[問題]PREP-T1-Q202 Q211 (DS)

[oneseven]

Q202
Each employee of Company Z is an employee of either Division X or Division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for Division X than for Division Z?

(1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for Division Z.
(2) More than half of the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.

Answer :D

有點被搞暈了 好複雜

Q211
If k is a positive integer, then 20k is divisible by how many different positive integers?

(1) k is a prime
(2) k = 7

Answer : B
為何條件一不充分呢 謝謝

[eced]

Q202 (你題目寫錯了,最後是Company Z)
設Full time的人為F, part time的為P
F(x)/P(x) > F(z)/P(z) ? F(z)=F(x)+F(y) ; P(z)=P(x)+P(y)
(1)F(y)/P(y) < F(z)/P(z)
=>[ F(z)-F(x)] / [ P(z)-P(x)] < F(z)/P(z)
=> P(z)F(z)-P(z)F(x) < F(z)P(z)-F(z)P(x)
=>F(x)/P(x) > F(z)/P(z)
sufficient

(2) F(x) > F(y); P(y) > P(x)
F(x)/P(x) > F(y)/P(y)=[F(z)-F(x)]/[P(z)-P(x)]
F(x)P(z)-F(x)P(x) > P(x)F(z)-P(x)F(x)
F(x)/P(x) > F(z)/P(z)

sufficient


Q211
(1)的問題在於 prime=2時
20k=40=2^3 * 5 因子共有4*2=8個
當prime為非2
20k=2^2 *5*k 因子共有3*2*2=12個

[chris8888]

Q202
設Full time的人為F, part time的為P
F(x)/P(x) > F(z)/P(z) ? F(z)=F(x)+F(y) ; P(z)=P(x)+P(y)
(1)F(y)/P(y) < F(z)/P(z)
=>[ F(z)-F(x)] / [ P(z)-P(x)] < F(z)/P(z)
=> P(z)F(z)-P(z)F(x) < F(z)P(z)-F(z)P(x)
=>F(x)/P(x) > F(z)/P(z)
sufficient

(2) F(x) > F(y); P(y) > P(x)
F(x)/P(x) > F(y)/P(y)=[F(z)-F(x)]/[P(z)-P(x)]
F(x)P(z)-F(x)P(x) > P(x)F(z)-P(x)F(x)
F(x)/P(x) > F(z)/P(z)

sufficient


解的好, 不過我自己剛剛算了一下, 仔仔細細的算, 發現就算已經知道方法, 也要花掉5-10分中才能確定自己整個過程沒出錯, 我真懷疑自己對於這種複雜的比大小可以在2分鐘內解出, 太誇張了, G妹這樣出題. 不是不會算, 是考試若第一次看見這種題型, 題目內容再改一下, 要在兩分鐘內做對實在是很誇張.

看題目, 了解題意
列出可能的式子
進行解題,
驗證錯誤
解出
這樣子只有2分鐘要克服這題. 難.

除非事先看過這題, 而且算得熟, 幾乎不需要判斷解法和找出思路, 直接計算改過的數據題目, 我想才可能.

[Huang Hsin-Yi]

Q211
(1)的問題在於 prime=2時
20k=40=2^3 * 5 因子共有4*2=8個當prime為非2
20k=2^2 *5*k 因子共有3*2*2=12個[/quote]

看得懂因為2而因數數量不同
不過為何是8和12

[Huang Hsin-Yi]

[aqboy[origen]]

40=2^3*5^1

40的因數個數=(2+1)*(1+1)=8

60=2^2*3^1*5^1

60的因數個數=(2+1)*(1+1)*(1+1)=12

記得這個是公式 , 不然也可以自己找 反正數字不大

[sunnyblue13s]

40=2^3*5^1

40的因數個數=(3+1)*(1+1)=8

60=2^2*3^1*5^1

60的因數個數=(2+1)*(1+1)*(1+1)=12

記得這個是公式 , 不然也可以自己找 反正數字不大

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我來幫你修正一下~~

[Philosophia]

Q211
(1)的問題在於 prime=2時
20k=40=2^3 * 5 因子共有4*2=8個
當prime為非2
20k=2^2 *5*k 因子共有3*2*2=12個

5 也是prime number
so (1) 可得到 8, 9, 12 三種不同答案
but (2) 12 factors only

ans B