Re: [計量] GWD 2-Q18

看板GMAT (GMAT入學考試)作者dounts (忘記過去)時間15年前 (2010/11/04 07:40)推噓1(1推 0噓 0→)

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※ 引述《creamz (creamz)》之銘言： : On a certain transatlantic crossing, 20 percent of a ship's passengers held : round-trip tickets and also took their cars abroad the ship. If 60 percent of : the passengers with round-trip tickets did not take their cars abroad the : ship, what percent of the ship's passengers held round-trip tickets? : A.33 1/3% : B.40% : C.50% : D.60% : E.66 2/3% : Ans: C 我們假設 # = number of people R = held tound-trip tickets C = took car abord C'= did not too car abord 從題目得知： #(R & C) / #(Total) = 20% #(R & C') / #(R) = 60% 求 #(R) / #(Total) = ? 我們變化第二句 #(R & C') / #(R) = 60% --> #(R & C) / #(R) = 1 - #(R & C') / #(R) = 1 - 60% = 40% #(R & C) / #(Total) ------------------- = #(R) / #(Total) = 20% / 40% = 50% #(R & C) / #(R) 簡單文字敘述持來回票又挾帶車子的人有 20% 在這持來回票之中沒有帶車子的有 60% 換句話說持來回票之中有帶車子的有 40% 不喜歡用什麼貝式定理的話可以把總數直接當成 100 人不過要記得搞清楚數字和 percent 所以持來回票又挾帶車子的有 20 人持來回票之中有帶車子的有 40% 所以這 20 人佔了持來回票的 40% ==> 持來回票的人有 50 人所以答案就是 50 % 這個問題討厭的就是要搞清楚分母分子在我指導美國人的時候 (雖然他也是很笨...呃說不太聰明好了) 他自己也是搞不清楚分子分母可見得 percent 的題目真的閱讀要仔細 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 69.86.175.71