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Finding Time (T)

Introduction

In simple interest problems, you are often given the principal (P), rate (R), and the total interest (SI) earned, and asked to calculate the time (T). This is useful in real-life situations such as knowing how long it takes for money to double at a given rate.

Pattern: Finding Time (T)

Pattern

Key concept: Rearrange SI = (P × R × T) / 100 to solve for T.

Formula:
T = (SI × 100) / (P × R)

Step-by-Step Example

Question

A sum of ₹5,000 earns a simple interest of ₹1,500 at 10% per annum. Find the time period.

Options:
A. 2 years
B. 3 years
C. 4 years
D. 5 years

Solution

  1. Step 1: Identify the given values

    P = 5,000, SI = 1,500, R = 10%.

  2. Step 2: Write the rearranged formula for time

    T = (SI × 100) / (P × R)

  3. Step 3: Substitute the known values

    T = (1,500 × 100) / (5,000 × 10)

  4. Step 4: Simplify numerator and denominator stepwise

    Numerator = 1,500 × 100 = 150,000
    Denominator = 5,000 × 10 = 50,000

  5. Step 5: Divide to obtain time

    T = 150,000 ÷ 50,000 = 3

  6. Final Answer:

    3 years → Option B

  7. Quick Check:

    Yearly SI = (5,000 × 10) / 100 = 500; for 3 years → 500 × 3 = 1,500 ✅

Quick Variations

1. If time is in months or days, convert to years (e.g., 18 months = 1.5 years).

2. Sometimes total SI is given for a fraction of time - rearrange carefully to avoid confusion.

3. Useful in “doubling time” problems when SI equals the principal.

Trick to Always Use

  • Step 1 → Convert time into years (if months/days given).
  • Step 2 → Apply formula: T = (SI×100) ÷ (P×R).
  • Step 3 → Verify by recalculating SI = (P×R×T)/100.

Summary

Summary

  • Use T = (SI × 100) / (P × R) to find time when P, R, and SI are known.
  • Always convert months/days into years before substituting into the formula.
  • Simplify numerator and denominator stepwise (compute SI×100 and P×R separately) to avoid arithmetic errors.
  • Quick-check by computing yearly SI = (P×R)/100 and multiplying by T to verify the given SI.

Example to remember:
P = ₹5,000, R = 10%, SI = ₹1,500 → T = 3 years.

Practice

(1/5)
1. A sum of ₹6000 earns a simple interest of ₹900 at 5% per annum. How long was the money invested?
easy
A. 2 years
B. 3 years
C. 4 years
D. 5 years

Solution

  1. Step 1: Identify given values

    P = 6000, SI = 900, R = 5%.

  2. Step 2: Apply the time formula

    T = (SI × 100) / (P × R).

  3. Step 3: Substitute the values

    T = (900 × 100) / (6000 × 5) = 90000 / 30000 = 3.

  4. Final Answer:

    3 years → Option B

  5. Quick Check:

    Yearly SI = (6000 × 5)/100 = 300; ×3 years = 900 ✅
Hint: Divide total SI by yearly SI to get time quickly.
Common Mistakes: Forgetting to multiply P and R in the denominator.
2. A principal of ₹2000 yields ₹200 as simple interest at a rate of interest of 5% per annum. For how long was the money lent?
easy
A. 1 year
B. 2 years
C. 3 years
D. 2.5 years

Solution

  1. Step 1: Identify given values

    P = 2000, SI = 200, R = 5%.

  2. Step 2: Apply the formula

    T = (200 × 100) / (2000 × 5) = 20000 / 10000 = 2.

  3. Final Answer:

    2 years → Option B

  4. Quick Check:

    Yearly SI = (2000 × 5)/100 = 100; ×2 = 200 ✅
Hint: Calculate SI×100 first, then divide by (P×R).
Common Mistakes: Mixing up percent decimals or skipping the ×100 factor.
3. A sum of ₹3000 yields a simple interest of ₹450 when invested at 6% per annum. For how many years was it invested?
easy
A. 2 years
B. 1.5 years
C. 2.5 years
D. 3 years

Solution

  1. Step 1: Identify given values

    P = 3000, SI = 450, R = 6%.

  2. Step 2: Use the formula for time

    T = (450 × 100) / (3000 × 6) = 45000 / 18000 = 2.5.

  3. Final Answer:

    2.5 years → Option C

  4. Quick Check:

    Yearly SI = (3000 × 6)/100 = 180; ×2.5 = 450 ✅
Hint: Multiply R×P first in denominator to simplify division.
Common Mistakes: Treating 2.5 as 25 or misplacing decimal points.
4. The simple interest on a sum at 8% per annum for a certain period is ₹640. If the principal is ₹2000, how long was the money invested?
medium
A. 2 years
B. 3 years
C. 4 years
D. 5 years

Solution

  1. Step 1: Identify given values

    P = 2000, SI = 640, R = 8%.

  2. Step 2: Apply formula

    T = (640 × 100) / (2000 × 8) = 64000 / 16000 = 4.

  3. Final Answer:

    4 years → Option C

  4. Quick Check:

    Yearly SI = (2000 × 8)/100 = 160; ×4 = 640 ✅
Hint: Compute (P×R) then divide SI×100 by that product.
Common Mistakes: Dividing by R only or forgetting to include time factor correctly.
5. A sum of ₹1200 earns a simple interest of ₹90 at a rate of 3.75% per annum. For how long was the money invested?
medium
A. 2 years
B. 1.5 years
C. 3 years
D. 2.5 years

Solution

  1. Step 1: Identify given values

    P = 1200, SI = 90, R = 3.75%.

  2. Step 2: Apply formula

    T = (90 × 100) / (1200 × 3.75) = 9000 / 4500 = 2.

  3. Final Answer:

    2 years → Option A

  4. Quick Check:

    Yearly SI = (1200 × 3.75)/100 = 45; ×2 = 90 ✅
Hint: Expect decimal rates; keep precision (e.g., 3.75) intact during division.
Common Mistakes: Rounding 3.75 too early or misreading months as years.

Mock Test

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