Simple and compound interest for SBI PO

Interest is the one quant topic every banking aspirant will use at work, and it shows up across the SBI PO paper: as a word problem in the prelims, as a statement in data sufficiency, as one side of a quantity comparison, and inside caselets about deposits and loans. The formulas are short. The marks go to candidates who can apply them without writing out long decimal multiplications.

This guide covers the formulas and why they work, the fraction method that makes compound interest fast, and the question types that go beyond the basics: CI–SI differences, changing rates, half-yearly compounding and equal instalments.

The formulas, and why they work

Simple interest (SI) is charged on the original principal only. Each year earns the same amount.

• SI = P × R × T ÷ 100.
• Amount = P + SI.

Compound interest (CI) is charged on the principal plus the interest already added. Each year's amount is the previous year's × (1 + R/100).

• Amount = P × (1 + R/100)ᵀ.
• CI = amount − P.

Why CI grows faster. In year 2, CI earns interest on year 1's interest as well. For two years at 10% on ₹10,000, SI is ₹2,000 and CI is ₹2,100. The extra ₹100 is 10% of year 1's interest of ₹1,000.

That idea gives the two difference formulas:

• CI − SI for 2 years = P × (R/100)².
• CI − SI for 3 years = P × (R/100)² × (3 + R/100).

The fraction method

Turn the rate into a fraction and each year becomes one multiplication by a simple ratio.

For 10%, the multipliers are worth memorising: 1.1, 1.21 and 1.331 for one, two and three years.

Worked examples

Example 1. Fractional rate. Find the CI on ₹16,000 at 12.5% a year for 2 years.

• 12.5% = 1/8, so each year multiplies by 9/8.
• Amount = 16,000 × 81/64 = 250 × 81 = 20,250.
• CI = ₹4,250.

Example 2. CI − SI for three years. Find the difference on ₹8,000 at 10% for 3 years.

• 8,000 × (0.1)² × 3.1 = 8,000 × 0.01 × 3.1 = ₹248.
• Check: SI = 2,400; CI = 8,000 × 0.331 = 2,648; the difference is 248.

Example 3. Principal from the difference. The CI − SI on a sum at 5% for 2 years is ₹50. Find the sum.

• P × (0.05)² = 50, so P = 50 ÷ 0.0025 = ₹20,000.

Example 4. Working back. A sum becomes ₹13,310 in 3 years at 10% compound interest. Find it.

• 13,310 ÷ 1.331 = ₹10,000.

Example 5. Doubling under SI. At what simple interest rate does a sum double in 8 years?

• Doubling means the interest equals the principal. P × R × 8 ÷ 100 = P, so R = 100 ÷ 8 = 12.5%.

Example 6. Multiples under CI. A sum doubles in 5 years at compound interest. In how many years will it become 8 times?

• Every 5 years it multiplies by 2. 8 = 2³, so it needs three such periods: 15 years.
• This works only for CI, because CI multiplies by the same factor in every equal period. Under SI, a sum that doubles in 5 years becomes 8 times in 35 years (it needs 7 times the principal as interest).

Example 7. Different rates each year. ₹10,000 is invested at 10% in the first year and 20% in the second, compounded annually. Find the CI.

• 10,000 × 1.1 × 1.2 = 13,200. CI = ₹3,200.

Example 8. Half-yearly compounding. What is the effective annual rate of 10% a year compounded half-yearly?

• Half-yearly means 5% for each half-year, twice a year.
• (1.05)² − 1 = 0.1025, so 10.25%.
• General rule: for half-yearly compounding, halve the rate and double the number of periods.

Example 9. Equal annual instalments. A loan of ₹21,000 at 10% compound interest is repaid in two equal annual instalments. Find each instalment.

• Each instalment, brought back to today, must add up to the loan: x ÷ 1.1 + x ÷ 1.21 = 21,000.
• Over a common denominator: x × (1.1 + 1) ÷ 1.21 = 21,000, so x = 21,000 × 1.21 ÷ 2.1 = ₹12,100.
• Check: after year 1 the balance is 23,100 − 12,100 = 11,000; after year 2 it is 12,100 − 12,100 = 0.

Reading what the question wants

If a question says only "compound interest" without the frequency, assume annual compounding, which is the usual convention in these papers.

Common mistakes

• Using SI logic for CI multiples. Doubling in 5 years means 8 times in 15 years under CI, but in 35 years under SI.
• Forgetting to halve the rate for half-yearly compounding, or forgetting to double the number of periods.
• Answering with the amount when the question asks for the interest, or the other way round.
• Rounding 1.21 or 1.331. Keep these exact; they make the final answer come out whole.

Practice set

1. Find the CI on ₹20,000 at 10% a year for 3 years.
2. Find the SI on ₹7,500 at 8% a year for 5 years.
3. Find the CI − SI on ₹12,000 at 5% a year for 2 years.
4. What is the effective annual rate of 8% a year compounded half-yearly?
5. A sum triples in 6 years at compound interest. In how many years will it become 9 times?
6. Find the CI on ₹15,000 at 20% a year for 1½ years, compounded half-yearly.
7. A loan of ₹33,100 at 10% compound interest is repaid in three equal annual instalments. Find each instalment.

Answers:

1. ₹6,620. 20,000 × 0.331.
2. ₹3,000. 7,500 × 8 × 5 ÷ 100.
3. ₹30. 12,000 × (0.05)² = 12,000 × 0.0025.
4. 8.16%. (1.04)² − 1 = 0.0816.
5. 12 years. 9 = 3², so two periods of 6 years.
6. ₹4,965. 10% for each of 3 half-years: 15,000 × 1.331 = 19,965; 19,965 − 15,000.
7. ₹13,310. x × (1/1.1 + 1/1.21 + 1/1.331) = 33,100. Over 1,331: x × (1,210 + 1,100 + 1,000) ÷ 1,331 = 33,100, so x = 33,100 × 1,331 ÷ 3,310 = 13,310.

What to do next

• Memorise the fraction table and the 10% multipliers (1.1, 1.21, 1.331, 1.4641).
• Solve five interest questions a day for a week using fractions only, no long decimals.
• See interest used as statements in data sufficiency and as quantities in quantity comparison.
• Revise profit, loss and discount, which uses the same multiplying-factor idea.