A summer academic camp has a total of 120 enrolled students. Every student participates in… — GMAT

A summer academic camp has a total of $120$ enrolled students. Every student participates in at least one of two workshops: Data Analysis or Public Speaking. Exactly $70$ students participate in Data Analysis and exactly $80$ students participate in Public Speaking. What is the average (arithmetic mean) test score of the students who participate ONLY in Public Speaking?

Statement (1): The average test score of all $120$ students in the camp is $84$ .
Statement (2): The average test score of the students who participate in BOTH workshops is $90$ , and the average test score of the students who participate ONLY in Data Analysis is $78$ .

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

Cevap

Both statements together are sufficient to uniquely determine the average score of the students participating only in Public Speaking, but neither statement alone is sufficient.

The correct response identifies that both statements together provide enough information to solve for the target average, whereas neither statement alone is sufficient. By deconstructing the overlapping sets into three distinct groups (Data Analysis only = 40, Both = 30, Public Speaking only = 50), the weighted average equation connects the overall average to the three subgroup averages. Statement (1) supplies only the overall average, leaving two unknown subgroup averages. Statement (2) supplies two subgroup averages, leaving the overall average unknown. Combining both statements yields a single equation with only one unknown (

A_{P\text{-only}} = 85.2

), producing a unique numerical solution.

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Rephrase the question stem by calculating the exact number of students in each mutually exclusive subset.

Using the overlapping sets formula

N(D \cup P) = N(D) + N(P) - N(D \cap P)

, we get

120 = 70 + 80 - N(D \cap P)

, so

N(D \cap P) = 30

students in both workshops. Therefore, students in Data Analysis only =

70 - 30 = 40

, students in Public Speaking only =

80 - 30 = 50

, and students in both =

30

Deconstructing the overlapping sets into three distinct, non-overlapping groups (Data Analysis only, Both, Public Speaking only) is essential for formulating a weighted average equation.

Formulate the weighted average equation relating all subset averages to the total average.

Total Score Sum =

120 \cdot A_{total} = 40 \cdot A_{D\text{-only}} + 30 \cdot A_{both} + 50 \cdot A_{P\text{-only}}

, where

A_{P\text{-only}}

is the target value.

The total sum of all scores is the weighted sum of scores from the three mutually exclusive subsets.

Evaluate Statement (1) independently.

Statement (1) gives

A_{total} = 84

. Substituting gives

120(84) = 10080 = 40 \cdot A_{D\text{-only}} + 30 \cdot A_{both} + 50 \cdot A_{P\text{-only}}

. Since

A_{D\text{-only}}

and

A_{both}

remain unknown,

A_{P\text{-only}}

cannot be uniquely solved.

One linear equation with three unknown variables does not yield a unique solution. Statement (1) alone is NOT sufficient.

Evaluate Statement (2) independently.

Statement (2) gives

A_{both} = 90

and

A_{D\text{-only}} = 78

. Substituting gives

120 \cdot A_{total} = 40(78) + 30(90) + 50 \cdot A_{P\text{-only}} = 5820 + 50 \cdot A_{P\text{-only}}

. Since

A_{total}

is unknown,

A_{P\text{-only}}

cannot be uniquely solved.

One linear equation with two unknown variables does not yield a unique solution. Statement (2) alone is NOT sufficient.

Evaluate Statement (1) and Statement (2) combined.

Combining both statements provides

A_{total} = 84

A_{both} = 90

, and

A_{D\text{-only}} = 78

. The equation becomes

120(84) = 40(78) + 30(90) + 50 \cdot A_{P\text{-only}}

, which simplifies to

10080 = 5820 + 50 \cdot A_{P\text{-only}} \implies 4260 = 50 \cdot A_{P\text{-only}} \implies A_{P\text{-only}} = 85.2

We have one linear equation with only one unknown variable, giving a single unique numerical value. Both statements together are SUFFICIENT.

Anahtar Kavram

Combining overlapping set cardinalities with group weighted averages in Data Sufficiency

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