Compound Interest for SSC CGL — Formulas, Shortcuts & Solved PYQs

Concept

Simple interest (साधारण ब्याज) pays you interest only on the original principal. Compound interest (चक्रवृद्धि ब्याज) pays you interest on the principal plus all the interest you have already earned. That single difference is why compounding is called "interest on interest" — and it is the reason every examiner loves it.

Here is a concrete analogy. Imagine you deposit ₹1,000 at 10% per annum. Under simple interest, every year you earn exactly ₹100 — always calculated on the original ₹1,000. Under compound interest, the first year you still earn ₹100, but now your balance is ₹1,100. In the second year, 10% is applied to ₹1,100, so you earn ₹110. That extra ₹10 is the "compounding effect." It looks small at 10% for 2 years, but over long periods or at high rates, the gap between CI and SI becomes significant.

In SSC CGL, compound interest questions appear almost every year in the Quantitative Aptitude section. The question types you will see are:

• Find the compound interest or final amount given P, r, n.
• Find the difference between CI and SI for 2 or 3 years.
• Doubling / tripling time problems.
• Half-yearly or quarterly compounding variations.
• Finding the original principal given the CI and rate.

The core formula is:

$A = P\left(1 + \frac{r}{100}\right)^n$

where $A$ is the final amount, $P$ is the principal, $r$ is the annual rate of interest, and $n$ is the number of years. The compound interest itself is $CI = A - P$.

You do not need calculus, you do not need logarithms (for standard CGL questions). What you need is fast arithmetic with a few power expansions memorised and two or three structural shortcuts locked in. That is exactly what this page gives you.

Deep Dive

The Core Formula and Its Variants

The standard formula is:

$A = P\left(1 + \frac{r}{100}\right)^n$

For half-yearly compounding, the rate is halved and the number of periods is doubled:

$A = P\left(1 + \frac{r}{200}\right)^{2n}$

For quarterly compounding:

$A = P\left(1 + \frac{r}{400}\right)^{4n}$

Look — the pattern is mechanical. Whatever the compounding frequency, divide the rate by the number of periods per year and multiply $n$ by the same number. SSC CGL almost always sticks to annual or half-yearly compounding, so memorise those two.

CI vs SI: The Difference Formulas

This is the single most tested sub-type. The difference between CI and SI for 2 years has a clean closed form:

$CI - SI = P \cdot \left(\frac{r}{100}\right)^2$

For 3 years, the difference is:

$CI - SI = P \cdot \left(\frac{r}{100}\right)^2 \cdot \left(3 + \frac{r}{100}\right)$

These are not approximations — they are exact. Derive them once, then just use them. Here is the 2-year derivation so you understand why it works:

$SI_2 = P \cdot \frac{2r}{100}$

$CI_2 = P\left[\left(1 + \frac{r}{100}\right)^2 - 1\right] = P\left[1 + \frac{2r}{100} + \frac{r^2}{10000} - 1\right] = P \cdot \frac{2r}{100} + P \cdot \frac{r^2}{10000}$

$CI_2 - SI_2 = P \cdot \frac{r^2}{10000} = P \cdot \left(\frac{r}{100}\right)^2$

The only term left after cancellation is the square of the rate on the principal — which is why it is so clean.

Successive Year Breakdown

For 2-year problems, breaking the interest year-by-year is often faster than expanding the full formula:

• Year 1 interest: $I_1 = P \times r / 100$
• New principal: $P_1 = P + I_1$
• Year 2 interest: $I_2 = P_1 \times r / 100$
• Total CI: $I_1 + I_2$

At 10% on ₹8,000: Year 1 interest = ₹800, new base = ₹8,800, Year 2 interest = ₹880. Total CI = ₹1,680. That took about 10 seconds — faster than computing $(1.1)^2$ and multiplying.

Doubling and Tripling Time

If a principal doubles in $t$ years at CI, it will be $2^k$ times the principal in $k \times t$ years. This is a pure exponent logic:

• Double in 5 years → 4 times ($2^2$) in 10 years → 8 times ($2^3$) in 15 years.

The trap variant: "double in 5 years at SI — how long to quadruple?" SI is linear, not exponential. Doubling under SI takes 5 years, so quadrupling takes exactly 15 years (not 10). Do not apply exponential logic to SI doubling questions.

Common Power Expansions Worth Memorising

| Factor | Value | |---|---| | $(1.05)^2$ | $1.1025$ | | $(1.05)^3$ | $1.157625$ | | $(1.08)^2$ | $1.1664$ | | $(1.10)^2$ | $1.21$ | | $(1.04)^2$ | $1.0816$ | | $(1.06)^3$ | $1.191016$ |

These six cover roughly 80% of SSC CGL compound interest questions. You do not need to derive them in the exam hall — just have them at your fingertips.

Effective Rate of Interest

When compounding is more frequent than annual, the effective annual rate is higher than the nominal rate. For half-yearly compounding at nominal rate $r$:

$r_{eff} = \left(1 + \frac{r}{200}\right)^2 - 1$

SSC CGL rarely asks this directly, but understanding it helps you cross-check answers in half-yearly compounding problems.

Memory Tricks & Shortcuts

Fast-Solving Framework

Read the question and immediately identify which of these four situations applies:

Situation 1 — Standard CI (given P, r, n, find CI or A):

• If $n = 2$: use year-by-year build. Do not expand the formula.
• If $n = 3$ and $r \in \{5\%, 6\%, 8\%, 10\%\}$: use the memorised $(1+r)^3$ value directly.
• If $r$ or $n$ is unusual: expand $(1 + r/100)^n$ step by step.

Situation 2 — CI minus SI:

• For 2 years: one-shot formula $P(r/100)^2$.
• For 3 years: $P(r/100)^2(3 + r/100)$.

Situation 3 — Doubling / tripling / 4x problems:

• Identify the base multiple (2, 4, 8, etc.), express as power of the doubling base, multiply the doubling time.

Situation 4 — Half-yearly or quarterly compounding:

• Perform the rate/period substitution first, then treat as a standard problem.

If the answer options are far apart, use estimation — compute to 2 decimal places and match. If options are close (like ₹5,408 vs ₹5,400 vs ₹5,416), you must compute exactly.

Solved PYQs

Why this question: Tests basic 2-year CI computation — the most common question type. Distinguishing CI from SI is the core trap here.

Solving path: Use year-by-year build. Year 1 interest on ₹8,000 at 10% = ₹800. New principal = ₹8,800. Year 2 interest = ₹880. Total CI = ₹800 + ₹880 = ₹1,680. Option D. Note that option B (₹1,600) is the SI trap — if you used $P \times r \times n / 100 = 8000 \times 10 \times 2 / 100$, you get ₹1,600. That is the most common error on this type.

Why this question: Tests finding the final amount (not just CI). Students sometimes confuse amount with interest.

Solving path: $(1.04)^2 = 1.0816$. Amount $= 5000 \times 1.0816 = ₹5,408$. Option D. Option C (₹5,400) is the SI amount trap: $5000 + 5000 \times 0.04 \times 2 = 5000 + 400 = 5400$. The compounding adds ₹8 extra — small but that is exactly what the options test.

Why this question: The CI-SI difference type. This is the purest test of whether you know the shortcut formula.

Solving path: Direct formula: $CI - SI = P(r/100)^2 = 10000 \times (5/100)^2 = 10000 \times 0.0025 = ₹25$. Option D. If you computed CI and SI separately: $SI = 10000 \times 5 \times 2 / 100 = ₹1000$; $CI = 10000 \times [(1.05)^2 - 1] = 10000 \times 0.1025 = ₹1025$; difference = ₹25. Same answer, 4 extra steps. The shortcut formula saves you in the exam hall.

Why this question: Tests 3-year compound interest with a non-round rate. Requires either the memorised $(1.06)^3$ or the chain multiplication approach.

Solving path: Use chain multiplication. Year 1: $5000 \times 1.06 = ₹5,300$. Year 2: $5300 \times 1.06 = ₹5,618$. Year 3: $5618 \times 1.06 = ₹5,955.08$. CI = $5955.08 - 5000 = ₹955.08$. Option D. Alternatively, $(1.06)^3 = 1.191016$, so $5000 \times 1.191016 = 5955.08$. Note option B (₹980.40) is a common distractor for aspirants who miscalculate the chain multiplication.

Why this question: The doubling-time problem type. Conceptual, no heavy computation needed.

Solving path: $P$ doubles in 5 years means after 5 years: $2P$. After 5 more years (10 total), the $2P$ doubles again to $4P$. Answer: 10 years. Option C. The logic is that at CI, equal time intervals produce equal multiplier effects — every 5 years the amount multiplies by 2. $4 = 2^2$, so you need 2 intervals of 5 years = 10 years. Do not add 5 + 5 = 10 years mechanically without understanding the exponential structure, or you will fail the "8 times" variant.

Common Mistakes

• Using SI formula on CI questions. The answer choices always include both the SI result and the CI result as options. If you compute $P \times r \times n / 100$, you will land on a distractor every time. Always check: is the question asking for compound interest?

• Confusing Amount with Compound Interest. CI = Amount - Principal. On ₹5,000 questions, answers like ₹5,408 (amount) and ₹408 (CI) are both calculable. Read the question word: "amount" vs "compound interest."

• Applying doubling-time logic to Simple Interest. If P doubles in t years under SI, it becomes 4P in 3t years (linear scaling), not 2t years (exponential). The doubling-chain shortcut is valid only for CI.

• Forgetting to adjust rate and period for half-yearly compounding. Half-yearly at 10% annual is not the same as annual at 10%. The effective half-yearly rate is 5%, applied 4 times over 2 years. Missing this substitution step leads to a lower (incorrect) answer.

• Rounding intermediate calculations. On 3-year problems, rounding after Year 2 before computing Year 3 interest introduces error. Keep at least 2 decimal places throughout the chain.

• Misreading "difference between CI and SI" as asking for the SI amount. The question asks for CI minus SI, a single number. Compute both separately and subtract — or use the direct difference formula. Do not report SI or CI alone.

Related Topics

• simple interest
• percentage
• ratio proportion
• time work
• profit loss
• successive discount

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