Simple and compound interest for SSC GD

Interest is the extra money paid for using someone else's money. A bank pays you interest on your savings; you pay interest on a loan. Interest questions in SSC GD come in a small number of forms, and each one has a formula you can learn in a day.

The topic also reuses percentage. Simple interest is a percentage of the same amount every year. Compound interest is successive percentage growth, like the population questions you may have met already.

The terms

Simple interest (SI)

In simple interest, the interest is the same every year, because it is always calculated on the original principal.

SI = P × R × T ÷ 100

Amount = P + SI

Rearranging the same formula gives the other three:

Doubling and tripling. A sum doubles when the interest equals the principal, so R × T = 100. It triples when the interest is twice the principal, so R × T = 200. At 10% a year, a sum doubles in 10 years and triples in 20.

Compound interest (CI)

In compound interest, each year's interest is added to the principal, and the next year's interest is calculated on the new total. So the interest grows every year.

Amount = P × (1 + R/100)ᵀ, where the bracket is multiplied by itself T times.

CI = Amount − P

For two or three years, working year by year is often quicker than the formula, and it is harder to slip.

Half-yearly compounding. If interest is added every six months, halve the rate and double the number of periods. ₹10,000 at 10% a year, compounded half-yearly for 1 year, means 5% for 2 periods: 10,000 × 1.05 × 1.05 = ₹11,025.

SI and CI compared

Shortcut for two years: CI − SI = P × (R/100)²

Solved examples

Example 1. Find the simple interest and amount on ₹12,000 at 7.5% a year for 4 years.

1. SI = 12,000 × 7.5 × 4 ÷ 100.
2. SI = 3,60,000 ÷ 100 = ₹3,600.
3. Amount = 12,000 + 3,600 = ₹15,600.

Example 2. What principal earns ₹1,260 simple interest in 3 years at 6% a year?

1. P = 1,260 × 100 ÷ (6 × 3).
2. P = 1,26,000 ÷ 18 = ₹7,000.

Example 3. ₹6,000 becomes ₹7,200 in 4 years at simple interest. Find the rate.

1. SI = 7,200 − 6,000 = ₹1,200.
2. R = 1,200 × 100 ÷ (6,000 × 4) = 1,20,000 ÷ 24,000 = 5%.

Example 4. A sum doubles in 8 years at simple interest. Find the rate, and the time in which it will triple.

1. Doubling: R × 8 = 100, so R = 12.5%.
2. Tripling needs R × T = 200, so T = 200 ÷ 12.5 = 16 years.

Example 5. Find the compound interest on ₹10,000 at 10% a year for 3 years.

1. Year 1: interest 1,000, amount 11,000.
2. Year 2: interest 1,100, amount 12,100.
3. Year 3: interest 1,210, amount 13,310.
4. CI = 13,310 − 10,000 = ₹3,310.

Example 6. The difference between CI and SI on a sum for 2 years at 8% is ₹32. Find the sum.

1. CI − SI = P × (8/100)² = P × 0.0064.
2. P × 0.0064 = 32, so P = 32 ÷ 0.0064 = ₹5,000.
3. Check: SI = 800; CI = 5,000 × 1.08 × 1.08 − 5,000 = 5,832 − 5,000 = 832. Difference 32.

Common mistakes

Practice set

1. Find the SI on ₹6,500 at 8% a year for 3 years.
2. Find the amount when ₹9,000 is lent at 5% simple interest for 4 years.
3. ₹5,000 becomes ₹6,500 in 5 years at simple interest. Find the rate.
4. In how many years will ₹4,800 earn ₹1,080 at 7.5% simple interest?
5. A sum doubles in 10 years at simple interest. Find the rate.
6. Find the CI on ₹15,000 at 10% a year for 2 years.
7. Find the amount on ₹4,000 at 5% compound interest for 3 years.
8. Find the difference between CI and SI on ₹12,000 at 5% for 2 years.

Answers:

1. 6,500 × 8 × 3 ÷ 100 = ₹1,560.
2. SI = 9,000 × 5 × 4 ÷ 100 = 1,800; amount = ₹10,800.
3. SI = 1,500; R = 1,500 × 100 ÷ (5,000 × 5) = 6%.
4. T = 1,080 × 100 ÷ (4,800 × 7.5) = 1,08,000 ÷ 36,000 = 3 years.
5. R × 10 = 100, so 10%.
6. 15,000 → 16,500 → 18,150; CI = ₹3,150.
7. 4,000 → 4,200 → 4,410 → ₹4,630.50.
8. 12,000 × (5/100)² = 12,000 × 0.0025 = ₹30.

What to do next

• Learn the SI formula and its three rearrangements.
• Work 10 CI questions year by year before you try the formula.
• Learn the two-year CI − SI shortcut and test it on two sums.
• Mix interest with profit and loss and percentage in a timed set of 20.