Simple Interest

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Simple Interest (SI) is a fundamental financial concept tested extensively in SSC CGL. When you borrow or lend money, interest is the extra amount paid over the principal. Simple Interest remains constant on the original principal throughout the loan period — unlike compound interest where interest accrues on accumulated interest.

Key Formula: $$SI = \frac{P \times R \times T}{100}$$

Where:

• P = Principal (the original amount borrowed or lent)
• R = Rate of interest per annum (in %)
• T = Time period (in years)

Amount Payable: $$A = P + SI = P\left(1 + \frac{RT}{100}\right)$$

Quick Facts:

• Interest is directly proportional to principal, rate, and time
• Double time means double interest (keeping P and R constant)
• Double rate means double interest (keeping P and T constant)
• SI for 1 year = PRT/100

⚡ SSC CGL Exam Tips:

• Always convert time to years (divide months by 12)
• If time is given in days, divide by 365 (or 366 for leap year)
• A common trick: “at the same rate” means rate doesn’t change -疑 Many questions ask for the sum when SI and rate are known for a certain period

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of Simple Interest through practice and examples.

Understanding Simple Interest with Worked Examples

Example 1: Basic SI Calculation A sum of ₹8,000 is lent at 5% per annum for 3 years. Find the Simple Interest.

$$SI = \frac{8000 \times 5 \times 3}{100} = \frac{120000}{100} = ₹1,200$$

Amount = 8000 + 1200 = ₹9,200

Example 2: Finding Principal The SI on a sum at 7% per annum for 2 years is ₹840. Find the principal.

$$P = \frac{SI \times 100}{R \times T} = \frac{840 \times 100}{7 \times 2} = \frac{84000}{14} = ₹6,000$$

Example 3: Finding Rate The SI on ₹5,000 for 4 years is ₹1,600. Find the rate per annum.

$$R = \frac{SI \times 100}{P \times T} = \frac{1600 \times 100}{5000 \times 4} = \frac{160000}{20000} = 8%$$

Example 4: Finding Time At what rate will ₹2,500 amount to ₹3,000 in 5 years?

SI = 3000 - 2500 = ₹500 $$R = \frac{500 \times 100}{2500 \times 5} = \frac{50000}{12500} = 4%$$

Types of SI Questions in SSC CGL:

Question Type	Approach
Find SI	Use $SI = \frac{PRT}{100}$ directly
Find Amount	Calculate SI first, then add to Principal
Find Principal	Rearrange: $P = \frac{SI \times 100}{R \times T}$
Find Rate	Rearrange: $R = \frac{SI \times 100}{P \times T}$
Find Time	Rearrange: $T = \frac{SI \times 100}{P \times R}$

Common Student Mistakes:

1. Forgetting to convert months to years (e.g., 8 months = 8/12 years)
2. Using wrong time period when interest is per month instead of per annum
3. Confusing SI with Compound Interest in calculation
4. Not dividing rate by number of periods when interest is calculated half-yearly or quarterly

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory with derivations, special cases, and previous year SSC CGL patterns.

Mathematical Derivation of Simple Interest

The concept of Simple Interest derives from the assumption that interest is calculated only on the original principal throughout the loan tenure. If P is lent at R% per annum for T years:

• Interest for 1st year = $\frac{P \times R \times 1}{100}$
• Interest for 2nd year = $\frac{P \times R \times 1}{100}$
• Interest for 3rd year = $\frac{P \times R \times 1}{100}$

Total SI for T years = $T \times \frac{PR}{100} = \frac{PRT}{100}$

Special Cases and Tricks:

1. Interest for fraction of year:

   • For 6 months: $T = \frac{6}{12} = \frac{1}{2}$ year
   • For 3 months: $T = \frac{3}{12} = \frac{1}{4}$ year

2. Rate per month: If annual rate is R%, monthly rate = R/12%

3. Sum becoming n times itself: If sum P becomes n times in T years at R% SI: $$nP = P + \frac{P \times R \times T}{100}$$ $$n - 1 = \frac{RT}{100}$$ $$T = \frac{100(n-1)}{R}$$

4. Difference between SI and CI: For 2 years at R% p.a., difference = $P \times \left(\frac{R}{10}\right)^2$ (when R is single digit)

Previous Year SSC CGL Patterns:

Questions from SSC CGL 2022-2023 on Simple Interest typically follow these patterns:

• Finding the principal when SI, rate, and time are given (moderate difficulty)
• Comparing SI and CI for the same principal and rate (moderate)
• Questions involving payments in installments (difficult)

SSC CGL 2022 Question: The SI on a sum of money for 3 years at 8% p.a. is ₹1,440. What sum is this if the rate is reduced by 2%?

Solution: Original SI = 1440, R = 8%, T = 3 $$P = \frac{1440 \times 100}{8 \times 3} = \frac{144000}{24} = ₹6,000$$ New rate = 6% $$New\ SI = \frac{6000 \times 6 \times 3}{100} = ₹1,080$$

SSC CGL 2023 Question: A sum at SI becomes 1.25 times of itself in T years. At what rate percent per annum will it become double in double the time?

Solution: $1.25 = 1 + \frac{RT}{100}$, so $\frac{RT}{100} = 0.25 = \frac{1}{4}$ When time doubles to 2T, we need SI to equal P (to become 2P): $$2 = 1 + \frac{R \times 2T}{100}$$ $$\frac{2RT}{100} = 1 \Rightarrow \frac{RT}{100} = \frac{1}{2}$$ Original had $\frac{RT}{100} = \frac{1}{4}$, so rate doubles: New R = 2R

Advanced Applications:

• SI is used in calculating installments where equal payments are made
• Fictitious principal concept in partnership problems
• Equated Monthly Installments (EMI) calculations often start with SI principles

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