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Difficulty: HardOverlapping Sets, Statistics, and Data Distributions

In a graduating class of 120120 students, each student participated in at least one of two extracurricular activities: the Science Club or the Debate Team. The mean score on a national mathematics exam for all students who participated in the Science Club was 8585, and the mean score for all students who participated in the Debate Team was 8080. What was the mean mathematics score for all 120120 students in the graduating class?

(1) Exactly 4040 students participated in both the Science Club and the Debate Team, and their mean mathematics score on the exam was 9090.
(2) The total number of students who participated in the Science Club was equal to the total number of students who participated in the Debate Team.

  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.
Both statements together are sufficient. Statement (1) establishes that 4040 students are in both activities with a mean score of 9090, leaving 8080 students in only one activity, but does not specify how those 8080 students are divided between the two clubs. Statement (2) specifies that the two club sizes are equal, which implies that the number of students participating only in Science equals the number participating only in Debate. Combining these facts determines that exactly 4040 students are in Science only, 4040 in Debate only, and 4040 in both, allowing the total score sum (96009600) and overall mean (8080) to be uniquely calculated.

Step-by-Step Solution

1
Set up the algebraic model for overlapping set counts and statistics sums.
Let aa be the number of students in Science Club only, bb be the number of students in Debate Team only, and cc be the number of students in both. a+b+c=120a + b + c = 120. Total sum of scores = 85(a+c)+80(b+c)−Sum(S∩D)85(a+c) + 80(b+c) - \text{Sum}(S \cap D).
Scores of students in the intersection are counted in both club averages, so subtracting the overlap sum prevents double counting.
2
Evaluate Statement (1) independently.
Statement (1) gives c=40c = 40 and Sum(S∩D)=40×90=3600\text{Sum}(S \cap D) = 40 \times 90 = 3600. Then a+b=80a + b = 80, and Total Sum = 85a+80b+3000=5a+940085a + 80b + 3000 = 5a + 9400. Since aa can vary from 00 to 8080, Total Sum is not unique.
Statement (1) alone is insufficient because aa remains a free variable.
3
Evaluate Statement (2) independently.
Statement (2) gives a+c=b+c  ⟹  a=ba + c = b + c \implies a = b. Without cc or intersection scores, Total Sum cannot be computed.
Statement (2) alone is insufficient.
4
Evaluate Statement (1) and Statement (2) combined.
From (1), a+b=80a + b = 80 and c=40c = 40. From (2), a=ba = b. Thus 2a=80  ⟹  a=402a = 80 \implies a = 40 and b=40b = 40. Substituting a=40a = 40 gives Total Sum = 5(40)+9400=96005(40) + 9400 = 9600. Overall mean = 9600/120=809600 / 120 = 80.
The combined system yields a single unique overall average score.

Key Concept

Weighted averages in overlapping sets using principle of inclusion-exclusion for statistical sums.
Estimated Time:2m 0s
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Topics

Overlapping Sets, Statistics, and Data Distributions

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