Simple Interest — Same duration at two rates; recover P and time: A certain sum becomes ₹ 600 at 6% p.a. SI in a certain time and ₹ 200 at 1% p.a. SI in the same time. Find the principal and the time.

Difficulty: Hard

Correct Answer: ₹ 120 and 66 2/3 yr

Explanation:


Introduction / Context:
Two simple-interest amounts with the same time but different rates provide two linear equations in principal P and time t. Solving them simultaneously yields both the initial sum and the duration, even when the time is unusually long.



Given Data / Assumptions:

  • A1 = P * (1 + 0.06 * t) = ₹ 600
  • A2 = P * (1 + 0.01 * t) = ₹ 200
  • Same t in both cases


Concept / Approach:
Divide equations or subtract to eliminate P or t. One direct approach is to solve as a 2×2 system for P and t in real numbers under SI linearity.



Step-by-Step Solution:

From A2: P = 200 / (1 + 0.01 t).Substitute in A1: 200 * (1 + 0.06 t) / (1 + 0.01 t) = 600 ⇒ (1 + 0.06 t) / (1 + 0.01 t) = 3.Cross-multiply: 1 + 0.06 t = 3 + 0.03 t ⇒ 0.03 t = 2 ⇒ t = 66.666... years.Then P = 200 / (1 + 0.6666...) = 200 / 1.6666... = ₹ 120.


Verification / Alternative check:

Check A1: 120 * (1 + 0.06 * 66.666...) = 120 * 5 = 600; A2: 120 * (1 + 0.01 * 66.666...) = 120 * 1.6666... = 200.


Why Other Options Are Wrong:

  • Other principals do not satisfy both amounts simultaneously for one common t.


Common Pitfalls:

  • Assuming realistic times; math permits long durations under SI.


Final Answer:
₹ 120 and 66 2/3 years.

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