Skip to content
Free shipping above ₹499
Oakspine Press

Simple and compound interest for IBPS PO

For a banking exam, interest is a must-know. SI from two amounts, CI year by year, the CI − SI shortcuts for two and three years, half-yearly and quarterly compounding, rates from two CI amounts and multiplying sums, with why each works, worked examples and practice.

3 Oct 2026 6 min read

In this guide
  1. Simple interest
  2. Compound interest
  3. Worked examples
  4. Common mistakes
  5. Practice set
  6. What to do next

Interest is the one arithmetic topic every bank officer uses at work, and IBPS PO asks it in several forms: a direct word problem in the prelims, one side of a quantity comparison, a data sufficiency stem, or a column in a mains DI table. The questions are rarely hard. They are lost to small slips: amount confused with interest, a rate not halved, a shortcut used where it doesn't apply.

The fix is to understand what each formula is doing. Simple interest grows by the same amount every year. Compound interest grows by the same percentage every year. Almost every shortcut follows from that one difference.

Simple interest

  • SI = P × R × T ÷ 100. Amount = P + SI.
  • SI is the same every year: P × R ÷ 100.

Why it works: interest is charged only on the original principal, so each year adds the same amount. Over T years, that is T times the yearly interest.

Two amounts, two times. If a sum becomes A₁ after t₁ years and A₂ after t₂ years at SI, the yearly interest is (A₂ − A₁) ÷ (t₂ − t₁). Then subtract t₁ years of interest from A₁ to get the principal.

Doubling at SI. A sum doubles when SI equals P, that is when R × T = 100. At 12.5% it takes 8 years; at 10% it takes 10 years.

Compound interest

  • Amount = P(1 + R/100)ⁿ. CI = Amount − P.
  • For small n, go year by year: add R% of the latest amount each year.

Why it works: each year's interest is added to the principal, so next year's interest is charged on a bigger sum. Each year multiplies the amount by (1 + R/100).

Shortcuts and the reasons behind them

ShortcutFormulaWhy
CI − SI, 2 yearsP(R/100)²In year 2, CI also earns interest on year 1's interest: P × R/100 × R/100
CI − SI, 3 yearsP(R/100)² × (3 + R/100)Interest on interest builds up in years 2 and 3
Half-yearly compoundingRate R/2, periods 2nInterest is added twice a year at half the annual rate
Quarterly compoundingRate R/4, periods 4nFour additions a year at a quarter of the rate
Rate from two consecutive CI amountsR = (A₂ − A₁) ÷ A₁ × 100The later amount is the earlier one multiplied by (1 + R/100)
Multiplying sums at CIIf P becomes k times in n years, it becomes k² times in 2n yearsEach block of n years multiplies by k

Worked examples

Example 1: A sum amounts to ₹7,200 in 2 years and ₹8,000 in 3 years at SI. Find the principal and the rate.

  • Yearly SI = 8,000 − 7,200 = 800.
  • P = 7,200 − 2 × 800 = ₹5,600.
  • R = 800 ÷ 5,600 × 100 = 14 2/7%, about 14.29%.

Example 2: Find the CI on ₹10,000 at 10% a year for 3 years, compounded annually.

  • 10,000 → 11,000 → 12,100 → 13,310.
  • CI = ₹3,310.

Example 3: The difference between CI and SI on a sum for 2 years at 10% is ₹60. Find the sum.

  • P × (0.1)² = 60, so P = 60 ÷ 0.01 = ₹6,000.

Example 4: Find the difference between CI and SI on ₹10,000 at 10% for 3 years.

  • Formula: 10,000 × 0.01 × 3.1 = ₹310.
  • Check: CI = 3,310 (Example 2), SI = 3,000. Difference 310.

Example 5: Find the CI on ₹8,000 at 10% a year, compounded half-yearly, for 1 year.

  • 5% for 2 half-years: 8,000 × 1.05 × 1.05 = 8,820.
  • CI = ₹820. Annual compounding would give ₹800.

Example 6: A sum amounts to ₹6,050 after 2 years and ₹6,655 after 3 years at CI. Find the rate and the principal.

  • R = (6,655 − 6,050) ÷ 6,050 × 100 = 605 ÷ 6,050 × 100 = 10%.
  • P = 6,050 ÷ 1.21 = ₹5,000.

Example 7: A sum doubles in 5 years at CI. In how many years will it become 8 times?

  • 8 = 2³, so three blocks of 5 years: 15 years.
  • At SI this would be wrong. Doubling in 5 years at SI means 20% a year, and 8 times needs SI = 7P, which takes 35 years.

Common mistakes

  • Giving the amount when the question asks for interest, or the reverse.
  • Forgetting to halve the rate (and double the periods) for half-yearly compounding.
  • Using the 2-year CI − SI shortcut for 3 years. Use the 3-year form.
  • Using the multiplying-sum rule at SI. It works only for compound interest.
  • Subtracting two CI amounts to get a year's interest, then dividing by P. For CI, divide by the earlier amount, not by P.

Practice set

  1. Find the SI on ₹12,000 at 7.5% a year for 4 years.
  2. Find the CI on ₹5,000 at 20% a year for 2 years, compounded annually.
  3. The difference between CI and SI for 2 years at 5% is ₹25. Find the principal.
  4. A sum becomes ₹4,400 in 2 years and ₹4,800 in 4 years at SI. Find the principal and the rate.
  5. Find the CI on ₹16,000 at 10% a year, compounded half-yearly, for 1 year.
  6. A sum amounts to ₹8,820 in 2 years and ₹9,261 in 3 years at CI. Find the rate and the principal.
  7. At what rate of SI does a sum triple in 6 years?
  8. Find the CI on ₹10,000 at 8% a year, compounded quarterly, for 6 months.

Answers:

  1. ₹3,600. 12,000 × 7.5 × 4 ÷ 100.
  2. ₹2,200. 5,000 × 1.44 = 7,200.
  3. ₹10,000. P × 0.0025 = 25.
  4. ₹4,000 at 5%. Yearly SI = 400 ÷ 2 = 200; P = 4,400 − 400 = 4,000; R = 200 ÷ 4,000 × 100.
  5. ₹1,640. 16,000 × 1.05² = 17,640.
  6. 5%; ₹8,000. 441 ÷ 8,820 = 0.05; 8,820 ÷ 1.1025 = 8,000.
  7. 33⅓%. Tripling means SI = 2P, so R × 6 = 200.
  8. ₹404. 2% a quarter for 2 quarters: 10,000 × 1.02² = 10,404.

What to do next

  • Write the shortcut table from memory once a day for a week, with the reason for each line.
  • Solve ten interest questions a day, alternating SI and CI, and check each answer against the year-by-year method.
  • Use interest questions in quantity comparison practice, where CI versus SI is a favourite pair.
  • Keep percentage factors sharp with the percentage guide.

A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Institute of Banking Personnel Selection website .

Get the next IBPS PO guide by email

New guides every week. No spam, unsubscribe any time.

Keep reading

IBPS POQuantitative AptitudeProfit, loss and discount for IBPS POMark-up and discount, "selling price of 20 equals cost price of 25", equal selling prices with profit and loss, false weights, buy-x-get-y-free offers and successive discounts. The multiplier method and why it works, eight worked examples and practice with solutions. 6 min read·2 Oct 2026IBPS POQuantitative AptitudePercentage for IBPS POPercentage sits under most of IBPS PO quant, from arithmetic to DI and quantity comparison. The base rule, fraction equivalents, reverse and successive changes, elections, exams, price and consumption, income and savings, with eight worked examples and practice. 6 min read·1 Oct 2026IBPS POQuantitative AptitudeQuantity comparison (Q1 vs Q2) for IBPS POWork out Quantity I and Quantity II, then say which is bigger, whether they are equal, or that no relation can be set. The answer options, a method that avoids wasted calculation, the unit and two-value traps, eight worked examples across arithmetic topics, and practice with solutions. 6 min read·30 Sept 2026IBPS POQuantitative AptitudeQuantitative data sufficiency for IBPS POYou don't have to find the answer, only decide whether the statements give enough information to find one. The standard options, a checking order, yes/no questions, the three-statement version, and the traps where even both statements are not enough, with worked examples and practice. 7 min read·29 Sept 2026