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Time Finding Problems

Introduction

In compound interest problems, sometimes you are asked to find how long it takes for a sum to grow to a certain amount at a given rate. These are known as Time Finding Problems. Understanding this pattern helps you reverse the compounding process to find the duration required.

Pattern: Time Finding Problems

Pattern

The key formula is: (1 + R/100)T = A / P

To find time (T), take logarithms on both sides:

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

This formula works for annual, half-yearly, or quarterly compounding when rate and time are expressed consistently.

Step-by-Step Example

Question

In how many years will ₹10,000 amount to ₹12,100 at 10% per annum, compounded annually?

Solution

  1. Step 1: List given values

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

  2. Step 2: Form the compound relation

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

  3. Step 3: Take logarithms to solve for T

    Take log on both sides: log(1.21) = T × log(1.10) → T = log(1.21) / log(1.10).

  4. Step 4: Compute T

    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:

    At 10% yearly, 10,000 → 11,000 → 12,100 in 2 years ✅

Quick Variations

1. When rate and time units differ (e.g., quarterly compounding), adjust rate and time accordingly.

2. Logarithmic formula works for fractional years as well.

3. Sometimes, approximate values can be checked using compound growth tables or calculators.

Trick to Always Use

  • Step 1: Write (1 + R/100)T = A/P.
  • Step 2: Take logarithms → T = log(A/P) / log(1 + R/100).
  • Step 3: Adjust R and T for compounding type (annual, half-yearly, quarterly).

Summary

Summary

  • Time-finding problems use the exponential relationship between amount, rate, and time.
  • Use T = log(A/P) / log(1 + R/100) for annual compounding.
  • For half-yearly or quarterly, replace R with R/n and T with nT as needed.
  • Always double-check results using successive growth or a quick verification step.

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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