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?
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A25%
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B33 1/3%
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C40%
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D50%
Answer
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%