More Questions from Unitary Method

If 10 spiders can catch 10 flies in 10 minutes (all working at the same constant rate), how many flies can 200 spiders catch in 200 minutes?

Aptitude Unitary Method Difficulty: Easy
Choose an option
  • A
    2000
  • B
    5000
  • C
    4000
  • D
    3000
  • E
    2500

Answer

Correct Answer: 4000

Explanation

Introduction: Catching flies here scales with the number of spiders and with time, assuming a constant per-spider catch rate. We infer the rate from the base scenario and then scale to the larger numbers using direct proportionality (spiders × minutes × rate per spider-minute).

Given Data / Assumptions:

  • 10 spiders catch 10 flies in 10 minutes.
  • All spiders have the same constant efficiency.
  • Find flies caught by 200 spiders in 200 minutes.

Concept / Approach: Let r be the number of flies caught by one spider per minute. Then total flies = spiders × minutes × r. Using the base case we find r, then apply it to the new case by multiplication without rounding since values are exact here.

Step-by-Step Solution:

Base: 10 = 10 * 10 * r → r = 10 / 100 = 0.1 flies per spider-minute New: Flies = 200 * 200 * 0.1 = 4000

Verification / Alternative check: Proportion method: New/Base = (200/10) * (200/10) = 20 * 20 = 400. Then 10 * 400 = 4000, same result.

Why Other Options Are Wrong: 2000, 3000, 5000, 2500 are not consistent with the direct quadratic scaling in spiders and minutes based on the stated rate.

Common Pitfalls: Treating the relationship as linear in one factor only (either spiders or minutes). The catch count scales with both.

Final Answer: 4000

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