2006-07-30 16:21:24 · 14 answers · asked by Bobby T 1 in Science & Mathematics ➔ Mathematics
10
give 70 figs to 70 people.
Then give 30 pears to everyone else.
You have 45 pears left so at least 45 have to have a fig and a pear.
Now give 55 plumbs to everyone that doesn't have a plumb and a pear.
You have to give the remaining 25 plumbs to people who already have a fig and a pear.
Now you can give 75 toys to everyone who doesn't have a plumb, pear, and fig.
The remaining 10 toys have to go to people with plums pears and figs.
In the end you have
10 with all 4
15 with a fig and pear and plumb
20 with a fig and pear and toy
25 with a fig and plumb and toy
30 with a pear and plumb and toy
2006-07-30 16:26:45 · answer #1 · answered by bogusman82 5 · 0⤊ 0⤋
First of all, Broncos is a jackass who doesn't understand the complexity of the problem.
Now...to the problem itself. To be honest, I'm not sure if I can give you the exact answer, but I think I can start down the right trail.
Let's say that the first 70 people own figs, and the last 75 own pears. That leaves an "overlap" of 45 people that own both of those. In other words, persons #1-#70 own figs and #26-#100 own pears, so only persons #26-#70 own both.
Now, let's say that we split the plums between the beginning and end of the line - persons #1-#40 and #61-#100 own plums. What we find now is that persons #41-#60 DON'T own plums, so that takes 20 more people out of the running, leaving us with 25 people who own all three: Persons #26-#40 (15) and #61-#70 (10).
So, we're left with the toys to disperse in such a way as to minimize the number of people with all four. Since we know we only have 25 that own all three, there are 75 people that we can give toys to before creating a situation where somebody has all four. So, if we give the toys to those 75 first, we are only left with 10 toys that must be given to some of the 25 people who already have the other three.
So, your answer would be 10 people.
There's probably a formula that could be used to figure this out, but I couldn't tell you what it would be.
LATE NOTE:
I see that John and Jim looked at it from the other side and figured it out more quickly. Nicely done.
2006-07-30 23:40:46 · answer #2 · answered by firemedicgm 4 · 0⤊ 0⤋
100-85=15 did not get toys. similarly, 20 did not get plums., 25 did not get pears and 30 did not get figs. So, 100-(15+20+25+30) must own all four. That is 10 only.
2006-07-31 07:55:26 · answer #3 · answered by sharanan 2 · 0⤊ 0⤋
10. 15 people don't own toys. 25 don't own plums, 25 don't own pears, 30 don't own figs. If there are no duplicates, that's a max of 90 who don't own something, so the min who own all four is 100 - 90.
2006-07-30 23:34:48 · answer #4 · answered by Jim H 3 · 0⤊ 0⤋
4 is the least but the least to be sure is 85+80+75+1=241
blah blah blah
241
2006-07-30 23:28:19 · answer #5 · answered by Croasis 3 · 0⤊ 0⤋
10
100-85=15
100-80=20
100-75=25
100-70=30
15+20+25+30=90
so 90 people can have 3.
2006-07-30 23:28:37 · answer #6 · answered by justnotright 4 · 0⤊ 0⤋
Zero
2006-07-30 23:28:18 · answer #7 · answered by Matthew S 2 · 0⤊ 0⤋
70 people?
2006-07-31 00:02:12 · answer #8 · answered by matt 3 · 0⤊ 0⤋
i guess 70. its been a while
2006-07-30 23:25:23 · answer #9 · answered by viper3ez 2 · 0⤊ 0⤋
get a ****** life loser whats 1 +1 fooled you huh
2006-07-30 23:24:00 · answer #10 · answered by Anonymous · 0⤊ 0⤋