一題JJ

[lisayang83]

The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.
The argument above is flawed because it fails to take into account
A.Changes in the population density of both Meadowbrook and Parkdale over the past four years.
B.How the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale
C.The ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale
D.The violent crime rates in Meadowbrook and Parkdale four years ago
E.How Meadowbrooks’ expenditures for crime prevention over the past four years compare to Parkdale’s expenditures.

請問是A嗎?
因為這樣可以知道人口?

[dibert8]

這題考點是"比較基準",選 (D)
(A) violent crime rate 無關於 population density
(B) violent crime rate 無關於 rate of population growth
(C) violent crime rate 無關於 ratio of nonviolent crimes
(E) violent crime rate 無關於 expenditures

[Huang Hsin-Yi]

我也認為是D
因為兩個城市犯罪率 比較值 一定要和他們以前的犯罪率比較,才能斷定哪個城市較安全
舉例
Meadowbrook 四年前是10%-10人 今年70%-70人 增加60%
Pardele 四年年是80%-80人 今年90%-90人 增加10%
結果是pardele較危險

[wonderland]

我也認為是D

因為兩個城市犯罪率 比較值 一定要和他們以前的犯罪率比較,才能斷定哪個城市較安全

舉例

Meadowbrook 四年前是10%-10人 今年70%-70人 增加60%

Pardele 四年年是80%-80人 今年90%-90人 增加10%

結果是pardele較危險

假設

Meadowbrook 四年前是10%-10人 今年70%-70人 增加60%

Pardele 四年年是30%-30人 今年40%-40人 增加10%

不就變成Meadowbrook比較危險了嗎？
不太懂為什麼這題沒有把基數"人口數"考慮進去，五個選項好像都選不下手

哪位可以解釋一下邏輯呢？