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标题: GWD-3-05 請問喔 [打印本页]

作者: xiao19870205 时间: 2009-9-29 07:08 标题: GWD-3-05 請問喔

Q5:

If (y+3)(y-1) – (y-2)(y-1) = r(y-1), what is the value of y?

(1) r² = 25

(2) r = 5

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.

Answer:

我的解題:

題目化簡 : (y+3)(y-1) – (y-2)(y-1) = r(y-1) => (y+3) - (y-2) = r

(1) r² = 25, 所以 r =5 or -5

當 r =5, 則可推出 y= -3 時會滿足

r = -5, 無答案

(2) r = 5 時, 則可推知 y= -3

因此我選 (B).

但答案是給 (E) 我不知道自己哪裡忽略了

請高手幫忙指導謝謝~~

作者: applebeer 时间: 2009-9-30 06:39

感觉应该是r不等于5时，才能推出y=1

如果r=5，y任何值都满足情况啊。

因为y不等于1时，有（y+3)-(y-2)=y+3-y+2=5=r

作者: jiroushiren 时间: 2009-9-30 08:37

LZ 的这步：“題目化簡 : (y+3)(y-1) – (y-2)(y-1) = r(y-1) => (y+3) - (y-2) = r ” 就已经中了GMAC的圈套了

推成这步就默认了 Y不等于1 忽略了 Y=1 的情况当Y+1时 r 为任意值所以答案为E

作者: xiao19870205 时间: 2009-9-30 20:14

感謝我弄懂了!

謝啦~

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