Composition percentage after removal: A box has 100 blue, 50 red, and 50 black balls. After removing 25% of the blue balls and 50% of the red balls, what percentage of the balls are black?

Difficulty: Medium

Correct Answer: 33 1/3%

Explanation:

Introduction / Context:When parts of some categories are removed from a collection, the total count changes and so do category percentages. We carefully recompute the new totals and then find the proportion of black balls in the updated mixture.Given Data / Assumptions:

  • Initial counts: Blue = 100, Red = 50, Black = 50.
  • Remove 25% of Blue: 0.25 * 100 = 25 removed.
  • Remove 50% of Red: 0.50 * 50 = 25 removed.
  • Black balls are unchanged.

Concept / Approach:Compute post-removal counts for each color and sum to get the new total. Then compute the percentage of black balls as (black / total) * 100%.

Step-by-Step Solution:

Blue after removal = 100 − 25 = 75.Red after removal = 50 − 25 = 25.Black remains = 50.New total = 75 + 25 + 50 = 150.Percentage black = 50 / 150 * 100% = 1/3 * 100% = 33 1/3%.

Verification / Alternative check:Fractional view: colors in ratio 75 : 25 : 50 = 3 : 1 : 2. Black is 2 parts out of 6 ⇒ 1/3 ⇒ 33 1/3%.

Why Other Options Are Wrong:

  • 25%: Would be true only if black were 1/4 of the new total, which it is not.
  • 40% or 50%: Overstate black’s presence relative to the recalculated total.

Common Pitfalls:Using the original total of 200 when calculating percentages after removal. Always recompute the total after changes.

Final Answer:

33 1/3%

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