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Numerical Ability · AFCAT

NA04 — Simple & Compound Interest

📈 Numerical Ability – NA04 AFCAT Level ☆ Low Priority

📌 AFCAT Focus 2022–2026: SI & CI together fetch 1–2 questions per paper. The most tested sub-topics are: finding principal/rate/time from SI, the difference between SI and CI for 2 years (= Pr²/100²), and population/depreciation as CI problems. Master the four SI variables and the two CI shortcuts — that covers 90% of AFCAT questions here.

1. Simple Interest (SI)

Fig. 1.1 — Simple Interest Triangle: Finding Any Variable from the Other Three

SI Core Formulas:
● SI = PRT/100 A = P + SI = P(1 + RT/100)
● If rate changes each year: SI = P(R₁+R₂+R₃...)/100
● SI doubles P when T = 100/R SI triples P when T = 200/R

✎ Worked Example — Finding Rate

Rs 5,000 amounts to Rs 6,500 in 3 years at SI. Find the rate of interest.

SI = Amount − Principal = 6500 − 5000 = Rs 1,500
R = 100 × SI / (P × T) = 100 × 1500 / (5000 × 3) = 1,50,000 / 15,000 = 10% p.a.

✔ Rate = 10% per annum

COMPOUND INTEREST

2. Compound Interest (CI)

Fig. 2.1 — Compound Interest: Compounding Frequencies & Key Formulas

🏭 Population Growth (CI Model)

Population after n years at growth rate r%:
Pₙ = P×(1 + r/100)ⁿ
If growth rate is different each year:
Pₙ = P×(1+r₁/100)(1+r₂/100)...
If rate is given per half-year, adjust accordingly
Present population given future: work backwards
P = Pₙ / (1 + r/100)ⁿ

🏭 Depreciation (CI Model)

Value after n years at depreciation rate r%:
Vₙ = V×(1 − r/100)ⁿ
Value decreases each year — use minus sign
Machine worth Rs 50,000 depreciates 10%/yr:
After 2 yrs: 50,000×(0.9)² = 50,000×0.81 = Rs 40,500
Mixed: appreciation then depreciation — multiply factors

✎ Worked Example — CI vs SI Difference

The CI on a sum at 10% p.a. for 2 years is Rs 210. Find the SI for the same sum, rate and period.

For 2 years: CI − SI = P(R/100)²
Let SI (2 yrs) = 2Pr/100 = 2P(10)/100 = P/5
CI = P[(1.1)²−1] = 0.21P = Rs 210 → P = Rs 1,000
SI = PRT/100 = 1000×10×2/100 = Rs 200
Check: CI−SI = 210−200 = 10 = 1000×(10/100)² = 1000×0.01 = 10 ✓

✔ SI = Rs 200

Topic CSI Applications — Instalments & LoansAFCAT Direct

Equal Instalments

If a loan of P is repaid in n equal annual instalments at r% SI per annum:
Instalment = P(1 + nr/100) / n
Total amount paid = n × Instalment

Instalment Example

Loan = Rs 10,000, 2 years, 10% SI p.a., paid in 2 equal instalments.
Amount after 2 yrs = 10,000 + 10,000×10×2/100 = Rs 12,000.
Each instalment = 12,000/2 = Rs 6,000

Buying on Credit

Cash price = C. Paid as down payment D + n instalments of I each at r% SI:
SI is charged on (C − D) for the instalment period.
Total paid = D + n×I. Extra paid (interest) = Total paid − C.

Topic DCI for 3 Years & DepreciationAFCAT Direct

CI for 3 years

CI (3 yrs) − SI (3 yrs) = SI × r(300+r) / 100²
where SI = PRT/100 for 3 years, r = rate per annum.
Simpler: A = P(1+r/100)³ — just substitute and compute.

Effective Rate

10% p.a. for 3 years CI: A = P×(1.1)³ = P×1.331 → Effective rate = 33.1%
20% p.a. for 2 years CI: Effective = 44%. Memorise common values.

Depreciation Formula

Value after n years = V × (1 − r/100)ⁿ
Depreciation reduces value; use minus sign (not plus).
The formula is identical to CI but with minus.

Appreciation

Value increases: V × (1 + r/100)ⁿ — same as CI formula. E.g. land value Rs 5 lakh appreciates 10% p.a. After 2 yrs: 5,00,000 × (1.1)² = Rs 6,05,000

✎ Worked Example — Depreciation

A car is purchased for Rs 8,00,000. It depreciates at 15% p.a. Find its value after 2 years.

Value after 2 years = P × (1 − r/100)ⁿ
= 8,00,000 × (1 − 15/100)²
= 8,00,000 × (0.85)²
= 8,00,000 × 0.7225 = Rs 5,78,000

✔ Value after 2 years = Rs 5,78,000

📐 Formula Sheet — NA04

Simple Interest

SI = PRT/100
A = P + SI = P(1 + RT/100)
P = 100×SI/(R×T)
R = 100×SI/(P×T) T = 100×SI/(P×R)

Compound Interest

Annual: A = P(1+R/100)ⁿ
Half-yearly: A = P(1+R/200)²ⁿ
Quarterly: A = P(1+R/400)⁴ⁿ
CI = A − P

SI vs CI Difference

2 years: CI−SI = P×R²/10,000
3 years: CI−SI = PR²(300+R)/10³×100³
CI > SI always (for same P,R,T>1yr)
For 1 year at same rate: CI = SI

Population & Depreciation

Growth: Pₙ = P(1+r/100)ⁿ
Depreciation: Vₙ = V(1−r/100)ⁿ
Mixed rates: multiply each factor
Find past value: divide by factor

Key SI Shortcuts

SI doubles: T = 100/R years
SI triples: T = 200/R years
If sum doubles in T yrs (SI): R = 100/T
Rate per annum always unless stated

Quick Recall

10% for 2 yrs CI: 21% effective
10% for 3 yrs CI: 33.1% effective
20% for 2 yrs CI: 44% effective
Quarterly 8% = Half-yearly 8.16%

📝 Topic-Wise PYQs — NA04

Q1. A sum of Rs 8,000 earns SI of Rs 2,000 in 4 years. Find the rate of interest per annum. AFCAT PYQ

(a) 5%(b) 6%(c) 6.25%(d) 8%

✔ Answer: (c) 6.25%

R = 100×SI/(P×T) = 100×2000/(8000×4) = 2,00,000/32,000 = 6.25%

Q2. Find the CI on Rs 10,000 at 10% p.a. for 2 years, compounded annually. AFCAT PYQ

(a) Rs 2,000(b) Rs 2,100(c) Rs 2,200(d) Rs 1,900

✔ Answer: (b) Rs 2,100

A = 10,000×(1.1)² = 10,000×1.21 = Rs 12,100. CI = 12,100−10,000 = Rs 2,100. Shortcut: SI for 2 yrs = Rs 2,000; extra = P(R/100)² = 10,000×0.01 = Rs 100. CI = 2000+100 = Rs 2,100.

Q3. A town’s population is 1,00,000. It increases at 10% p.a. What will the population be after 3 years? AFCAT PYQ

(a) 1,31,000(b) 1,33,100(c) 1,30,000(d) 1,21,000

✔ Answer: (b) 1,33,100

Pₙ = 1,00,000×(1.1)³ = 1,00,000×1.331 = 1,33,100. Population growth = CI model.

Q4. The difference between CI and SI on Rs 5,000 at 8% p.a. for 2 years is: ⚡ Tricky

(a) Rs 30(b) Rs 32(c) Rs 40(d) Rs 80

✔ Answer: (b) Rs 32

CI−SI = P(R/100)² = 5000×(8/100)² = 5000×64/10,000 = 5000×0.0064 = Rs 32.

Q5. At what rate of SI will a sum triple itself in 25 years? ⚡ Tricky

(a) 6%(b) 8%(c) 10%(d) 4%

✔ Answer: (b) 8%

Sum triples → SI = 2P (since A = 3P). R = 100×SI/(P×T) = 100×2P/(P×25) = 200/25 = 8%. Shortcut: SI triples when T = 200/R, so R = 200/25 = 8%.

🧠 Quick Memory Chart — NA04

📈 SI Formulas

SI = PRT/100
A = P + SI
Doubles: T = 100/R yrs
Triples: T = 200/R yrs
SI same every year
Calculated on original P

📈 CI Formulas

Annual: P(1+R/100)ⁿ
Half-yearly: R÷2, T×2
Quarterly: R÷4, T×4
CI > SI always (T>1)
2-yr diff: PR²/10,000
10% for 2yr = 21% effective

📈 Population & Depn.

Growth: P(1+r/100)ⁿ
Depreciation: V(1−r/100)ⁿ
Depn: use − sign
Mixed rates: multiply factors
Same formula as CI

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