Simple and Compound Interest questions in CDS are formula-driven and fast — once you know which formula applies, calculation is straightforward. The distinction between SI and CI, the half-yearly/quarterly adjustments, and the two-year/three-year difference formulas account for virtually all questions in this chapter.
📌 CDS exam focus:(1) Finding P, R, or T when three of the four SI variables are given; (2) CI with half-yearly and quarterly compounding — adjusting rate and time; (3) Difference between CI and SI for exactly 2 years (formula shortcut); (4) Population growth and machinery depreciation modelled as CI; (5) Instalment problems on loans.
Topics at a Glance
① Simple Interest
SI = PRT/100; finding P, R, T
② Compound Interest
Annual, half-yearly, quarterly
③ SI vs CI Difference
2-year and 3-year shortcut formulas
④ Population & Depreciation
Growth and decay as CI applications
⑤ Instalments
Equal instalment on loans
⑥ Effective Rate
Equivalent annual rate for non-annual compounding
1. Simple Interest
1.1
Formula, Variations & Key Applications
SI grows linearly — same interest every period
⚡ Simple Interest — All Variations
SI = P × R × T / 100
A = P + SI = P(1 + RT/100)
Finding unknowns:
P = 100 × SI / (R × T)
R = 100 × SI / (P × T)
T = 100 × SI / (P × R)
When a sum doubles at SI: T = 100/R years
When a sum triples at SI: T = 200/R years
When a sum becomes n times: T = (n−1) × 100/R years
If different rates for different years:
SI = P(R₁ + R₂ + R₃)/100 (for T₁=T₂=T₃=1 year each)
Worked Example
A sum of Rs 5,000 amounts to Rs 6,500 in 4 years at SI. Find the rate of interest.
SI = A − P = 6500 − 5000 = Rs 1500. R = 100 × SI / (P × T) = 100 × 1500 / (5000 × 4) = 1,50,000/20,000 = 7.5% p.a.
2. Compound Interest
2.1
Annual, Half-Yearly & Quarterly Compounding
Adjust rate and time for non-annual compounding — most common error source
⚡ Compound Interest Formulas
A = P(1 + R/100)ⁿ [annual compounding, n years]
CI = A − P
Half-yearly compounding: A = P(1 + R/200)²ⁿ
→ rate becomes R/2, periods become 2n
Quarterly compounding: A = P(1 + R/400)⁴ⁿ
→ rate becomes R/4, periods become 4n
When rates differ each year:
A = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)
Effective annual rate (for half-yearly at R%):
Effective = [(1 + R/200)² − 1] × 100
Memorise: For CDS, when "compounded half-yearly" — always halve the rate and double the time. This is the most tested adjustment.
2.2
SI vs CI Difference — Shortcut Formulas
These shortcuts save 90 seconds per question — must memorise
⚡ CI − SI Difference Shortcuts
For 2 years:
CI − SI = P(R/100)² = P × R² / 10000
For 3 years:
CI − SI = P(R/100)² × (R/100 + 3)
= PR²(300 + R) / 1000000
Also useful for 2 years:
CI = SI + P(R/100)²
If SI for 2 years = S, then CI = S + S²/(4P) [when rate is same]
The 2-year difference formula is tested almost every CDS exam. Memorise: CI − SI = P(R/100)².
Worked Example — 2-Year Difference
The difference between CI and SI on Rs 8,000 for 2 years is Rs 20. Find rate of interest.
CI − SI = P(R/100)² → 20 = 8000 × R²/10000 → R² = 20 × 10000/8000 = 25 → R = 5% p.a.
3. Population & Depreciation
3.1
CI Applied to Real-World Growth and Decay
Same formula — just change the sign for decay
👥 Population Growth
Population after n years: P(1 + r/100)ⁿ
Population n years ago: P_now / (1 + r/100)ⁿ
Different rates each year: P × ∏(1 + rᵢ/100)
Increase in one year, decrease in another: use ± appropriately
CDS often asks: given current population, find future or past
🔧 Depreciation of Assets
Value after n years: V(1 − r/100)ⁿ
Value n years ago: V_now / (1 − r/100)ⁿ
Combined: appreciates at a%, depreciates at b% → V(1+a/100)(1−b/100)
Machines, vehicles, buildings use this model
Rate is always given as % per year unless otherwise stated
4. Instalment Problems
4.1
Equal Instalments on Loans at Simple Interest
Appears as 1–2 questions — use backward calculation
⚡ Instalment Calculation (SI)
If loan amount P is to be repaid in n equal annual instalments
at rate R% SI per annum:
Each instalment x satisfies:
x + x×(n−1)R/100 + x + x×(n−2)R/100 + ... + x = P + P×n×R/100
Simplified (equal annual instalments):
x = P(100 + nR) / [n(100 + (n−1)R/2 × ... )]
Most CDS questions give n=2 instalments — solve by:
P + SI on P for n yrs = sum of instalments + SI on each instalment
Work backwards from the last instalment.
For 2-instalment problems: P(1 + 2R/100) = x(1 + R/100) + x. Solve for x.
Worked Example — 2 Equal Instalments
A sum of Rs 2,600 is to be paid back in 2 equal annual instalments at 26% SI. Find each instalment.
Let instalment = x. After year 1: x pays off → remaining = 2600 + SI(1yr) − x = 2600(1.26) − x = 3276 − x.
This remaining must equal x after another year of interest: (3276 − x)(1.26) = x
3276 × 1.26 = x + 1.26x → 4127.76 = 2.26x → x = 4127.76/2.26 = Rs 1827. Alternatively: Use x = P×100(100+R) / [100×(100+R) + 100²] for 2 instalments at R%.
📋 TOPIC-WISE PYQ
SI & CI — CDS Questions
Q1. At what rate of SI will a sum of money triple itself in 20 years?
(a) 8% (b) 10% (c) 12% (d) 15%
Answer: (b) 10%
Sum triples → SI = 2P. T = (n−1)×100/R → 20 = (3−1)×100/R → R = 200/20 = 10%.
Q2. What is the compound interest on Rs 10,000 at 10% per annum for 3 years?
🧩 T1. SI on a sum for 2 years is Rs 240 and CI for the same sum, rate and time is Rs 252. Find rate and principal.
Solution: R = 10%, P = Rs 1,200.
CI − SI = 252 − 240 = Rs 12 = P(R/100)². SI for 1 year = 240/2 = Rs 120 = P×R/100.
So P×R/100 = 120 → PR = 12,000. Also P×R²/10000 = 12 → PR × R/10000 = 12 → 12,000 × R/10000 = 12 → R = 10%.
P = 12,000/R = 12,000/10 = Rs 1,200.
🧩 T2. A machine depreciates at 10% per year. After how many years will its value be less than half the original?
Solution: 7 years.
Value after n years = V×(0.9)ⁿ. Need (0.9)ⁿ < 0.5.
(0.9)¹=0.9, (0.9)²=0.81, (0.9)³=0.729, (0.9)⁴=0.6561, (0.9)⁵=0.5905, (0.9)⁶=0.5314, (0.9)⁷=0.4783 < 0.5.
Answer: 7 years.
🧩 T3. On what sum does the difference between CI and SI for 3 years at 5% amount to Rs 76.25?
Solution: Rs 10,000.
CI − SI (3yr) = P×R²(300+R)/10⁶ = P×25×305/10⁶ = P×7625/10⁶.
76.25 = P×7625/10⁶ → P = 76.25×10⁶/7625 = Rs 10,000.
📐 Formula Sheet — MC04
Simple Interest
SI = PRT/100
A = P + SI = P(1+RT/100)
Doubles: T=100/R; Triples: T=200/R
n times: T=(n−1)×100/R
Compound Interest
A = P(1+R/100)ⁿ (annual)
Half-yearly: P(1+R/200)²ⁿ
Quarterly: P(1+R/400)⁴ⁿ
Different rates: P(1+R₁/100)(1+R₂/100)...
CI−SI Difference
2 years: P(R/100)²
3 years: PR²(300+R)/10⁶
CI always ≥ SI
Equal only at T=1 year
Growth & Decay
Growth: P(1+r/100)ⁿ
Decay/Depreciation: V(1−r/100)ⁿ
Past value: divide by factor
Different rates: multiply factors
Key Values (10%)
SI 2yr: 20%; CI 2yr: 21%
SI 3yr: 30%; CI 3yr: 33.1%
Diff 2yr at 10%: P/100
Instalment (2 equated)
P(1+2R/100) = x(1+R/100)+x
Solve for x (each instalment)
Work backwards from final instalment
⚡ Quick Revision Booster — MC04
SI Shortcuts
Doubles in 100/R years
Triples in 200/R years
SI=240 for 2yr → SI/yr=120
CI Adjustments
Half-yearly: R→R/2, n→2n
Quarterly: R→R/4, n→4n
CI always > SI (T>1)
2-Year Difference
CI−SI = P(R/100)²
Know this cold!
3yr: PR²(300+R)/10⁶
Effective Rate
10% half-yearly → 10.25% effective
= (1+R/200)²−1 × 100%
Growth/Decay
Growth: × (1+r/100) per year
Decay: × (1−r/100) per year
🚨 Key Traps
Half-yearly → adjust R AND n
CI−SI formula: P(R/100)² not P×R/100
SI flat; CI exponential
✏️ PRACTICE EXERCISE
Test Yourself — MC04
E1. The simple interest on a sum for 5 years at 8% p.a. is Rs 2,400. Find the principal.
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