Simple and compound interest for CDS

Interest questions look like banking but are really percentage in disguise. Simple interest adds the same amount every year. Compound interest multiplies by the same factor every year. Once you see those two pictures clearly, the formulas stop being things to memorise and become things you can rebuild in ten seconds.

CDS papers tend to ask a few interest questions: a direct calculation, a CI minus SI difference, a "doubles in so many years" pattern, or a question that gives two amounts and asks for the rate. This guide covers each with the reasoning behind it.

Simple interest: a straight line

Simple interest is charged only on the original principal. Every year adds the same interest.

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

Because each year adds the same amount, two amounts at different times tell you the yearly interest directly. That is the key to most SI word problems.

For doubling and multiplying at simple interest:

• A sum doubles when the total interest equals the principal, so R × T = 100.
• A sum becomes n times when the interest is (n − 1) times the principal, so R × T = (n − 1) × 100.

Compound interest: a multiplier

Compound interest is charged on the principal plus the interest already added. Each year multiplies the amount by (1 + R/100).

• A = P × (1 + R/100)ⁿ
• CI = A − P

If the rate changes from year to year, multiply the separate factors: at 10% and then 20%, the multiplier is 1.1 × 1.2 = 1.32.

Half-yearly compounding: halve the rate and double the number of periods. At 10% a year compounded half-yearly for 1 year, the multiplier is 1.05² = 1.1025. Quarterly: divide the rate by 4 and multiply the periods by 4.

Why CI and SI differ, and by how much

In the first year, CI and SI are the same, because both are charged on P alone. The gap starts in the second year, when compound interest also charges interest on the first year's interest.

• Two years: CI − SI = P × (R/100)². This is simply the interest on the first year's interest.
• Three years: CI − SI = P × (R/100)² × (3 + R/100).

Doubling at compound interest

If a sum doubles in t years at compound interest, it becomes 4 times in 2t years, 8 times in 3t years, and 2ᵏ times in kt years. Why: each block of t years multiplies by 2, and the blocks multiply together. This only works for compound interest. At simple interest the growth is a straight line, so a sum that doubles in 5 years becomes 3 times in 10 years, not 4 times.

Quick comparison

Worked questions

Question 1: A sum at simple interest amounts to ₹7,800 in 3 years and to ₹9,000 in 5 years. Find the sum and the rate.

• The extra 2 years add 9,000 − 7,800 = 1,200, so one year's interest is ₹600.
• Three years' interest = 1,800, so P = 7,800 − 1,800 = ₹6,000.
• R = 600 ÷ 6,000 × 100 = 10%.

Question 2: Find the compound interest on ₹16,000 for 1 year at 10% a year, compounded half-yearly.

• Rate per half-year = 5%, number of periods = 2.
• A = 16,000 × 1.05² = 16,000 × 1.1025 = 17,640.
• CI = ₹1,640. Compounded yearly, it would have been ₹1,600.

Question 3: On a certain sum, simple interest for 2 years is ₹400 and compound interest for 2 years at the same rate is ₹420. Find the rate and the sum.

• One year's interest = 400 ÷ 2 = 200.
• CI − SI = 20 is the interest on ₹200 for one year, so R = 20 ÷ 200 × 100 = 10%.
• P = 200 ÷ 0.1 = ₹2,000.

Question 4: Find the difference between CI and SI on ₹8,000 for 3 years at 5% a year.

• Formula: 8,000 × (0.05)² × (3 + 0.05) = 8,000 × 0.0025 × 3.05 = ₹61.
• Check: CI = 8,000 × (1.157625 − 1) = 1,261 and SI = 1,200. The difference is 61.

Question 5: A sum amounts to ₹8,820 in 2 years and ₹9,261 in 3 years at compound interest. Find the rate and the sum.

• The third year's interest = 9,261 − 8,820 = 441, earned on 8,820.
• R = 441 ÷ 8,820 × 100 = 5%.
• P = 8,820 ÷ 1.05² = 8,820 ÷ 1.1025 = ₹8,000.

Question 6: A sum becomes 3 times itself in 8 years at simple interest. In how many years will it become 5 times itself?

• Tripling means interest = 2 × P, so R × 8 = 200 and R = 25%.
• Five times means interest = 4 × P, so 25 × T = 400 and T = 16 years.

Numbers worth knowing

Squares and cubes of common multipliers save a lot of time.

Practice set

1. Find the simple interest on ₹6,000 at 4% a year for 5 years.
2. Find the compound interest on ₹10,000 at 10% a year for 3 years.
3. The difference between CI and SI on a sum for 2 years at 10% a year is ₹50. Find the sum.
4. A sum doubles in 4 years at compound interest. In how many years will it become 16 times?
5. At what rate of simple interest does a sum double in 8 years?
6. Find the compound interest on ₹12,500 at 8% a year for 2 years.
7. ₹5,000 is invested at 10% in the first year and 20% in the second year, compounded yearly. Find the compound interest.
8. Find the compound interest on ₹20,000 for 1 year at 8% a year, compounded half-yearly.

Answers

1. ₹1,200. 6,000 × 4 × 5 ÷ 100.
2. ₹3,310. 10,000 × (1.331 − 1).
3. ₹5,000. P × (0.1)² = 50, so P = 50 ÷ 0.01.
4. 16 years. 16 = 2⁴, so 4 × 4 years.
5. 12.5%. R × 8 = 100.
6. ₹2,080. 12,500 × 1.1664 = 14,580, and 14,580 − 12,500 = 2,080.
7. ₹1,600. 5,000 × 1.1 × 1.2 = 6,600.
8. ₹1,632. 4% for 2 half-years: 20,000 × 1.0816 = 21,632.

What to do next

• Learn the multiplier table for 5%, 10%, 20% and 8%
• Solve five CI − SI questions using only the "interest on interest" idea
• Time yourself on the interest questions from the last five CDS papers
• Revise percentage and profit and loss together with this topic, since all three use multipliers