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Difficulty: MediumOverlapping Sets, Statistics, and Data Distributions

In a financial firm, a total of 100100 portfolio audits were conducted last year. Each audit was classified as High-Risk, International, or both. Exactly 6060 audits were classified as High-Risk, and exactly 5050 audits were classified as International. What was the average (arithmetic mean) duration, in days, of the 100100 portfolio audits?

(1) The average duration of the High-Risk audits was 1414 days, and the average duration of the International audits was 1818 days.
(2) The average duration of the audits classified as both High-Risk and International was 2020 days.

  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, but NEITHER statement ALONE is sufficient.
The correct answer states that both statements together are sufficient, but neither statement alone is sufficient. By set theory, the number of audits in both categories is 60+50−100=1060 + 50 - 100 = 10. The total sum of days across all 100100 audits satisfies Stotal=SHigh-Risk+SInternational−SBothS_{\text{total}} = S_{\text{High-Risk}} + S_{\text{International}} - S_{\text{Both}}. Statement (1) gives SHigh-Risk=60×14=840S_{\text{High-Risk}} = 60 \times 14 = 840 and SInternational=50×18=900S_{\text{International}} = 50 \times 18 = 900, yielding Stotal=1740−SBothS_{\text{total}} = 1740 - S_{\text{Both}}, which is insufficient alone. Statement (2) gives SBoth=10×20=200S_{\text{Both}} = 10 \times 20 = 200, which is insufficient alone. Combining both gives Stotal=1740−200=1540S_{\text{total}} = 1740 - 200 = 1540 days, allowing us to find a unique overall mean of 15.415.4 days.

Step-by-Step Solution

1
Determine set sizes using Principle of Inclusion-Exclusion.
Total N=100N = 100. N(H)=60N(H) = 60, N(I)=50N(I) = 50. Since every audit is in HH, II, or both, N(H∪I)=100=60+50−N(H∩I)  ⟹  N(H∩I)=10N(H \cup I) = 100 = 60 + 50 - N(H \cap I) \implies N(H \cap I) = 10.
Identify the count of audits in the overlapping set (both High-Risk and International).
2
Formulate total duration sum in terms of set sums.
Stotal=SH+SI−SH∩IS_{\text{total}} = S_H + S_I - S_{H \cap I}, where SHS_H is sum of High-Risk durations, SIS_I is sum of International durations, and SH∩IS_{H \cap I} is sum of durations in both categories.
Double counting principle applies to the sum of values across sets.
3
Evaluate Statement (1) alone.
SH=60×14=840S_H = 60 \times 14 = 840 days, SI=50×18=900S_I = 50 \times 18 = 900 days. Stotal=840+900−SH∩I=1740−SH∩IS_{\text{total}} = 840 + 900 - S_{H \cap I} = 1740 - S_{H \cap I}.
Since SH∩IS_{H \cap I} is unknown, StotalS_{\text{total}} cannot be calculated. Statement (1) alone is insufficient.
4
Evaluate Statement (2) alone.
SH∩I=10×20=200S_{H \cap I} = 10 \times 20 = 200 days.
Without SHS_H or SIS_I, StotalS_{\text{total}} cannot be calculated. Statement (2) alone is insufficient.
5
Evaluate Statements (1) and (2) together.
Stotal=1740−200=1540S_{\text{total}} = 1740 - 200 = 1540 days. Mean duration = 1540/100=15.41540 / 100 = 15.4 days.
A unique value for the overall mean is obtained. Both statements together are sufficient.

Key Concept

Overlapping Sets and Weighted Averages (Sum Double Counting)
Estimated Time:2m 0s
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Topics

Overlapping Sets, Statistics, and Data Distributions

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