TrollMan Posted January 19, 2012 Report Share Posted January 19, 2012 There were 100 people present at a baseball-card show: 59 were men 72 were football- and baseball-card collectors 81 were college graduates 89 were right-handed What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates? Show your reasoning Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 19, 2012 Report Share Posted January 19, 2012 There were 100 people present at a baseball-card show: 59 were men 72 were football- and baseball-card collectors 81 were college graduates 89 were right-handed What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates? Show your reasoning Am I missing something here? Isn't this simple arithmetic (deduction actually)? It's 31, right? I didn't show my reasoning because for some reason I think I've reasoned incorrectly. That is a reasonable assumption based on reasonable doubt. Come on, let's be reasonable, if for no other reason than ... that's it, it is unreasonable to go on with these puns (i.e. no rhyme nor reason). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 19, 2012 Report Share Posted January 19, 2012 Oops, I just voted this down by mistake. Sorry Trollman. It was a slip of the click. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 19, 2012 Report Share Posted January 19, 2012 There are 41 women. (100-59) If all the women were dual card collectors, that means minimally 31 men are such collectors. (72-41) If all women are college grads, then minimally 40 men are college grads. (81-41) If all women are right handed, then minimally 48 men are right handed. (89-41) If all left handed men are college grads, then minimally 29 men are right handed college grads. (40-(59-48)) So we have 31 men who are dual card collectors and 29 men who are right handed college grads. The minimum overlap of these two sets within a total of 59 men is 1. So the answer is 1. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 19, 2012 Report Share Posted January 19, 2012 This looks suspiciously like homework... Quote Link to comment Share on other sites More sharing options...
0 TrollMan Posted January 19, 2012 Author Report Share Posted January 19, 2012 (edited) well, IQ154, with a name like mine who wouldnt downrate me Homework? Are you saying you didnt enjoy this puzzle!? sditty, good job. Edited January 20, 2012 by TrollMan Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 There were 100 people present at a baseball-card show: 59 were men 72 were football- and baseball-card collectors 81 were college graduates 89 were right-handed What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates? Show your reasoning There are 41 women. Assuming all women are part of all groups. The least number of men that satisfy all the criteria is min of ((72-41), (81-41),(89-41)). Which makes it 31 men who were both football- and baseball-car collectors and right-handed college graduates. Quote Link to comment Share on other sites More sharing options...
0 bhramarraj Posted January 20, 2012 Report Share Posted January 20, 2012 (edited) No. of total women = 100 - 59 = 41 If all the 41 women had been football- & baseball-card collectors, then least no. of men who are football & baseball collectors = 72 - 41 = 31. Similarly least no. of men who are college graduates = 81 - 41 = 40. ------do------------------------------- right handed = 89 - 41 = 48 Then from above it is clear that 31 is the least no. of men who are both football- & baseball- card collectors and righthanded college graduates. Edited January 20, 2012 by bhramarraj Quote Link to comment Share on other sites More sharing options...
0 hhh3 Posted January 20, 2012 Report Share Posted January 20, 2012 (edited) 31 Edited January 20, 2012 by hhh3 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 59 men from 100 leaves 41 persons present who are not men (but may be women or children). Assuming each of the 41 (women or children) were card collectors, college graduates and right-handed, then 72 - 41 = 31 card collectors who were men, 81 - 41 = 40 college graduates were men, and 89 - 41 = 48 men were right-handed. If 27 men were right-handed college graduates but not card collectors, 20 men were right-handed card collectors but not college graduates, 10 men were card collectors and college graduates but not right-handed, and 1 man who was a college graduate but neither a card collector nor right-handed, then there would be 1 man who was a right-handed card collector and college graduate -- the minimum possible number (given the givens). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 I looked at it backwards. A maximum of 28 men could be non-card collectors, a maximum of 19 men (who could all be different than the 28 non-card collectors) could be non-graduates, and a maximum of 11 men (who could all be different than the 28 non-card men and 19 non-grad men) could be left-handed. 28 + 19 + 11 = 58. So at least one man must be a right-handed card collecting graduate. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 Would it not be asier to think that the MINIMUM number of right handed college grads, that are dual card collectors, would be 1. This is because the MINIMUM number of people in total can only be one. It easier to think first without using maths. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 Ok, so. The lease number to have EVERY condition. So because ALL conditions need to be met, we are adding finding out how many DONT have degrees, are NOT right handed, and DONT collect both and subtracting them from 100. if 59 are men the minimum number of be men is 59. Both men and baseball and football card collectors, so: 59 - (100-72) = 31 (the minumum number who are both men and collect BOTH kinds of cards) Men, collectors of both cards and college graduates: 59 - (100-72) - (100-81*grads*) = 12 (The minumum number of men who are collectors of BOTH cards AND grads) Men who collect both cards, are grads AND right handed: 59 - (100-72) - (100 - 81) - (100 - 89) = 1 So what I am saying is that 41 people were NOT men 28 people were NOT collectors of both 19 people were NOT graduates 11 people were NOT right handed And they are not all the same people so 41 + 28 + 19 + 11 = 99 (the possible number of people who dont meet ALL the conditions) Therefor 100 - 99 = 1 (only one person must be all 4 of those conditions), Though most of them fulfil 3 of the 4 criteria in this scenario. Statisitacically the avererage answer is 30.62 people fulfilled all of those conditions on the day, but as we know, statistics is the science of guessing so for sure its possible it could have been 59 of them (the lowest single condition needed). If the question was OR not and, it becomes a whole lot harder... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 if all the women were collectors of both kinds of cards, then there would only be 31 men who were collectors of both types of cards if all the women were right handed, there would be 48 right handed men if all the women were college graduates, there would be 40 college graduate men within the 48 men that were right handed, there would only have to be 20 who were collectors of both types and 29 who were college graduates if the 11 left handed men met those two conditions due to the fact that 20 + 29 is only 49, only one man has to meet all three conditions for the statistics to be satisfied Therefore, the answer is 1 Quote Link to comment Share on other sites More sharing options...
0 Peekay Posted January 20, 2012 Report Share Posted January 20, 2012 The best way to do this is to take the Negative of the stated conditions: Not Men = 100-59 = 41 Not FBCC = 100-72 = 28 Not CG = 100-81 = 19 Not RH = 100-89 = 11 So, a total of 99 people might NOT be have one or more of the above characteristics. Which leaves us with a MINIMUM number of 1 person who will definitely have the above characteristics. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 <p>There is 59 men, out of a 100 people would leave 41 woman. For each scenario the amount of woman has to be subtracted to give the least amount of men in each scenario the results being 31 least amount of men for card collectors, 40 least amount of men for college grads, 48 least amount of men right handed. The question ask for the least amount of men between the 2 categories of football and baseball car collectors and right handed men, which is comparing the results of 31 and 48. One might concluded that 31 one is the least amount, but 31 only works for the collectors not the right handed category, so 48 is answer.</p> Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 Wow, I can't believe the number of wrong answers on here Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted January 20, 2012 Report Share Posted January 20, 2012 Yes, that shows that it's a good puzzle--it was clearly stated, but still people got different answers. A lot of us learned something here. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 20, 2012 Report Share Posted January 20, 2012 There were 59 men, Let us assume there were 28 men who were not sports card collectors, leaving 31 men Of the remaining men, 19 were not college graduates, leaving just 12 men 11 of the remaining men happened to be left handed, leaving a minimum of 1 man who met none of the other criteria Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 21, 2012 Report Share Posted January 21, 2012 (edited) Not men = 100 - 59 = 41 Least no of men who are card collectors = 72 - 41 = 31 Out of these 31 men, there may be (100 - 81 = 19) non-college grads; thus least no of men who are card collectors and college grads = 31 - 19 = 12 Out of these 12 men, there may be (100 - 89 = 11) who are left-handed; thus least no of men who are card collectors and college grads and are right-handed = 12 - 11 = 1 Edited January 21, 2012 by Sheila Chong Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 21, 2012 Report Share Posted January 21, 2012 59 men, 41 women 72 are card collectors. Worst case, assume all 41 women are card collectors. So, 31 men are card collectors and 28 are not. 81 are college graduates. Worst case, all women and non-card collectors are college graduates. So, 41+28=69 graduates are either women or dont collect cards. Remaining 12 (81-69) are graduate men who collect cards. 89 are right handed. Worst case, consider remaining 88 (100-12) are right handed. So, only one out of 12 is a graduate man who collects cards and is right handed. Answer: At least 1. Is this correct? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 21, 2012 Report Share Posted January 21, 2012 Wow, guess when you get old the brain doesn't work so well.....this is my first try at this so, maybe in about 50 more I might get it. Quote Link to comment Share on other sites More sharing options...
0 TrollMan Posted January 21, 2012 Author Report Share Posted January 21, 2012 (edited) ^ slow and steady wins the race =) 59 men, 41 women 72 are card collectors. Worst case, assume all 41 women are card collectors. So, 31 men are card collectors and 28 are not. 81 are college graduates. Worst case, all women and non-card collectors are college graduates. So, 41+28=69 graduates are either women or dont collect cards. Remaining 12 (81-69) are graduate men who collect cards. 89 are right handed. Worst case, consider remaining 88 (100-12) are right handed. So, only one out of 12 is a graduate man who collects cards and is right handed. Answer: At least 1. Is this correct? Yes, you got it! Edited January 21, 2012 by TrollMan Quote Link to comment Share on other sites More sharing options...
Question
TrollMan
There were 100 people present at a baseball-card show:
59 were men
72 were football- and baseball-card collectors
81 were college graduates
89 were right-handed
What is the least number of men who were both football- and baseball-car collectors and right-handed college graduates?
Show your reasoning
Link to comment
Share on other sites
22 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.