关于GMAT数学题的问题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?在一次横渡大西洋中,20%的乘客有round-trip然后带着他们的车上船.如果60%的乘客有round-trip没有带车上船,问:有多少百分比的乘客有round-trip?a.33.3333333% b.40% c.50% d.60% e.66.6666666666666%机经上的一道题

问题描述:

关于GMAT数学题的问题
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?
在一次横渡大西洋中,20%的乘客有round-trip然后带着他们的车上船.如果60%的乘客有round-trip没有带车上船,问:有多少百分比的乘客有round-trip?
a.33.3333333% b.40% c.50% d.60% e.66.6666666666666%
机经上的一道题~~~做不来
Freiheit1218提到:持有往返票(round-trip)并带着车的占总乘客数的20%;所有持有往返票的乘客中,没带车的占60%。
那么也就是说:持有往返票并带着车的乘客占总乘客数的(1-60%)
也就是说题目其实在问实际乘船人数和理论乘船人数的百分比是伐!?

看不懂

20%/(1-60%)=50%

c1.理解题意条件:a船上所有乘客中的百分之二十持有双程票并带自己的车上船b持有双程票的所有乘客中的百分之六十没有带自己的车上船提问:船上所有乘客中的百分之几的持有双程票?2.分析由条件b,推得所有持有双程票的...

翻译的也不大对头,应该是:持有往返票(round-trip)并带着车的占总乘客数的20%;所有持有往返票的乘客中,没带车的占60%。
那么也就是说:持有往返票并带着车的乘客占总乘客数的(1-60%)
所以:
20%除以(1-60%)=50%
选C