Compound Interest — Advanced Exam Problems

🎯 Learning Objective

Reverse-engineer rate, sum and installments from CI clues — consecutive amounts, CI-SI gaps and the k-times ladder.

💡 Concept

• Consecutive-amount pattern: A₃/A₂ = 1 + r/100 — one division reveals the rate
• CI − SI for 2 years is interest on year-1's interest → r = 200 × (CI − SI)/SI
• k-times ladder: doubling in n years means 4× in 2n and 8× in 3n — powers stack at CI
• Equal installments at CI: bring every installment to present value and equate the total to the loan
• Every reverse CI question is factor-hunting — find (1 + r/100) hiding inside a ratio

🧮 Key Formulas

Consecutive amounts: A(n+1)/A(n) = 1 + r/100

>

2 years: r = 200 × (CI − SI)/SI

>

k× in n years → kᵐ× in m·n years

>

Installments: x/(1 + r/100) + x/(1 + r/100)² + … = Loan

✏️ Easy Example

Q. A sum doubles in 5 years at compound interest. In how many years will it become 8 times?

1. At CI, growth multiplies — each doubling acts on the already-doubled amount
2. 8 = 2 × 2 × 2 = 2³, so becoming 8 times needs three doublings
3. Each doubling takes the same 5 years at CI → 3 × 5 = 15

Answer: 15 years

🇮🇳 Real-Life Example

Every EMI quote from a bank is the installment maths of example three — present values equated to your loan.

📝 Exam-Level Example

Q. A sum of money at compound interest amounts to ₹4,840 in 2 years and ₹5,324 in 3 years. Find the rate of interest and the sum.

1. The jump from year 2 to year 3 is ONE year of interest earned on ₹4,840 — consecutive CI amounts differ by exactly one growth factor
2. Interest of that year = 5324 − 4840 = ₹484
3. Rate = (484/4840) × 100 = 10% — the base is the year-2 amount, because CI grows on the running balance
4. Peel two years off the ₹4,840: P = 4840 ÷ (1.1)² = 4840 ÷ 1.21
5. P = ₹4,000; check: 4000 × 1.331 = 5324 ✓

Answer: Rate = 10%, Sum = ₹4,000

📝 Exam-Level Example

Q. The simple interest on a sum for 2 years is ₹800 and the compound interest on the same sum for 2 years at the same rate is ₹832. Find the rate and the sum.

1. Year-1 interest is identical in both systems — SI's yearly slice = 800/2 = ₹400
2. CI's extra ₹32 (= 832 − 800) is interest earned ON that ₹400 during year 2 — the only place CI differs
3. Rate = (32/400) × 100 = 8% per annum
4. Now use the yearly slice to find P: 8% of P = 400
5. P = 400 × 100/8 = ₹5,000; check: 5000 × 1.08² = 5832 → CI = 832 ✓

Answer: Rate = 8%, Sum = ₹5,000

📝 Exam-Level Example

Q. A loan of ₹8,400 at 10% per annum compound interest is to be cleared in two equal annual installments. Find the value of each installment.

1. Each installment x is paid at a year-end; discount each back to today — money paid later is worth less now
2. Present value of installment 1 = x/1.1; of installment 2 = x/(1.1)² = x/1.21
3. Their present values must add up to the loan: x/1.1 + x/1.21 = 8400
4. Combine over 1.21: x(1.1 + 1)/1.21 = 8400 → 2.1x = 8400 × 1.21 = 10164
5. x = 10164/2.1 = ₹4,840
6. Verify forward: 8400 × 1.1 = 9240, pay 4840 → 4400 left; 4400 × 1.1 = 4840, final payment clears it ✓

Answer: ₹4,840

🪄 Memory Trick

k-times CI ladder: 2× in n years → 4× in 2n, 8× in 3n. Powers stack — never use SI's linear adding here.

⚠️ Common Mistakes

• ❌ Applying the SI multiple rule ((k−1) extra principals) to CI — CI stacks powers instead
• ❌ Dividing the year-2-to-year-3 jump by the PRINCIPAL instead of the year-2 amount when finding r
• ❌ Setting each CI installment as loan/2 plus interest — only present-value discounting works

🏆 Exam Tips

• ✅ Consecutive amounts like 5324/4840 always hand you 1 + r/100 directly
• ✅ Memorise 1.1² = 1.21, 1.1³ = 1.331, 1.05² = 1.1025 — they unlock reverse questions
• ✅ Verify installments forward: grow, pay, grow, pay — the balance must land on zero

📌 Summary

• Consecutive CI amounts → divide to get the rate
• CI − SI (2 yr) = one year's interest on year-1's interest
• Multiples stack as powers: 2× in n → 8× in 3n
• Installments: discount to present value, sum equals loan