Railway Exams Maths Simple Interest: Easy Concepts, Formulas & Examples

Simple Interest is a fundamental concept in arithmetic and an important topic frequently covered in Railway Exams quantitative aptitude sections. It refers to the interest calculated only on the initial principal amount, which remains constant throughout the investment or loan period. Understanding Simple Interest concepts, formulas, shortcut methods, and practical applications can help candidates solve Railway Exams Maths questions more quickly and accurately.

Railway Exams Maths - Simple Interest vs. Compound Interest

The primary difference between Simple Interest (SI) and Compound Interest (CI) lies in how interest is calculated:

• Simple Interest: The principal remains constant, and interest is not charged on previously accrued interest.

• Compound Interest: Interest is charged on interest. Consequently, the principal does not remain constant as accrued interest is added to the principal for subsequent calculations.

Key Terminology in Simple Interest

In Simple Interest (SI), interest is calculated only on the original principal amount and does not get added back to the principal for future interest calculation. Understanding these terms is important for Simple Interest calculations. 

• Principal (P): Also known as मूलधन, this is the initial amount of money. The principal is always 100% of itself.

• Rate (R): The annual interest rate (दर).

• Time (T): The duration for which the money is borrowed or lent (समय), typically in years.

• Simple Interest (SI): The calculated interest (साधारण ब्याज).

• Amount (AMT): Also known as मिश्रधन, this is the total sum of money, including principal and interest. Amount = Principal + Interest.

Fundamental Simple Interest Formula & Related Derivations

The primary formula for calculating Simple Interest is:

SI = (P × R × T) / 100

Where:

• SI = Simple Interest

• P = Principal

• R = Rate (per annum)

• T = Time (in years)

From this, other variables can be derived:

• Rate (R) = (SI × 100) / (P × T)

• Time (T) = (SI × 100) / (P × R)

• Principal (P) = (SI × 100) / (R × T)

For situations where the Principal is constant, simple interest (SI) can be directly calculated as the product of Rate and Time as a percentage of the principal (SI % = Rate × Time).

Simple Interest Calculation Applications

Calculating Simple Interest

Problem: Find the Simple Interest on ₹4500 at 5% per annum for 3 years.

Solution:

Using SI = (P × R × T) / 100:

SI = (4500 × 5 × 3) / 100 = 45 × 15 = ₹675.

Alternatively, total interest rate = 5% × 3 years = 15%. SI = 15% of ₹4500 = ₹675.

Problem: Principal ₹2400, Rate 12.5%, Time 3 years. Calculate SI.

Solution:

Using fractional value of 12.5% as 1/8:

SI = P × (Fractional Rate) × T = 2400 × (1/8) × 3 = 300 × 3 = ₹900.

Calculating Principal from Simple Interest

Problem: Rate 7%, Time 3 years, Simple Interest ₹630. Find the Principal.

Solution:

Using Principal (P) = (SI × 100) / (R × T):

P = (630 × 100) / (7 × 3) = 63000 / 21 = ₹3000.

Alternatively, total interest rate = 7% × 3 = 21%. If 21% = ₹630, then 1% = ₹30. Since Principal is 100%, P = 100 × ₹30 = ₹3000.

Time Conversion for Simple Interest

Problem: Calculate Simple Interest on ₹7200 at 3 3/4% per annum for 8 months.

Memory Tip: When converting from a smaller unit to a larger unit (e.g., months to years), divide.

Solution:
Rate (R) = 3 3/4% = 15/4%.
Time (T) = 8 months = 8/12 years = 2/3 years.
SI = (P × R × T) / 100 = (7200 × (15/4) × (2/3)) / 100 = ₹180.

Calculating Total Amount Payable

Problem: Raghu took a loan of ₹1500 for 3 years at 10% Simple Interest. Find the total Amount to pay back.

Solution:

SI = (1500 × 10 × 3) / 100 = ₹450.

Amount = Principal + SI = 1500 + 450 = ₹1950.

Calculating Rate from Principal and Amount Ratio

Problem: The ratio of Principal and Amount after 3 years is 20:29. Find the annual rate of interest.

Solution:

If P=₹20, AMT=₹29, then SI = AMT - P = 29 - 20 = ₹9.

Rate (R) = (SI × 100) / (P × T) = (9 × 100) / (20 × 3) = 900 / 60 = 15%.

Simple Interest for Multiple Loans/Deposits

Problem: Sanjay borrowed ₹900 at 4% and ₹1100 at 5% for the same duration. Total interest paid was ₹364. Find the time period.

Solution:

Let T be the time.

Total SI = SI1 + SI2

364 = (900 × 4 × T) / 100 + (1100 × 5 × T) / 100

364 = 36T + 55T = 91T

T = 364 / 91 = 4 years.

Problem: Person deposits ₹5000 for 4 years and ₹6000 for 3 years at the same rate. Total interest received is ₹2470. Find the annual rate.

Solution:

Let R be the rate.

Total SI = SI1 + SI2

2470 = (5000 × R × 4) / 100 + (6000 × R × 3) / 100

2470 = 200R + 180R = 380R

R = 2470 / 380 = 247 / 38 = 13 / 2 = 6.5%.

Simple Interest: "Times" or "Multiples" Questions

This common problem type involves a sum of money becoming a multiple of itself over time.

Direct Formula for "Times" Questions

When a sum of money becomes 'n' times itself, the Simple Interest (SI) earned is (n - 1) times the Principal.

A universal formula to quickly find either Rate (R) or Time (T) is:

(n - 1) × 100 = R × T

• Where 'n' is the number of times the money becomes itself (e.g., triples means n=3).

• R is the Rate of Interest (in percentage).

• T is the Time (in years).

Problem: What is the rate of simple interest at which a sum of money triples itself in 50 years?

Solution:

n = 3, T = 50 years.

(3 - 1) × 100 = R × 50

2 × 100 = 50R

200 = 50R

R = 200 / 50 = 4%.

Examples of "Times" Questions

Problem 1: Find the rate at which a sum of money becomes five times itself in 8 years.

Solution: n = 5, T = 8. (5 - 1) × 100 = R × 8 => 400 = 8R => R = 50%.

Problem 2: In how many years will a sum of money triple itself at 25% per annum?

Solution: n = 3, R = 25. (3 - 1) × 100 = 25 × T => 200 = 25T => T = 8 years.

"Two Times" Questions: Changing Time or Rate

When the rate (R) is constant, the ratio of (n-1) to Time (T) remains constant.

(n₁ - 1) / T₁ = (n₂ - 1) / T₂

Problem: A sum of money triples itself in 15 years. In how much time will it become five times itself?

Solution:

Scenario 1: n₁ = 3, T₁ = 15.

Scenario 2: n₂ = 5, T₂ = ?

(3 - 1) / 15 = (5 - 1) / T₂

2 / 15 = 4 / T₂

2T₂ = 60 => T₂ = 30 years.

Memory Tip: For "Two Times" problems, you can compare (n-1) directly with Time.

Effect of Rate Increase on Final Amoun

Problem: ₹800 becomes ₹920 in 3 years. If the rate is increased by 3% per annum, what will the amount be in 3 years?

Memory Tip: Additional/reduced interest is always calculated on the original Principal.

Solution:
Original final amount = ₹920.
Increased rate = 3% per annum.
Total percentage increase in interest over 3 years = 3% × 3 = 9%.

Additional Simple Interest = 9% of original Principal (₹800) = (9/100) × 800 = ₹72.
New final amount = Original final amount + Additional Interest = 920 + 72 = ₹992.

Effect of Rate Decrease on Final Amount

Problem: ₹900 becomes ₹1260 in 4 years. If the rate is decreased by 2% per annum, what will the amount be in 4 years?

Solution:
Original final amount = ₹1260.
Decreased rate = 2% per annum.
Total percentage decrease in interest over 4 years = 2% × 4 = 8%.
Reduced Simple Interest = 8% of original Principal (₹900) = (8/100) × 900 = ₹72.
New final amount = Original final amount - Reduced Interest = 1260 - 72 = ₹1188.Finding Principal from Additional Interest

Problem: A sum is invested for 4 years. If invested at a 3% higher rate, it would have yielded ₹1440 more interest. What is the original sum (principal)?

Solution:
Increased rate = 3% per annum.
Time = 4 years.
Total percentage increase in SI = 3% × 4 = 12%.
This 12% increase corresponds to ₹1440.
If 12% of Principal = ₹1440, then 1% = 1440 / 12 = ₹120.
Principal (100%) = 120 × 100 = ₹12,000.

Finding Principal from Amounts at Different Times

This advanced problem type leverages the fact that in simple interest, the interest earned per year on a constant principal is always the same.

Problem: A sum of money amounts to ₹840 in 3 years and ₹1200 in 7 years at simple interest. What is the value of the principal?

Solution:

1. Amount after 3 years (A₃) = ₹840

2. Amount after 7 years (A₇) = ₹1200

3. Interest earned in (7 - 3) = 4 years = A₇ - A₃ = 1200 - 840 = ₹360.

4. Annual Simple Interest = ₹360 / 4 years = ₹90 per year.

5. Simple Interest for 3 years (SI₃) = ₹90/year × 3 years = ₹270.

6. Principal (P) = A₃ - SI₃ = 840 - 270 = ₹570.
   (Verification: SI for 7 years = 90 × 7 = 630. P = 1200 - 630 = ₹570).

Problem: A sum of money, at Simple Interest, becomes ₹1020 in 5 years and ₹1200 in 8 years. What is the Principal?

1. Interest earned in (8 - 5) = 3 years = 1200 - 1020 = ₹180.

2. Annual Simple Interest = ₹180 / 3 years = ₹60 per year.

3. Simple Interest for 5 years = ₹60/year × 5 years = ₹300.

4. Principal = Amount in 5 years - SI for 5 years = 1020 - 300 = ₹720.

The Allegation Method in Simple Interest

The Allegation Method, also known as Bandhan ka Rule, is a powerful technique used when two different quantities or rates are combined to yield a single average quantity or rate. It helps determine the ratio in which two components are mixed.

Memory Tip: Use the Allegation Method when two components, each with its own value (e.g., interest rate), are combined to yield an overall average value, and you need to determine the proportional contribution of each component.

Allegation Method Applications

Problem 1: Loan from Two Banks
A person takes a loan of ₹10,000. Part is at 8% p.a., rest at 10% p.a. Total annual interest is ₹950. Find the loan amount from the first bank.

Solution:

1. Overall average annual rate (R_avg) = (Total Interest / Total Principal) × 100
   R_avg = (950 / 10000) × 100 = 9.5%.

2. Apply Allegation Method:
   Bank 1 (8%) Bank 2 (10%) \ / \ / 9.5% (Overall) / \ / \ (10 - 9.5) = 0.5 : (9.5 - 8) = 1.5
   Ratio of principals (Bank 1 : Bank 2) = 0.5 : 1.5 = 1 : 3.

3. Total ratio parts = 1 + 3 = 4. Total loan = ₹10,000.
   One part = 10000 / 4 = ₹2500.
   Loan from Bank 1 = 1 part = ₹2500.

Problem 2: Investment with Monthly Interest
Mr. Datta deposited ₹3,00,000 partly at 10% p.a. and partly at 6% p.a. If his monthly interest income is ₹2,000, find the difference between his deposits.

1. Annual Interest = Monthly Interest × 12 = 2000 × 12 = ₹24,000.

2. Overall Annual Rate = (24000 / 300000) × 100 = 8%.

3. Apply Allegation Method:
   Post Office (10%) Bank (6%) \ / \ / 8% (Overall) / \ / \ (8 - 6) = 2 : (10 - 8) = 2
   Ratio of deposits (Post Office : Bank) = 2 : 2 = 1 : 1.

4. Since the ratio is 1:1, deposits are equal. The difference between deposits is zero.

Problem 3: Finding Principal for Multiple Deposits
Seema deposits the same amount of money into two different banks that offer Simple Interest at rates of 10% and 12% respectively. After 3 years, she receives a total interest of ₹3300. What was the total amount she deposited?

Solution:

Memory Tip: For Simple Interest, SI % of Principal = Rate × Time.

1. Interest percentage from Bank 1 = 10% × 3 years = 30%.

2. Interest percentage from Bank 2 = 12% × 3 years = 36%.

3. Total interest percentage = 30% + 36% = 66%.

4. This 66% corresponds to the total interest of ₹3300.

5. If 66% of P = ₹3300, then 1% = 3300 / 66 = ₹50.

6. Principal (100% per bank) = 50 × 100 = ₹5000.
   
   

Since she deposited the same amount in two banks, her total deposit is 2 × ₹5000 if the question implies 'P' is deposit in each bank. However, if the question implies P is the 'total amount deposited' distributed equally, then the total amount she deposited is ₹5000 if this 'P' is the effective total principal that yields 66% interest. Based on the previous derivation (P = 3300 * 100 / 66 = 5000), it implies the total principal across both.