本日小編給眾人匯集整頓了GMAT測驗大牛的溫習辦法履歷談，教給眾人若何用代數思惟來解題，同窗們可用來參考!
OG已刷到第五遍(照樣第六遍?忘了)，忽然有點恍然大悟的快活。決議一點一點記載下來，迎接評論辯論和分享!
一、Main Conclusion題
曩昔繞得好暈啊，各類辦法和邏輯，然則感到對我都不太實用，有點像去強迫按那些大牛們的思緒做題，然則明顯曉得這不是本身。。。
本日開端看CR OG12 準確的句子，發明亮點了!!!並且把OG裏main conclusion的題全殺了一遍，全中!!除一個66題大概稍隱晦，其余都對了。
先說道理再給例子眾人本身溫習嘗嘗。
1、道理：MC題型我發明根本出題人是如許做的：究竟1、究竟2、究竟3(龐雜一點的會有究竟3)，而謎底一樣平常模式便是：把究竟1、2代入究竟3，可以得出的那句話。
發明沒!!相似解代數方程式!!便是如許!!別不信任，我把OG標題粘給眾人看。
2、舉例：(標題來自OG12 CR)
101. For a trade embargo制止收支口岸 against a particular country to succeed, a high degree of both international accord協定，同等 and ability to prevent goods from entering or leaving that country must be sustained. A total blockade封閉 of Patria’s ports is necessary to an embargo, but such an action would be likely to cause international discord辯論 over the embargo.
The claims above, if true, most strongly support which of the following conclusions?
(D) Any trade embargo against Patria would be likely to fail at some time.
剖析：P1給出了embargo倆前提。P2給出了P國不克不及同時具有著倆。C：P2≠P1。便是謎底D!
102. Theater Critic: The play La Finestrina, now at Central Theater, was written in Italy in the eighteenth century. The director claims that this production is as similar to the original production as is possible in a modern theater . Although the actor who plays Harlequin the clown gives a performance very reminiscent表示的，令人想起…的 of the twentieth-century American comedian Groucho Marx, Marx’s comic style was very much within the comic acting tradition that had begun in sixteenth-century Italy.
The considerations given best serve as part of an argument that
(D) the performance of the actor who plays Harlequin in La Finestrina does not serve as evidence against the director’s claim
剖析：
P1 LF是18世紀意大利寫的
P2 導演說LF只管即便還原原著時期
P3 LF裏的腳色H讓人想到20世紀的作品GM，然則GM是16世紀風
P2=P1，P3=1，以是!!P3=P2結論! 便是謎底D!
19. A computer equipped with signature-recognition software, which restricts access to a computer to those people whose signatures are on file , identifies a person's signature by analyzing not only the form of the signature but also such characteristics as pen pressure and signing speed. Even the most adept forgers cannot duplicate復制、完整同樣的 all of the characteristics the program analyzes.
Which of the following can be logically concluded from the passage above?
(C) Nobody can gain access to a computer equipped with the software solely by virtue of skill at forging signatures.
剖析：
配景 sr軟件經由過程署名限定進口
P1 sr軟件道理：不但是form，另有pressure和speed
P2 最牛的都不克不及復制同樣的署名
這題更簡略，直接把P2代入P1便是結論，謎底C!
43. Laws requiring the use of headlights during daylight hours can prevent automobile collisions . However, since daylight visibility is worse in countries farther from the equator, any such laws would obviously be more effective in preventing collisions in those countries. In fact. the only countries that actually have such laws are farther from the equator than is the continental United States.
Which of the following conclusions could be most properly drawn from the information given above?
(E) Daytime headlight laws would probably do less to prevent automobile collisions in the continental United States than they do in the countries that have the laws.
剖析：
究竟1：Hdattack collision→究竟2：離得越遠，laws越有用→究竟3：比us離得遠的國度有laws。
把究竟1和究竟2代入究竟3，便是謎底!
LZ在刷第一遍第二遍乃至第三遍的時刻，43題和102題一向半知半解!如今終究清楚多了!你會發明，GMAC出題的思緒大部門便是：先給一個普適性的道理或究竟，再給一個個案的小例子而且告知你它一些特色(平日跟前面的道理有關的，要信任CR題幹沒有空話!)，然後Main Conclusion便是像代數題同樣，把特定的x值代入y方程，y的成果便是謎底!
如許解題必定不會偏!由於y方程裏沒有的，必定不是謎底!!
(二)Weaken題型
實在作為童貞座且是女士，weaken題我照樣挺有自負的(自黑)。但是一向位於黑一題算一題，本身能想到的attack有謎底就可以做對，想不到的然則謎底列出來了的就要費頭腦。顯著是缺少科學邏輯思惟憑履歷的搞法。沒方法月尾就要測驗了時光緊急，童貞座另一大亮點本性要閃光了——做總結!
代數思惟再一次救了時光段考期緊的我。空話不說!切入正題。
1、道理：
GMAC的weaken題型先輩們大多半牛人都總結了，最經典確當然是helr的6大分類：類比、統計、計劃、相幹因果、果因、因果。這個確切最先給了我很大啟示!其時我就在斟酌這些分類終極在邏輯鏈裏若何去應用呢?照樣線性代數!!
Y=ax+b
又在刷OG，逐漸發明：GMAC對CR weaken的出題思緒—— 題幹告知你，此題的變量有x和y，結論是x變更就能夠得出y變更。若何去weaken這個結論?
請眾人腦補Y=ax+b方程式，你就會看到，如x的變更可以直接反響到y對應變更，有三個方面——x自己、a和b。
x天然不說，直接否認x和y的幹系，太直白太露骨。然則要防備GMAC的變體情勢題(背面我列的第一個例子)
a：代表了跟x相幹。最簡略的懂得莫過於，題幹的情況產生了變更，致使a1和a2不相稱，以是x變更時，y不必定能得出!
b：代表了傳中的圈外人。最簡略的懂得(也是GMAC最愛出的範例)，y結論的產生，可不多是b釀成的?
以是全部weaken題型便是在探求a和b，由於分歧於main conclusion之處在於，weaken題型裏一樣平常結論是交卸給你的。
那末，必定要記著：由於CR是沒法在幾句話裏做到窮舉的，以是必定有破綻可以給你捉住!要有這個信念。
來幾道OG標題給眾人加深下懂得：
104. Although the discount stores in Goreville’s central shopping district are expected to close within five years as a result of competition from a SpendLess discount department store that just opened, those locations will not stay vacant for long. In the five years since the opening of Colson’s, a non-discount department store, a new store has opened at the location of every store in the shopping district that closed because it could not compete with Colson’s.
Which of the following, if true, most seriously weakens the argument?
(B) Increasingly, the stores that have opened in the central shopping district since Colson’s opened have been discount stores .
剖析：這題我做錯過好幾回，在CD上翻過很多多少帖子想弄明確，但照樣暈!發明壯大的代數思惟再一次辦理了我T^T
P1 S扣頭店開張，致使G區的其他扣頭店都關門，在將來五年。
C 這些關掉的門面其實不會空良久
P2 一種non-discount的C店開了，new store已在PK不贏C店而倒閉的門店上又開了起來
Y=vacant，X=開的新店。本題的X只有兩種大概=discount store和non-discount store
P1告知咱們的是，當X=discount store時，Y建立。Y1=A×X1+B。
P2告知咱們的是，X已又來了，然則不曉得是否是discount!成果GMAC竟然就得出了Y必定不可立!是否是很扯!
題目問甚麽情形下，Y會持續vacant?固然是新的X持續屈服P1的方程時!也便是X2=X1的時刻!沒錯只有謎底B!
總結：必定要找準X和Y的變量究竟是誰，然後就看它七十二變!
上面這題偏難，然則很沖動我終究想明確了。來道經典的、更常常會考的、簡略的例子：
114. Guidebook writer: I have visited hotels throughout the country and have noticed that in those built before 1930 the quality of the original carpentry work is generally superior to that in hotels built afterward. Clearly carpenters working on hotels before 1930 typically worked with more skill, care, and effort than carpenters who have worked on hotels built subsequently.
Which of the following, if true, most seriously weakens the guidebook writer’s argument?
(D) The better the quality of original carpentry in a building, the less likely that building is to fall into disuse and be demolishe
剖析：這題是典範的a變量產生轉變。
配景：all hotels visited(樣本已充分，不要想weaken作者這一點)
P1 quality(1930前)>quality(1930後)
C skill&effort(1930前)>skill&effort(1930後)
方程式就釀成了：skill&effort=a×quality+b
想weaken，x和y的幹系就從a和b入手。
D選項：a不同樣!a(1930前)≠a(1930後)——a是demolish(假如1930前的hotel是demolish了，那末x的影響因子a就不相稱了!)
以是原方程式不可立。即skill&effort(1930前)不必定>skill&effort(1930後)
再來一道b不同樣的，這類從圈外人找緣故原由的題最常考，不外如今看來也最簡略了。弄個輕微龐雜點的，也便是x是由多個身分組成的：
111. The difficulty with the proposed high-speed train line is that a used plane can be bought for one-third the price of the train line, and the plane, which is just as fast, can fly anywhere. The train would be a fixed linear system, and we live in a world that is spreading out in all directions and in which consumers choose the freewheel systems (cars, buses, aircraft), which do not have fixed routes. Thus a sufficient market for the train will not exist.
Which of the following, if true, most severely weakens the argument presented above?
(C) Planes are not a free-wheel system because they can fly only between airports, which are less convenient for consumers than the high speed train’s stations would be.
P1 price：train=1/3的plane
P2 speed和規模 plane>train
P3 行駛路線：train是牢固的，然則cars、buses、plane都是freewheel
y=(speed+price+free wheel)×a +b
C 由於train的speed+price+free wheel都是最弱，以是train的y也最弱
這個找weaken就很輕易了!只要(speed+price+free wheel)隨意率性一個train不是最弱、大概train的b比他們大，train就有了勝出的上風!
謎底C: b=convenient，且plane的free wheel和train同樣弱!雙響炮啊，以是y不可立!
數學萬歲!!!T^T 我崇敬全部數學成就好的娃，我也要盡力。
市場價：HK$ 110
