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Time Finding Problems

Introduction

Compound Interest problems இல், சில சமயம் ஒரு தொகை குறிப்பிட்ட amount ஆக வளர எவ்வளவு காலம் எடுத்துக் கொள்கிறது என்பதை கண்டறிய சொல்லப்படும். இவை Time Finding Problems என்று அழைக்கப்படுகின்றன. இந்த pattern ஐ புரிந்துகொண்டால், compounding process ஐ reverse செய்து தேவையான duration ஐ எளிதாகக் கண்டறியலாம்.

Pattern: Time Finding Problems

Pattern

Key formula: (1 + R/100)T = A / P

Time (T) ஐ காண, இரு பக்கங்களிலும் logarithms எடுக்கவும்:

T = [log(A/P)] / [log(1 + R/100)]

Rate மற்றும் time ஒரே unit இல் (annual, half-yearly, quarterly) வெளிப்படுத்தப்பட்டால், இந்த formula அனைத்துக்கும் பொருந்தும்.

Step-by-Step Example

Question

வருடத்திற்கு 10% வட்டியில் (annual compounding), ₹10,000 தொகை ₹12,100 ஆக மாற எத்தனை years ஆகும்?

Solution

  1. Step 1: Given values

    P = ₹10,000, A = ₹12,100, R = 10%.

  2. Step 2: Compound relation அமைத்தல்

    (1 + R/100)T = A/P → (1.10)T = 12,100 / 10,000 = 1.21.

  3. Step 3: Logarithms பயன்படுத்தி T ஐ காண்க

    log(1.21) = T × log(1.10) → T = log(1.21) / log(1.10).

  4. Step 4: T கணக்கிடல்

    log(1.21) = 0.0828; log(1.10) = 0.0414 → T = 0.0828 / 0.0414 = 2 years.

  5. Final Answer:

    Time = 2 years

  6. Quick Check:

    10% yearly → 10,000 → 11,000 → 12,100 (2 years) ✅

Quick Variations

1. Rate மற்றும் time units வேறுபட்டால் (உதா., quarterly compounding), அதற்கு ஏற்ப rate மற்றும் time ஐ adjust செய்யவும்.

2. Logarithmic formula fractional years க்கும் பொருந்தும்.

3. சில நேரங்களில் compound growth tables அல்லது calculators மூலம் approximate check செய்யலாம்.

Trick to Always Use

  • Step 1: (1 + R/100)T = A/P என்று எழுதுங்கள்.
  • Step 2: Logarithms எடுத்து → T = log(A/P) / log(1 + R/100).
  • Step 3: Compounding type (annual, half-yearly, quarterly) படி R மற்றும் T ஐ adjust செய்யுங்கள்.

Summary

Summary

  • Time-finding problems இல் amount, rate, time இடையிலான exponential relationship பயன்படுத்தப்படுகிறது.
  • Annual compounding க்கு T = log(A/P) / log(1 + R/100) பயன்படுத்தவும்.
  • Half-yearly அல்லது quarterly க்கு, தேவையான இடங்களில் R ஐ R/n ஆகவும் T ஐ nT ஆகவும் மாற்றவும்.
  • எப்போதும் successive growth அல்லது quick verification மூலம் answer ஐ double-check செய்யுங்கள்.

Practice

(1/5)
1. In how many years will ₹4,000 amount to ₹5,324 at 10% per annum, compounded annually?
easy
A. 3 years
B. 2 years
C. 4 years
D. 1 year

Solution

  1. Step 1: List given values

    P = ₹4,000; A = ₹5,324; R = 10% p.a.

  2. Step 2: Form the compound equation

    (1 + R/100)^T = A/P → (1.10)^T = 5,324 / 4,000 = 1.331.

  3. Step 3: Identify matching power

    (1.10)^3 = 1.331 → T = 3 years.

  4. Final Answer:

    3 years → Option A.

  5. Quick Check:

    4,000 × (1.10)^3 = 4,000 × 1.331 = 5,324 ✅
Hint: Compute A/P and check if it matches a known power of (1 + R/100).
Common Mistakes: Taking logarithms unnecessarily when the growth factor is an exact power.
2. In how many years will ₹5,000 amount to ₹5,512.50 at 5% per annum, compounded annually?
easy
A. 1 year
B. 2 years
C. 2.5 years
D. 3 years

Solution

  1. Step 1: Note given values

    P = ₹5,000; A = ₹5,512.50; R = 5% p.a.

  2. Step 2: Form the equation

    (1.05)^T = 5,512.50 / 5,000 = 1.1025.

  3. Step 3: Recognize exact power

    (1.05)^2 = 1.1025 → T = 2 years.

  4. Final Answer:

    2 years → Option B.

  5. Quick Check:

    5,000 × (1.05)^2 = 5,512.50 ✅
Hint: Compute A/P and see if it equals (1 + R/100)^n for an integer n.
Common Mistakes: Mistaking the simple-interest increase for compound growth.
3. Find the time required for ₹6,000 to amount to ₹6,365.40 at 6% per annum, compounded half-yearly.
easy
A. 0.5 years
B. 1.5 years
C. 1 year
D. 2 years

Solution

  1. Step 1: List given values

    P = ₹6,000; A = ₹6,365.40; R = 6% p.a.; n = 2 (half-yearly).

  2. Step 2: Convert to per-period rate

    Rate per period = 6/2 = 3% → factor = 1.03; total periods = 2T.

  3. Step 3: Form the equation

    (1.03)^{2T} = 6,365.40 / 6,000 = 1.0609.

  4. Step 4: Identify matching power

    (1.03)^2 = 1.0609 → 2T = 2 → T = 1 year.

  5. Final Answer:

    1 year → Option C.

  6. Quick Check:

    6,000 × (1.03)^2 = 6,365.40 ✅
Hint: Convert R and T into per-period terms (rate/n and nT).
Common Mistakes: Using annual rate directly without converting for half-yearly compounding.
4. In how many years will ₹10,000 amount to ₹13,000 at 10% per annum, compounded annually? (Give answer to two decimal places.)
medium
A. 2.00 years
B. 2.50 years
C. 3.00 years
D. 2.75 years

Solution

  1. Step 1: Identify given values

    P = ₹10,000; A = ₹13,000; R = 10% p.a.

  2. Step 2: Form the growth equation

    (1.10)^T = 13,000 / 10,000 = 1.3.

  3. Step 3: Solve using logarithms

    T = ln(1.3) / ln(1.10) ≈ 0.262364 / 0.095310 = 2.75274 → T ≈ 2.75 years.

  4. Final Answer:

    2.75 years → Option D.

  5. Quick Check:

    (1.10)^{2.75274} ≈ 1.3000 → 10,000 × 1.3000 ≈ 13,000 ✅
Hint: Use T = ln(A/P) ÷ ln(1 + R/100).
Common Mistakes: Rounding logs too early, causing inaccurate T.
5. Find the time required for ₹8,000 to amount to ₹10,000 at 8% per annum, compounded half-yearly. (Round to two decimals.)
medium
A. 2.84 years
B. 2.50 years
C. 3.00 years
D. 2.25 years

Solution

  1. Step 1: List given values

    P = ₹8,000; A = ₹10,000; R = 8% p.a.; n = 2 (half-yearly).

  2. Step 2: Convert to per-period rate

    Rate per half-year = 8/2 = 4% → factor = 1.04; total periods = 2T.

  3. Step 3: Form the equation

    (1.04)^{2T} = 10,000 / 8,000 = 1.25.

  4. Step 4: Solve using logarithms

    2T = ln(1.25)/ln(1.04) → T = [ln(1.25) / ln(1.04)] / 2 ≈ 5.689431 / 2 = 2.84472 → T ≈ 2.84 years.

  5. Final Answer:

    2.84 years → Option A.

  6. Quick Check:

    (1.04)^{2×2.84472} ≈ 1.25 → 8,000 × 1.25 = 10,000 ✅
Hint: Use T = ln(A/P) ÷ [n × ln(1 + R/(100n))].
Common Mistakes: Using annual logs instead of per-period logs.

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