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Basic CI Formula Application

Introduction

Compound Interest (CI) aptitude exams में सबसे ज़्यादा पूछे जाने वाले topics में से एक है। यह बताता है कि पैसा कैसे बढ़ता है जब ब्याज न सिर्फ़ original amount (principal) पर, बल्कि समय-समय पर जुड़े हुए interest पर भी लगाया जाता है। Basic formula समझने से आप किसी भी CI problem को जल्दी और accurately solve कर सकते हैं।

Pattern: Basic CI Formula Application

Pattern

मुख्य concept यह है: Compound Interest principal और accumulated interest - दोनों पर लगता है।

Main formulas:
Amount (A) = P × (1 + R/100)T
Compound Interest (CI) = A - P

जहाँ:
P = Principal amount, R = प्रति वर्ष ब्याज दर, T = समय (सालों में)।

Step-by-Step Example

Question

₹5,000 पर 8% प्रति वर्ष की दर से 3 साल के लिए compound interest निकालें।

Solution

  1. Step 1: Given values पहचानें

    Principal (P) = ₹5,000, Rate (R) = 8%, Time (T) = 3 years.

  2. Step 2: Amount का formula लगाएँ

    A = P × (1 + R/100)T
    = 5,000 × (1 + 8/100)3
    = 5,000 × (1.08)3

  3. Step 3: Value calculate करें

    (1.08)3 = 1.2597
    A = 5,000 × 1.2597 = ₹6,298.50

  4. Step 4: Compound Interest निकालें

    CI = A - P = 6,298.50 - 5,000 = ₹1,298.50

  5. Final Answer:

    Compound Interest = ₹1,298.50

  6. Quick Check:

    1 साल बाद → ₹5,400; 2 साल बाद → ₹5,832; 3 साल बाद → ₹6,298.50 ✅

Quick Variations

1. कभी-कभी आपको Total Amount निकालने को कहा जाता है - ऐसे में सीधे A निकालें।

2. अगर compounding half-yearly या quarterly हो तो R और T को उसी हिसाब से adjust करें।

3. CI किसी भी currency या percentage पर लागू होता है - concept हमेशा same रहता है।

Trick to Always Use

  • Step 1: Question से P, R और T सही से पहचानें।
  • Step 2: Annual compounding के लिए A = P(1 + R/100)T formula लगाएँ।
  • Step 3: Amount में से Principal घटाकर CI निकालें।
  • Step 4: छोटे T के लिए successive percentage increase से cross-check कर लें।

Summary

Summary

Basic CI Formula Application pattern में:

  • Amount = P(1 + R/100)T मुख्य formula है।
  • Compound Interest = Amount - Principal।
  • जब तक कुछ और न कहा जाए, annual compounding ही मानें।
  • Short time period वाले questions में तेजी से check करने के लिए successive year growth देखें।

Practice

(1/5)
1. Find the compound interest on ₹8,000 at 10% per annum for 2 years.
easy
A. ₹1,680.00
B. ₹1,600.00
C. ₹1,700.00
D. ₹1,800.00

Solution

  1. Step 1: Identify given values

    P = ₹8,000, R = 10% p.a., T = 2 years.

  2. Step 2: Apply formula for Amount

    A = P × (1 + R/100)^T = 8,000 × (1 + 10/100)^2 = 8,000 × (1.1)^2.

  3. Step 3: Compute

    (1.1)^2 = 1.21 → A = 8,000 × 1.21 = ₹9,680.00.

  4. Final Answer:

    CI = A - P = 9,680.00 - 8,000.00 = ₹1,680.00 → Option A.

  5. Quick Check:

    Year 1 interest = 10% of 8,000 = 800; principal for year 2 = 8,800; year 2 interest = 10% of 8,800 = 880; total interest = 800 + 880 = ₹1,680.00 ✅
Hint: Compute A = P(1 + R/100)^T, then subtract P to get CI.
Common Mistakes: Using SI formula (P×R×T/100) instead of compounding year-by-year.
2. What will be the compound interest on ₹10,000 at 5% per annum for 2 years?
easy
A. ₹1,050.00
B. ₹1,025.00
C. ₹1,000.00
D. ₹1,100.00

Solution

  1. Step 1: Identify given values

    P = ₹10,000, R = 5% p.a., T = 2 years.

  2. Step 2: Apply CI formula

    A = 10,000 × (1 + 5/100)^2 = 10,000 × (1.05)^2.

  3. Step 3: Compute

    (1.05)^2 = 1.1025 → A = 10,000 × 1.1025 = ₹11,025.00.

  4. Final Answer:

    CI = A - P = 11,025.00 - 10,000.00 = ₹1,025.00 → Option B.

  5. Quick Check:

    10,000 → 10,500 (after 1 year) → 11,025 (after 2 years) → CI = 11,025 - 10,000 = ₹1,025.00 ✅
Hint: Raise (1 + R/100) to T first, then multiply by P for A.
Common Mistakes: Rounding intermediate values too early; leads to small errors.
3. Find the compound interest on ₹6,000 at 12% per annum for 2 years.
easy
A. ₹1,440.20
B. ₹1,500.60
C. ₹1,526.40
D. ₹1,520.80

Solution

  1. Step 1: Identify given values

    P = ₹6,000, R = 12% p.a., T = 2 years.

  2. Step 2: Apply CI formula

    A = 6,000 × (1 + 12/100)^2 = 6,000 × (1.12)^2.

  3. Step 3: Compute

    (1.12)^2 = 1.2544 → A = 6,000 × 1.2544 = ₹7,526.40 → CI = 7,526.40 - 6,000.00 = ₹1,526.40.

  4. Final Answer:

    Compound Interest = ₹1,526.40 → Option C.

  5. Quick Check:

    Year 1 interest = 12% of 6,000 = 720 → principal becomes 6,720; Year 2 interest = 12% of 6,720 = 806.40; total interest = 720 + 806.40 = ₹1,526.40 ✅
Hint: Compute year-by-year interest if powers are hard; add yearly interests to confirm.
Common Mistakes: Forgetting to include the first year's interest when calculating the second year's base.
4. Find the compound interest on ₹15,000 at 9% per annum for 3 years.
medium
A. ₹4,425.44
B. ₹4,185.44
C. ₹4,167.44
D. ₹4,050.44

Solution

  1. Step 1: Identify given values

    P = ₹15,000, R = 9% p.a., T = 3 years.

  2. Step 2: Apply CI formula

    A = 15,000 × (1 + 9/100)^3 = 15,000 × (1.09)^3.

  3. Step 3: Compute

    (1.09)^2 = 1.1881; (1.09)^3 = 1.1881 × 1.09 = 1.295029 → A = 15,000 × 1.295029 ≈ ₹19,425.44 → CI = 19,425.44 - 15,000.00 = ₹4,425.44.

  4. Final Answer:

    Compound Interest = ₹4,425.44 → Option A.

  5. Quick Check:

    Year-wise: 15,000 → 16,350.00 → 17,811.50 → 19,425.44; CI = 19,425.44 - 15,000 = ₹4,425.44 ✅
Hint: Use (1 + R/100)^3 values remembered or compute sequentially: multiply by (1 + R/100) three times.
Common Mistakes: Using truncated values for (1 + R/100)^3 too early, causing small rounding errors.
5. The compound interest on ₹20,000 at 10% per annum for 3 years will be approximately:
medium
A. ₹6,100.00
B. ₹6,200.00
C. ₹6,300.00
D. ₹6,620.00

Solution

  1. Step 1: Identify given values

    P = ₹20,000, R = 10% p.a., T = 3 years.

  2. Step 2: Apply formula for Amount

    A = 20,000 × (1 + 10/100)^3 = 20,000 × (1.1)^3.

  3. Step 3: Compute

    (1.1)^2 = 1.21; (1.1)^3 = 1.331 → A = 20,000 × 1.331 = ₹26,620.00 → CI = 26,620.00 - 20,000.00 = ₹6,620.00.

  4. Final Answer:

    Compound Interest = ₹6,620.00 → Option D.

  5. Quick Check:

    Successive amounts: 20,000 → 22,000 → 24,200 → 26,620; CI = 26,620 - 20,000 = ₹6,620.00 ✅
Hint: Use successive multiplication by (1 + R/100) for each year to avoid power calculations if needed.
Common Mistakes: Using simple interest formula or stopping compounding one year early.

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