn a multi-step calculation into a single multiplication and are among the most heavily tested shortcuts in judiciary and CLAT-PG aptitude.

How do I adjust the formula for half-yearly or quarterly compounding?

Divide the annual rate by the number of compoundings per year and multiply the number of years by that same count. Half-yearly gives A = P(1 + R/200)²ⁿ and quarterly gives A = P(1 + R/400)⁴ⁿ. So 10% per annum compounded half-yearly means 5% applied twice a year, not 10% applied once.

If a sum doubles in 5 years under compound interest, when does it become 8 times?

Use the power law: doubling means the multiplier is 2 over 5 years, so 2^k times is reached in 5k years. Since 8 = 2³, the sum becomes 8 times in 5 × 3 = 15 years. Note this multiplicative rule applies only to compound interest; under simple interest, doubling obeys the additive relation R × T = 100.

Why are simple and compound interest equal in the first year?

In the first year there is no previously accumulated interest for compounding to act on, so interest is computed on the original principal in both regimes and the two are identical. The divergence begins only from the second year, which is why every CI minus SI shortcut starts at n = 2 with the term P(R/100)².

How do equal-installment compound-interest problems work?

Each installment is discounted back to the loan date and the present values must sum to the loan. For a loan A repaid in n equal annual installments x at rate R, A equals the sum of x/(1+R/100)^k for k from 1 to n. For example, Rs 1,820 repaid in two equal installments at 20% gives x/1.2 + x/1.44 = 1820, so each installment is Rs 1,200.