[計量] OG的168與175題

看板GMAT (GMAT入學考試)作者 (大寶與困寶)時間16年前 (2008/07/22 16:35), 編輯推噓1(1013)
留言14則, 1人參與, 最新討論串1/1
題目是這樣的 168.An airline passenger is planning a trip that involves three connecting flights that leaves from Airports A, B, and C,respectively. The first flight leaves Airport A every hour, beginning at 8:00 a.m., and arrives at Airport B 2 and 1/2 hours later. The second flight leaves Airport B every 20 minutes, beginning at 8:00 a.m., and arrives at Airport C 1 and 1/6 hours later. The third flight leaves Airport C every hour, beginning at 8:45 a.m. What is the least total amount of time tha passenger must spend between flights if all flights keep to theur schedules? Ans:1 hr 15 minutes 這題其實題目有點看不太懂..外加解答看的霧煞煞.. 另一題175. A square countertop has a square tile inlay in the center, leaving an untiles strip of uniform width around the tile. If the ratio of the tiled area to the untiled area is 25 to 39, which of the following could be the width, in inches , of the strip? Ans:1.3.4 這題也是題目看不太懂@@....麻煩大家了 謝謝:) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 123.193.252.16

07/22 16:53, , 1F
168. A->B->C 三段航程聯票 就是A出發到B B出發到C
07/22 16:53, 1F

07/22 16:54, , 2F
請問旅客所花費的"候機"時間?
07/22 16:54, 2F

07/22 16:55, , 3F
航站A 每小時一班飛機飛B
07/22 16:55, 3F

07/22 16:56, , 4F
航站B 每20分鐘一班飛機飛C
07/22 16:56, 4F

07/22 16:59, , 5F
假設蝙蝠俠 在航站A搭乘早上8:00AM經過2 1/2小時到航B
07/22 16:59, 5F

07/22 17:00, , 6F
他到達航站B 時間為10:30 所以蝙蝠俠比須等待 10分鐘
07/22 17:00, 6F

07/22 17:00, , 7F
搭乘10:40的飛機到航站C
07/22 17:00, 7F

07/22 17:02, , 8F
蝙蝠俠花了1 1/6到達航站C 到達航站C時間為11:50
07/22 17:02, 8F

07/22 17:02, , 9F
小時
07/22 17:02, 9F

07/22 17:04, , 10F
然後蝙蝠俠必須等12:45離開C航廈的班機
07/22 17:04, 10F

07/22 17:05, , 11F
因為C航站從 8:45起 每一小時發一班飛機
07/22 17:05, 11F

07/22 17:05, , 12F
所以蝙蝠俠必須等待 55分鐘的候機時間~
07/22 17:05, 12F

07/22 17:06, , 13F
所以 10 + 55 = 65 = 1 hour 5 minutes
07/22 17:06, 13F

07/22 17:08, , 14F
*補充* A->B->C 三段航程聯票=A到B B到C C再到某地!
07/22 17:08, 14F
文章代碼(AID): #18XPlPD3 (GMAT)
文章代碼(AID): #18XPlPD3 (GMAT)