Why does money grow on its own?

Ever wondered why a tiny amount in a bank can turn into a much bigger sum without you doing anything? That's the magic of compound interest – the interest you earn on both the original money (principal) and the interest that’s already been added.

In simple words, compound interest is when your money earns interest, and that interest itself starts earning interest. Imagine planting a seed that grows a new branch every year, and each branch also sprouts leaves that keep growing.

What is Compound Interest?

Compound interest is the extra amount you get when interest is calculated on the principal (the original sum) plus any interest that’s already been added. Think of it like a snowball rolling downhill: it picks up more snow (interest) as it rolls, and the bigger it gets, the faster it gathers even more.

Compound Interest Formula Explained

The standard formula that the UP Board expects you to use is:

A = P (1 + r/n)^(n·t)

  • A: Amount of money you’ll have after the time period (principal + interest).
  • P: Principal, the original sum you start with.
  • r: Annual interest rate expressed as a decimal (so 5% becomes 0.05).
  • n: Number of times interest is added (compounded) each year. If it’s yearly, n=1; half‑yearly, n=2; quarterly, n=4; monthly, n=12.
  • t: Time the money is left to grow, in years.

Notice how the term (1 + r/n) is raised to the power n·t. That power tells the computer (or your brain) how many times the interest‑adding step happens.

Step‑by‑Step Method to Solve Compound Interest Problems

When you see a question, follow this simple checklist:

  1. Read the problem carefully and write down the values of P, r, n, and t.
  2. Convert the percentage rate into a decimal.
  3. Plug the numbers into the formula.
  4. Do the arithmetic inside the parentheses first, then raise to the power.
  5. Round the final answer to the required number of decimal places.
graph TD A[Identify Principal (P)] --> B[Identify Rate (R) & Time (T)] --> C[Choose Compounding Frequency (n)] --> D[Apply Formula A = P(1 + R/n)^(nT)] --> E[Calculate & Round]

Worked Example 1 – Yearly Compounding

Problem: A student deposits ₹5,000 in a bank that offers 5% interest per year, compounded annually. How much will be in the account after 3 years?

Solution:

  1. P = 5,000
  2. r = 5% = 0.05
  3. n = 1 (interest added once a year)
  4. t = 3 years

Plug into the formula:

A = 5,000 × (1 + 0.05/1)^(1 × 3) = 5,000 × (1.05)^3.

Calculate (1.05)^3 ≈ 1.157625.

Finally, A ≈ 5,000 × 1.157625 = ₹5,788.13.

So after three years the amount becomes about ₹5,788.

Worked Example 2 – Half‑Yearly Compounding

Problem: Rahul saves ₹2,000 at an 8% annual rate, compounded half‑yearly. Find the amount after 2 years.

Solution:

  1. P = 2,000
  2. r = 8% = 0.08
  3. n = 2 (twice a year)
  4. t = 2 years

Formula:

A = 2,000 × (1 + 0.08/2)^(2 × 2) = 2,000 × (1 + 0.04)^4 = 2,000 × (1.04)^4.

(1.04)^4 ≈ 1.169859.

A ≈ 2,000 × 1.169859 = ₹2,339.72.

Hence Rahul will have roughly ₹2,340 after two years.

Quick Summary Table

SymbolMeaningExample Value
PPrincipal (initial amount)₹5,000
rAnnual interest rate (decimal)0.05 for 5%
nNumber of compounding periods per year1 for yearly, 2 for half‑yearly
tTime in years3 years
AAmount after t years₹5,788.13

Common Mistakes to Avoid

  • Forgetting to change the percent into a decimal.
  • Mixing up the compounding frequency (n) with the time period.
  • Skipping the parentheses – always compute 1 + r/n before raising to the power.

📝 Likely Exam Questions

  1. Question: A sum of ₹10,000 is invested at 6% per annum, compounded quarterly. Find the amount after 1.5 years.
    Answer: n=4, r=0.06, t=1.5 → A = 10,000(1+0.06/4)^(4×1.5) = 10,000(1.015)^6 ≈ ₹10,938.
  2. Question: If ₹7,500 grows to ₹9,000 in 2 years at a rate compounded annually, what is the rate?
    Answer: Use A = P(1+r)^t → 9,000 = 7,500(1+r)^2 → (1+r)^2 = 1.2 → 1+r = √1.2 ≈ 1.095 → r ≈ 9.5% per annum.
  3. Question: Find the compound interest earned on ₹4,000 at 10% per annum, compounded half‑yearly, for 3 years.
    Answer: n=2, r=0.10, t=3 → A = 4,000(1+0.10/2)^(2×3)=4,000(1.05)^6≈4,000×1.3401=₹5,360.4. Interest = A‑P = ₹1,360.4.
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