求助10月JJ 56

bobomomo

56. 郑和号横越大西洋，20%的乘客是既有返程票又有带车上船，知道有返程票的乘客中60%没有带车上船。问：一共多少乘客买了返程票？

解法：有返程票又没有带车上船的人占总乘客的百分比为X，则X/60%=20%/40%，X=30%，故买了返程票的乘客（包括带车和没带车上船的）共有30%+20%=50%

(有返程票无带车/总的数量)/(无带车的/有返程票的)=(有返程票有带车/总的数量)/(有带车的/有返程票的) - 这是狗狗的解释

但我还是不太清楚 这T 想的头都快炸了 笨死了 还是想不出来 这种T 有什么 很明确的思路在里面吗？

horsefish

20%X=(1-60%)Y>>>y/x=50%; X:所有passengers, Y: round-trip

tiger1234

20% of ppl have round tickets and cars (we will call them rc), and we know that out of all the ppl who have round tickets(we will call them r) 60% of r did not bring their cars(we will call these ppl rnc);

so for all those ppl who have round tickets, they either brought their cars (rc) or they did not (rnc), so rc + rnc = r, and we know that rnc = 60% * r = .6r so we know that rc = .4 * r, and we also know that 20% of the ppl on the ship are ppl with round tickets and cars(rc) so we know that rc = .2 of overall ppl on the ship, since rc = .4r as well => .4r = .2 of overall ppl on the ship => r = 1/2 of the overall ppl on the ship

I can't type Chinese right now, sorry about that

blackhorse

如果再解释简单点，就是你要明白这句话的意思“20%的乘客是既有返程票又有带车上船，知道有返程票的乘客中60%没有带车上船”

它的意思就是：有返程票的乘客中60%没有带车上船，所以 带票且带车的人＝带票总数的（1－60％）＝40％。

总共有20％的人带票且带车，所以带票的人占总数的＝20％/40%＝50％。

这样明白了吗？

bobomomo

谢谢马鱼 JJ 谢谢 TIGER 谢谢黑马

早上一起来 什么都没做 就想看这T 终于弄明白了

谢谢