Compound interest for RRB NTPC: formula, shortcuts and practice

Compound interest looks harder than simple interest because of the power in the formula. In NTPC questions, though, the numbers are chosen so that you rarely need more than two or three years of calculation. Most questions can be done with the year-by-year method or one of three short formulas.

The topic is also a percentage topic in disguise. Population growth, depreciation of a machine, and successive percentage increases all use the same multiplier. Learn it here and you have it for percentage too.

The formula, and why it has a power

• Amount = P × (1 + R/100)ⁿ
• CI = Amount − P

Here n is the number of compounding periods, usually years.

Why a power: at the end of each year, the interest is added to the principal. The next year's interest is then R% of this larger sum. So each year multiplies the money by the same factor (1 + R/100). After n years you have multiplied by it n times.

The year-by-year method

For 2 or 3 years, calculating one year at a time is often faster than the formula, and it shows exactly where the extra interest comes from.

Example: ₹10,000 at 10% a year.

In year 1, CI and SI are equal. From year 2, CI earns extra because it is also earning on earlier interest.

Fraction rates speed things up. 12.5% = 1/8, so each year multiplies by 9/8. ₹6,400 at 12.5% for 2 years: 6,400 × 9/8 × 9/8 = 8,100, so the CI is ₹1,700.

Effective rates. 10% for 2 years is a total rise of 10 + 10 + 1 = 21% (the successive-change rule a + b + ab/100). 5% for 2 years is 10.25%. 8% for 2 years is 16.64%.

CI minus SI: the gap formulas

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

Why the 2-year formula works: in the first year, CI and SI earn the same. In the second year, SI earns R% of P again, but CI also earns R% on the first year's interest, which was P × R/100. That extra is P × (R/100)², and it is the whole gap.

Half-yearly and quarterly compounding

• Half-yearly: rate per period = R/2, number of periods = 2n.
• Quarterly: rate per period = R/4, number of periods = 4n.

Why: "10% a year compounded half-yearly" means interest of 5% is added every six months. The bank pays half the rate twice as often.

The "becomes k times" rule

If a sum becomes k times in n years at compound interest, it becomes k² times in 2n years and k³ times in 3n years.

Why: every n years the money is multiplied by k. Two such stretches multiply it by k × k.

This rule does not hold for simple interest. A sum that doubles in 5 years at SI becomes 3 times in 10 years, not 4 times.

Worked questions at NTPC level

Q1. Find the difference between compound and simple interest on ₹5,000 at 10% for 2 years, and for 3 years.

2 years: 5,000 × (10/100)² = 5,000 × 0.01 = ₹50.
3 years: 5,000 × 0.01 × (3 + 0.1) = 50 × 3.1 = ₹155.
Check for 3 years: CI = 5,000 × 1.331 − 5,000 = 1,655; SI = 1,500; gap 155.

Q2. Find the compound interest on ₹4,000 at 10% a year, compounded half-yearly, for 1 year.

Rate per half-year = 5%, periods = 2. Amount = 4,000 × 1.05 × 1.05 = 4,410.
CI = ₹410. (Yearly compounding would give ₹400; the extra ₹10 is interest on the first half-year's interest.)

Q3. A sum doubles in 5 years at compound interest. In how many years will it become 8 times?

8 = 2³, so 3 × 5 = 15 years.

Q4. Find the compound interest on ₹5,000 for 2 years when the rate is 10% in the first year and 20% in the second.

Amount = 5,000 × 1.1 × 1.2 = 6,600. CI = ₹1,600.
The formula with a single R does not apply; multiply by each year's factor.

Q5. The difference between compound and simple interest on a sum for 2 years at 8% is ₹32. Find the sum.

P × (8/100)² = 32, so P × 0.0064 = 32, and P = ₹5,000.
Check: CI = 5,000 × 1.1664 − 5,000 = 832; SI = 800; gap 32.

Q6. A sum amounts to ₹8,820 in 2 years and ₹9,261 in 3 years at compound interest. Find the rate and the principal.

The third year's interest is 9,261 − 8,820 = 441, earned on 8,820. Rate = 441 ÷ 8,820 × 100 = 5%.
Principal = 8,820 ÷ (1.05)² = 8,820 ÷ 1.1025 = ₹8,000.

Common mistakes

• Using the SI formula for a CI question, or the other way round.
• Forgetting to halve the rate and double the periods in half-yearly questions.
• Giving the amount when the question asks for interest.
• Using the two-amounts SI method for CI. At CI, the difference between consecutive years' amounts is one year's interest on the earlier amount, not on the principal.
• Applying the k-times rule to SI.

Practice set

1. Find the compound interest on ₹5,000 at 10% for 2 years.
2. Find the amount on ₹20,000 at 5% for 2 years.
3. Find the difference between CI and SI on ₹8,000 at 5% for 2 years.
4. A sum doubles in 4 years at compound interest. In how many years will it become 4 times?
5. Find the compound interest on ₹1,000 at 20% a year, compounded half-yearly, for 1 year.
6. Find the compound interest on ₹10,000 at 10% for 3 years.
7. A sum amounts to ₹6,050 in 2 years and ₹6,655 in 3 years at compound interest. Find the rate and the principal.
8. Find the compound interest on ₹6,250 at 8% for 2 years.

Answers:

1. ₹1,050. 5,000 × 1.21 = 6,050.
2. ₹22,050. 20,000 × 1.1025.
3. ₹20. 8,000 × (5/100)² = 8,000 × 0.0025.
4. 8 years. 4 = 2², so 2 × 4.
5. ₹210. 10% per half-year for 2 periods: 1,000 × 1.21 = 1,210.
6. ₹3,310. 10,000 × 1.331 = 13,310.
7. 10% and ₹5,000. 6,655 − 6,050 = 605, and 605 ÷ 6,050 = 10%; 6,050 ÷ 1.21 = 5,000.
8. ₹1,040. 6,250 × 1.1664 = 7,290.

What to do next

• Learn the four powers in the exam angle box and the effective 2-year rates for 5%, 10% and 20%.
• Solve 15 CI questions this week using the year-by-year method, then redo them with the formula to compare speed.
• Practise ten CI − SI questions until the 2-year formula is automatic.
• Revise simple interest alongside, since the two are often compared in one question.