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Time Finding Problems

Introduction

Compound interest problems में कई बार पूछा जाता है कि किसी राशि को एक निश्चित rate पर बढ़कर किसी तय amount तक पहुँचने में कितना समय लगेगा। इन्हें Time Finding Problems कहा जाता है। यह pattern compounding को reverse करके required time निकालने में मदद करता है।

Pattern: Time Finding Problems

Pattern

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

Time (T) निकालने के लिए दोनों तरफ logarithm लेते हैं:

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

यह formula annual, half-yearly, या quarterly compounding-सभी में काम करता है, बस rate और time को consistent units में रखना जरूरी है।

Step-by-Step Example

Question

₹10,000 कितने साल में बढ़कर ₹12,100 हो जाएगा, यदि rate 10% प्रति वर्ष हो (annual compounding)?

Solution

  1. Step 1: Given values

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

  2. Step 2: Compound relation

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

  3. Step 3: Log लें

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

  4. Step 4: T निकालें

    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 साल में) ✅

Quick Variations

1. Rate और time अलग units में हों (जैसे quarterly), तो पहले units adjust करें।

2. Logarithmic formula fractional years के लिए भी perfectly काम करता है।

3. कभी-कभी approximate values को growth tables या calculator से verify किया जा सकता है।

Trick to Always Use

  • Step 1: (1 + R/100)T = A/P लिखें।
  • Step 2: Log लें → T = log(A/P) / log(1 + R/100)।
  • Step 3: Compounding type (annual, half-yearly, quarterly) के हिसाब से R और T adjust करें।

Summary

Summary

  • Time-finding problems amount, rate और time के exponential relation पर आधारित होते हैं।
  • Annual compounding के लिए formula: T = log(A/P) / log(1 + R/100)।
  • Half-yearly या quarterly compounding के लिए R को R/n और T को nT में बदलें।
  • हमेशा quick verification से answer confirm करें।

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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