SSC CGL Preparation – Day 18

Quantitative Aptitude – Simple Interest and Compound Interest (Advanced)

✅ 1. Key Concepts Recap

🔹 Simple Interest (SI)

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

Where:

• PPP = Principal
• RRR = Rate of interest per annum
• TTT = Time (in years)

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🔹 Compound Interest (CI)

🧮 When Interest is Compounded Annually

$$\text{CI} = P \left(1 + \frac{R}{100} \right)^T – P$$

🧮 When Compounded Half-Yearly

• Rate = R/2, Time = 2T

$$\text{CI} = P \left(1 + \frac{R}{2 \times 100} \right)^{2T} – P$$

🧮 When Compounded Quarterly

• Rate = R/4, Time = 4T

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🔹 Difference between CI and SI for 2 years

$$\text{Difference} = P \times \left( \frac{R}{100} \right)^2$$

(for 2 years only)

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✅ 2. Special Cases & Shortcuts

🔸 Case 1: Time = 2 years

$$\text{CI – SI} = \frac{P \cdot R^2}{100^2}$$

🔸 Case 2: Amount is given

$$\text{Amount} = \text{Principal} + \text{Interest} \Rightarrow P = \frac{A}{\left(1 + \frac{R}{100} \right)^T}$$​

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✅ 3. Questions You Can Expect in SSC CGL

Type	Example
SI calculation	Find SI on ₹6000 at 10% for 3 years
CI calculation	CI on ₹8000 at 5% for 2 years
Difference b/w SI and CI	What is the difference between SI and CI on ₹1000 at 10% for 2 years?
Time or Rate	If CI on ₹2000 becomes ₹2420 in 2 years, find rate
Compounded semi-annually	Slightly advanced, useful for Tier 2

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✅ 4. Examples

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Q1. What is the Simple Interest on ₹10,000 at 8% per annum for 3 years?

Solution: $$\text{SI} = \frac{P \times R \times T}{100} = \frac{10000 \times 8 \times 3}{100} = ₹2400$$

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Q2. What is the Compound Interest on ₹5000 at 10% per annum for 2 years?

Solution: $$\text{CI} = 5000 \left(1 + \frac{10}{100} \right)^2 – 5000 = 5000 \times (1.1)^2 – 5000 = 5000 \times 1.21 – 5000 = ₹6050 – ₹5000 = ₹1050$$

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Q3. Find the difference between CI and SI on ₹8000 at 5% p.a. for 2 years.

Solution:
Use shortcut: $$\text{Difference} = \frac{P \cdot R^2}{100^2} = \frac{8000 \cdot 25}{10000} = ₹20$$

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Q4. Find the CI on ₹16000 at 20% p.a. compounded annually for 3 years.

Solution:
Use: $$\text{CI} = 16000 \left(1 + \frac{20}{100} \right)^3 – 16000 = 16000 \times (1.2)^3 = 16000 \times 1.728 = ₹27648 \Rightarrow \text{CI} = 27648 − 16000 = ₹11648$$

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Q5. In how many years will ₹1000 amount to ₹1210 at 10% p.a. compound interest?

Solution: $$\text{Amount} = P \left(1 + \frac{R}{100} \right)^T \Rightarrow 1210 = 1000 \times (1.1)^T \Rightarrow (1.1)^T = 1.21 \Rightarrow T = 2 \text{ years}$$

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✅ 5. Important Tips for SSC CGL

• Use Difference formula only for 2 years.
• Avoid step-by-step power multiplication: memorize squares of decimals like:
  • (1.1)² = 1.21
  • (1.2)² = 1.44, (1.2)³ = 1.728
  • (1.05)² = 1.1025
• When time or rate is missing, always apply logics using basic CI/Amount formula.
• In exam, CI questions often appear in MCQ format with difference in SI & CI.