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题库 / GMATLA-PS-32
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?
~$33\frac{1}{3}$~%
40%
50%
60%
~$66\frac{2}{3}$~%
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看清楚题目的表达，第一个条件是and，就是总乘客里的20%带车往返，第二个条件是有往返票的乘客有60%是不带车的，所以往返乘客中带车的的比例是40%，由此可知，往返票占总数的50%
总人数*20% （往返票且带车）=40%*往返票人员 （带车），那么往返票人员是总人员的一半
If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, so 40 percent took
0.2x=0.4y y/x=0.5
If 60 percent of the passengers with round-trip tickets ，一定要注意of后面接的群体。
理解题目！
写错，是a/x=0.2/0.4=50%
设总乘客数为x,持往返票的人数为a,则有0.2x+0.6a=a,则有a/x=0.2/0.5=50%
ratio=20%/（1-60%）
错因：掉进坑，没明白题意
【韦恩图】
20%的ship passengers,60%的round-trip passengers注意题目的坑
60 percent of the passengers（ with round-trip tickets）括号里是修饰成分修饰passengers，搞错了
20%的ship passengers,60%的round-trip passengers注意题目的坑
找出一个群体(round trip ticket w/ cars)能用不同未知数代表的，它占人口总数20%,同时占round trip人口总数 (1-60%)，等式成立
40%* R=20*all passengers
要看清分母是谁！
令y=with round trip ticket的，a=whole people； 所以roundtrip ticket with car 0.4y=0.2a； 所以y=0.5a， 所以就是50%