Compound interest for AFCAT: multipliers, CI–SI difference and half-yearly compounding

Compound interest looks harder than simple interest because of the power in the formula. In practice, AFCAT questions keep the time short (usually two or three years) and the rates friendly, so every question can be done with a couple of multiplications. The real skill is recognising which shortcut the question is built around: the CI–SI difference, half-yearly compounding, or two amounts in consecutive years.

This guide explains the formula as a chain of percentage changes, which is what it is, and then works through each question type.

The formula, and why it is just successive percentages

Amount = P × (1 + R/100)ⁿ, where n is the number of years. Compound interest = amount − P.

Each year, the balance grows by R%, so it is multiplied by (1 + R/100). After n years it has been multiplied n times. That is the whole formula.

At 10% for 2 years, the multiplier is 1.1 × 1.1 = 1.21. So the sum grows by 21% in two years, not 20%. The extra 1% is the interest earned on the first year's interest. This is the successive-change rule from the percentage chapter: 10 + 10 + (10 × 10)/100 = 21.

Multipliers worth memorising

The effective two-year growth is 2R + R²/100. At 20% that is 40 + 4 = 44%.

The CI–SI difference

For the same P and R, compound and simple interest are equal in year one. They differ from year two, because compound interest also earns interest on the first year's interest.

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

Why the two-year rule works: in year two, compound interest earns R% on the first year's interest, which was P × R/100. So the extra is P × R/100 × R/100.

Compounding more often than once a year

If interest is compounded half-yearly, halve the rate and double the number of periods. Quarterly: quarter the rate and multiply the periods by four.

₹8,000 at 10% a year compounded half-yearly for 1 year is 5% for 2 periods: 8,000 × 1.05 × 1.05 = ₹8,820. Compounded yearly it would be ₹8,800. The more often interest is added, the more it earns.

Doubling and "n times"

If a sum becomes k times itself in t years at compound interest, it becomes k² times in 2t years, k³ times in 3t years, and so on. The same multiplier repeats every t years.

So a sum that doubles in 5 years becomes 4 times in 10 years and 8 times in 15 years. Compare simple interest, where it would become 4 times in 15 years.

Worked examples

Example 1 (basic CI). Find the CI on ₹10,000 at 10% a year for 2 years, compounded yearly.

• Amount = 10,000 × 1.21 = ₹12,100.
• CI = ₹2,100.

Example 2 (CI–SI for three years). Find the difference between CI and SI on ₹8,000 at 5% a year for 3 years.

• Shortcut: 8,000 × (0.05)² × (3 + 0.05) = 8,000 × 0.0025 × 3.05 = ₹61.
• Check: CI = 8,000 × 1.157625 − 8,000 = ₹1,261. SI = 8,000 × 5 × 3 ÷ 100 = ₹1,200. Difference = ₹61.

Example 3 (principal from the difference). The difference between CI and SI on a sum for 2 years at 8% is ₹32. Find the sum.

• P × (0.08)² = 32, so P × 0.0064 = 32.
• P = ₹5,000.

Example 4 (two consecutive amounts). A sum amounts to ₹6,050 in 2 years and ₹6,655 in 3 years at compound interest. Find the rate and the principal.

• The third year turned ₹6,050 into ₹6,655: 6,655 ÷ 6,050 = 1.1, so R = 10%.
• P = 6,050 ÷ 1.21 = ₹5,000.

Example 5 (rates that change). ₹20,000 is invested at 10% in the first year and 20% in the second, compounded yearly. Find the CI.

• Amount = 20,000 × 1.1 × 1.2 = ₹26,400.
• CI = ₹6,400.

Example 6 (half-yearly). Find the CI on ₹16,000 at 10% a year for 1½ years, compounded half-yearly.

• 5% per half-year, 3 periods.
• Amount = 16,000 × 1.157625 = ₹18,522.
• CI = ₹2,522.

Common mistakes

• Giving the amount when CI is asked, or the other way round. Read the last line of the question twice.
• Adding rates across years. 10% for 2 years is 21%, not 20%.
• Treating "doubles in 5 years" as linear. At compound interest, 4 times takes 10 years; at simple interest, 15.
• Long multiplication under pressure. Use the table, or multiply step by step: 10% of 12,100 is 1,210, so year three takes it to 13,310.

Practice set

1. Find the CI on ₹6,250 at 4% a year for 2 years.
2. Find CI − SI on ₹10,000 at 5% a year for 2 years.
3. Find the amount on ₹1,000 at 10% a year for 3 years.
4. At 20% a year compounded yearly, by what percentage does a sum grow in 2 years?
5. A sum amounts to ₹17,640 in 2 years at 5% compound interest. Find the sum.
6. A sum triples in 4 years at compound interest. In how many years will it become 9 times?
7. A town of 50,000 people grows at 4% a year. What is its population after 2 years?
8. Find the amount on ₹8,000 at 10% a year for 1 year, compounded half-yearly.

Answers:

1. ₹510. 6,250 × 1.0816 = 6,760, minus 6,250.
2. ₹25. 10,000 × (0.05)² = 10,000 × 0.0025.
3. ₹1,331. 1,000 × 1.331.
4. 44%. 1.2 × 1.2 = 1.44.
5. ₹16,000. 17,640 ÷ 1.1025.
6. 8 years. 9 = 3², so twice the tripling time.
7. 54,080. 50,000 × 1.0816.
8. ₹8,820. 5% for 2 half-years: 8,000 × 1.1025.

What to do next

• Memorise the 5%, 10% and 20% rows of the multiplier table.
• Practise ten CI–SI difference questions, including reverse ones where the difference is given.
• Revise simple interest alongside, since many options mix the two, then move on to time and work.