GMAT邏輯推理對很多同學(xué)來說都是一大難點，只有掌握好GMAT邏輯技巧才能從容應(yīng)對GMAT邏輯題庫。下面請隨小編一起看看邏輯備考推理兩大怪題解析，歡迎閱讀！
題一
In Patton City, days are categorized as having heavy rainfall (more than two inches),moderate rainfall (more than one inch, but no more than two inches), light rainfall (at least a trace, but no more than one inch), or no rainfall. In 1990, there were fewer days with light rainfall than in 1910 and fewer with moderate rainfall, yet total rainfall for the year was 20 percent higher in 1990 than in 1910.
If the statements above are true, then it is also possible that in Patton City
A. the number of days with heavy rainfall was lower in 1990 than in 1910
B. the number of days with some rainfall, but no more than two inches, was the same in 1990 as in 1910
C. the number of days with some rainfall, but no more than two inches, was higher in 1990 than in 1910
D. the total number of inches of rain that fell on days with moderate rainfall in 1990 was more than twice what it had been in 1910
E. the average amount of rainfall per month was lower in 1990 than in 1910
已知：heavy rainfall > 2 inches; 2>=moderate rainfall > 1; 0 <="1"> 且 1990年小雨和中雨的天數(shù)都比1910少，然而1990年全年的降雨量還比1910多，那就只能是1990年的大雨天數(shù)比1910多，一般我們都會這樣的預(yù)估，但選項恰恰沒有該項。迷茫中，看到了D，誤以為找到了救星，正好落在了出題者的陷阱之中。
細看該題所問“If the statements a above are true, then it is also possible that in Patton City”，原來并非是由原文推出選項的導(dǎo)出題，而是一道怪題;只要是“可能的”項就是正確的。換句話說，依據(jù)原文，有四個選項是絕不可能的，只有一個是可能的。
那再看D，在1990年中雨天數(shù)比1910少的情況下，其降雨量可以超過1910的2倍嗎?絕不可能 ------ 因為2>=moderate rainfall > 1，即最大的中雨量不超過最小的中雨量的2倍，也就是：即便1990的中雨天每次都是最大量，而1910的中雨量每次都是最小量，那么D都不能成立。
而B與C明顯與已知條件矛盾(“1990年小雨和中雨的天數(shù)都比1910少”)
而E與“1990年全年的降雨量還比1910多”沖突看來也就A有希望了;但A與我們的預(yù)估相反。
但A的確是可能的，這一點人間惆悵客和其樓上的新浪網(wǎng)友都有很好的解釋。說個最極端的情況，1990年只下了一天雨，那是一場萬年不遇的億寸豪雨，其降雨量可以遠遠超過1910年的天天小雨、中雨或so so 大雨。一天就超過365天這有什么不可能的呢?
其實這題一點也不難，就是怪，思路逆向而走，在狹縫中尋找可能空間;看到這里，各位看官霎時領(lǐng)悟，真可謂：“忽報人間曾伏虎，淚飛頓作傾盆雨”
題二
In the United States， vacationers account for more than half of all visitors to what are technically called “pure aquariums” but for fewer than one quarter of all visitors to zoos， which usually include a “zoo aquarium” of relatively modest scope.
Which of the following， if true， most helps to account for the difference described above between visitors to zoos and visitors to pure aquariums?
(A) In cities that have both a zoo and a pure aquarium， local residents are twice as likely to visit the aquarium as they are to visit the zoo.
(B) Virtually all large metropolitan areas have zoos，whereas only a few large metropolitan areas have pure aquariums.
(C) Over the last ten years，newly constructed pure aquariums have outnumbered newly established zoos by a factor of two to one.
(D) People who visit a zoo in a given year are two times more likely to visit a pure aquarium that year than are people who do not visit a zoo.
(E) The zoo aquariums of zoos that are in the same city as a pure aquarium tent to be smaller than the aquariums of zoos that have no pure aquarium nearby.
從問題中的“helps to account for”可以明確該題是解釋題，所以要懷著讀出“奇怪或矛盾”的目的去閱讀原文(絕不是懷著奇怪的心態(tài)去閱讀)
讀完原文發(fā)現(xiàn)，一方面vacationer占了參觀pure aquariums人數(shù)的一半，而同時只占了參觀zoo人數(shù)的不到四分之一。這是怎么回事呢?尤其是zoo中還有zoo aquarium呢。這題讀完，本身覺得也沒啥矛盾，就是心里覺得怪怪的。(注意原文中account for 意為“占”，而問題中的意為“解釋”)
初步預(yù)估，大概是vacationers都是純爺們，特愛看純水族，不是很愛看ZOO中的水族(畢竟scope不如純的全面)
如果這樣想，很多人就會青睞D項;D說人民群眾參觀ZOO的可能性是參觀純水族的兩倍多，好像解釋了;仔細一想不對啊，這和vacationer占參觀人數(shù)的比例毫無關(guān)系啊;同學(xué)們，分子分母思維法，不可只想分子，更不可只想分母啊。
該題的核心是vacationer;這是原文的軌道，切不可脫離，那vacationer的參觀者和其它參觀者有啥子區(qū)別呢? 恩，vacationer，旅游度假者也，多是去外地觀光，而其它參觀者當(dāng)然是當(dāng)?shù)鼐用窳恕Ｒ粋�旅游觀光者到外地一般會看什么景點呢?當(dāng)然是自己本地所沒有的。所以嘛，這道題的答案就已經(jīng)浮出了水面。
應(yīng)該是 B (有純水族的城市少，所以每到這個城市的vacationer要去純水族的可能性就比去參觀ZOO的可能性大的多，畢竟幾乎所有城市都有ZOO;而當(dāng)?shù)赜慰瓦@種偏愛就不明顯;分子分母思維法，定要牢記)待憶~思盈的分析已經(jīng)接近真理，望再接再厲 ”。