Say that 70% have lost an eye, 75% an ear, 80% an arm, 85% a leg. What percentage, at least, must have lost all four?Found this one in Herstein's Abstract Algebra. pic wikimedia
Monday, June 8, 2009
Subscribe to:
Post Comments (Atom)
The Unofficial Tunxis Math Department Blog
Say that 70% have lost an eye, 75% an ear, 80% an arm, 85% a leg. What percentage, at least, must have lost all four?Found this one in Herstein's Abstract Algebra. pic wikimedia
4 comments:
I did this one visually and came out with 10%.
Number of people with NO eye injury 30%
Number of people with NO ear injury 25%
Maximum number of people with NEITHER head injury 55%
Number of people with NO arm injury 20%
Number of people with NO leg injury 15%
Maximum number of people with NEITHER appendage injury 35%
Maximum number of people with NONE of the four injuries 55%+35% = 90%
Minimum number of people with ALL four injuries 100%-90%=10%
Assume, w/o loss of generality, that we are dealing w/ 100 people.
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.
Let this number of people be called u. Let v stand for the number of people that have four wounds.
3u + v = 310
or
u = (310-v)/3
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.
Here's Ken's visual solution. I did the problem exactly the same way
Post a Comment