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II．        工程问题
【怎么读】：读出模型的几个元素——
工程问题通用模型：
1/a + 1/b +1/c+…… = 1/t
原理：
ｔ×（va+vb+……）=T（工作总量）

105.     11075-!-item-!-187;#058&007490
Two water pumps, working simultaneously at their respective constant rates, took exactly 4 hours to fill a certain swimming pool.  If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

(A) 5
(B)  
(C)  
(D) 6
(E)  

5.         1353-!-item-!-187;#058&000886
Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours.  If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate[l15] ?

(A) 25%
(B)
(C) 50%
(D)
(E) 75%

179. 17563-!-item-!-187;#058&011544
Working independently at their respective constant rates, pumps X and Y took 48 minutes to fill an empty tank with water. What fraction of the water in the full tank came from pump X ? (1) Working alone at its constant rate, pump X would have taken 80 minutes to fill the tank with water.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank with water.

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179. 17750-!-item-!-187;#058&011285
A manufacturer conducted a survey to determine how many people buy products P and Q. [l7] What fraction of the people surveyed said that they buy neither product P nor product Q ?
(1)1/3 of the people surveyed said that they buy product P but not product Q.
(2) 1/2 of the people surveyed said that they buy product Q.

202. 19528-!-item-!-187;#058&012279
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 Company Z[l8]  ?
(1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for Company Z.
(2) More than half of the full-time employees of Company Z are employees of Division X[l9] , and more than half of the part-time employees of Company Z are employees of Division Y[l10] .

212. 20491-!-item-!-187;#058&012823
This morning, a certain sugar container was full. Since then some of the sugar from this container was used to make cookies. If no other sugar was removed from or added to the container, by what percent did the amount of sugar in the container decrease[l11] ?
(1) The amount of sugar in the container after making the cookies would need to be increased by 30 percent to fill the container.
(2) Six cups of sugar from the container were used to make the cookies.

85.       9649-!-item-!-187;#058&007038
Of the 800 employees of Company X, 70 percent have been with the company for at least ten years.  If y of these "long-term" members were to retire and no other employee changes were to occur, what value of y would reduce the percent of "long-term" employees in the company to 60 percent?[l12]

(A) 200
(B)  160
(C)  112
(D)  80
(E)  56

88.       9789-!-item-!-187;#058&007104
In May Mrs. Lee's earnings were 60 percent of the Lee family's total income.  In June Mrs. Lee earned 20 percent more than in May. [l13]  If the rest of the family's income was the same both months, then, in June, Mrs. Lee's earnings were approximately what percent of the Lee family's total income?

(A) 64%
(B) 68%
(C) 72%
(D) 76%
(E) 80%

176. 17259-!-item-!-187;#058&011402
In 1995 Division A of Company X had 4,850 customers. If there were 86 service errors in Division A that year, what was the service-error rate, in number of service errors per 100 customers[l14] , for Division B of Company X in 1995 ?
(1) In 1995 the overall service-error rate for Divisions A and B combined was 1.5 service errors per 100 customers.
(2) In 1995 Division B had 9,350 customers, none of whom were customers of Division A.

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IV．     附录：

附件1： 理解问题专题集锦
【注】：以下题目均来源于PREP破解1和2，下面只分析读题，有兴趣解题的同学可以查阅破解版相应题目的答案。

I．          比例问题、比较问题+集合问题（含浓度问题）
【怎么读】：
1. 比例问题：读出分子分母
1）遇到percentage和fraction：注意跟在这两个词后的of后的部分（包含其修饰语）就是分母；
2）ratio：ratio of A to B——A是分子，B是分母
2. 集合：读出分类标准，并且能够通过英文表述准确识别是否是全集、有没有交集或并集
1）一个标准分类的：通常用韦恩图或公式直接求解
2）两个标准分类的：通常用二维表解决较为简单明了
3. 比较问题：准确读出比较对象

96.       10591-!-item-!-187;#058&007292
According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of water to make orange juice.[l1]   How many 12-ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice[l2] ?

(A)  25
(B)  34
(C)  50
(D)  67
(E)  100

142.     14580-!-item-!-187;#058&010013
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present.[l3]   If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase

164.     17177-!-item-!-187;#058&011044
A certain scholarship committee awarded scholarships in the amounts of $1,250, $2,500, and $4,000[l4] .  The committee awarded twice as many $2,500 scholarships as $4,000 scholarships, and it awarded three times as many $1,250 scholarships as $2,500 scholarships. [l5] If a total of $37,500 was awarded in $1,250 scholarships, how many $4,000 scholarships were awarded?

(A)  5
(B)  6
(C)  9
(D) 10
(E) 15

165. 16466-!-item-!-187;#058&010795
Of the 1,400 college teachers surveyed, 42 percent said that they considered engaging in research an essential goal. How many of the college teachers surveyed were women?
(1) In the survey, 36 percent of the men and 50 percent of the women said that they considered engaging in research an essential goal.[l6]  
(2) In the survey, 288 men said that they considered engaging in research an essential goal.
【解】：A
1）破题：
A）理解的关键：见批注
B）破题：这是一个标准的二维表问题
2）思路：见下表
        E         E取非         T
F         50%x;700                  X
M         36*(1400-x);288                  1400-x
T         *42%                  1400
【NOTE】：黑色字为题干已知；绿色字为条件1已知；红色字为条件2已知。

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（4）重视讨论帖：自己先看，通过整理机井+看讨论后依然不明白的，放到答疑专区，并并写清楚您的问题是什么，我们好有的放矢地解决。

（5）考前把自己最后的错题分类总结快速看一遍：早上考的，头晚系统看看；下午考的，第二天上午看看。
主要看什么：自己之前解决的难题，简单的就不用看了；有争议的也看看：不是记住答案，而是告诉自己，遇到这些要小心，这是没有确定的，需要我自己好好思考的，不要考场上乱点。
这一遍会让机经带给你的解题感觉在脑中清晰和系统化，有助于临阵不乱。

4. 考场上：不要去想GMAC是怎么评分的，一切都是未知和徒劳，专气致柔如婴儿是在做好上述准备后获得高分的唯一要诀。

5. 对于知识基础扎实、题目阅读理解顺畅、pace很好的同学：
备考前把OG的math review过一遍，直接进入JJ整理（当月或前月均可，经验反复证明：机经对于前面月份相同题目的重复经常存在），在此基础上，在上述4步中挑选你认为有必要的步骤进行补充训练即可。

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【注意几点】：
A） 马虎问题：
我觉得比较好的：整理JJ的工作可以和做PREP同步进行。所以，就把自己这段时间以来的PREP的错题（之前已经分类）中马虎问题集中起来做一遍，总结出一种对GMAT数学prefer的一些马虎点的感觉，然后再拿出JJ中的马虎问题看2遍，记住自己容易出错的地方
B）理解问题：
做PREP时就要注意积累一些自己不熟悉的表达，做JJ时遇到理解问题，继续积累收藏，考前看看就OK了。
C）思维陷阱：以JJ为纲，按知识点整理出一类题自己怎么爱犯错的。如果基础再差一点儿，可以加上PREP的相同类型错误的题目一起整理。考前反复看几遍，把这种敏感融入到题感之中。（比如，概率中的有没有放回，是否是独立事件；余数问题：）。
D）悬而未决：
有争议的、大家都没辙的难题——放了，依靠考场灵感。
E）知识点遗忘：看看自己薄弱的知识点在哪一部分，有针对性地补——博森的数学书“知识点讲解”部分或任意下载一些整理好的数学知识点中文总结即可，OG的math review是基础仅仅提供广度和系统性的了解，但在这个范围内各考点的深度需要利用中文资料自己进一步弥补，建议使用好google和维基百科等网络工具来查漏补缺。

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【经验】：考场上，特别是有JJ时，很可能做得比较好，此时会遇到一些新的难题，这些题虽然陌生，一定逃不出你整理过的某个知识点，运用做JJ时给你的感觉和思路，一定可以激发平时没有的考场灵感，这就是数学和语文最大的不同。
C）提高考场效率：一定是建立在水到渠成的基础上。不是直接背了就去，除非心态极佳不怕变体或记忆力超群且识别变体的能力超群。
（3）目标：最后上考场的感觉是JJ的思路已经进入了骨髓，觉得是不上51都难的感觉。
（4）如何使用相应月份的JJ：
1）自己先做一遍：这一遍就是找出自己的错题，同时按如下两条线进行分类
2）无需按照PS或DS来分。
两条线——
A）错误类型：
a）马虎问题和理解问题：
前者：如果时间不够充裕的，就按我后面说的方式进行，我后面会专门讲数学的备考
后者：基础不好且有时间的可以看看红书；没有时间的就按照我说的备考进行——积累和熟练为主。
b）思维陷阱：
c）知识点遗忘：单独一部分进行
d）DS充分性把握不好
e）悬而未决：有争议的；大家都搞不定的难题
f）记住答案的：有些题目需要大量计算的，可以先用计算器算好了，直接背，不是刻意，多看几遍一定就印象深刻了。
B）知识点类型：我在10.16的答疑部分的讲义有所演示如何进行这部分分类

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2.     理解能力训练、pace训练和知识点遗忘的克服：PREP数学的破解（博森小班专区资料下载会有题目和部分讲解的下载）
（1）      资料：PREP破解 DS和PS——最接近实战的英文题
（2）      怎么练：
A） 每天37道，自己组合成套题配合着语文和作文来做，计时。
B） 每天做完以后：总结错题，猜的但猜对了的不管。
按照错误类型进行总结：马虎问题、思维陷阱、理解问题、知识点遗忘。
尤其是知识点遗忘部分，建议把相应知识点的题目放在一起，积累到一定程度以后，会发现这一类知识点的出题思路和解题思路。
C） 定时复习。
（3）      练到什么程度：当发现3个标准的题感有了，就可以酌量减少，一周做2套什么的维持下感觉，根据自己的模考计划而定。
哪部分差做哪部分，时间紧的建议做完DS 2个practice，PS practice1。
3.     思维训练：机经
（1）      资料：当月JJ；if没有，就用上月JJ
（2）      对待JJ的心态：
不是背答案，因为考场压力可能迫使你完全忘记了答案。平时反复整理JJ的目的是让你实现以下几个方面：
A）找到自己知识点的漏洞：练习方法我下面会讲
B）熟悉GMAC近期出题的思路方向：JJ的趋势较红书时代有所变化，这就是为什么用JJ的参考价值更大。以这个出题方向为纲引导你形成一种思想套路之类的东西，也就是：看到这类题就会自然而然地产生一个思考的方向，且这个方向90%以上是准确的。

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GMAT数学备考的三个层次：
1. 基础训练与错误点的寻找：OG
1）OG review + DS和PS题目开始前的guide（DS是P273-276：了解出题规则），只有知己知彼才能百战百胜。
2）OG DS和PS——DS160道，PS250道左右。分别把后面的100道做一遍，总结出自己的理解问题，这是基础积累过程；
A）如果各做100道：一次做完；如果都做完，就分2次完成，两次的时间间隔不能太长，DS和PS搭配着来
B）错题总结：按照错误类型总结——
a）思维陷阱
b）马虎问题
c）理解问题：并把阻碍理解的表达积累起来
e）“DS的充分性”没有把握好

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Eg：
相关考题（ID 2426）：
If s4v3x7 < 0, is svx < 0?
(1)     v < 0
(2)     x > 0
            
          Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
            
          Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
            
          BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
            
          EACH statement ALONE is sufficient.
            
          Statements (1) and (2) TOGETHER are NOT sufficient.
【解析】：E
本题知道v或x都没有用，关键是s，因为题干已经可以确定vx<0了，所以，最终取决于s与0的关系，每个选项单独以及结合起来都得不到s的信息，所以对于问题依然可能回答yes也可能回答no，因此无论如何都无法充分。

III．    GMAT数学备考策略

首先：水平测试——
用PREP模考软件数学部分测试一下：
1）  如果能够保证错误在3个以内且不超时、pace稳定，就可以直接完成下述三个层次的第三个，大考前需要看一遍OG review，即可。
2）  如果错误率超过1）的标准，或做题感觉很艰难（包括pace不稳定、读题困难、题目思路把握不准确，etc），就按照如下三个层次来，在进行过程中，可以不断用I的三个标准来判断自己的进步和水平现状，当找到相应感觉以后，就可以进入下一个层次。

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5．    解题过程不必要的繁复，浪费时间：该类问题常见于DS题
【根源】：没有搞清楚DS“充分性”的含义！主要有以下两大类（其他占比很小的零碎题型就不总结了）——
(1) what 型：只有能够求出唯一值来回答what才叫充分！
Eg:
What is the value of the integer n ?
(1)
(2)
            
          Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
            
          Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
            
          BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
            
          EACH statement ALONE is sufficient.
            
          Statements (1) and (2) TOGETHER are NOT sufficient.
【解析】：B
条件1单独不充分，因为求出的n有2个值。
条件2单独充分：能求出一个唯一值（可以用试算）n=3.
【注意】：如果条件2的方程的解>1个，则看1）+2）取交集后是否能够确定唯一值，如果可以，充分选C！
（2）“Is”型：能够确定回答yes或确定回答no都叫充分；可能回答yes也可能回答no叫不充分。