चक्रवृद्धि ब्याज

हिंदी में पढ़ें

Compound Interest — The Eighth Wonder That Works For (or Against) You

What You Will Learn

By the end of this chapter, you will be able to:

• Compute compound interest for any rate, time, and compounding frequency — yearly, half-yearly, quarterly.
• Handle the classic exam traps: different rates for different years, fractional time periods, and the CI–SI difference.
• Use shortcut formulas (like the CI–SI difference for 2 and 3 years) to solve questions in under 30 seconds.
• Decide instantly which formula applies — amount, interest, effective rate, or growth/depreciation.

The Hook: One Grain of Rice That Emptied a Kingdom

Legend says a wise man asked a king for a modest reward: one grain of rice on the first square of a chessboard, two on the second, four on the third — doubling each time. The king laughed. By square 64, the total was more rice than the world has ever produced.

That is compounding. Growth that feeds on its own growth. Albert Einstein allegedly called compound interest "the eighth wonder of the world" — and whether or not he said it, every bank, every loan shark, and every exam paper setter behaves as if he did.

If you have ever felt that compound interest questions are "just formula plugging," you are half right — and that half is exactly where examiners set their traps. Let's fix that.

1. The Core Idea: Interest on Interest

In simple interest (SI), interest is calculated only on the original principal, every single year. The principal never grows.

In compound interest (CI), the interest earned in each period is *added* to the principal, and the next period's interest is calculated on this larger amount. Your money earns money, and then that money earns money.

Example. ₹10,000 at 10% per annum for 2 years:

• SI = 10,000 × 10 × 2 / 100 = ₹2,000
• CI: Year 1 interest = ₹1,000 → new principal ₹11,000. Year 2 interest = 10% of 11,000 = ₹1,100. Total CI = ₹2,100.

The extra ₹100 is "interest on interest" — 10% of the first year's ₹1,000. That tiny gap is the entire soul of this chapter.

🧠 Active Recall: Without looking back — why is CI for 2 years always more than SI for 2 years at the same rate? What exactly does the extra amount represent?

2. The Master Formula

When interest is compounded annually:

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

where A = amount, P = principal, R = rate % per annum, T = time in years. Then:

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

Think of it this way: every year your money gets multiplied by the growth factor (1 + R/100). At 10%, the factor is 1.1. At 20%, it is 1.2. Three years at 10% means multiplying by 1.1 three times: 1.331. So ₹1,000 becomes ₹1,331 — CI = ₹331.

🧠 Active Recall: What is the growth factor at 25% per annum? What is the amount of ₹8,000 in 2 years at 25% compounded annually? (Hint: factor = 1.25, and 1.25² = 1.5625.)

3. Compounding More Than Once a Year

Banks rarely compound only once a year. The rule is simple:

> Divide the rate, multiply the time.

• Half-yearly: rate becomes R/2, time becomes 2T.
  $$A = P\left(1 + \frac{R/2}{100}\right)^{2T}$$
• Quarterly: rate becomes R/4, time becomes 4T.
  $$A = P\left(1 + \frac{R/4}{100}\right)^{4T}$$

Example. ₹10,000 at 10% p.a. for 1 year, compounded half-yearly: factor per half-year = 1.05, applied twice → 1.1025. Amount = ₹11,025, CI = ₹1,025 — versus ₹1,000 with annual compounding. More frequent compounding = more interest, always.

🧠 Active Recall: For the same principal, rate, and time, rank these from least to most interest: annual, quarterly, half-yearly compounding. Why?

4. The CI–SI Difference: The Examiner's Favourite

This is where SSC CGL, IBPS PO, and CSAT live. Memorise these two results cold.

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

Equivalently: the difference is simply interest on the first year's interest = (R/100) × (SI of one year).

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

Example. The difference between CI and SI on a sum for 2 years at 10% is ₹50. Find the sum.
50 = P × (0.1)² = P/100 → P = ₹5,000. Ten seconds, no options needed.

🧠 Active Recall: CI–SI difference for 2 years at 5% is ₹25. What is the principal? (Use P × (1/20)² = 25.)

5. The Traps: Different Rates, Fractional Time

Different rates for different years (R₁, R₂, R₃): multiply the factors directly.
$$A = P\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)$$

Fractional time (e.g., 2⅓ years at 12%): compound for the whole years, then apply *simple* interest for the fraction on the amount so far.
$$A = P\left(1+\frac{R}{100}\right)^2 \left(1+\frac{R/3}{100}\right)$$

Effective rate for 2 years: two successive years at R% give an overall rate of
$$2R + \frac{R^2}{100}\%$$
At 10%: 20 + 1 = 21% — this is the "a + b + ab/100" successive-change formula in disguise.

🧠 Active Recall: A sum grows at 10% in year 1 and 20% in year 2. What is the total percentage growth? (1.1 × 1.2 = 1.32 → 32%, not 30%!)

6. Growth and Depreciation — Same Formula, New Costume

• Population growth / appreciation: A = P(1 + R/100)ᵀ
• Depreciation (machines, vehicles): A = P(1 − R/100)ᵀ
• Finding the past: if population is P *now* and has grown at R% for T years, the population T years ago was P ÷ (1 + R/100)ᵀ.

A machine worth ₹1,00,000 depreciating at 10% p.a. is worth 1,00,000 × 0.9² = ₹81,000 after 2 years. Same maths, minus sign.

7. The Mnemonic: "R-T Twins"

> "Compounding more often? R and T are TWINS — Rate gets Thin, Time gets Tall."

Half-yearly → R halves, T doubles. Quarterly → R quarters, T quadruples. If you remember the twins move in opposite directions, you will never mix up the formula again.

Second memory hook for the 2-year difference: "The gap is the square of the rate." Difference = P × (R/100)². Square = 2 years. Cube-ish (with the 3 + R/100 twist) = 3 years.

8. A Worked Advanced Example

Question. A sum becomes ₹13,310 in 3 years at 10% p.a. compounded annually. Find the sum, and the CI–SI difference for the same period.

• 1.1³ = 1.331, so P = 13,310 / 1.331 = ₹10,000.
• CI = ₹3,310. SI = 10,000 × 10 × 3/100 = ₹3,000.
• Difference = ₹310. Check with the formula: 10,000 × 0.01 × (3.1) = ₹310. ✔

Notice how recognising 1.331 = 1.1³ turned a "hard" question into a 15-second one. Learn the common powers: 1.1² = 1.21, 1.1³ = 1.331, 1.2² = 1.44, 1.05² = 1.1025, 1.25² = 1.5625.

Summary Table

📌 Review this chapter after 2 days. Redo the CI–SI difference derivations and the common powers (1.1³, 1.2², 1.05²) from a blank page — if you can reproduce them cold, this chapter is permanently yours.