Compound Interest questions with answers and concepts

Learn compound interest with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

116 practice questionsPage updated 11 Nov 2026Reviewed by AISEA Editorial DeskPractice questions

Practise this concept

Compound interest is interest calculated on the accumulated amount, so each period can earn interest on earlier interest as well as on the original principal. In aptitude exams, first identify the principal, nominal annual rate, compounding frequency and number of periods. Convert the rate and time to one compounding period, apply the growth factor repeatedly, and subtract the principal only when the question asks for compound interest rather than the final amount.

Key formulas and rules

8 rules

Annual compounding: $A = P (1 + \frac{R}{100})^{T}$ , where $T$ is the number of years.

Compounding $n$ times per year: $A = P (1 + \frac{R}{100 \times n})^{n \times T}$ .

Compound interest: $C I = A - P = P [(1 + \frac{R}{100})^{T} - 1]$ .

Present principal: $P = \frac{A}{( 1 + \frac{R}{100} ) ^{T}}$ .

Effective annual rate: $EAR = [(1 + \frac{R}{100 \times n})^{n} - 1] \times 100%$ .

Depreciation: $Value = P (1 - \frac{R}{100})^{T}$ .

Difference between CI and SI (2 years): $C I - S I = P (\frac{R}{100})^{2}$ .

Difference between CI and SI (3 years): $C I - S I = P (\frac{R}{100})^{2} (3 + \frac{R}{100})$ .

Key concepts

Separate principal, amount and interest

The principal P is the starting sum. The amount A is the principal plus all accumulated interest. Compound interest CI is A − P. Many mistakes come from calculating A correctly and then reporting it as CI, or from subtracting the principal twice.

Convert the nominal rate and frequency

If an annual rate R is compounded twice a year, each half-year uses R / 2 and there are 2 periods per year. Quarterly uses R / 4 and four periods per year; monthly uses R / 12 and twelve periods per year. The rate per period and the number of periods must change together.

Use the compound growth factor

Each period multiplies the current amount by one plus the rate per period. After several periods, the same multiplier is applied repeatedly, which is why compound growth accelerates. Do not add the same interest to the original principal for every period; that is simple interest.

Effective annual rate versus nominal rate

A quoted annual rate may be nominal when interest is credited more than once a year. The effective annual rate is the actual one-year growth after all subperiods compound. For positive rates, more frequent compounding gives a higher effective rate than the same nominal rate compounded annually.

Solve for time or rate from powers

When the final amount is a familiar multiple of the principal, match the growth factor to powers such as 1.1, 1.05 or 1.025. If the exact power is not evident, use logarithms or a calculator only when allowed. Verify the result by substituting it back into the compound formula.

Handle partial years explicitly

For half-yearly or quarterly compounding, a period such as 1.5 years converts naturally to 3 half-year periods. For annual compounding over a fractional year, do not silently choose a convention: the question must specify whether the fractional part earns simple interest, is ignored or is treated by another rule. State the convention before calculating.

Find interest for a particular period

The interest in the third year is calculated on the amount at the end of the second year, not on the original principal. First find the opening balance of the requested period, then multiply by the rate for that period. This is different from total compound interest across all years.

Compare compound and simple interest

For the same principal, rate and time, compound interest is usually higher because interest is reinvested. For two years at an annual rate, the difference is P × (R / 100)². For longer periods, calculate both amounts rather than extending the two-year shortcut.

Apply successive increases and decreases

Population growth, inflation and depreciation use successive multiplicative factors. An increase of 10% followed by a decrease of 10% gives 1.1 × 0.9 = 0.99 of the starting value, not the original value. For depreciation, use one minus the rate as the factor in each period.

Do not confuse deposits with instalments

The single-deposit compound formula applies to one principal left invested. If equal payments are made at different dates, each payment earns for a different number of periods, so timing matters. Use an annuity or present-value method when the question asks for regular instalments or payments that discharge a debt.

Check units and rounding

Use a decimal rate when the formula expects a decimal, or use the percentage form with the divisor 100, but do not mix them. Keep intermediate decimals until the final step. Check that the final amount is greater than the principal for a positive rate, and that a rounded option is close to the unrounded result.

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Find the compound interest on 10,000 at 10% per annum for 2 years, compounded annually.

Solution

Step 1: Calculate the accumulated amount.

A = P (1 + \frac{R}{100})^{n} = 10, 000 (1 + \frac{10}{100})^{2} = 10, 000 \times (1.1)^{2} = 12, 100

Step 2: Compute compound interest.

CI = A - P = 12, 100 - 10, 000 = 2, 100

Answer: 2,100.

Example 2

Find the difference between compound interest and simple interest on 5,000 for 2 years at 8% per annum.

Solution

Step 1: Apply the 2-year difference formula.

CI - SI = P (\frac{R}{100})^{2}

Step 2: Substitute $P = 5, 000$ and $R = 8$ .

Difference = 5, 000 \times (\frac{8}{100})^{2} = 5, 000 \times \frac{64}{10 , 000} = \frac{320 , 000}{10 , 000} = 32

Answer: 32.

Example 3

In how many years will 8,000 amount to 9,261 at 5% per annum compound interest?

Solution

Step 1: Set up the growth equation.

\frac{A}{P} = (1 + \frac{R}{100})^{n}

\frac{9 , 261}{8 , 000} = (1 + \frac{5}{100})^{n} = (\frac{21}{20})^{n}

Step 2: Recognize powers:

(\frac{21}{20})^{3} = \frac{9 , 261}{8 , 000} ⟹ n = 3 years

Answer: 3 years.

Example 4

What is the effective annual rate for a nominal rate of 20% per annum compounded half-yearly?

Solution

Step 1: Calculate rate per conversion period.

R_{half} = \frac{20%}{2} = 10%

Step 2: Calculate effective annual yield.

R_{eff} = (1 + \frac{10}{100})^{2} - 1 = (1.1)^{2} - 1 = 1.21 - 1 = 0.21 = 21%

Answer: 21% per annum.

Example 5

12,500 amounts to 15,125 in 2 years at compound interest compounded annually. Find the annual rate.

Solution

Step 1: Growth ratio for 2 years:

(1 + \frac{R}{100})^{2} = \frac{15 , 125}{12 , 500} = 1.21

Step 2: Take square root:

1 + \frac{R}{100} = 1.21 = 1.1 ⟹ \frac{R}{100} = 0.1 ⟹ R = 10%

Answer: 10% per annum.

Example 6

Find the amount when 20,000 is invested at 8% per annum compounded quarterly for 1 year.

Solution

Step 1: Determine quarterly parameters.
- Periodic rate: $R_{q} = \frac{8%}{4} = 2%$
- Number of quarters: $n = 1 \times 4 = 4$

Step 2: Compute compound amount.

A = 20, 000 \times (1 + 0.02)^{4} = 20, 000 \times (1.02)^{4} \approx 20, 000 \times 1.082432 = 21, 648.64

Answer: 21,648.64.

Example 7

Find the compound interest on 25,000 for 1.5 years at 10% per annum compounded half-yearly.

Solution

Step 1: Parameters for half-yearly compounding:
- Periodic rate: $R_{h} = \frac{10%}{2} = 5%$
- Periods: $n = 1.5 \times 2 = 3$

Step 2: Compute amount and interest.

A = 25, 000 \times (1.05)^{3} = 25, 000 \times 1.157625 = 28, 940.63

CI = 28, 940.63 - 25, 000 = 3, 940.63

Answer: Approximately 3,940.63.

Example 8

The compound interest on a sum for 2 years at 5% per annum is 512.50. Find the principal.

Solution

Step 1: Express CI in terms of $P$ .

CI = P [(1 + \frac{5}{100})^{2} - 1] = P [(1.05)^{2} - 1] = P [1.1025 - 1] = 0.1025 P

Step 2: Solve for $P$ .

0.1025 P = 512.50 ⟹ P = \frac{512.50}{0.1025} = 5, 000

Answer: 5,000.

Example 9

A machine worth 100,000 depreciates at 10% per annum. What will be its book value after 2 years?

Solution

Step 1: Apply depreciation formula.

V = P (1 - \frac{R}{100})^{n}

V = 100, 000 \times (1 - 0.10)^{2} = 100, 000 \times (0.90)^{2} = 100, 000 \times 0.81 = 81, 000

Answer: 81,000.

Example 10

A population increases by 10% in the first year and decreases by 10% in the second year. If it starts at 100,000, what is the population after 2 years?

Solution

Step 1: Apply consecutive annual growth factors.
- After Year 1: $100, 000 \times 1.10 = 110, 000$
- After Year 2: $110, 000 \times 0.90 = 99, 000$

Step 2: Net change:

Net Factor = (1 + 0.10) (1 - 0.10) = 1 - (0.10)^{2} = 0.99 ⟹ 99% of 100, 000 = 99, 000

Answer: 99,000.

Example 11

6,000 is invested for 2 years at 10% per annum. How much larger is the compound interest when compounded half-yearly rather than annually?

Solution

Step 1: Compounded annually ( $n = 2, R = 10%$ ):

A_{annual} = 6, 000 \times (1.10)^{2} = 7, 260 ⟹ CI_{annual} = 1, 260

Step 2: Compounded half-yearly ( $n = 4, R = 5%$ ):

A_{half} = 6, 000 \times (1.05)^{4} \approx 6, 000 \times 1.215506 = 7, 293.04

CI_{half} = 7, 293.04 - 6, 000 = 1, 293.04

Step 3: Difference:

Δ CI = 1, 293.04 - 1, 260 = 33.04

Answer: Approximately 33.04.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

Rs. 12500 amounts to Rs. 15125 in 2 years at compound interest. What is the rate of interest?

Answer: B. 10%

Explanation

$12500 x (1 + r /100)^{2} = 15125$
$(1 + r /100)^{2} = 15125/12500 = 1.21$
$1 + r /100 = 1.1$
r = 10%

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

The compound interest on a sum for 2 years at 10% per annum is Rs. 3150. What is the simple interest for the same period?

Rs. 3000
Rs. 3150
Rs. 3300
Rs. 3500

Answer: A. Rs. 3000

Explanation

CI = P[(1.1)^2 - 1] = 0.21P = 3150
$P = 3150/0.21 = 15000$
$S I = (15000 \times 10 \times 2) /100 = 3000$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

The compound interest on Rs. 40000 for 2 years is Rs. 8400. What is the rate of interest?

Answer: C. 10%

Explanation

Amount = 40000 + 8400 = 48400
$40000 x (1 + r /100)^{2} = 48400$
$(1 + r /100)^{2} = 48400/40000 = 1.21$
$1 + r /100 = 1.1$
r = 10%

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Write the principal, nominal rate, compounding frequency and total periods before calculating.
Divide the annual rate by the number of compounding periods and multiply the years by that same number.
Subtract the original principal only when the question asks for compound interest rather than amount.
For a particular year's interest, use the amount at the start of that year as the base.
Use effective annual rate when comparing different compounding frequencies over one year.
Treat growth and depreciation as repeated multiplication, not repeated addition.
State the convention for a fractional year under annual compounding before solving.
Do not use the one-deposit formula for regular instalments; payment timing changes the calculation.
Keep exact intermediate values and round only the final answer.

Sources and review notes

Reviewed by AISEA Editorial Desk on 2 Sep 2026.

The page was written against the Compound Interest question bank and covers annual, half-yearly, quarterly and monthly compounding, effective rates, unknown variables, depreciation and successive changes.
Worked examples were recalculated independently and checked for period conversion, amount-versus-interest distinction, rate bases, growth factors and final rounding.
The page explicitly separates single-deposit compounding from instalments and calls out the convention required for fractional periods under annual compounding.

Compound InterestOpenStax · Checked 2 Sep 2026Supports: compound-interest growth factors, future value, present value and compounding frequency
Time Value of Money BasicsOpenStax · Checked 2 Sep 2026Supports: interest on accumulated interest and the effect of repeated periods on future value
Exponential FunctionsOpenStax · Checked 2 Sep 2026Supports: exponential growth and repeated multiplicative factors in compound-interest models

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Work through 116 questions with explanations after each answer.

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Next topic: Permutation and Combination

Compound Interest questions with answers and concepts

116 practice questionsPage updated 11 Nov 2026Reviewed by AISEA Editorial DeskPractice questions

Practise this concept

Key formulas and rules

8 rules

Annual compounding: $A = P (1 + \frac{R}{100})^{T}$ , where $T$ is the number of years.

Compounding $n$ times per year: $A = P (1 + \frac{R}{100 \times n})^{n \times T}$ .

Compound interest: $C I = A - P = P [(1 + \frac{R}{100})^{T} - 1]$ .

Present principal: $P = \frac{A}{( 1 + \frac{R}{100} ) ^{T}}$ .

Effective annual rate: $EAR = [(1 + \frac{R}{100 \times n})^{n} - 1] \times 100%$ .

Depreciation: $Value = P (1 - \frac{R}{100})^{T}$ .

Difference between CI and SI (2 years): $C I - S I = P (\frac{R}{100})^{2}$ .

Difference between CI and SI (3 years): $C I - S I = P (\frac{R}{100})^{2} (3 + \frac{R}{100})$ .

Key concepts

Separate principal, amount and interest

Convert the nominal rate and frequency

Use the compound growth factor

Effective annual rate versus nominal rate

Solve for time or rate from powers

Handle partial years explicitly

Find interest for a particular period

Compare compound and simple interest

Apply successive increases and decreases

Do not confuse deposits with instalments

Check units and rounding

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Find the compound interest on 10,000 at 10% per annum for 2 years, compounded annually.

Solution

Step 1: Calculate the accumulated amount.

A = P (1 + \frac{R}{100})^{n} = 10, 000 (1 + \frac{10}{100})^{2} = 10, 000 \times (1.1)^{2} = 12, 100

Step 2: Compute compound interest.

CI = A - P = 12, 100 - 10, 000 = 2, 100

Answer: 2,100.

Example 2

Find the difference between compound interest and simple interest on 5,000 for 2 years at 8% per annum.

Solution

Step 1: Apply the 2-year difference formula.

CI - SI = P (\frac{R}{100})^{2}

Step 2: Substitute $P = 5, 000$ and $R = 8$ .

Difference = 5, 000 \times (\frac{8}{100})^{2} = 5, 000 \times \frac{64}{10 , 000} = \frac{320 , 000}{10 , 000} = 32

Answer: 32.

Example 3

In how many years will 8,000 amount to 9,261 at 5% per annum compound interest?

Solution

Step 1: Set up the growth equation.

\frac{A}{P} = (1 + \frac{R}{100})^{n}

\frac{9 , 261}{8 , 000} = (1 + \frac{5}{100})^{n} = (\frac{21}{20})^{n}

Step 2: Recognize powers:

(\frac{21}{20})^{3} = \frac{9 , 261}{8 , 000} ⟹ n = 3 years

Answer: 3 years.

Example 4

What is the effective annual rate for a nominal rate of 20% per annum compounded half-yearly?

Solution

Step 1: Calculate rate per conversion period.

R_{half} = \frac{20%}{2} = 10%

Step 2: Calculate effective annual yield.

R_{eff} = (1 + \frac{10}{100})^{2} - 1 = (1.1)^{2} - 1 = 1.21 - 1 = 0.21 = 21%

Answer: 21% per annum.

Example 5

12,500 amounts to 15,125 in 2 years at compound interest compounded annually. Find the annual rate.

Solution

Step 1: Growth ratio for 2 years:

(1 + \frac{R}{100})^{2} = \frac{15 , 125}{12 , 500} = 1.21

Step 2: Take square root:

1 + \frac{R}{100} = 1.21 = 1.1 ⟹ \frac{R}{100} = 0.1 ⟹ R = 10%

Answer: 10% per annum.

Example 6

Find the amount when 20,000 is invested at 8% per annum compounded quarterly for 1 year.

Solution

Step 1: Determine quarterly parameters.
- Periodic rate: $R_{q} = \frac{8%}{4} = 2%$
- Number of quarters: $n = 1 \times 4 = 4$

Step 2: Compute compound amount.

A = 20, 000 \times (1 + 0.02)^{4} = 20, 000 \times (1.02)^{4} \approx 20, 000 \times 1.082432 = 21, 648.64

Answer: 21,648.64.

Example 7

Find the compound interest on 25,000 for 1.5 years at 10% per annum compounded half-yearly.

Solution

Step 1: Parameters for half-yearly compounding:
- Periodic rate: $R_{h} = \frac{10%}{2} = 5%$
- Periods: $n = 1.5 \times 2 = 3$

Step 2: Compute amount and interest.

A = 25, 000 \times (1.05)^{3} = 25, 000 \times 1.157625 = 28, 940.63

CI = 28, 940.63 - 25, 000 = 3, 940.63

Answer: Approximately 3,940.63.

Example 8

The compound interest on a sum for 2 years at 5% per annum is 512.50. Find the principal.

Solution

Step 1: Express CI in terms of $P$ .

CI = P [(1 + \frac{5}{100})^{2} - 1] = P [(1.05)^{2} - 1] = P [1.1025 - 1] = 0.1025 P

Step 2: Solve for $P$ .

0.1025 P = 512.50 ⟹ P = \frac{512.50}{0.1025} = 5, 000

Answer: 5,000.

Example 9

A machine worth 100,000 depreciates at 10% per annum. What will be its book value after 2 years?

Solution

Step 1: Apply depreciation formula.

V = P (1 - \frac{R}{100})^{n}

V = 100, 000 \times (1 - 0.10)^{2} = 100, 000 \times (0.90)^{2} = 100, 000 \times 0.81 = 81, 000

Answer: 81,000.

Example 10

A population increases by 10% in the first year and decreases by 10% in the second year. If it starts at 100,000, what is the population after 2 years?

Solution

Step 1: Apply consecutive annual growth factors.
- After Year 1: $100, 000 \times 1.10 = 110, 000$
- After Year 2: $110, 000 \times 0.90 = 99, 000$

Step 2: Net change:

Net Factor = (1 + 0.10) (1 - 0.10) = 1 - (0.10)^{2} = 0.99 ⟹ 99% of 100, 000 = 99, 000

Answer: 99,000.

Example 11

6,000 is invested for 2 years at 10% per annum. How much larger is the compound interest when compounded half-yearly rather than annually?

Solution

Step 1: Compounded annually ( $n = 2, R = 10%$ ):

A_{annual} = 6, 000 \times (1.10)^{2} = 7, 260 ⟹ CI_{annual} = 1, 260

Step 2: Compounded half-yearly ( $n = 4, R = 5%$ ):

A_{half} = 6, 000 \times (1.05)^{4} \approx 6, 000 \times 1.215506 = 7, 293.04

CI_{half} = 7, 293.04 - 6, 000 = 1, 293.04

Step 3: Difference:

Δ CI = 1, 293.04 - 1, 260 = 33.04

Answer: Approximately 33.04.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

Rs. 12500 amounts to Rs. 15125 in 2 years at compound interest. What is the rate of interest?

Answer: B. 10%

Explanation

$12500 x (1 + r /100)^{2} = 15125$
$(1 + r /100)^{2} = 15125/12500 = 1.21$
$1 + r /100 = 1.1$
r = 10%

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

The compound interest on a sum for 2 years at 10% per annum is Rs. 3150. What is the simple interest for the same period?

Rs. 3000
Rs. 3150
Rs. 3300
Rs. 3500

Answer: A. Rs. 3000

Explanation

CI = P[(1.1)^2 - 1] = 0.21P = 3150
$P = 3150/0.21 = 15000$
$S I = (15000 \times 10 \times 2) /100 = 3000$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

The compound interest on Rs. 40000 for 2 years is Rs. 8400. What is the rate of interest?

Answer: C. 10%

Explanation

Amount = 40000 + 8400 = 48400
$40000 x (1 + r /100)^{2} = 48400$
$(1 + r /100)^{2} = 48400/40000 = 1.21$
$1 + r /100 = 1.1$
r = 10%

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Write the principal, nominal rate, compounding frequency and total periods before calculating.
Divide the annual rate by the number of compounding periods and multiply the years by that same number.
Subtract the original principal only when the question asks for compound interest rather than amount.
For a particular year's interest, use the amount at the start of that year as the base.
Use effective annual rate when comparing different compounding frequencies over one year.
Treat growth and depreciation as repeated multiplication, not repeated addition.
State the convention for a fractional year under annual compounding before solving.
Do not use the one-deposit formula for regular instalments; payment timing changes the calculation.
Keep exact intermediate values and round only the final answer.

Sources and review notes

Reviewed by AISEA Editorial Desk on 2 Sep 2026.

The page was written against the Compound Interest question bank and covers annual, half-yearly, quarterly and monthly compounding, effective rates, unknown variables, depreciation and successive changes.
Worked examples were recalculated independently and checked for period conversion, amount-versus-interest distinction, rate bases, growth factors and final rounding.
The page explicitly separates single-deposit compounding from instalments and calls out the convention required for fractional periods under annual compounding.

Compound InterestOpenStax · Checked 2 Sep 2026Supports: compound-interest growth factors, future value, present value and compounding frequency
Time Value of Money BasicsOpenStax · Checked 2 Sep 2026Supports: interest on accumulated interest and the effect of repeated periods on future value
Exponential FunctionsOpenStax · Checked 2 Sep 2026Supports: exponential growth and repeated multiplicative factors in compound-interest models

Ready to test your understanding?

Work through 116 questions with explanations after each answer.

Start practice

Next topic: Permutation and Combination

Compound InterestOpenStax · Checked 2 Sep 2026Supports: compound-interest growth factors, future value, present value and compounding frequency

Time Value of Money BasicsOpenStax · Checked 2 Sep 2026Supports: interest on accumulated interest and the effect of repeated periods on future value

Exponential FunctionsOpenStax · Checked 2 Sep 2026Supports: exponential growth and repeated multiplicative factors in compound-interest models

Key formulas and rules

Key concepts

Separate principal, amount and interest

Convert the nominal rate and frequency

Use the compound growth factor

Effective annual rate versus nominal rate

Solve for time or rate from powers

Handle partial years explicitly

Find interest for a particular period

Compare compound and simple interest

Apply successive increases and decreases

Do not confuse deposits with instalments

Check units and rounding

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Example 5

Solution

Example 6

Solution

Example 7

Solution

Example 8

Solution

Example 9

Solution

Example 10

Solution

Example 11

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Sources and review notes

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Separate principal, amount and interest

Convert the nominal rate and frequency

Use the compound growth factor

Effective annual rate versus nominal rate

Solve for time or rate from powers

Handle partial years explicitly

Find interest for a particular period

Compare compound and simple interest

Apply successive increases and decreases

Do not confuse deposits with instalments

Check units and rounding

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Example 5

Solution

Example 6

Solution

Example 7

Solution

Example 8

Solution

Example 9

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Example 10

Solution

Example 11

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