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Difficulty: MediumOverlapping Sets, Statistics, and Data Distributions

A high-tech manufacturing firm produced a batch of 150150 electronic assemblies. Each assembly in the batch underwent standard quality testing by Inspector Alpha, Inspector Beta, or both. Exactly 9090 assemblies were tested by Inspector Alpha and 105105 assemblies were tested by Inspector Beta. What was the average (arithmetic mean) testing duration per assembly for the entire batch of 150150 assemblies?

(1) The average testing duration for assemblies tested only by Inspector Alpha was 1212 minutes, and the average testing duration for assemblies tested only by Inspector Beta was 1818 minutes.
(2) The average testing duration for assemblies tested by both Inspector Alpha and Inspector Beta was 2525 minutes.

Which of the following statements provides sufficient information to answer the question?

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Both statements together are sufficient to determine a unique average duration, but neither statement alone is sufficient.
The correct response identifies that both statements combined provide all required subgroup averages. Using the overlapping set counts established in the rephrased question stem (4545 Alpha-only, 6060 Beta-only, and 4545 Both), the three subgroup averages supplied across both statements allow for a unique calculation of the overall weighted average.

Step-by-Step Solution

1
Apply the Principle of Inclusion-Exclusion to find the number of assemblies in each mutually exclusive subgroup.
Total=N(Alpha)+N(Beta)−N(Both)  ⟹  150=90+105−N(Both)  ⟹  N(Both)=45\text{Total} = N(\text{Alpha}) + N(\text{Beta}) - N(\text{Both}) \implies 150 = 90 + 105 - N(\text{Both}) \implies N(\text{Both}) = 45. Thus, N(Alpha only)=90−45=45N(\text{Alpha only}) = 90 - 45 = 45, N(Beta only)=105−45=60N(\text{Beta only}) = 105 - 45 = 60, and N(Both)=45N(\text{Both}) = 45.
Before evaluating the statements, rephrasing the stem by partitioning the set into three non-overlapping groups simplifies the weighted average equation.
2
Evaluate Statement (1) alone.
Statement (1) provides Mean(Alpha only)=12\text{Mean}(\text{Alpha only}) = 12 minutes and Mean(Beta only)=18\text{Mean}(\text{Beta only}) = 18 minutes. However, Mean(Both)\text{Mean}(\text{Both}) remains unknown. The overall sum of testing durations cannot be determined. Statement (1) is NOT sufficient.
We cannot compute a weighted mean of three groups if one group's mean is completely missing.
3
Evaluate Statement (2) alone.
Statement (2) provides Mean(Both)=25\text{Mean}(\text{Both}) = 25 minutes. However, Mean(Alpha only)\text{Mean}(\text{Alpha only}) and Mean(Beta only)\text{Mean}(\text{Beta only}) are unknown. Statement (2) is NOT sufficient.
Without the average durations for the single-inspector groups, the total testing duration cannot be computed.
4
Evaluate Statements (1) and (2) combined.
Combining both statements gives all three subgroup means: 1212 minutes for 4545 assemblies, 1818 minutes for 6060 assemblies, and 2525 minutes for 4545 assemblies. Total testing duration =45(12)+60(18)+45(25)=540+1080+1125=2745= 45(12) + 60(18) + 45(25) = 540 + 1080 + 1125 = 2745 minutes. The overall average duration is 2745150=18.3\frac{2745}{150} = 18.3 minutes. Statements (1) and (2) together are SUFFICIENT.
Knowing all subgroup sizes and all subgroup means enables exact calculation of the overall mean.

Key Concept

Overlapping Sets and Weighted Averages in Data Sufficiency
Estimated Time:2m 0s
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Topics

Overlapping Sets, Statistics, and Data Distributions

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