Each of the 100 employees at Company K works in Division X, Division Y, or both divisions.… — GMAT

Each of the $100$ employees at Company K works in Division X, Division Y, or both divisions. The arithmetic mean age of the employees in Division X is $35$ years, and the arithmetic mean age of the employees in Division Y is $45$ years. Is the arithmetic mean age of all $100$ employees at Company K greater than $40$ years?

(1) Exactly $40$ employees work in Division X and exactly $70$ employees work in Division Y.
(2) The arithmetic mean age of the employees who work in both Division X and Division Y is $40$ years.

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.Answer
D
EACH statement ALONE is sufficient.
E
Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Both statements together are sufficient to answer the question definitively, but neither statement alone is sufficient.

Combining both statements establishes the exact sizes of all three disjoint subsets (Division X only = 30, Division Y only = 60, both divisions = 10) and the average value of the overlapping subset (40 years). Calculating the total age sum gives 4,150 years, resulting in an exact overall mean of 41.5 years. This provides a definitive 'Yes' answer to whether the mean is greater than 40 years.

Step-by-Step Solution

Formulate the algebraic expressions for the total sum of ages.

Let

x

be the number of employees in Division X only,

y

be the number in Division Y only, and

z

be the number in both divisions.

x + y + z = 100

. Let

S_x, S_y, S_z

be the sum of ages of employees in Division X only, Division Y only, and both divisions, respectively. Then

S_x + S_z = 35(x + z)

and

S_y + S_z = 45(y + z)

. The total sum of ages of all

100

employees is

S_{total} = S_x + S_y + S_z = 35x + 45y + 80z - S_z

Because employees in both divisions contribute to the averages of both Division X and Division Y, simply adding

35(x+z)

and

45(y+z)

counts

S_z

twice.

Evaluate Statement (1) independently.

Statement (1) states

x + z = 40

and

y + z = 70

. Since

x + y + z = 100

, we find

z = (40 + 70) - 100 = 10

x = 30

, and

y = 60

. Thus,

S_{total} = 35(40) + 45(70) - S_z = 4550 - S_z

. The overall mean age is

\frac{4550 - S_z}{100} = 45.5 - \frac{S_z}{100}

. Depending on the value of

S_z

(the sum of ages of the

10

overlap employees), the overall mean can be greater than

40

or less than or equal to

40

Without knowing

S_z

or the average age of the overlap group, the overall mean cannot be uniquely bounded. Thus, Statement (1) alone is NOT sufficient.

Evaluate Statement (2) independently.

Statement (2) states

\frac{S_z}{z} = 40 \implies S_z = 40z

. Substituting into

S_{total}

gives

S_{total} = 35(x+z) + 45(y+z) - 40z = 35x + 45y + 40z

. The overall mean age is

\frac{35x + 45y + 40z}{x + y + z}

. The condition

\frac{35x + 45y + 40z}{x + y + z} > 40

simplifies to

35x + 45y > 40x + 40y \iff 5y > 5x \iff y > x

Statement (2) provides no information about whether

y > x

(whether more employees work exclusively in Division Y than in Division X). Thus, Statement (2) alone is NOT sufficient.

Evaluate Statements (1) and (2) together.

From Statement (1),

x = 30

y = 60

, and

z = 10

. From Statement (2),

S_z = 40(10) = 400

. Since

y = 60 > x = 30

, the condition

y > x

holds. Substituting these values into

S_{total}

yields

S_{total} = 35(30) + 45(60) + 40(10) = 1050 + 2700 + 400 = 4150

. The arithmetic mean age of all

100

employees is

\frac{4150}{100} = 41.5

years, which is strictly greater than

40

Combining both statements yields a unique and definitive 'Yes' answer to the question stem.

Key Concept

Weighted averages in overlapping sets with double-counted sums

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