Simple and Compound Interest — SSC CGL Quantitative Aptitude

SI grows in a straight line — the same interest every year. CI grows by successive percentages — each year's interest earns interest. Every exam question lives in that gap: the difference formulas, doubling times, and half-yearly conversions. Work in multipliers ($\times 1.1$ per 10% year) and CI stops being a formula to fear.

1. What SSC actually asks

Tier 1: 1–2 Q · Tier 2: 2–3 Q. The recurring types: straight SI computation, rate/time recovery ("doubles in 8 years"), CI for 2–3 years, CI−SI difference (a formula question in disguise), half-yearly compounding, population/depreciation growth, and "becomes $k$ times" doubling chains.

2. Simple interest — the straight line

$SI=\frac{P\times R\times T}{100},A=P+SI$

• Same interest every year: if a sum earns ₹360/year, three years earn ₹1080. Most SI questions are one division away once you extract the per-year interest.
• Doubles in $T$ years → $R=\frac{100}{T}$%. (Triples → $R=\frac{200}{T}$%: the interest equals $2P$.)
• Becomes $\frac{6}{5}$ of itself in 3 years → interest = $\frac{1}{5}P$ over 3 years → $R=\frac{20}{3}\%\approx 6\frac{2}{3}\%$.

3. Compound interest — successive multipliers

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

Think in multipliers: 10% p.a. for 2 years = $\times 1.1\times 1.1=\times 1.21$ → CI on ₹10,000 = ₹2100 (not ₹2000 — the extra ₹100 is interest-on-interest).

Conversion rules:

• Half-yearly: halve the rate, double the periods ($10\%$ yearly → $5\%$ × 2 halves).
• Quarterly: quarter the rate, ×4 periods.
• Population growth / depreciation are CI in costume: growth multiplies by $(1+r)$, depreciation by $(1-r)$.

4. The CI−SI difference formulas (guaranteed marks)

For the same P, R, T:

$\text{2 years: }CI-SI=P{\left(\frac{R}{100}\right)}^2$

$\text{3 years: }CI-SI=P{\left(\frac{R}{100}\right)}^2\left(3+\frac{R}{100}\right)$

P = ₹8000, R = 5%, 2 years: difference $=8000\times 0.0025=$ ₹20. These two lines convert a five-minute computation into fifteen seconds — and the question appears almost every year.

5. Doubling logic

• SI: doubling time $=\frac{100}{R}$ years; each further $P$ of interest takes the same time again (2× → 3× takes the same years as 1× → 2×).
• CI: doubling chains. If a sum doubles in 4 years, it quadruples in 8, becomes 8× in 12, 16× in 16 years — count the doublings: $16=2^4$.
• Rule of 72 (estimate): doubling years $\approx \frac{72}{R}$ — good for sanity checks, not exact answers.

6. Solved PYQ-style examples

Q1. SI on ₹5000 at 8% for 3 years? Solution. $\frac{5000\times 8\times 3}{100}=$ ₹1200.

Q2. A sum doubles itself in 8 years at simple interest. The rate is… Solution. Interest = P over 8 years → $R=\frac{100}{8}=$ 12.5%.

Q3. CI on ₹8000 at 10% p.a. for 1 year, compounded half-yearly? Solution. 5% per half-year, 2 periods: $8000\times 1.05^2=8820$ → CI = ₹820 (₹20 more than yearly's ₹800).

Q4. Difference between CI and SI on ₹10,000 at 10% for 3 years? Solution. $P{\left(\frac{R}{100}\right)}^2\left(3+\frac{R}{100}\right)=10000\times 0.01\times 3.1=$ ₹310.

Q5. A sum doubles in 4 years at compound interest. In how many years does it become 16 times? Solution. $16=2^4$ → four doublings → $4\times 4=$ 16 years.

7. Exam protocol

1. SI questions: extract the per-year interest first; almost everything is one division after that.
2. CI: write the multiplier chain ($\times 1.1\times 1.1$), never expand the binomial by hand.
3. CI−SI difference asked? Formula, fifteen seconds, move on.
4. Half-yearly/quarterly: convert rate and periods before computing anything.
5. "Becomes $k$ times" at CI: count doublings ($k=2^n$); at SI: count how many extra principals of interest.