thecoolguy 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
Is B the answer?
From a weighted average, one can determine an overall average.
if we know that there are twice as many houses in the "x" pool, then we know that "x" will have twice as much weight in the overall average.
In this case, since the ratio of x:y is 2:1, x will have 2/3 weight and y will have 1/3 weight. So, we could set up the equation:
overall average = 2/3 (212000) + 1/3 (194000)
So, statement (2) is sufficient.
Statement (1), on the other hand, doesn't help us assess the relative weight of x and y (e.g. we could have 35 houses in the x pool and 1 in the y pool or we could have 1 in the x pool and 35 in the y pool, giving us radically different answers to the question).
(2) is suff and (1) is insuff: choose (b).
