sondenso wrote:

If the average (arithmetic mean) of the assessed values

of x houses is $212,000 and the average of the assessed

values of y other houses is $194,000, what is the average

of the assessed values of the x+y houses?

(1) x+y=36

(2) x=2y

OA is wrong!

\(SumX/X = 212\) ==> \(SumX = 212X\)

\(SumY/Y = 194\) ==> \(SumY = 194Y\)

Question:\((\frac{SumX + SumY}{X+Y})\)?

Question:\((\frac{212X + 194Y}{X+Y})\)?

(1) \(X+Y=36\)

Substitute into question:

Question:\((\frac{212X + 194Y}{36})\)?

Insuficient

(2) X=2Y

Substitute into question

Question:\((\frac{212(2Y) + 194Y}{2Y+Y)})\)?

Question:\((\frac{618Y}{3Y})\)?

Question:\((\frac{618}{3})\)?

Question:\((206)\)?

Wait a second-- do you know what \(206\) is? Of course we do... Sufficient!

Final answer, \(B\).

_________________

Hades