Simple Interest and Compound Interest Questions for SSC CGL: Quick Answer
SSC CGL interest questions test five ideas: SI = P × R × T / 100, Amount under CI = P × (1 + R/100)ⁿ, the CI minus SI difference for 2 and 3 years, half-yearly compounding (halve the rate, double the time) and equal instalments. Learn these five, and almost every SI/CI question in Tier-I and Tier-II can be solved in under a minute. Below are 15 fully worked examples in the current SSC style, then a practice set with answers.
Key takeaway: For 2 years, CI − SI = P × (R/100)². For 3 years, CI − SI = P × (R/100)² × (3 + R/100). These two shortcuts alone solve about a third of SSC interest questions.
Every answer on this page was checked with exact fraction arithmetic in Python before publishing.
Simple Interest and Compound Interest Formulas for SSC CGL
| Concept | Formula | Use it when |
|---|---|---|
| Simple Interest | SI = P × R × T / 100 | Interest is on the principal only |
| SI Amount | A = P + SI = P(1 + RT/100) | "Sum amounts to" with SI |
| CI Amount (yearly) | A = P(1 + R/100)ⁿ | Interest is added every year |
| Compound Interest | CI = A − P | Question asks for interest, not amount |
| Half-yearly compounding | Rate = R/2, Time = 2n | "Compounded half-yearly" |
| Quarterly compounding | Rate = R/4, Time = 4n | "Compounded quarterly" |
| Different rates each year | A = P(1 + R₁/100)(1 + R₂/100)… | Rate changes year to year |
| CI − SI (2 years) | P(R/100)² | Difference for 2 years |
| CI − SI (3 years) | P(R/100)²(3 + R/100) | Difference for 3 years |
| Rate from 2 successive CI amounts | R = (Aₙ₊₁ − Aₙ) / Aₙ × 100 | Amounts for consecutive years given |
| SI to multiply the sum k times | T = (k − 1) × 100 / R | "Doubles / triples in … years" (SI) |
| CI: sum becomes x times in n years | Becomes xᵐ times in m × n years | "Becomes 4 times… 16 times" (CI) |
Fraction shortcuts that save time in SSC CGL
| Rate | As a fraction | Multiplier for 1 year |
|---|---|---|
| 4% | 1/25 | 26/25 |
| 5% | 1/20 | 21/20 |
| 8% | 2/25 | 27/25 |
| 10% | 1/10 | 11/10 |
| 12.5% | 1/8 | 9/8 |
| 20% | 1/5 | 6/5 |
| 25% | 1/4 | 5/4 |
Pick principals that are powers of the denominator (for example 15,625 = 25³ at 4%) and the CI arithmetic becomes exact.
Simple Interest Questions for SSC CGL with Solutions
Example 1: Find the simple interest
Q. Find the simple interest on ₹8,000 at 7½% per annum for 3 years.
Solution: SI = 8,000 × 7.5 × 3 / 100 = ₹1,800.
Example 2: Find the rate of simple interest
Q. A sum of ₹6,000 amounts to ₹7,680 in 4 years at simple interest. Find the rate.
Solution: SI = 7,680 − 6,000 = ₹1,680. R = 1,680 × 100 / (6,000 × 4) = 7%.
Example 3: Doubling and tripling at simple interest
Q. A sum doubles itself in 8 years at simple interest. In how many years will it become three times?
Solution: Doubling means interest = 100% of P in 8 years, so R = 100/8 = 12.5%. To triple, interest must be 200% of P: T = 200 / 12.5 = 16 years.
Shortcut: T = (k − 1) × 100 / R, so time is proportional to (k − 1). Doubling (k − 1 = 1) takes 8 years, so tripling (k − 1 = 2) takes 16 years.
Example 4: Dividing a sum at two SI rates
Q. ₹7,000 is divided into two parts. One part is lent at 8% and the other at 10% simple interest. The total yearly interest is ₹640. Find the two parts.
Solution: Let the 8% part be a. Then 8a/100 + 10(7,000 − a)/100 = 640, so 8a + 70,000 − 10a = 64,000, giving a = 3,000. Parts: ₹3,000 at 8% and ₹4,000 at 10%.
Check: 240 + 400 = 640. ✓
Compound Interest Questions for SSC CGL with Solutions
Example 5: Basic compound interest
Q. Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.
Solution: A = 10,000 × (11/10)² = 12,100. CI = ₹2,100.
Example 6: Compound interest at 4% for 3 years
Q. Find the amount and compound interest on ₹15,625 at 4% per annum for 3 years.
Solution: A = 15,625 × (26/25)³. Since 25³ = 15,625, this is simply 26³ = ₹17,576. CI = 17,576 − 15,625 = ₹1,951.
Example 7: Compounded half-yearly
Q. Find the compound interest on ₹16,000 for 1 year at 10% per annum, compounded half-yearly.
Solution: Rate per half-year = 5%, number of half-years = 2. A = 16,000 × (21/20)² = 17,640. CI = ₹1,640.
Note: Yearly compounding would give only ₹1,600, so half-yearly compounding adds ₹40 here.
Example 8: Effective rate of half-yearly compounding
Q. What is the effective annual rate when 10% per annum is compounded half-yearly?
Solution: (1.05)² − 1 = 0.1025 → 10.25%.
Example 9: Different rates in different years
Q. Find the compound interest on ₹25,000 for 2 years if the rate is 10% in the first year and 20% in the second year.
Solution: A = 25,000 × (11/10) × (6/5) = 33,000. CI = ₹8,000.
Example 10: Rate from two successive amounts
Q. A sum at compound interest amounts to ₹7,260 in 2 years and ₹7,986 in 3 years. Find the rate and the principal.
Solution: Interest in the 3rd year = 7,986 − 7,260 = 726. R = 726 / 7,260 × 100 = 10%. P = 7,260 / (1.1)² = 7,260 / 1.21 = ₹6,000.
Example 11: Becoming 4 times and 16 times at compound interest
Q. At compound interest, a sum becomes 4 times in 6 years. In how many years will it become 16 times?
Solution: 16 = 4². If it becomes 4 times in 6 years, it becomes 4² times in 2 × 6 = 12 years.
CI and SI Difference Questions for SSC CGL
Example 12: Principal from CI − SI for 2 years
Q. The difference between compound interest and simple interest on a sum for 2 years at 8% per annum is ₹48. Find the sum.
Solution: CI − SI = P(R/100)² → 48 = P × (8/100)² = P × 64/10,000. P = 48 × 10,000 / 64 = ₹7,500.
Example 13: CI − SI for 3 years
Q. Find the difference between CI and SI on ₹5,000 for 3 years at 10% per annum.
Solution: Shortcut: P(R/100)²(3 + R/100) = 5,000 × 0.01 × 3.1 = ₹155.
Check: CI = 5,000 × 1.331 − 5,000 = 1,655; SI = 1,500; difference = 155. ✓
Example 14: Rate and principal from 2-year SI and CI
Q. The simple interest on a sum for 2 years is ₹1,500 and the compound interest for the same period at the same rate is ₹1,575. Find the rate and the sum.
Solution: SI for one year = 750. The extra ₹75 in CI is interest on the first year's interest: R = 75 / 750 × 100 = 10%. P = 750 × 100 / 10 = ₹7,500.
Check: CI on 7,500 at 10% for 2 years = 7,500 × 1.21 − 7,500 = 1,575. ✓
Compound Interest Instalment Questions for SSC CGL
Example 15: Two equal annual instalments
Q. A loan is repaid in two equal annual instalments of ₹12,100 each, at 10% compound interest per annum. Find the loan amount.
Solution: The loan equals the present value of the instalments: 12,100/1.1 + 12,100/1.21 = 11,000 + 10,000 = ₹21,000.
How to Solve SI and CI Questions Quickly in SSC CGL
- Read the compounding period first. Half-yearly or quarterly changes both the rate and the number of periods.
- Convert the rate to a fraction. 10% → 11/10, 5% → 21/20, 4% → 26/25.
- Check the options for the CI − SI shortcut whenever "difference" appears.
- For "amounts in n and n + 1 years", the rate is the growth between the two amounts.
- For instalments, divide each instalment by the multiplier raised to its year and add.
- Verify with a quick check: CI is always more than SI for the same P, R and T when T > 1.
Common Mistakes in SI and CI Questions
| Mistake | Correct approach |
|---|---|
| Using 10% for 1 year when compounding is half-yearly | Use 5% for 2 half-years |
| Answering the amount when the question asks for interest | CI = A − P |
| Applying the 2-year CI − SI formula to 3 years | Use P(R/100)²(3 + R/100) |
| Adding rates in "different rates" questions (10% + 20% = 30%) | Multiply: (1.1)(1.2) = 1.32 |
| Treating "doubles in 8 years" at CI like SI | At CI, 4 times takes 16 years, not 24 |
SI and CI Questions for SSC CGL: Practice Set with Answers
- Find the SI on ₹12,500 at 6% per annum for 2½ years.
- Find the CI on ₹20,000 at 5% per annum for 2 years.
- Find the difference between CI and SI on ₹10,000 for 2 years at 12%.
- At what rate of SI will a sum become 7/4 of itself in 5 years?
- A sum amounts to ₹9,680 in 2 years at 10% CI. Find the sum.
- At CI, a sum becomes 3 times in 5 years. In how many years will it become 27 times?
- The SI on a sum is 16/25 of the sum, and the rate per cent equals the number of years. Find the rate.
- A loan of ₹4,200 is repaid in two equal annual instalments at 10% CI. Find each instalment.
Answers and explanations
- ₹1,875. 12,500 × 6 × 2.5 / 100.
- ₹2,050. 20,000 × (21/20)² = 22,050; 22,050 − 20,000.
- ₹144. 10,000 × (0.12)².
- 15%. Interest = 3/4 of P = 75% in 5 years → 15% per year.
- ₹8,000. 9,680 / 1.21.
- 15 years. 27 = 3³, so 3 × 5 years.
- 8%. R × R / 100 = 16/25 → R² = 64 → R = 8.
- ₹2,420. 4,200 = x/1.1 + x/1.21 → x = 4,200 × 1.21 / 2.1 = 2,420.
Simple Interest and Compound Interest FAQs for SSC CGL
How many SI and CI questions come in SSC CGL Tier 1?
SSC does not publish a topic-wise count. Interest questions appear regularly in the Quantitative Aptitude section, usually one or two per shift, so it is a high-return topic.
What is the formula for the difference between CI and SI for 2 years?
CI − SI = P × (R/100)². For 3 years it is P × (R/100)² × (3 + R/100).
How do I solve half-yearly compound interest questions?
Halve the annual rate and double the number of years, then use A = P(1 + r/100)ⁿ with the new values.
Is simple interest or compound interest more important for SSC CGL?
Both appear, but compound interest questions (especially CI − SI and instalments) are more common in recent papers and take more practice.
What is the fastest way to calculate compound interest for 3 years?
Use the fraction multiplier. At 10%, multiply by 1.331; at 5%, by 1.157625; at 4%, choose a principal like 15,625 so that 15,625 × (26/25)³ = 17,576 is exact.
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