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A list of 18 numbers is arranged from the least to the [#permalink]
01 Oct 2003, 22:36
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
2. A list of 18 numbers is arranged from the least to
the greatest. The average of the first n numbers in the list is x
and the average of the remaining numbers is y. If the average of x
and y is equal to the average of all 18 numbers, which of the following
msut be true:
A) n = 1
B) n = 9
C) n = 17
D) 1 < n < 9
E) 9 < n < 17
i admit that i'm having a hard time wrapping my brain around this.
Of course, if they are 18 consecutive numbers, then the average of all would be the one in the middle, and the average of the first 9 would be the middle of the first 9, and the average of the second 9 would be the middle of the second 9, and the average of those two would be the middle of everything, since consecutives just always work out so nicely like that.
My problem to understand is what if they aren't consecutive. Would this still be possible?
But it doesn't matter, because I've just shown that it is true that this could come up 9, and there's only one answer in the whole list that includes 9, and that's B. D and E both exclude 9.
So I'm going with B, having not tried the non-consecutive option.