Simple Interest — Advanced Exam Problems

🎯 Learning Objective

Crack two-amount, rate-equals-time and borrow-lend SI questions using the straight-line timeline.

💡 Concept

• Two-amount pattern: (A₂ − A₁) over the extra years = fixed yearly SI; subtract back to reach P
• Rate numerically equal to time → SI = PR²/100, solved by a square root
• Borrow at R₁, lend at R₂ → yearly gain = the spread (R₂ − R₁) on the whole principal
• Different rates for different periods → add each period's PRT/100 slab separately
• SI never bends — every hard question is a straight-line timeline of equal yearly jumps

🧮 Key Formulas

Yearly SI = (A₂ − A₁)/(t₂ − t₁)

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P = A₁ − t₁ × yearly SI

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R = T → SI = PR²/100

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Arbitrage gain = P × (R₂ − R₁) × T/100

✏️ Easy Example

Q. A sum amounts to ₹6,200 in 2 years and ₹7,400 in 5 years at simple interest. Find the yearly interest.

1. The amount grew from 6200 to 7400 between year 2 and year 5 — that growth is pure interest of 3 years
2. 3 years' SI = 7400 − 6200 = ₹1,200
3. SI is equal every year, so yearly interest = 1200/3 = ₹400

Answer: ₹400 per year

🇮🇳 Real-Life Example

Money lenders who borrow from banks at 10% and lend onward at 24% live entirely on the spread — the maths of example three.

📝 Exam-Level Example

Q. A sum amounts to ₹5,800 in 2 years and ₹7,000 in 5 years at simple interest. Find the sum and the rate of interest.

1. Amount gap = 7000 − 5800 = ₹1,200 across 5 − 2 = 3 years — under SI this gap is pure interest, no compounding
2. Yearly SI = 1200/3 = ₹400
3. Walk back to the start: P = (amount after 2 years) − (2 years' interest) = 5800 − 800 = ₹5,000
4. For the rate, ask: ₹400 is what per cent of ₹5,000 for one year?
5. R = (400/5000) × 100 = 8% per annum

Answer: Sum = ₹5,000, Rate = 8%

📝 Exam-Level Example

Q. A sum of ₹1,600 earns ₹1,024 as simple interest at a rate numerically equal to the time in years. Find the rate.

1. Let rate = R% and time = R years — the question says they are numerically equal, so one unknown is enough
2. SI = P × R × T/100 = 1600 × R × R/100 = 16R²
3. Set the earning equal: 16R² = 1024
4. R² = 1024/16 = 64
5. R = √64 = 8 → rate 8% per annum (and time 8 years)

Answer: 8% per annum (time 8 years)

📝 Exam-Level Example

Q. A man borrows ₹12,000 at 4% p.a. simple interest and immediately lends the whole amount at 6.25% p.a. Find his total gain in 3 years.

1. He pays 4% but earns 6.25% on the SAME ₹12,000 — his real earning is the spread, 6.25 − 4 = 2.25% per year
2. Yearly gain = 2.25% of 12000 = (2.25 × 12000)/100 = ₹270
3. SI is flat, so 3 years' gain = 270 × 3 — no compounding, just multiply
4. Total gain = ₹810

Answer: ₹810

🪄 Memory Trick

Two amounts given? Gap ÷ year-gap = yearly SI; peel back to P; rate = yearly SI/P × 100. Three moves finish every such question.

⚠️ Common Mistakes

• ❌ Compounding the gap — under SI the amount grows in a straight line, so the gap is plain interest
• ❌ Taking the 2-year amount itself as P instead of stripping the interest back out
• ❌ Using the full lending rate (6.25%) instead of the spread (2.25%) for arbitrage gains

🏆 Exam Tips

• ✅ R = T questions always end in a square root — expect R² equal to a clean square
• ✅ Draw the timeline P → A₁ → A₂ with equal yearly jumps before touching formulas
• ✅ Arbitrage in one multiplication: spread × principal × years ÷ 100

📌 Summary

• Two amounts → gap is pure SI; walk the straight line back to P
• R = T → SI = PR²/100, answer via square root
• Borrow low, lend high → gain on the spread only
• Everything stays linear — that is the power of SI