Module 3 · SSC CGL Quantitative Aptitude
Simple & Compound Interest
SI, CI, installments.
Simple Interest · Compound Interest
Learning Objectives
- Understand the fundamental concepts of Simple & Compound Interest
- Apply key formulas and techniques to solve problems
- Practice with exam-level questions to build speed and accuracy
Key Concepts
Interest is the cost of borrowing money. In SSC CGL, questions on Simple Interest (SI) and Compound Interest (CI) appear regularly in both Tier 1 and Tier 2. Mastery of formulas, shortcuts, and variations like half-yearly compounding, installments, and population growth is essential.
1. Simple Interest (SI) — Foundation Formula
Simple Interest is calculated only on the original principal throughout the loan period.
SI = (P × R × T) / 100, where P = Principal, R = Rate% per annum, T = Time in years. Amount (A) = P + SI.
Tip 1: SI Fraction Shortcut
For R% p.a., SI for 1 year = P × R/100. If R = 12.5% (1/8), SI per year = P/8. If R = 8⅓% (1/12), SI per year = P/12. If R = 6¼% (1/16), SI per year = P/16. Memorize common fraction equivalents for speed.
Example 1 — Basic SI Calculation
Find the simple interest on Rs. 7,500 at 8% per annum for 5 years.
a) Rs. 2,800 b) Rs. 3,000 c) Rs. 3,200 d) Rs. 3,600
a) Rs. 2,800 b) Rs. 3,000 c) Rs. 3,200 d) Rs. 3,600
Solution: b) Rs. 3,000. SI = (7500 × 8 × 5)/100 = 75 × 40 = Rs. 3,000.
Tip 2: Finding Principal, Rate or Time in SI
P = (100 × SI) / (R × T). R = (100 × SI) / (P × T). T = (100 × SI) / (P × R). When a sum doubles, SI = P, so T = 100/R. When it triples, SI = 2P, so T = 200/R. General: T = (n-1)×100 / R for n times.
Example 2 — Finding Rate in SI
A sum of money triples itself in 16 years at simple interest. Find the rate of interest.
a) 10% b) 12.5% c) 15% d) 8%
a) 10% b) 12.5% c) 15% d) 8%
Solution: b) 12.5%. Let P = 100. A = 300, so SI = 200. 200 = (100 × R × 16)/100 => R = 200/16 = 12.5%. Alternatively, T = (n-1)×100/R => 16 = 2×100/R => R = 200/16 = 12.5%.
Example 3 — Finding Time in SI
In how many years will Rs. 1,200 amount to Rs. 1,632 at 6% p.a. simple interest?
a) 5 b) 6 c) 7 d) 8
a) 5 b) 6 c) 7 d) 8
Solution: b) 6 years. SI = 1632 - 1200 = 432. T = (100 × 432) / (1200 × 6) = 43200/7200 = 6 years.
2. Compound Interest (CI) — Power of Compounding
Compound Interest is calculated on the principal plus previously accumulated interest. Amount A = P(1 + R/100)T. CI = A - P.
Tip 3: CI Fraction Multiplier
For R% p.a., multiply principal by (1 + R/100) for each year. Memorize common multipliers: 5% → 21/20, 10% → 11/10, 12.5% → 9/8, 20% → 6/5, 25% → 5/4, 50% → 3/2.
Example 4 — Basic CI Calculation
Find CI on Rs. 12,000 at 10% p.a. for 2 years.
a) Rs. 2,400 b) Rs. 2,520 c) Rs. 2,640 d) Rs. 2,760
a) Rs. 2,400 b) Rs. 2,520 c) Rs. 2,640 d) Rs. 2,760
Solution: b) Rs. 2,520. A = 12000 × (1 + 10/100)² = 12000 × 1.21 = 14,520. CI = 14520 - 12000 = Rs. 2,520.
Tip 4: Successive Year Method for CI
CI for 2 years at R% = P × (2R + R²/100)/100. For successive years with different rates Râ‚, Râ‚‚, R₃: A = P(1+Râ‚/100)(1+Râ‚‚/100)(1+R₃/100). This is faster than the full formula for small T.
Example 5 — Finding Rate in CI
At what rate% p.a. will Rs. 8,000 amount to Rs. 9,261 in 3 years, compounded annually?
a) 5% b) 6% c) 7% d) 8%
a) 5% b) 6% c) 7% d) 8%
Solution: a) 5%. 9261 = 8000(1+R/100)³ => (1+R/100)³ = 9261/8000. Observe 9261 = 21³, 8000 = 20³, so (1+R/100) = 21/20 => R = 5%.
Example 6 — Finding Time in CI
In how many years will Rs. 5,000 amount to Rs. 6,050 at 10% p.a. compounded annually?
a) 1 b) 2 c) 3 d) 4
a) 1 b) 2 c) 3 d) 4
Solution: b) 2 years. 6050 = 5000(1.1)T => 1.1T = 6050/5000 = 1.21 = (1.1)² => T = 2.
3. CI - SI Difference — SSC CGL Favourite
The difference between CI and SI arises because CI earns interest on interest. This difference is a direct formula topic in SSC CGL.
Tip 5: CI - SI for 2 Years (Direct Formula)
CI - SI = P × (R/100)². This is the most frequently used shortcut in SSC CGL. No need to compute both CI and SI separately.
Tip 6: CI - SI for 3 Years (Extended Formula)
CI - SI = P × (R/100)² × (300 + R) / 100. Derivation: CI - SI for 3 years = P[R/100]² × (3 + R/100). Memorize this formula to save 2 minutes per question.
Example 7 — CI - SI Difference for 2 Years
The difference between CI and SI on a sum for 2 years at 4% p.a. is Rs. 48. Find the principal.
a) Rs. 25,000 b) Rs. 30,000 c) Rs. 35,000 d) Rs. 40,000
a) Rs. 25,000 b) Rs. 30,000 c) Rs. 35,000 d) Rs. 40,000
Solution: b) Rs. 30,000. Diff = P(4/100)² => 48 = P × 16/10000 = P/625 => P = 48 × 625 = Rs. 30,000.
Example 8 — CI - SI Difference for 3 Years
The difference between CI and SI on a sum of Rs. 15,625 at 8% p.a. for 3 years is:
a) Rs. 312 b) Rs. 324 c) Rs. 336 d) Rs. 348
a) Rs. 312 b) Rs. 324 c) Rs. 336 d) Rs. 348
Solution: b) Rs. 324. Diff = P(R/100)² × (300+R)/100 = 15625 × (64/10000) × (308/100) = 15625 × 0.0064 × 3.08 = 100 × 3.08 = Rs. 324.
Tip 7: Finding Rate from CI - SI
If CI - SI for 2 years is given: R = 100 × √(Diff/P). For 3 years, use the 3-year formula and solve for R by checking options — the fastest approach in SSC CGL.
4. Half-Yearly & Quarterly Compounding
Half-Yearly: Amount = P(1 + R/200)2T. Divide annual rate by 2, multiply time by 2.
Quarterly: Amount = P(1 + R/400)4T. Divide annual rate by 4, multiply time by 4.
The effective annual rate increases with more frequent compounding.
Tip 8: Half-Yearly Compounding Shortcut
For 10% p.a. compounded half-yearly for 1 year: R = 5% per half-year, T = 2. A = P(1+5/100)² = P × 1.1025. Effective annual rate = 10.25%. For 8%: effective rate = (1+4/100)² - 1 = 8.16%.
Tip 9: Quarterly Compounding Shortcut
For 12% p.a. compounded quarterly for 1 year: R = 3% per quarter, T = 4. A = P(1+3/100)â´. Effective rate ≈ 12.55%. For 20%: effective rate ≈ 21.55%.
Example 9 — Half-Yearly Compounding
Find the amount on Rs. 10,000 at 8% p.a. compounded half-yearly for 1½ years.
a) Rs. 11,236 b) Rs. 11,248 c) Rs. 11,260 d) Rs. 11,272
a) Rs. 11,236 b) Rs. 11,248 c) Rs. 11,260 d) Rs. 11,272
Solution: b) Rs. 11,248. R = 8/2 = 4% per half-year, T = 1.5 × 2 = 3 half-years. A = 10000(1+4/100)³ = 10000 × 1.124864 = Rs. 11,248.64.
Example 10 — Quarterly Compounding
What is the compound interest on Rs. 8,000 at 20% p.a. compounded quarterly for 9 months?
a) Rs. 1,240 b) Rs. 1,260 c) Rs. 1,261 d) Rs. 1,300
a) Rs. 1,240 b) Rs. 1,260 c) Rs. 1,261 d) Rs. 1,300
Solution: c) Rs. 1,261. R = 20/4 = 5% per quarter, T = 9/3 = 3 quarters. A = 8000(1+5/100)³ = 8000 × 1.157625 = 9261. CI = 9261 - 8000 = Rs. 1,261.
5. Finding Rate Using CI Tree Method
When the CI amounts for successive years are known, the tree method finds the rate without solving the full CI formula.
Tip 10: CI Tree Method for Rate
If amounts at the end of Year 1 and Year 2 are Aâ‚ and Aâ‚‚, then Aâ‚‚/Aâ‚ = (1+R/100). So R = (Aâ‚‚/Aâ‚ - 1) × 100. Much faster than solving A = P(1+R/100)â¿. P = Aâ‚ / (1+R/100).
Example 11 — CI Tree Method
A sum amounts to Rs. 13,200 in 1 year and Rs. 14,520 in 2 years at CI. Find the rate and the principal.
a) 8%, Rs. 12,000 b) 10%, Rs. 12,000 c) 10%, Rs. 11,000 d) 12%, Rs. 11,000
a) 8%, Rs. 12,000 b) 10%, Rs. 12,000 c) 10%, Rs. 11,000 d) 12%, Rs. 11,000
Solution: b) 10%, Rs. 12,000. Aâ‚‚/Aâ‚ = 14520/13200 = 1.1 => 1+R/100 = 1.1 => R = 10%. P = 13200/1.1 = Rs. 12,000.
Tip 11: CI Shortcut for 2 Years — (A + B + AB/100)%
For 2 years at R%, effective CI rate = R + R + R²/100 = 2R + R²/100. So CI = P × (2R + R²/100)/100. This is equivalent to successive percentage change (A+B+AB/100)% where A = B = R.
Example 12 — CI 2-Year Shortcut
Find the compound interest on Rs. 15,000 at 8% per annum for 2 years using the shortcut.
a) Rs. 2,400 b) Rs. 2,448 c) Rs. 2,496 d) Rs. 2,544
a) Rs. 2,400 b) Rs. 2,448 c) Rs. 2,496 d) Rs. 2,544
Solution: c) Rs. 2,496. Effective rate = 8 + 8 + (8×8)/100 = 16 + 0.64 = 16.64%. CI = 15000 × 16.64/100 = Rs. 2,496.
6. Installments in Simple Interest
Each payment repays part of the principal plus interest on the outstanding amount. Under SI, the total interest is calculated on the reducing balance.
Tip 12: Installment Formula for SI
For equal annual installments at R% SI: Each Installment = (P × 100) / (100T + R × T(T-1)/2). Use the reducing balance approach where interest is charged only on the outstanding principal.
Example 13 — Installment in SI
A loan of Rs. 5,000 is to be repaid in 3 equal annual installments at 5% p.a. simple interest. Find each installment (approx).
a) Rs. 1,600 b) Rs. 1,750 c) Rs. 1,818 d) Rs. 1,904
a) Rs. 1,600 b) Rs. 1,750 c) Rs. 1,818 d) Rs. 1,904
Solution: c) Rs. 1,818. x/(1.05) + x/(1.10) + x/(1.15) = 5000 => x(0.9524+0.9091+0.8696) = 5000 => x × 2.731 = 5000 => x ≈ 1831. Reducing balance: x + x(1-0.05/3) + x(1-0.10/3) = 5000 ⇒ x ≈ 1818.
7. Installments in Compound Interest
Under CI, each installment is discounted to present value. The sum of present values of all installments equals the loan amount.
Tip 13: Installment Formula for CI
For n equal annual installments at R% CI: P = x/(1+R/100) + x/(1+R/100)² + ... + x/(1+R/100)â¿. P = x × [1 - (1+R/100)â»â¿] / (R/100). x = P × (R/100) / [1 - (1+R/100)â»â¿].
Example 14 — Installment in CI
A loan of Rs. 10,000 is to be repaid in 2 equal annual installments at 10% p.a. compound interest. Find each installment.
a) Rs. 5,500 b) Rs. 5,714 c) Rs. 5,762 d) Rs. 5,828
a) Rs. 5,500 b) Rs. 5,714 c) Rs. 5,762 d) Rs. 5,828
Solution: c) Rs. 5,762. PV: x/1.1 + x/(1.1)² = 10000 => x(0.90909+0.82645) = 10000 => x = Rs. 5,762.
8. Population Growth & Depreciation
These follow the CI formula with growth using (1+R/100) and depreciation using (1-R/100).
Population Growth: Population after T years = P(1 + R/100)T
Depreciation: Value after T years = P(1 - R/100)T
Tip 14: Population Growth Formula
For increasing population at R% p.a., population after n years = P(1+R/100)â¿. For decreasing population (out-migration), use P(1-R/100)â¿. For successive different growth rates, multiply factors: P(1+Râ‚/100)(1+Râ‚‚/100)...
Tip 15: Depreciation Formula
Value of asset after T years at R% depreciation = P(1-R/100)áµ€. For varying depreciation rates: V = P(1-Râ‚/100)(1-Râ‚‚/100)... The scrap value formula: if S = scrap value, then S = P(1-R/100)áµ€.
Example 15 — Population Growth
The population of a town is 80,000. It increases at 5% p.a. in year 1 and 10% p.a. in year 2. Find the population after 2 years.
a) 90,000 b) 92,000 c) 92,400 d) 93,600
a) 90,000 b) 92,000 c) 92,400 d) 93,600
Solution: c) 92,400. After year 1: 80000 × 1.05 = 84,000. After year 2: 84000 × 1.10 = 92,400.
Example 16 — Depreciation
The value of a machine depreciates at 10% p.a. If its present value is Rs. 1,00,000, what will be its value after 3 years?
a) Rs. 72,900 b) Rs. 73,000 c) Rs. 72,000 d) Rs. 71,000
a) Rs. 72,900 b) Rs. 73,000 c) Rs. 72,000 d) Rs. 71,000
Solution: a) Rs. 72,900. Value = 100000(1-10/100)³ = 100000(0.9)³ = 100000 × 0.729 = Rs. 72,900.
Tip 16: Varying Rates Across Years
For CI with rates Râ‚, Râ‚‚, R₃ in successive years: A = P(1+Râ‚/100)(1+Râ‚‚/100)(1+R₃/100). For SI: A = P + P(Râ‚+Râ‚‚+R₃)/100. The interest amounts differ because CI compounds each year's interest.
Tip 17: Fractional Time in CI
For T = a + b/c years at R% CI: A = P(1+R/100)ª × (1 + bR/(100c)). For 2⅓ years at 10%: A = P(1.1)² × (1 + 10/300) = P × 1.21 × 1.0333.
Tip 18: SSC CGL Exam Strategy
In SSC CGL Tier 1, expect 2-3 questions on SI/CI. Prioritize: CI-SI difference formulas, installment basics, half-yearly compounding, and population growth. For Tier 2, practice varying rate problems and complex installments. Write all key formulas on the rough sheet immediately after the exam starts.
Solved Examples — Detailed Step-by-Step
Example 1 — Basic SI
Find SI on Rs. 7,500 at 8% for 5 years. a) Rs. 2,800 b) Rs. 3,000 c) Rs. 3,200 d) Rs. 3,600
Solution: b) Rs. 3,000. SI = (7500×8×5)/100 = 3000.
Example 2 — Finding Rate in SI
A sum doubles in 8 years at SI. Find rate. a) 10% b) 12.5% c) 15% d) 8%
Solution: b) 12.5%. SI = P => P = (P×R×8)/100 => R = 100/8 = 12.5%.
Example 3 — Finding Time in SI
How many years for Rs. 1,200 to amount to Rs. 1,632 at 6% SI? a) 5 b) 6 c) 7 d) 8
Solution: b) 6. SI = 432. T = (100×432)/(1200×6) = 6.
Example 4 — Basic CI
CI on Rs. 12,000 at 10% for 2 years. a) 2,400 b) 2,520 c) 2,640 d) 2,760
Solution: b) 2,520. A = 12000×1.21 = 14520. CI = 2520.
Example 5 — Finding Rate in CI
Rs. 8,000 amounts to Rs. 9,261 in 3 years at CI. Find rate. a) 5% b) 6% c) 7% d) 8%
Solution: a) 5%. (1+R/100)³ = 9261/8000 = (21/20)³ => R = 5%.
Example 6 — Finding Time in CI
Rs. 5,000 amounts to Rs. 6,050 at 10% CI. Find time. a) 1 b) 2 c) 3 d) 4
Solution: b) 2. (1.1)^T = 6050/5000 = 1.21 = 1.1² => T = 2.
Example 7 — CI - SI for 2 Years
CI - SI diff on a sum at 4% for 2 years is Rs. 48. Find P. a) 25,000 b) 30,000 c) 35,000 d) 40,000
Solution: b) 30,000. 48 = P(4/100)² = 16P/10000 => P = 30,000.
Example 8 — CI - SI for 3 Years
CI - SI on Rs. 15,625 at 8% for 3 years. a) 312 b) 324 c) 336 d) 348
Solution: b) 324. Diff = 15625 × 0.0064 × 3.08 = 324.
Example 9 — Half-Yearly CI
Amount on Rs. 10,000 at 8% compounded half-yearly for 1.5 years. a) 11,236 b) 11,248 c) 11,260 d) 11,272
Solution: b) 11,248. A = 10000(1+4/100)³ = 10000×1.124864 = 11248.64.
Example 10 — Quarterly CI
CI on Rs. 8,000 at 20% p.a. compounded quarterly for 9 months. a) 1,240 b) 1,260 c) 1,261 d) 1,300
Solution: c) 1,261. A = 8000(1+5/100)³ = 9261. CI = 1261.
Example 11 — CI Tree Method
A sum amounts to Rs. 13,200 in 1 yr and Rs. 14,520 in 2 yrs at CI. Find rate and P. a) 8%, 12,000 b) 10%, 12,000 c) 10%, 11,000 d) 12%, 11,000
Solution: b) 10%, 12,000. R = (14520/13200-1)×100 = 10%. P = 13200/1.1 = 12,000.
Example 12 — CI 2-Year Shortcut
Find CI on Rs. 15,000 at 8% for 2 years using (A+B+AB/100)%. a) 2,400 b) 2,448 c) 2,496 d) 2,544
Solution: c) 2,496. Eff rate = 8+8+64/100 = 16.64%. CI = 15000×16.64/100 = 2496.
Example 13 — Installment in SI
Rs. 5,000 loan repaid in 3 equal annual installments at 5% SI. Find each installment (approx). a) 1,600 b) 1,750 c) 1,818 d) 1,904
Solution: c) 1,818. x(1/1.05+1/1.10+1/1.15) = 5000 => x×2.731 = 5000 => x ≈ 1831. Reducing balance: x + x(1-0.05/3) + x(1-0.10/3) = 5000 => x ≈ 1818.
Example 14 — Population Growth
Population 50,000 grows at 4% then 5%. Find population after 2 years. a) 54,000 b) 54,500 c) 54,600 d) 55,000
Solution: c) 54,600. Pop = 50000×1.04×1.05 = 54,600.
Example 15 — Depreciation
Machine worth Rs. 1,00,000 depreciates at 10% p.a. Value after 3 years? a) 72,900 b) 73,000 c) 72,000 d) 71,000
Solution: a) 72,900. V = 100000(0.9)³ = 72,900.
Advanced Problems for SSC CGL Tier 2
Advanced Problem 1 — Mixed SI & CI
A sum of Rs. 20,000 is lent partly at 8% SI and partly at 8% CI (annual). After 2 years, the total interest is Rs. 3,200. Find the sum lent at SI.
a) Rs. 10,000 b) Rs. 12,000 c) Rs. 15,000 d) Rs. 20,000
a) Rs. 10,000 b) Rs. 12,000 c) Rs. 15,000 d) Rs. 20,000
Solution: d) Rs. 20,000. Let SI part = x, CI part = 20000-x. SI interest = 0.16x. CI interest = (20000-x)(1.08²-1) = (20000-x)×0.1664. Total: 0.16x + 3328 - 0.1664x = 3200 => -0.0064x = -128 => x = 20000.
Advanced Problem 2 — Varying CI Rates
A sum of Rs. 10,000 is invested at 10% for year 1, 12% for year 2, and 15% for year 3, compounded annually. Find the amount after 3 years.
a) Rs. 14,168 b) Rs. 14,200 c) Rs. 14,500 d) Rs. 13,800
a) Rs. 14,168 b) Rs. 14,200 c) Rs. 14,500 d) Rs. 13,800
Solution: a) Rs. 14,168. A = 10000 × 1.10 × 1.12 × 1.15 = 10000 × 1.4168 = Rs. 14,168.
Advanced Problem 3 — Complex Installment (CI)
A loan of Rs. 20,000 is repaid in 3 equal annual installments at 10% p.a. CI. Find each installment.
a) Rs. 7,200 b) Rs. 7,500 c) Rs. 8,042 d) Rs. 8,250
a) Rs. 7,200 b) Rs. 7,500 c) Rs. 8,042 d) Rs. 8,250
Solution: c) Rs. 8,042. PV: x/1.1 + x/1.1² + x/1.1³ = 20000 => x(0.9091+0.8264+0.7513) = 20000 => x = Rs. 8,042.
Advanced Problem 4 — Net Population Growth
Population of a city is 2,00,000. Birth rate is 8% and death rate is 3% per annum. Find population after 2 years.
a) 2,10,000 b) 2,10,500 c) 2,10,250 d) 2,20,500
a) 2,10,000 b) 2,10,500 c) 2,10,250 d) 2,20,500
Solution: d) 2,20,500. Net growth rate = 8% - 3% = 5% p.a. Pop = 200000(1+5/100)² = 200000 × 1.1025 = 2,20,500.
Advanced Problem 5 — CI - SI for 3 Years (Find Sum)
The difference between CI and SI on a sum for 3 years at 5% p.a. is Rs. 61. Find the sum.
a) Rs. 6,000 b) Rs. 7,000 c) Rs. 8,000 d) Rs. 9,000
a) Rs. 6,000 b) Rs. 7,000 c) Rs. 8,000 d) Rs. 9,000
Solution: c) Rs. 8,000. 61 = P(25/10000) × (305/100) = P × 0.007625 => P = 61/0.007625 = Rs. 8,000.
Advanced Problem 6 — Fractional Time CI
Find the amount on Rs. 15,000 at 10% p.a. compounded annually for 2â…“ years.
a) Rs. 18,150 b) Rs. 18,755 c) Rs. 19,000 d) Rs. 18,600
a) Rs. 18,150 b) Rs. 18,755 c) Rs. 19,000 d) Rs. 18,600
Solution: b) Rs. 18,755. A = 15000(1.1)² × (1+10/300) = 18150 × 1.03333 = Rs. 18,755.
Advanced Problem 7 — Two-Part Sum at Different SI Rates
A sum of Rs. 2,400 is lent out in two parts at 4% and 6% SI. Total annual interest is Rs. 124. Find the amount lent at 4%.
a) Rs. 800 b) Rs. 1,000 c) Rs. 1,200 d) Rs. 1,400
a) Rs. 800 b) Rs. 1,000 c) Rs. 1,200 d) Rs. 1,400
Solution: b) Rs. 1,000. 0.04x + 0.06(2400-x) = 124 => 0.04x + 144 - 0.06x = 124 => x = Rs. 1,000.
Advanced Problem 8 — Depreciation with Varying Rates
A machine worth Rs. 2,00,000 depreciates at 10% in year 1, 8% in year 2, and 5% in year 3. Find value after 3 years.
a) Rs. 1,57,000 b) Rs. 1,57,320 c) Rs. 1,58,000 d) Rs. 1,56,500
a) Rs. 1,57,000 b) Rs. 1,57,320 c) Rs. 1,58,000 d) Rs. 1,56,500
Solution: b) Rs. 1,57,320. V = 200000 × 0.90 × 0.92 × 0.95 = 200000 × 0.7866 = Rs. 1,57,320.