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If the average (arithmetic mean) of the assessed values of x houses is $212,000 and the average of the assessed...

GMAT Data Sufficiency : (DS) Questions

Source: Mock
Data Sufficiency
DS - Statistics
MEDIUM
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Notes
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If the average (arithmetic mean) of the assessed values of \(\mathrm{x}\) houses is \($212,000\) and the average of the assessed values of \(\mathrm{y}\) other houses is \($194,000\), what is the average of the assessed values of the \(\mathrm{x + y}\) houses?

  1. \(\mathrm{x + y = 36}\)
  2. \(\mathrm{x = 2y}\)
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.
C
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D
EACH statement ALONE is sufficient.
E
Statements (1) and (2) TOGETHER are not sufficient.
Solution

Understanding the Question

We need to find the average assessed value of all houses combined, given:

  • x houses have an average value of $212,000
  • y houses have an average value of $194,000

To calculate this combined average, we must determine the weighted average. This requires knowing either the exact values of x and y, or their ratio. Without this information, we cannot determine how much weight each group contributes to the overall average.

Key insight: For sufficiency, we need information that allows us to calculate one unique average value for all \(\mathrm{x + y}\) houses.

Analyzing Statement 1

Statement 1: \(\mathrm{x + y = 36}\)

This tells us there are 36 houses total, but doesn't specify how many belong to each group. Let's test whether different distributions yield different averages:

Test Case 1: Equal distribution (\(\mathrm{x = 18, y = 18}\))

  • Total value = 18 × $212,000 + 18 × $194,000 = $7,308,000
  • Average = $7,308,000 ÷ 36 = $203,000

Test Case 2: Unequal distribution (\(\mathrm{x = 24, y = 12}\))

  • Total value = 24 × $212,000 + 12 × $194,000 = $7,416,000
  • Average = $7,416,000 ÷ 36 = $206,000

Since different distributions of the 36 houses produce different average values ($203,000 vs $206,000), we cannot determine a unique answer.

Statement 1 is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Now we forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: \(\mathrm{x = 2y}\)

This establishes a fixed 2:1 ratio between the two groups. Let's see if this ratio alone determines the average:

Using \(\mathrm{x = 2y}\):

  • Total houses = \(\mathrm{x + y = 2y + y = 3y}\)
  • Total value = (2y × $212,000) + (y × $194,000) = $424,000y + $194,000y = $618,000y
  • Average = $618,000y ÷ 3y = $618,000 ÷ 3 = $206,000

Notice that y cancels out completely! This means regardless of the actual number of houses, as long as the ratio is 2:1, the average will always be $206,000.

[STOP - Sufficient!]

Statement 2 is sufficient.

This eliminates choices C and E.

The Answer: B

Statement 2 alone provides the ratio needed to calculate a unique weighted average, while Statement 1 alone does not.

Answer Choice B: "Statement 2 alone is sufficient, but Statement 1 alone is not sufficient."

Answer Choices Explained
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.
C
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D
EACH statement ALONE is sufficient.
E
Statements (1) and (2) TOGETHER are not sufficient.
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