Simple Interest questions with answers and concepts

Learn simple 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.

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

Practise this concept

Simple interest is calculated only on the original principal for the full stated period. It grows linearly with principal, rate and time, so each year at a fixed rate contributes the same interest. In aptitude questions, first separate principal, interest and final amount, then put the rate and time in compatible units before solving for the unknown.

Key formulas and rules

9 rules

Simple interest: $S I = \frac{P \times R \times T}{100}$ , when $R$ is a percentage rate per annum and $T$ is in years.

Amount: $A = P + S I = P (1 + \frac{R \times T}{100})$ .

Rate: $R = \frac{100 \times S I}{P \times T}$ .

Time: $T = \frac{100 \times S I}{P \times R}$ .

Principal: $P = \frac{100 \times S I}{R \times T}$ .

For $m$ months: $S I = \frac{P \times R \times m}{100 \times 12}$ .

If $A = k P$ : $S I = (k - 1) P$ and $R = \frac{100 ( k - 1 )}{T}$ .

Difference in interest: For the same $P$ and $T$ , $Δ S I = \frac{P \times Δ R \times T}{100}$ .

Annual interest: If amounts at two times use the same rate, $Annual Interest = \frac{A _{2} - A _{1}}{T _{2} - T _{1}}$ .

Key concepts

Identify P, R, T, SI and A

P is the original principal, R is the annual rate expressed as a percentage, T is time in years, SI is interest alone and A is the final amount. The central relationship is $A = P + S I$ . Write these five quantities in a small list before substituting numbers so that amount is not mistaken for interest.

Simple interest versus compound interest

Under simple interest, every period's interest is based on the original principal. Under compound interest, later interest can be based on a principal that includes earlier interest. Use the simple-interest formula only when the question explicitly says simple interest or gives a context that defines it; do not compound the amount between periods.

Keep rate and time compatible

An annual percentage rate needs time in years. Convert months to months divided by 12, and follow any stated day-count convention for days. Do not use 8 months as $T = 8$ in the annual formula. If a rate is already written as a decimal, use $S I = P$ × r × T instead of dividing that decimal by 100 again.

Solve for the missing variable

Find SI first when the amount is given: $S I = A$ − P. Then rearrange the formula for the unknown principal, rate or time. Keep the percentage form of the rate in the formula with the divisor 100, or convert R to a decimal and remove that divisor, but never mix the two conventions.

Use amount differences to remove the principal

With simple interest, the difference between amounts at two dates equals the interest for the time gap. Divide that difference by the number of years to get one year's interest. Subtract the interest accumulated before the earlier date from the earlier amount to recover the principal, then use the principal to find the rate if needed.

Handle multiple rates on one principal

If the same principal earns simple interest at one rate for one period and another rate for a later period, calculate each period separately on the original principal and add the interest. If different portions are invested at different rates, use a weighted sum of their interests and make the portions add to the total principal.

Use multiples of the principal

If a sum doubles, the interest equals one principal. If it triples, the interest equals two principals. More generally, if the amount is k times the principal, the interest is (k − 1) times the principal. This turns many rate and time questions into a direct proportion without choosing an arbitrary principal.

Compare rates, returns and lending gains

A change in annual rate changes simple interest by P × rate difference × T / 100. For a borrower who lends the same money onward, gain is the interest earned minus the interest paid. Compare the same principal and period, and do not report the final amount as the interest or gain.

Check units, rounding and reasonableness

Interest should be zero when time or rate is zero, and it should scale linearly with each of P, R and T. Check that the final amount is P + SI, that a rate is a percentage rather than a decimal, and that rounding is applied only after the exact calculation. If an option is approximate, verify the exact value before choosing it.

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

Worked examples

Example 1

Find the simple interest on 5,000 at 8% per annum for 3 years.

Solution

Step 1: Apply the standard simple interest formula.

SI = \frac{P \times R \times T}{100}

Step 2: Substitute $P = 5, 000$ , $R = 8$ , and $T = 3$ .

SI = \frac{5 , 000 \times 8 \times 3}{100} = 50 \times 24 = 1, 200

Answer: 1,200.

Example 2

A sum of 6,000 amounts to 7,200 in 4 years at simple interest. Find the annual rate.

Solution

Step 1: Compute interest earned.

SI = A - P = 7, 200 - 6, 000 = 1, 200

Step 2: Solve for rate $R$ .

R = \frac{100 \times SI}{P \times T} = \frac{100 \times 1 , 200}{6 , 000 \times 4} = \frac{120 , 000}{24 , 000} = 5%

Answer: 5% per annum.

Example 3

What principal will amount to 7,000 in 4 years at 10% per annum simple interest?

Solution

Step 1: Relate Amount and Principal.

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

7, 000 = P (1 + \frac{10 \times 4}{100}) = P (1 + 0.40) = 1.40 P

Step 2: Solve for $P$ .

P = \frac{7 , 000}{1.40} = 5, 000

Answer: 5,000.

Example 4

A sum amounts to 5,400 in 3 years and 6,200 in 5 years at simple interest. Find the principal.

Solution

Step 1: Calculate interest earned over $(5 - 3) = 2 years$ .

Interest for 2 years = 6, 200 - 5, 400 = 800

Interest per year = \frac{800}{2} = 400

Step 2: Calculate principal by subtracting 3 years' interest from 3-year amount.

P = 5, 400 - (3 \times 400) = 5, 400 - 1, 200 = 4, 200

Answer: 4,200.

Example 5

Find the simple interest on 12,000 for 8 months at 15% per annum.

Solution

Step 1: Convert time period to years.

T = \frac{8}{12} = \frac{2}{3} years

Step 2: Calculate interest.

SI = \frac{12 , 000 \times 15 \times \frac{2}{3}}{100} = \frac{12 , 000 \times 10}{100} = 1, 200

Answer: 1,200.

Example 6

At what annual rate will a principal double itself in 8 years at simple interest?

Solution

Step 1: For a principal to double, $SI = P$ .

P = \frac{P \times R \times 8}{100} ⟹ 1 = \frac{8 R}{100} ⟹ R = \frac{100}{8} = 12.5%

Answer: 12.5% per annum.

Example 7

15,000 is split between investments at 8% and 12% per annum simple interest. If the total annual interest is 1,440, how much is invested at 8%?

Solution

Step 1: Let the amount at 8% be $x$ , and the amount at 12% be $(15, 000 - x)$ .

Step 2: Set up the annual interest equation.

\frac{8 x}{100} + \frac{12 ( 15 , 000 - x )}{100} = 1, 440

8 x + 180, 000 - 12 x = 144, 000

4 x = 36, 000 ⟹ x = 9, 000

Answer: 9,000.

Example 8

12,000 is invested for 2 years at 8% per annum and for the next 2 years at 12% per annum simple interest. Find the final amount, assuming interest is calculated on the original principal throughout.

Solution

Step 1: Calculate interest across both periods.

SI_{1} = \frac{12 , 000 \times 8 \times 2}{100} = 1, 920

SI_{2} = \frac{12 , 000 \times 12 \times 2}{100} = 2, 880

Total SI = 1, 920 + 2, 880 = 4, 800

Step 2: Total final amount:

A = 12, 000 + 4, 800 = 16, 800

Answer: 16,800.

Example 9

The difference between the simple interest from two banks on 5,000 for 4 years is 200. Find the difference between their annual interest rates.

Solution

Step 1: Express difference in SI in terms of rate difference $Δ R$ .

Δ SI = \frac{P \times Δ R \times T}{100}

200 = \frac{5 , 000 \times Δ R \times 4}{100} = 200Δ R ⟹ Δ R = 1%

Answer: 1 percentage point per annum.

Example 10

A person borrows 5,000 at 4% per annum simple interest for 2 years and lends it at 6% for the same period. Find his net gain.

Solution

Step 1: Calculate net interest gain using rate spread $(6% - 4% = 2%)$ .

Gain = \frac{5 , 000 \times ( 6 - 4 ) \times 2}{100} = \frac{5 , 000 \times 4}{100} = 200

Answer: 200.

Representative solved questions

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

Question 1

A sum of money at simple interest amounts to 7020 in 5 years and 7830 in 8 years. Find the principal.

5000
5400
6000
5670

Answer: D. 5670

Explanation

SI for 3 years = 7830 - 7020 = 810
SI for 1 year = 810/3 = 270
SI for 5 years = 270 x $5 = 1350$
Principal = 7020 - 1350 = 5670

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

A sum of 8000 becomes 9600 in 4 years at simple interest. In how many years will it become 11200?

8 years
6 years
7 years
10 years

Answer: A. 8 years

Explanation

SI in 4 years = 9600 - 8000 = 1600
Rate = (1600 x 100)/(8000 x 4) = 5%
For amount 11200, $S I = 11200 - 8000 = 3200$
Time = (3200 x 100)/(8000 x 5) = 8 years

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

A sum of money triples itself in 20 years at simple interest. In how many years will it become 5 times itself?

30 years
35 years
40 years
50 years

Answer: C. 40 years

Explanation

Triples in 20 years means $S I = 2 P$ in 20 years
Rate = (2P x 100)/(P x 20) = 10%
To become 5 times, $S I = 4 P$
Time = (4P x 100)/(P x 10) = 40 years

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 P, R, T, SI and A separately before substituting values.
If amount is given, calculate SI as amount minus principal before finding the rate or time.
Convert months to years for an annual rate and do not divide a decimal rate by 100 twice.
For amount differences at two dates, divide by the time gap to find the interest per year.
For split investments, make the portions add to the total principal and use a weighted interest equation.
Under simple interest, later periods use the original principal; adding interest to the principal changes the problem to compound interest.
Use the doubling and tripling shortcuts when the amount is a multiple of the principal.
Check that amount equals principal plus interest and apply rounding only at the end.

Sources and review notes

Reviewed by AISEA Editorial Desk on 2 Sep 2026.

The page was written against the Simple Interest question bank and covers direct calculations, unknown variables, partial years, amount differences, changing rates, split investments and lending gains.
Worked examples were recalculated independently and checked for annual-rate units, principal-versus-amount distinction, linear interest treatment and final rounding.
The page explicitly separates simple interest from compounding and flags the original-principal assumption when rates change across periods.

Solve Simple Interest ApplicationsOpenStax · Checked 2 Sep 2026Supports: simple-interest variables, the SI formula and solving for principal, rate and time
Simple InterestOpenStax · Checked 2 Sep 2026Supports: future amount under simple interest and time conversions for months and days
Solve a Formula for a Specific VariableOpenStax · Checked 2 Sep 2026Supports: rearranging formulas to solve for an unknown variable

Ready to test your understanding?

Work through 115 questions with explanations after each answer.

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Next topic: Compound Interest

Simple Interest questions with answers and concepts

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

Practise this concept

Key formulas and rules

9 rules

Simple interest: $S I = \frac{P \times R \times T}{100}$ , when $R$ is a percentage rate per annum and $T$ is in years.

Amount: $A = P + S I = P (1 + \frac{R \times T}{100})$ .

Rate: $R = \frac{100 \times S I}{P \times T}$ .

Time: $T = \frac{100 \times S I}{P \times R}$ .

Principal: $P = \frac{100 \times S I}{R \times T}$ .

For $m$ months: $S I = \frac{P \times R \times m}{100 \times 12}$ .

If $A = k P$ : $S I = (k - 1) P$ and $R = \frac{100 ( k - 1 )}{T}$ .

Difference in interest: For the same $P$ and $T$ , $Δ S I = \frac{P \times Δ R \times T}{100}$ .

Annual interest: If amounts at two times use the same rate, $Annual Interest = \frac{A _{2} - A _{1}}{T _{2} - T _{1}}$ .

Key concepts

Identify P, R, T, SI and A

Simple interest versus compound interest

Keep rate and time compatible

Solve for the missing variable

Use amount differences to remove the principal

Handle multiple rates on one principal

Use multiples of the principal

Compare rates, returns and lending gains

Check units, rounding and reasonableness

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

Worked examples

Example 1

Find the simple interest on 5,000 at 8% per annum for 3 years.

Solution

Step 1: Apply the standard simple interest formula.

SI = \frac{P \times R \times T}{100}

Step 2: Substitute $P = 5, 000$ , $R = 8$ , and $T = 3$ .

SI = \frac{5 , 000 \times 8 \times 3}{100} = 50 \times 24 = 1, 200

Answer: 1,200.

Example 2

A sum of 6,000 amounts to 7,200 in 4 years at simple interest. Find the annual rate.

Solution

Step 1: Compute interest earned.

SI = A - P = 7, 200 - 6, 000 = 1, 200

Step 2: Solve for rate $R$ .

R = \frac{100 \times SI}{P \times T} = \frac{100 \times 1 , 200}{6 , 000 \times 4} = \frac{120 , 000}{24 , 000} = 5%

Answer: 5% per annum.

Example 3

What principal will amount to 7,000 in 4 years at 10% per annum simple interest?

Solution

Step 1: Relate Amount and Principal.

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

7, 000 = P (1 + \frac{10 \times 4}{100}) = P (1 + 0.40) = 1.40 P

Step 2: Solve for $P$ .

P = \frac{7 , 000}{1.40} = 5, 000

Answer: 5,000.

Example 4

A sum amounts to 5,400 in 3 years and 6,200 in 5 years at simple interest. Find the principal.

Solution

Step 1: Calculate interest earned over $(5 - 3) = 2 years$ .

Interest for 2 years = 6, 200 - 5, 400 = 800

Interest per year = \frac{800}{2} = 400

Step 2: Calculate principal by subtracting 3 years' interest from 3-year amount.

P = 5, 400 - (3 \times 400) = 5, 400 - 1, 200 = 4, 200

Answer: 4,200.

Example 5

Find the simple interest on 12,000 for 8 months at 15% per annum.

Solution

Step 1: Convert time period to years.

T = \frac{8}{12} = \frac{2}{3} years

Step 2: Calculate interest.

SI = \frac{12 , 000 \times 15 \times \frac{2}{3}}{100} = \frac{12 , 000 \times 10}{100} = 1, 200

Answer: 1,200.

Example 6

At what annual rate will a principal double itself in 8 years at simple interest?

Solution

Step 1: For a principal to double, $SI = P$ .

P = \frac{P \times R \times 8}{100} ⟹ 1 = \frac{8 R}{100} ⟹ R = \frac{100}{8} = 12.5%

Answer: 12.5% per annum.

Example 7

15,000 is split between investments at 8% and 12% per annum simple interest. If the total annual interest is 1,440, how much is invested at 8%?

Solution

Step 1: Let the amount at 8% be $x$ , and the amount at 12% be $(15, 000 - x)$ .

Step 2: Set up the annual interest equation.

\frac{8 x}{100} + \frac{12 ( 15 , 000 - x )}{100} = 1, 440

8 x + 180, 000 - 12 x = 144, 000

4 x = 36, 000 ⟹ x = 9, 000

Answer: 9,000.

Example 8

12,000 is invested for 2 years at 8% per annum and for the next 2 years at 12% per annum simple interest. Find the final amount, assuming interest is calculated on the original principal throughout.

Solution

Step 1: Calculate interest across both periods.

SI_{1} = \frac{12 , 000 \times 8 \times 2}{100} = 1, 920

SI_{2} = \frac{12 , 000 \times 12 \times 2}{100} = 2, 880

Total SI = 1, 920 + 2, 880 = 4, 800

Step 2: Total final amount:

A = 12, 000 + 4, 800 = 16, 800

Answer: 16,800.

Example 9

The difference between the simple interest from two banks on 5,000 for 4 years is 200. Find the difference between their annual interest rates.

Solution

Step 1: Express difference in SI in terms of rate difference $Δ R$ .

Δ SI = \frac{P \times Δ R \times T}{100}

200 = \frac{5 , 000 \times Δ R \times 4}{100} = 200Δ R ⟹ Δ R = 1%

Answer: 1 percentage point per annum.

Example 10

A person borrows 5,000 at 4% per annum simple interest for 2 years and lends it at 6% for the same period. Find his net gain.

Solution

Step 1: Calculate net interest gain using rate spread $(6% - 4% = 2%)$ .

Gain = \frac{5 , 000 \times ( 6 - 4 ) \times 2}{100} = \frac{5 , 000 \times 4}{100} = 200

Answer: 200.

Representative solved questions

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

Question 1

A sum of money at simple interest amounts to 7020 in 5 years and 7830 in 8 years. Find the principal.

5000
5400
6000
5670

Answer: D. 5670

Explanation

SI for 3 years = 7830 - 7020 = 810
SI for 1 year = 810/3 = 270
SI for 5 years = 270 x $5 = 1350$
Principal = 7020 - 1350 = 5670

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

A sum of 8000 becomes 9600 in 4 years at simple interest. In how many years will it become 11200?

8 years
6 years
7 years
10 years

Answer: A. 8 years

Explanation

SI in 4 years = 9600 - 8000 = 1600
Rate = (1600 x 100)/(8000 x 4) = 5%
For amount 11200, $S I = 11200 - 8000 = 3200$
Time = (3200 x 100)/(8000 x 5) = 8 years

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

A sum of money triples itself in 20 years at simple interest. In how many years will it become 5 times itself?

30 years
35 years
40 years
50 years

Answer: C. 40 years

Explanation

Triples in 20 years means $S I = 2 P$ in 20 years
Rate = (2P x 100)/(P x 20) = 10%
To become 5 times, $S I = 4 P$
Time = (4P x 100)/(P x 10) = 40 years

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 P, R, T, SI and A separately before substituting values.
If amount is given, calculate SI as amount minus principal before finding the rate or time.
Convert months to years for an annual rate and do not divide a decimal rate by 100 twice.
For amount differences at two dates, divide by the time gap to find the interest per year.
For split investments, make the portions add to the total principal and use a weighted interest equation.
Under simple interest, later periods use the original principal; adding interest to the principal changes the problem to compound interest.
Use the doubling and tripling shortcuts when the amount is a multiple of the principal.
Check that amount equals principal plus interest and apply rounding only at the end.

Sources and review notes

Reviewed by AISEA Editorial Desk on 2 Sep 2026.

The page was written against the Simple Interest question bank and covers direct calculations, unknown variables, partial years, amount differences, changing rates, split investments and lending gains.
Worked examples were recalculated independently and checked for annual-rate units, principal-versus-amount distinction, linear interest treatment and final rounding.
The page explicitly separates simple interest from compounding and flags the original-principal assumption when rates change across periods.

Solve Simple Interest ApplicationsOpenStax · Checked 2 Sep 2026Supports: simple-interest variables, the SI formula and solving for principal, rate and time
Simple InterestOpenStax · Checked 2 Sep 2026Supports: future amount under simple interest and time conversions for months and days
Solve a Formula for a Specific VariableOpenStax · Checked 2 Sep 2026Supports: rearranging formulas to solve for an unknown variable

Ready to test your understanding?

Work through 115 questions with explanations after each answer.

Start practice

Next topic: Compound Interest

Solve Simple Interest ApplicationsOpenStax · Checked 2 Sep 2026Supports: simple-interest variables, the SI formula and solving for principal, rate and time

Simple InterestOpenStax · Checked 2 Sep 2026Supports: future amount under simple interest and time conversions for months and days

Solve a Formula for a Specific VariableOpenStax · Checked 2 Sep 2026Supports: rearranging formulas to solve for an unknown variable

Key formulas and rules

Key concepts

Identify P, R, T, SI and A

Simple interest versus compound interest

Keep rate and time compatible

Solve for the missing variable

Use amount differences to remove the principal

Handle multiple rates on one principal

Use multiples of the principal

Compare rates, returns and lending gains

Check units, rounding and reasonableness

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

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

Identify P, R, T, SI and A

Simple interest versus compound interest

Keep rate and time compatible

Solve for the missing variable

Use amount differences to remove the principal

Handle multiple rates on one principal

Use multiples of the principal

Compare rates, returns and lending gains

Check units, rounding and reasonableness

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

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.