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 Answer: 我选B。答案是C,我觉得B对。
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 Answer: 我选B。答案是C,我觉得B对。
-- by 会员 茶色的秋天 (2011/10/27 18:55:13)
这题应该是B 因为 average of X + Y houses is: (x * 21200 + y *19400) / (x+y) (1) only know x + y but not the individual values of x and y, so insufficient (2) x = 2y ==> (2y*21200+y*19400)/(3y) y可以约掉, 所以可以 所以答案是B
If X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2)XY/Y2-Y<1 Answer: 答案是E,我选的是B
-- by 会员 茶色的秋天 (2011/10/27 18:56:03)
我觉得答案应该是E
we want to know whether x<y, we know both x and y are greater than 1 (not necessarily integer)
(1)x^2/(XY+X) < 1 ==> x < y + 1 we cannot get relation of x and y since: when x = 1.5, y = 1.1, y+1 = 2.1, 可以 (x > y) when x = 1.5, y = 2, y+1 = 3, 可以, (x < y) 同一个条件2个结果,所以 insufficient (2) (XY/Y^2)-Y<1 ==> x < y (y+1) when x = 1.5, y = 1.6, y (y+1) = 4.16 可以,(x < y) when x = 2, y = 1.5, y (y+1) = 3.75 可以, (x>y) 同一个条件2个结果, insufficient (1)+(2) x < y + 1 x < y (y+1) (2)其实跟一差不多,多乘以一个大于1的数,没有任何新的信息,所以insufficient 所以 E