Simple and Compound Interest

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What: Simple Interest (SI) and Compound Interest (CI) — the two ways money grows over time. Plus the standard CDS variants: half-yearly and quarterly compounding, difference between SI and CI for 2 and 3 years, and instalment problems.
Why it matters: CDS papers from 2000 to 2023 carry 3–5 questions per sitting on this single chapter — among the highest-weighted arithmetic heads.
Key fact: For 2 years, the difference between CI and SI on principal $P$ at rate $r\%$ is exactly $P (r/100)^2$. Memorising this single identity solves at least one CDS problem every year.

Simple and Compound Interest is the financial backbone of CDS arithmetic. The formulas are tight, the problem types repeat, and the highest-marks questions test pattern recognition more than computation. Master the four core formulas and the SI-vs-CI difference identities and you will pick up 3–5 marks per paper.

This page is grounded in CDS Previous Year Questions across 2000–2023, plus NCERT Class 8 commercial arithmetic chapters. Pair with Percentage (rates are percentages) and Ratio and Proportion (instalment problems).

What This Topic Covers

CDS scope: (1) simple interest — $SI = Prt/100$; (2) compound interest — $A = P(1 + r/100)^t$; (3) compounding frequency — half-yearly, quarterly, monthly; (4) SI vs CI difference — exact formulas for 2 and 3 years; (5) instalments — equal payments with interest; and (6) word applications — growth, depreciation, ratio of amounts, finding rate or time.

Why This Topic Matters

3–5 CDS questions per paper, with clean numerical setups.
The CI − SI two-year identity is a 5-second formula that solves dozens of variants.
Concepts here connect directly to Percentage (population growth) and any growth/decay problem.

Exam Pattern & Weightage

Year / Paper	No.	Subtopics Tested
2007-I/II	3	SI basic, CI 2-year, rate finding
2008-I/II	3	CI half-yearly, difference SI-CI
2010-I/II	4	SI/CI comparison, instalment
2011-I/II	3	CI 3-year, rate equation
2012-I/II	4	Difference CI-SI, half-yearly
2013-I/II	4	Ratio of amounts, doubling time
2014-I/II	3	Multiple rate periods, CI compound
2015-I/II	3	SI vs CI, half-yearly
2018-II	2	Doubling, tripling time
2019-II	2	Mixed SI/CI, instalment
2021-I/II	3	Difference identity, rate finding
2022-I / 2023-I	3	Compounding frequency variants

⚡ CDS Alert

"Doubling time" at rate $r\%$ by compound interest: $t = \log 2 / \log(1 + r/100)$. For simple interest doubling: $t = 100/r$ years. At 10% CI, money doubles in ~7.27 years; at 10% SI, in exactly 10 years.

Core Concepts

Simple Interest

Simple Interest Formula $$SI = \frac{P \cdot r \cdot t}{100}, \qquad A = P + SI = P\left(1 + \frac{rt}{100}\right)$$

Where $P$ is principal, $r$ is rate per annum (in percent), $t$ is time in years, $A$ is the final amount.

Compound Interest

Compound Interest Formula $$A = P\left(1 + \frac{r}{100}\right)^t, \qquad CI = A - P$$

For other compounding frequencies, divide the rate and multiply the period:

Half-Yearly Compounding $$A = P\left(1 + \frac{r/2}{100}\right)^{2t}$$

Quarterly Compounding $$A = P\left(1 + \frac{r/4}{100}\right)^{4t}$$

Difference Between CI and SI

This is the high-yield CDS identity. For the same $P, r, t$:

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

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

⚠ Common Trap

"Compound interest for 2 years at $r\%$ is some value. What is the principal?" CDS expects you to use the difference identity backward, not compute CI step by step. Try $P (r/100)^2 = \text{given diff}$ as the first move.

Doubling / Tripling Time

For SI: time to double = $100/r$ years; to triple = $200/r$; to become $n$ times = $(n-1) \cdot 100 / r$.
For CI: time to double satisfies $(1 + r/100)^t = 2$, so $t = \log 2 / \log(1 + r/100)$. At 5%, doubling = 14.21 years; at 10%, 7.27 years; at 20%, 3.80 years.

Instalment Problems

A loan of $L$ is to be repaid in $n$ equal annual instalments with interest at $r\%$. Each instalment equals:

Annual Equal Instalment (CI basis) $$\text{Instalment} = \frac{L \cdot r/100}{1 - (1 + r/100)^{-n}}$$

CDS variants are simpler — usually 2-year or 3-year loans where direct equation works.

Worked Examples

Example 1 — SI Basic (2007-II)

Q: Find the simple interest on ₹5000 at 8% per annum for 3 years.

Apply $SI = Prt/100 = 5000 \cdot 8 \cdot 3 / 100 = 1200$.
Answer: ₹1200.

Example 2 — CI 2 Years (2008-I)

Q: Compute the compound interest on ₹8000 at 10% per annum for 2 years.

Apply $A = P(1 + r/100)^t = 8000 \cdot (1.10)^2 = 8000 \cdot 1.21 = 9680$.
CI $= A - P = 9680 - 8000 = 1680$.

Example 3 — CI − SI Difference (2012-II)

Q: The difference between CI and SI on a sum for 2 years at 10% per annum is ₹50. Find the principal.

Apply the 2-year identity: $CI - SI = P (r/100)^2$.
$50 = P \cdot (0.1)^2 = 0.01 P \implies P = 5000$.

Example 4 — Doubling Time SI (2018-II)

Q: In how many years will a sum double itself at 8% simple interest?

Time to double on SI: $t = 100/r = 100/8 = 12.5$ years.

Example 5 — Half-Yearly Compounding (2008-II)

Q: Find the CI on ₹10000 at 8% per annum compounded half-yearly for 1 year.

Half-yearly rate = 4%; periods = 2.
$A = 10000 \cdot (1.04)^2 = 10000 \cdot 1.0816 = 10816$. CI = 816.

Example 6 — Find Rate (2011-II)

Q: A sum becomes 1.21 times itself in 2 years on CI. Find the rate.

$(1 + r/100)^2 = 1.21 \implies 1 + r/100 = 1.10 \implies r = 10\%$.

Example 7 — Mixed SI/CI (2010-I)

Q: The SI for 2 years at 5% is ₹500. Find the CI for the same period and same rate.

From SI: $Prt/100 = 500 \Rightarrow P \cdot 0.05 \cdot 2 = 500 \Rightarrow P = 5000$.
CI: $A = 5000 \cdot (1.05)^2 = 5000 \cdot 1.1025 = 5512.50$; CI = 512.50.
Alternatively: difference identity $CI - SI = P(r/100)^2 = 5000 \cdot 0.0025 = 12.50$, so CI = 512.50.

How CDS Tests This Topic

Six recurring archetypes: (1) plain SI computation, (2) plain CI computation (especially 2 or 3 years), (3) difference between CI and SI (use the identity), (4) doubling/tripling time, (5) half-yearly or quarterly compounding, (6) finding the rate or time when amount and principal are given.

Exam Shortcuts (Pro-Tips)

Shortcut 1 — CI − SI Identity for 2 Years

2-Year Difference $$CI - SI = P (r/100)^2$$

Shortcut 2 — Sum Becomes $k$ Times

For SI, $A/P = 1 + rt/100$. For "sum becomes $k$ times", $rt = 100(k-1)$. For CI, $(1 + r/100)^t = k$ — usually solve by recognising $(1 + r/100)$ as a clean number like 1.05, 1.10, 1.20.

Shortcut 3 — Rule of 72 (Approximate Doubling)

At CI, money roughly doubles in $72/r$ years. At 6% CI, ~12 years; at 9% CI, ~8 years. Not exact but good for fast elimination.

Shortcut 4 — Recognise the Year Pattern

$(1.10)^2 = 1.21,\; (1.10)^3 = 1.331,\; (1.05)^2 = 1.1025,\; (1.20)^2 = 1.44,\; (0.9)^3 = 0.729$. Memorise these — CDS uses them in nearly every CI question.

Shortcut 5 — Half-Yearly Half the Rate, Double the Time

Whenever you see "compounded half-yearly", just halve the rate and double the time. "Compounded quarterly" → quarter the rate, quadruple the time.

Common Question Patterns

Pattern 1 — Plain SI / CI Computation

Apply the standard formula. CDS usually gives a clean rate (5%, 8%, 10%) and 2 or 3 years.

Pattern 2 — Find Principal from Difference

"CI − SI for 2 years at $r\%$ is given." Apply $P (r/100)^2 = $ difference. Solve for $P$.

Pattern 3 — Compounding Frequency

Half-yearly, quarterly, monthly. Adjust rate and time. Watch out for "annual nominal rate" vs effective rate.

Pattern 4 — Doubling / Tripling Time

SI: $t = (k-1) \cdot 100/r$. CI: $(1 + r/100)^t = k$.

Pattern 5 — Mixed SI / CI / Comparison

Same principal, same rate, same time — compare SI and CI. The difference identity is the fastest path.

Preparation Strategy

Week 1. Master SI and CI formulas. Drill 15 problems on plain computation, then 10 on rate-finding and time-finding. Memorise the standard powers: $(1.05)^2, (1.10)^2, (1.10)^3, (1.20)^2$.

Week 2. Master the CI − SI difference identities for 2 and 3 years. Drill half-yearly and quarterly compounding. Layer in Percentage for rate fluency and Ratio for instalment problems.

Mock testing. Take timed CDS papers. Track whether you spot the difference identity opportunity quickly enough — many candidates compute CI fully when one line of the identity would suffice. Use CDS mock tests for speed practice.

Drill SI and CI in Real Time

CDS mocks with the difference identity, half-yearly compounding, and doubling-time problems. Six archetypes — six formulas — reflex.

Start Free Mock Test

Frequently Asked Questions

What is the difference between Simple and Compound Interest?

SI is calculated only on the original principal each year. CI is calculated on the new amount (principal + accumulated interest) each year, so interest "grows on interest". Over more than one year, CI always exceeds SI for the same $P, r, t$.

What is the 2-year CI − SI identity?

For the same $P, r$ and 2 years: $CI - SI = P (r/100)^2$. This single line solves dozens of CDS variants. Example: at $P = 5000, r = 10\%$, $CI - SI = 5000 \cdot 0.01 = 50$.

What is the 3-year CI − SI identity?

$CI - SI = P (r/100)^2 (3 + r/100)$. At small $r$, the $r/100$ term is negligible and the difference is roughly $3 \cdot P (r/100)^2$, three times the 2-year difference.

How do I handle half-yearly compounding?

Halve the rate and double the time. So 10% per annum compounded half-yearly for 2 years becomes 5% per half-year for 4 half-years. Use $A = P(1.05)^4$. Similarly for quarterly: quarter the rate, quadruple the time.

How long does money take to double?

On SI at rate $r\%$: $t = 100/r$ years. So 8% SI doubles in 12.5 years. On CI: $t = \log 2 / \log(1 + r/100)$. At 10% CI, ~7.27 years. The Rule of 72 ($t \approx 72/r$) is a quick approximation for CI doubling.

How do I find the rate when amount and principal are known?

For SI: $r = (A - P) \cdot 100 / (Pt)$. For CI: $(1 + r/100)^t = A/P$ — take the $t$-th root, subtract 1, multiply by 100. CDS chooses values where $A/P$ is a clean power like 1.21, 1.331, 1.44.

Which CDS Maths topics connect to SI and CI?

Percentage supplies rate fluency. Ratio and Proportion appears in instalment problems. The compounding formula is structurally the same as population growth / depreciation problems in Percentage.