Compound Interest
Compound Interest
Under compound interest (CI) the interest of each period is added to the principal, so the next period's interest is earned on a bigger amount. Every CI question is therefore a product of factors (1 + R/100). NTPC asks two- and three-year CI, half-yearly and quarterly compounding, "doubles in n years", amounts in consecutive years, different rates in different years and fractional years.
1. Formula Box
A = P(1 + R/100)ⁿ CI = A − P
Half-yearly: rate R/2, periods 2n Quarterly: rate R/4, periods 4n
Different rates each year: A = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)
Fractional time (2 1/2 years, annual compounding): A = P(1 + R/100)² × (1 + R/200)
| Rate | 2-year factor | 3-year factor | CI % for 2 years | CI % for 3 years |
|---|---|---|---|---|
| 5% | 1.1025 | 1.157625 | 10.25% | 15.7625% |
| 10% | 1.21 | 1.331 | 21% | 33.1% |
| 20% | 1.44 | 1.728 | 44% | 72.8% |
| 4% | 1.0816 | 1.124864 | 8.16% | 12.4864% |
| 12% | 1.2544 | 1.404928 | 25.44% | 40.4928% |
Two-year CI percent = 2R + R²/100 (same as successive percentage change).
2. Shortcut Rules
| Situation | Rule | Example |
|---|---|---|
| Becomes n times in t years | Becomes nᵏ times in k × t years | Doubles in 6 y, 8 times (2³) in 18 y |
| Amounts after n and n + 1 years | Rate = (A₂ − A₁)/A₁ × 100 | 8,820 then 9,261: 441/8,820 = 5% |
| Interest of the kth year alone | P × (1 + R/100)^(k − 1) × R/100 | 3rd year at 10%: 0.121P |
| CI for consecutive years | Each year's CI = previous year's CI × (1 + R/100) | 1,000, 1,100, 1,210 at 10% |
✗ Doubles in 6 years so 4 times in 12 years, 8 times in 24 years | ✓ CI multiplies: 4 = 2², so 12 years; 8 = 2³, so 18 years
✗ 10% half-yearly for 1 year = 10% annual | ✓ Half-yearly uses 5% twice: 1.05² = 1.1025, so 10.25%
हिंदी नोट: चक्रवृद्धि ब्याज में हर वर्ष का गुणांक (1 + R/100) गुणा होता है। अर्धवार्षिक में दर आधी और अवधियाँ दुगुनी, त्रैमासिक में दर एक-चौथाई और अवधियाँ चार गुनी हो जाती हैं।
Exam Pointer: Verified NTPC patterns: annual CI for 2 or 3 years and the rate from two amounts (June 2022 CBT-2, August 2025 UG CBT-1), half-yearly and quarterly compounding (March 2021, June 2022 CBT-2), doubling time (January 2021), amounts in consecutive years and the interest of a single year (June 2022 CBT-2, August 2025 UG CBT-1), and different rates or fractional years (March 2021, June 2022 CBT-2).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Annual compounding for 2 or 3 years, and the rate from the amount
[PYQ: NTPC CBT-2 17-Jun-2022 Shift-2 | NTPC UG CBT-1 13-Aug-2025 Shift-3]
EXAM LEVEL
Q. Find the compound interest on ₹8,000 at 10% per annum for 3 years, compounded annually.
A = 8,000 × 1.1³ = 8,000 × 1.331 = ₹10,648. CI = ₹2,648.
Answer: ₹2,648
EXAMATLAS LEVEL
Q. ₹6,400 amounts to ₹7,225 in 2 years, compounded annually. Find the rate.
(1 + R/100)² = 7,225/6,400 = (85/80)², so 1 + R/100 = 85/80 = 1.0625 and R = 6.25%. Spotting the perfect squares 85² and 80² avoids any long division.
Answer: 6.25%
Pattern 2: Half-yearly and quarterly compounding
[PYQ: NTPC CBT-1 8-Mar-2021 Shift-1 | NTPC CBT-2 13-Jun-2022 Shift-2]
EXAM LEVEL
Q. Find the compound interest on ₹16,000 for 1 year at 10% per annum compounded half-yearly.
Rate 5% per half-year, 2 periods: A = 16,000 × 1.05² = 16,000 × 1.1025 = ₹17,640. CI = ₹1,640 (₹40 more than the ₹1,600 SI).
Answer: ₹1,640
EXAMATLAS LEVEL
Q. Find the compound interest on ₹40,000 for 9 months at 8% per annum compounded quarterly.
Rate per quarter = 2%, number of quarters = 3. A = 40,000 × 1.02³ = 40,000 × 1.061208 = ₹42,448.32. CI = ₹2,448.32.
Answer: ₹2,448.32
Pattern 3: A sum becomes n times under CI
[PYQ: NTPC CBT-1 12-Jan-2021 Shift-1]
EXAM LEVEL
Q. A sum doubles in 6 years at compound interest. In how many years will it become 8 times?
8 = 2³, so it takes 3 doubling periods: 18 years.
Answer: 18 years
EXAMATLAS LEVEL
Q. A sum becomes 4 times itself in 6 years at compound interest. In how many years will it become 32 times?
4 = 2² in 6 years means it doubles every 3 years. 32 = 2⁵ needs 5 doublings: 15 years. Answering 48 years (6 × 32/4) uses plain proportion, and 62 years uses SI logic (31P of interest at 0.5P per year); both are wrong under CI.
Answer: 15 years
Pattern 4: Amounts in consecutive years, and the interest of one particular year
[PYQ: NTPC CBT-2 17-Jun-2022 Shift-1 | NTPC UG CBT-1 19-Aug-2025 Shift-1]
EXAM LEVEL
Q. A sum at compound interest amounts to ₹8,820 in 2 years and ₹9,261 in 3 years. Find the rate and the sum.
The third year's interest is 9,261 − 8,820 = ₹441, earned on ₹8,820, so R = 441/8,820 × 100 = 5%. P = 8,820/1.05² = 8,820/1.1025 = ₹8,000.
Answer: 5%; ₹8,000
EXAMATLAS LEVEL
Q. The interest earned during the 3rd year on a sum at 10% compound interest is ₹1,452. Find the sum.
At the start of year 3 the amount is P × 1.1² = 1.21P, and the 3rd year's interest is 10% of that = 0.121P. So 0.121P = 1,452 and P = ₹12,000. Using 0.1P (simple interest thinking) gives ₹14,520, the trap.
Answer: ₹12,000
Pattern 5: Different rates in different years, and fractional years
[PYQ: NTPC CBT-1 13-Mar-2021 Shift-1 | NTPC CBT-2 14-Jun-2022 Shift-1]
EXAM LEVEL
Q. Find the compound interest on ₹20,000 if the rate is 5% for the first year and 8% for the second year.
A = 20,000 × 1.05 × 1.08 = ₹22,680, CI = ₹2,680.
Answer: ₹2,680
EXAMATLAS LEVEL
Q. Find the compound interest on ₹12,000 for 2 1/2 years at 10% per annum compounded annually.
Two full years: 12,000 × 1.21 = 14,520. For the last half-year, interest at 10% × 1/2 = 5% on 14,520: 14,520 × 1.05 = ₹15,246. CI = ₹3,246. Using 1.1^2.5 is the trap; the fractional year earns simple interest on the amount reached.
Answer: ₹3,246
60-Second Revision
- A = P(1 + R/100)ⁿ; two-year CI % = 2R + R²/100.
- Half-yearly: R/2 and 2n periods; quarterly: R/4 and 4n periods.
- n times in t years means nᵏ times in kt years.
- Consecutive amounts: rate = difference ÷ earlier amount; kth-year interest = P(1 + R/100)^(k − 1) × R/100.
- Different rates: multiply factors; fractional year: last part at proportional simple rate.