Simple and Compound Interest is one of the highest-scoring arithmetic chapters in CDS and OTA Maths, with two to four direct questions almost every attempt. The good news is that the whole topic runs on a handful of formulas. In this Cavalier lesson you will learn each formula, when to use it, and how to clear the half-yearly and rate-change traps that catch most candidates.
Why interest questions are easy marks
Interest is the money paid for the use of borrowed money. The amount borrowed is the principal (P), the time for which it is used is the time (T or n), and the percentage charged per year is the rate (R). The total returned at the end is the amount (A). These four quantities appear in every single interest problem, so memorising what each letter stands for is the first step.
Almost every CDS interest question simply asks you to plug numbers into one of four or five formulas. Because the calculations are short and the formulas are fixed, this is a chapter where a prepared candidate can score full marks in under five minutes. There is no theory to interpret, no diagram to draw, and the arithmetic is rarely heavy.
The examiners test you in three ways: a direct plug-in question, a reverse question where the interest is given and you must find P, R or T, and a comparison question pitting simple interest against compound interest. Once you can recognise which of these three a problem belongs to, choosing the right formula becomes automatic.
The basic relationship is always Amount = Principal + Interest, so A = P + I. Every interest formula is just a way of finding I or A.
The Simple Interest formula
In Simple Interest (SI), interest is calculated only on the original principal for the entire period. The principal never changes, so the interest earned each year is exactly the same. If a sum earns ₹300 in the first year, it earns ₹300 in every later year too. This makes simple interest grow in a straight line.
SI = (P × R × T) ÷ 100
Amount A = P + SI = P(1 + RT÷100)
Here R is the rate per annum (%) and T is time in years.
Because simple interest per year is constant, the total interest is just that yearly figure multiplied by the number of years. This is why the formula has T sitting outside any power. From this single relation you can rearrange to find any unknown:
- P = (100 × SI) ÷ (R × T)
- R = (100 × SI) ÷ (P × T)
- T = (100 × SI) ÷ (P × R)
If a sum becomes n times itself under simple interest in T years at rate R, then n − 1 = (R × T) ÷ 100. This one line solves most “doubles/triples in how many years” questions instantly.
The Compound Interest formula
In Compound Interest (CI), the interest of each period is added to the principal, so the next period earns interest on a larger base. This is “interest on interest”, and it is the way real bank deposits and loans actually work.
Amount A = P (1 + R÷100)n
CI = A − P = P[(1 + R÷100)n − 1]
where n is the number of times interest is compounded.
When compounding is yearly, n is simply the number of years. The difference from simple interest grows steadily because each year you earn on a bigger amount. After the first year the base is P(1 + R÷100); after the second it is P(1 + R÷100)2; and so on, which is exactly how the power n in the formula arises.
A useful way to picture it: year one earns the same as simple interest, but year two earns interest on the principal plus year one's interest. That tiny extra slice, compounded again and again, is what separates the two methods over long periods.
For the first year (or first period), SI and CI are always equal, because there is no accumulated interest yet to compound. They start to differ only from the second period onwards.
Half-yearly and quarterly compounding
This is the single biggest trap in CDS interest questions. When interest is compounded more than once a year, you must adjust both the rate and the time.
Half-yearly: rate becomes R÷2 per half-year, time becomes 2T half-years.
A = P (1 + R÷200)2T
Quarterly: rate becomes R÷4, time becomes 4T quarters.
A = P (1 + R÷400)4T
Candidates halve the rate but forget to double the time (or vice versa). For 2 years at 10% compounded half-yearly the exponent is 4, not 2, and the rate per period is 5%, not 10%.
For example, ₹8000 at 10% per annum compounded half-yearly for 1 year gives A = 8000(1 + 5÷100)2 = 8000 × 1.1025 = ₹8820. Compare this with yearly compounding, which would give 8000 × 1.10 = ₹8800. The extra ₹20 is the reward for compounding twice instead of once in the same year, which proves that more frequent compounding always produces a larger amount.
The same logic extends to monthly compounding, where the rate becomes R÷12 and the time becomes 12T. CDS rarely goes beyond quarterly, but the rule is identical: divide the rate by the number of periods per year and multiply the time by the same number.
The SI vs CI difference shortcuts
CDS frequently gives you the difference between CI and SI on the same sum and asks for P or R. Memorise these two results and you can answer in seconds.
For 2 years (same P and R): Difference = P (R÷100)2
For 3 years: Difference = P (R÷100)2 × (3 + R÷100)
The 2-year difference equals the interest-on-interest of the first year's interest, which is why it is exactly P(R/100)2. Think of it this way: in year two, compound interest earns an extra slice equal to R% of the first year's interest, and the first year's interest is PR÷100, so the extra slice is (R÷100) of (PR÷100), which simplifies to P(R÷100)2.
The 3-year formula looks heavier but is just the 2-year difference scaled by the factor (3 + R÷100). Many candidates write it as P(R÷100)2(300 + R)÷100 to avoid decimals; both forms give the same value. Keep whichever version you find easier to recall under exam pressure.
If a question gives the 2-year difference and the rate, find the principal directly: P = Difference × (100÷R)2. No long multiplication needed.
Growth, depreciation and population sums
The compound interest formula is really a compound growth formula, so the same maths covers population growth, machine depreciation and price changes that appear in CDS.
Growth / appreciation: Final = Initial (1 + R÷100)n
Depreciation / decay: Final = Initial (1 − R÷100)n
- Population increasing at R% per year → use the (1 + R÷100) form.
- Value of a machine or vehicle falling at R% per year → use the (1 − R÷100) form.
- If the rate differs each year, multiply the factors: A = P(1 + R1÷100)(1 + R2÷100)…
- If a quantity grows in some years and falls in others, mix plus and minus factors accordingly — for instance a value that rises 10% then falls 10% ends at P × 1.10 × 0.90 = 0.99P, a net 1% loss.
That last point is a favourite trap: a rise of x% followed by a fall of x% never returns you to the original value. The successive-change formula for a rise of a% and a fall of b% is a net change of (a − b − ab÷100)%, which the (1 + a÷100)(1 − b÷100) factors reproduce automatically.
To find a value two years ago at a known present value, divide instead of multiply: Past = Present ÷ (1 + R÷100)n.
Equal instalments on a loan
Some CDS papers ask for the equal annual instalment that clears a loan. Each instalment is discounted back to the present using the compound factor.
For two equal annual instalments X repaying a present sum P at rate R:
P = X ÷ (1 + R÷100) + X ÷ (1 + R÷100)2
For three instalments you simply add a third term with the cube in the denominator. Solve for X once P and R are known. The idea is that the borrower's repayments, when each is brought back to today's value, must add up to exactly the amount borrowed. These appear less often than basic SI and CI, so master the core formulas first and treat instalments as a bonus topic.
A quick sanity check: if the rate were zero, the two-instalment formula would give P = X + X = 2X, meaning each instalment is simply half the loan. A non-zero rate makes each instalment slightly larger than half, because the lender also charges for the delay. Recognising this pattern helps you reject absurd options in a multiple-choice paper.
Worked example: SI and CI together
A sum of ₹12,000 is invested at 10% per annum. Find the simple interest and the compound interest (compounded annually) for 2 years, and the difference between them.
The shortcut and the long method agree, confirming ₹120. In the exam, the shortcut alone would have given the answer in one step.
Common mistakes to avoid
Most lost marks in this chapter come from a few repeatable slips. Watch for these in every question.
- Using the SI formula when CI is asked (or the reverse). Read the word “compound” carefully.
- Forgetting to convert months into years: 9 months = 9÷12 = 0.75 year = 3÷4 year.
- For half-yearly compounding, changing the rate but not the time.
- Reporting the amount A when only the interest was asked, or vice versa.
- Mixing up principal and amount when reversing a CI calculation.
Underline what is asked — SI, CI, A, P, R or T — before you start. Half of the wrong options in MCQs are the “other” quantity, planted to catch the hurried reader.
Previous-year style question
Q. The difference between the compound interest and the simple interest on a certain sum at 10% per annum for 2 years is ₹50. What is the sum?
Answer: Using Difference = P(R÷100)2, we get 50 = P × (10÷100)2 = P × 0.01. So P = 50 ÷ 0.01 = ₹5000. The sum is ₹5000.
Notice how the 2-year difference shortcut turns a wordy problem into a single division. This exact pattern — give the difference and rate, ask for the sum — has appeared repeatedly in CDS papers, so it is worth drilling until it is automatic.
Quick revision
- SI = PRT÷100; Amount = P + SI.
- CI: A = P(1 + R÷100)n; CI = A − P.
- Half-yearly: rate → R÷2, time → 2T. Quarterly: R÷4, 4T.
- SI = CI for the first period only.
- 2-year difference = P(R÷100)2; 3-year difference = P(R÷100)2(3 + R÷100).
- Depreciation uses (1 − R÷100)n.
- Convert months to years and read whether SI, CI, A or P is wanted.
Drill ten mixed SI and CI problems from the Cavalier topic-wise PYQ set until each formula is reflex. That is enough to bank every interest mark in your CDS or OTA paper.
Frequently asked questions
When are Simple Interest and Compound Interest equal?
They are equal only for the first year (or first compounding period), because there is no accumulated interest yet to compound. From the second period onwards, CI exceeds SI.
How do I handle half-yearly compounding in CDS questions?
Halve the annual rate and double the number of years. So R becomes R÷2 per half-year and time T years becomes 2T periods: A = P(1 + R÷200) raised to the power 2T.
What is the fastest way to find the sum from the SI-CI difference?
For 2 years use Difference = P(R÷100) squared, so P = Difference multiplied by (100÷R) squared. This avoids any long compound-interest calculation.
Does the compound interest formula work for depreciation too?
Yes. For a value falling at R% per year, use Final = Initial (1 − R÷100) raised to the power n. It is the same formula with a minus sign instead of plus.
How many marks does this topic carry in CDS Maths?
Simple and Compound Interest typically gives two to four direct questions per CDS or OTA paper, making it one of the most reliable scoring areas in arithmetic.
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