PP1+PP2 數學完整詳解版-2009.08.19更新 (更新者: sunnyblue13s)

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PP1+PP2 數學完整詳解版-2009.08.19更新 (更新者: sunnyblue13s)

文章larry.tw » 2008-07-04 14:18

由於下載系統的問題, 我們遺失了重要的最新版PP

目前, 版大:sunnyblue13s 擁有最初的原始版本

DS1 - V1.1 / DS2 - V1.0
PS1 - V1.0 / PS2 - V1.0

有需要的版友, 請直接下載.

也請有回版的版友, 倘若擁有pp1-PS v2.0 / pp2-DS v1.2, 也煩請提供

感謝各位的支持

sunnyblue13s筆

PP1+PP2作者群:
pauline_chiou, william0625, gman, yuma168 ,larry.tw, Huang Hsin-Yi, Penda, Engedi, mauricew, charlie6989, abc123, sunnyblue13s, ruru.lu
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最後由 larry.tw 於 2008-11-19 00:09 編輯,總共編輯了 14 次。
larry.tw
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Re: PP1+PP2 數學完整詳解版 -- 勘誤表

文章larry.tw » 2008-09-13 13:49

[Philosophia ]

PP2-PS-Q176
x=4, y=-6, S is positive, 應是在第四象限,而非第二象限

PP2-DS-Q68
基本公式或應是 gross profit = revenue - expenses
如此可推 (1) and (2) revenue = 4 gross profit   
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Re: PP1+PP2 數學完整詳解版 -- 勘誤表

文章larry.tw » 2008-09-18 12:27

PP2-DS-135

we need to find the value of P-T

St 1 tells us P+Q+R+S = 388250*4
P = 388250*4-Q-R-S
But since it doesnt tell us anything about T its INSUFF

St 2 tells us Q+R+S+T = 383000*4
T=383000*4-Q-R-S
but since it doesnt tell us anything about P its INSUFF

Combining them we get
P = 388250*4-Q-R-S
T=383000*4-Q-R-S

Subtract the two statements we get P-T = (388250*4)-(383000*4)
Hence the answer is C
larry.tw
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Re: PP1+PP2 數學完整詳解版 -- 勘誤表

文章larry.tw » 2008-09-18 13:20

PP2-DS-155

"500 is the multiple of 100 that's closest to x": this means that, of all multiples of 100, 500 is closest to x.
in other words, x is closer to 500 than to 400 or 600 (or to any other multiple of 100, for that matter). what does this mean?
it means that 450 < x < 550. (note that this is almost the same thing as "x rounded to the nearest hundred produces 500", except for the piddling exception of 450.)

(另一種說法:For 500 to be the multiple of 100 closest to x, x has to be between 450 and 550. If x is less than 450 e.g 438 then the multiple of 100 closest to x will be 400 and not 500. Similarly if x is greater than 550 e.g. 578 then the multiple of 100 closest to x will be 600 and not 500. )


the second half of the problem statement, similarly interpreted, gives the following: 350 < y < 450.

(1)
in combination with the question stem, this means that 450 < x < 500 (and we still know that 350 < y < 450).
therefore, 800 < x + y < 950.
this isn't good enough to answer the question; the numbers between 800 and 850 are closer to 800, the numbers between 850 and 950 are closer to 900, and 850 is equidistant from 800 and 900.
insufficient.

(2)
in combination with the question stem, this means that 350 < y < 400 (and we still know that 450 < x < 550).
therefore, 800 < x + y < 950, exactly the same inequality that holds for statement (1).
insufficient.

(together)
now we know that 450 < x < 500 and 350 < y < 400.
this means that 800 < x + y < 900.
still not good enough to answer the question; the numbers between 800 and 850 are closer to 800, the numbers between 850 and 900 are closer to 900, and 850 is equidistant from 800 and 900.
insufficient.

ans = e
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Re: PP1+PP2 數學完整詳解版 -- 勘誤表

文章larry.tw » 2008-09-20 13:01

PP2-DS-184

題目應該就是問 K能不能等於2^r ?
(1)的條件 k/(2^6)=整數(設X)
所以k=(2^6)X 當X等於2的倍數 則條件成立
反之不成立 因此 條件一不充分
(2)的條件 K不能被大於一的任何奇數整除
換句話說 K只能被2的倍數整除 也就是說K一定是2的倍數且只有1和2的因數
所以條件2充分
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Re: PP1+PP2 數學完整詳解版 -- 勘誤表

文章larry.tw » 2008-09-22 12:58

PP2-DS-192

Add info:
(1) + (2) => y - 2 < x < 2y - 6
y - 2 < 2y - 6, y > 4 => x > 2
=> xy > 0
=> (C)
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Re: PP1+PP2 數學完整詳解版 -- 勘誤表

文章iamsmall » 2008-11-09 11:46

一戰的時候準備數學,發現PP數學難度很接近實戰,只是因為時間不夠加上很多題目不完整,所以沒能完全做完。現在準備二戰,想要讓數學可以更穩,所以花了一些時間先把PP1 PS 不完整的題目補上。

[文件]: PP1-PS
[題號]:
[內容]:補上Q5,6,11,13,14,15,17,18,25,28,29,31,37,38,46,47,50,55,56,59,61,62,63,67,72,78,80,83,87,91,92,93,97,
102,103,107,112,113,114,119,120,121,125,128,129,131,133,150,158,162,168,
173,176,177,179,182,187,191,190,193,等共60題,尚餘24題未齊,請大家繼續加油

http://www.4shared.com/file/71669175/e7 ... 1-V20.html

另外CD那邊有人把PP的DS圖形題都補齊了,
大家可以去下載
http://forum.chasedream.com/dispbbs.asp ... n=0&Star=1
檔案我也一併放在這邊
http://www.4shared.com/file/70351665/8f4a3e6c/ds1.html
http://www.4shared.com/file/70352297/f36044bd/ds2.html
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Re: PP1+PP2 數學完整詳解版-2009.08.19更新 (更新者: sunnyblue13s)

文章leafer » 2019-04-14 23:27

謝謝分享
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