Re: [計量] gwd ps

看板GMAT作者 (柏拉圖的永恆...)時間16年前 (2008/08/29 14:46), 編輯推噓1(101)
留言2則, 2人參與, 最新討論串4/4 (看更多)
※ 引述《flac (老獅子)》之銘言: : 請教一題GWD數學 : If x, y, and k are positive numbers such that [x/(x+y)](10) + [y/(x+y)](20) = k : and if x < y, which of the following could be the value of k? : A. 10 : B. 12 : C. 15 : D. 18 : E. 30 : 答案為D : 謝謝 原式 = 10X 20Y 10X+20Y 10X+10Y 10Y 10Y ----- + ----- = --------- = -------- + ----- = 10 + ----- = K X+Y X+Y X+Y X+Y X+Y X+Y We get 10Y ----- = K-10 X+Y (A) if K = 10 , then Y = 0 , which contradicts with the premise " y is positive " (B) if K = 12 , then 10Y ----- = 2 , and we get X = 4Y , which contradicts X+Y with the premise " X<Y " (C) if K = 15 , then 10Y ----- = 5 , and we get X = Y , which contradicts X+Y with the premise " X<Y " (D) if K = 18 , then 10Y ----- = 8 , and we get Y = 4X , it could be the answer X+Y (E) if K = 30 , then 10Y ----- = 20 , and we get Y = -2X , which contradicts X+Y with the premise " X and Y are both positive " Then obviously, (D) is the best answer. # -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.31.205.1

08/29 14:57, , 1F
感激啦 很清楚 大家都挺利害的
08/29 14:57, 1F

08/29 23:20, , 2F
老實說 這個推論想法比較好
08/29 23:20, 2F
文章代碼(AID): #18jvj6TR (GMAT)
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文章代碼(AID): #18jvj6TR (GMAT)