tag:blogger.com,1999:blog-4679715190635795973.post1807687890278713310..comments2018-09-13T20:33:45.972-07:00Comments on y of x: Gruesome, but FunUnknownnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4679715190635795973.post-66543041604980688732009-06-12T13:07:33.454-07:002009-06-12T13:07:33.454-07:00Here's Ken's visual solution. I did the pr...<a href="http://picasaweb.google.com/lh/photo/0_nEPkPQwuPUwn2xYQINNw?authkey=Gv1sRgCJuZsvPIkKiJEg&feat=directlink" rel="nofollow">Here's</a> Ken's visual solution. I did the problem exactly the same wayHMhttps://www.blogger.com/profile/12859709233583742208noreply@blogger.comtag:blogger.com,1999:blog-4679715190635795973.post-19473354375918806112009-06-11T03:48:40.378-07:002009-06-11T03:48:40.378-07:00Assume, w/o loss of generality, that we are dealin...Assume, w/o loss of generality, that we are dealing w/ 100 people.<br /><br />Since .7 have wound #1, .75 #2, .8 #3 & .85 #4, there are 70+75+80+85 = 310 wounds. The way to minimize the number who have all four wounds is to "give" three wounds to as many people as possible.<br /><br />Let this number of people be called u. Let v stand for the number of people that have four wounds.<br /><br />3u + v = 310<br /><br />or<br /><br />u = (310-v)/3<br /><br />The smallest v that generates an integral u (which also must be <= 100) is 10. So, 10 out of the 100, i.e. 10%, have all four wounds.Tinyc Timhttps://www.blogger.com/profile/01754672611765154678noreply@blogger.comtag:blogger.com,1999:blog-4679715190635795973.post-25103348850417741312009-06-10T10:14:38.530-07:002009-06-10T10:14:38.530-07:00Number of people with NO eye injury 30%
Number of ...Number of people with NO eye injury 30%<br />Number of people with NO ear injury 25%<br />Maximum number of people with NEITHER head injury 55%<br /><br />Number of people with NO arm injury 20%<br />Number of people with NO leg injury 15%<br />Maximum number of people with NEITHER appendage injury 35%<br /><br />Maximum number of people with NONE of the four injuries 55%+35% = 90%<br />Minimum number of people with ALL four injuries 100%-90%=10%Ken Colangelonoreply@blogger.comtag:blogger.com,1999:blog-4679715190635795973.post-90685440380565964022009-06-10T09:21:34.170-07:002009-06-10T09:21:34.170-07:00I did this one visually and came out with 10%.I did this one visually and came out with 10%.Ken Colangelonoreply@blogger.com