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Zorluk: ZorOverlapping Sets, Statistics, and Data Distributions

At a technical conference, a total of 150150 software engineers attended at least one of two technical sessions: System Architecture or Distributed Systems. Exactly 9090 engineers attended the System Architecture session. If the mean years of experience for all 150150 engineers combined was 88 years, what was the mean years of experience for the engineers who attended ONLY the Distributed Systems session?

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

Cevap

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
The correct answer is Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Rephrasing the question stem reveals that the total group of 150 engineers is divided into two disjoint subsets: the 90 engineers in the System Architecture session and the 60 engineers who attended ONLY the Distributed Systems session. Because the total experience of all 150 engineers is fixed at 1200 years (150 × 8), knowing the mean experience of the 90 System Architecture engineers in Statement (1) allows calculation of their total experience sum (90 × 7.5 = 675), which directly yields the remaining experience sum (1200 - 675 = 525) and mean (525 / 60 = 8.75) for the engineers in ONLY the Distributed Systems session. Statement (2) gives information about the overlap group but leaves the experience sum of the engineers in ONLY the System Architecture session unknown.

Adım Adım Çözüm

1
Rephrase the question stem using set relationships and statistics formulas.
Let SS be System Architecture attendees (S=90|S| = 90) and DD be Distributed Systems attendees. The total combined group is SDS \cup D with SD=150|S \cup D| = 150. The number of engineers attending ONLY Distributed Systems is DS=SDS=15090=60|D \setminus S| = |S \cup D| - |S| = 150 - 90 = 60.
Since every attendee is in at least one session, the total group consists of all attendees in SS plus those in DSD \setminus S, which are mutually disjoint sets.
2
Express the total sum of experience and set up the target equation.
Total combined experience sum Ttotal=150×8=1200T_{\text{total}} = 150 \times 8 = 1200. Since SS and DSD \setminus S partition the entire population, Ttotal=TS+TDST_{\text{total}} = T_S + T_{D \setminus S}, where TST_S is the sum of experience of all 9090 engineers in SS. Therefore, TDS=1200TST_{D \setminus S} = 1200 - T_S, and the target mean is TDS60=1200TS60\frac{T_{D \setminus S}}{60} = \frac{1200 - T_S}{60}.
Finding the mean experience for DSD \setminus S depends entirely on finding the total experience sum TST_S of the 9090 engineers in System Architecture.
3
Evaluate Statement (1): The mean years of experience for the engineers who attended the System Architecture session was 7.57.5 years.
TS=90×7.5=675T_S = 90 \times 7.5 = 675. Then TDS=1200675=525T_{D \setminus S} = 1200 - 675 = 525. Target mean =52560=8.75= \frac{525}{60} = 8.75 years. Statement (1) is SUFFICIENT.
Statement (1) directly gives the mean of set SS, allowing exact computation of TST_S and thus the target mean.
4
Evaluate Statement (2): Exactly 4040 engineers attended BOTH sessions, and their mean years of experience was 99 years.
This gives SD=40|S \cap D| = 40 and sum TSD=40×9=360T_{S \cap D} = 40 \times 9 = 360. Set SS is divided into SDS \setminus D (size 5050) and SDS \cap D (size 4040). TS=TSD+360T_S = T_{S \setminus D} + 360. Since TSDT_{S \setminus D} remains unknown, TST_S cannot be determined. Statement (2) is INSUFFICIENT.
Without the experience sum or mean of the engineers who attended ONLY System Architecture, we cannot determine TST_S.

Anahtar Kavram

Partitioning combined sets in weighted averages and Data Sufficiency rephrasing
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