Theory & Concepts
SSC CGL Compound Interest Questions, Formulas & Short Tricks
Prepare Compound Interest for SSC CGL with formulas, short tricks, solved examples, practice questions, PYQs, and free PDF notes for faster exam preparation.
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25 min readDifficulty: Advanced
Compound interest (ci) is often considered the most challenging topic in the arithmetic section. Unlike simple interest, where interest is calculated only on the principal, ci involve "interest on interest."
In the Ssc cgl exam, questions often involve messy calculations or compounding periods like 10 months or quarterly intervals. Mastering successive percentage shortcuts is the only way to solve these within 45 seconds.
Learning path
- The master ci formula
- Successive percentage tricks
- CI - SI difference rules
- 10 standard solved problems
- Compounding periods logic
1. Fundamental formulas
Total amount (a)
Where:
= Principal
= Rate per annum
= Number of years / Compounding periods
2. The successive percentage trick
Instead of the power formula, use the successive increase method for 2 and 3 years to save time.
Equivalent rate for 2 years
Example: For 10% rate, 2-year CI = 10 + 10 + 100/100 = 21%.
3. CI - SI difference (high yield)
For 2 years:
For 3 years:
4. Solved examples
Question 01Standard pattern
Calculate the compound interest on rs. 10,000 for 2 years at 10% per annum.
rs. 2,100
rs. 2,000
rs. 2,200
rs. 2,050
Correct answer: a) rs. 2,100
Solution
Equivalent rate for \( 10\% \) for 2 years = \( 10 + 10 + \frac{100}{100} = 21\% \).
CI = \( 21\% \text{ of } 10,000 \).
Calculation: \( \frac{21}{100} \times 10,000 = rs. 2,100 \).
Question 02Standard pattern
What is the difference between the compound interest and simple interest on rs. 5,000 for 2 years at 8% per annum?
rs. 32
rs. 40
rs. 28
rs. 35
Correct answer: a) rs. 32
Solution
Difference formula (2 years): \( D = P \times \left( \frac{R}{100} \right)^2 \).
Substitute: \( 5000 \times \left( \frac{8}{100} \right)^2 \).
Calculation: \( 5000 \times \frac{64}{10000} = \frac{5000 \times 64}{10000} \).
Final value = \( 32 \text{ units} \implies rs. 32 \).
Question 03Standard pattern
In what time will rs. 1,000 become rs. 1,331 at 10% per annum compound interest?
3 years
2 years
4 years
2.5 years
Correct answer: a) 3 years
Solution
Equation: \( 1331 = 1000 \times (1.1)^n \).
Ratio: \( \frac{1331}{1000} = (1.1)^n \).
\( 1.331 = (1.1)^n \).
Recognize: \( (1.1)^3 = 1.331 \).
Result = \( 3 \text{ years} \).
Question 04Standard pattern
A sum of money doubles itself at compound interest in 15 years. It will become eight times in:
45 years
30 years
60 years
75 years
Correct answer: a) 45 years
Solution
If money increases by factor \( 2 \) in \( 15 \) years.
To become \( 8 \) times: \( 8 = 2^3 \).
Time taken = \( 3 \times 15 \text{ years} \).
Result = \( 45 \text{ years} \).
Question 05Standard pattern
Find the compound interest on rs. 16,000 for 9 months at 20% per annum, interest being compounded quarterly.
rs. 2,522
rs. 2,400
rs. 2,640
rs. 2,100
Correct answer: a) rs. 2,522
Solution
New Rate (Quarterly) = \( \frac{20}{4} = 5\% \).
New Time (Quarterly) = \( 9 \text{ months} = 3 \text{ periods} \).
Equivalent rate for \( 5\% \) per 3 periods: \( 15.7625\% \).
CI = \( 16000 \times 15.7625\% = 2522 \).
Result = \( rs. 2,522 \).
Question 06Standard pattern
A sum of money placed at compound interest doubles itself in 5 years. In how many years will it amount to sixteen times?
20 years
15 years
25 years
10 years
Correct answer: a) 20 years
Solution
Base growth: Double ( \( 2 \) times) in \( 5 \) years.
Target growth: \( 16 \) times.
Express target as base power: \( 16 = 2^4 \).
Calculation: \( 4 \times 5 \text{ years} = 20 \text{ years} \).
Question 07Standard pattern
What sum of money will amount to rs. 665.50 in 3 years at 10% per annum compound interest?
rs. 500
rs. 450
rs. 600
rs. 550
Correct answer: a) rs. 500
Solution
Equivalent rate for \( 10\% \) for 3 years: \( 33.1\% \).
Equation: \( 1.331 \times P = 665.50 \).
Calculation: \( P = \frac{665.50}{1.331} \).
Result = \( rs. 500 \).
Question 08Standard pattern
The difference between SI and CI for 3 years at 10% is rs. 31. Find the principal.
rs. 1,000
rs. 1,500
rs. 1,200
rs. 2,000
Correct answer: a) rs. 1,000
Solution
Use \( D_3 \) formula: \( P \left( \frac{R}{100} \right)^2 \times (3 + \frac{R}{100}) = 31 \).
Substitute \( R = 10 \): \( P \times \frac{1}{100} \times 3.1 = 31 \).
Calculation: \( P \times 0.031 = 31 \implies P = 1000 \).
Result = \( rs. 1,000 \).
Question 09Standard pattern
At what rate percent compound interest per annum will rs. 400 amount to rs. 441 in 2 years?
5%
4%
8%
10%
Correct answer: a) 5%
Solution
Ratio: \( \frac{A}{P} = \frac{441}{400} \).
Equation: \( (1 + r)^2 = \frac{441}{400} \).
Root: \( 1 + r = \sqrt{\frac{441}{400}} = \frac{21}{20} \).
Rate: \( r = \frac{21}{20} - 1 = \frac{1}{20} = 5\% \).
Question 010Standard pattern
Find the compound interest on rs. 8,000 at 15% per annum for 2 years 4 months, compounded annually.
rs. 3,109
rs. 3,000
rs. 3,200
rs. 2,900
Correct answer: a) rs. 3,109
Solution
Amount after 2 years = \( 8000 \times (1.15)^2 = 10580 \).
Rate for rest 4 months (1/3 year) = \( \frac{15}{3} = 5\% \).
Interest for last period = \( 5\% \text{ of } 10580 = 529 \).
Total interest = \( (10580 - 8000) + 529 = 3109 \).
Result = \( rs. 3,109 \).
5. Strategy errors to avoid
Error 01Compounding period error: Not dividing the rate for half-yearly or quarterly questions.
Error 02Direct summation: Thinking that 10% CI for 2 years is 20% (it is 21%).
Error 03Formula fatigue: Using the full power formula for simple 2-year problems instead of successive %.
Error 04Base error: Calculating the third year interest on the original principal.
