Guest Posted March 31, 2010 Report Share Posted March 31, 2010 (edited) 90 logicians went to a restaurant. 3 of them had a pork pie, beer, and spotted dick. 24 of them had a pork pie; 5 of them had a pork pie and beer; 33 had beer; 10 had a beer and spotted dick; 38 had spotted dick; 8 had spotted dick and a pork pie. How many of the logicians had nothing? Edited March 31, 2010 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 14 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 90 logicians went to a restaurant. 3 of them had a pork pie, beer, and spotted dick. 24 of them had a pork pie; 5 of them had a pork pie and beer; 33 had beer; 10 had a beer and spotted dick; 38 had spotted dick; 8 had spotted dick and a pork pie. How many of the logicians had nothing? ------------------------------------------------------------------------------------------ 24 pork pie + 33 beer + 38 spotted dick = 95 = (pork only) + (beer only) + (spot only) + 2*((pork/spot) + (pork/beer) + (beer/spot)) + 3*(pork/beer/spot) 95 - 5 pork/beer - 10 beer/spot - 8 spot/pork = 72 = (pork only) + (beer only) + (spot only) + (pork/spot) + (pork/beer) + (beer/spot) 72 + 3 pie/beer/spot = 75= (pork only) + (beer only) + (spot only) + (pork/spot) + (pork/beer) + (/beer/spot) + (pork/beer/spot) 90 -75 = 15 who had nothing Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 Number that had something = PP + B + SD - (PP and B) - (PP and SD) - (B and SD) + (PP and B and SD) = 24 + 33 + 38 - 5 - 8 - 10 + 3 = 75. So 90-75 = 15 had nothing. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 I solved this by using Venn Diagrams - stuff I learned when I was about 13. It was called 'modern mathematics' then. My answer unsurprisingly agrees with the above correspondents (except the first one) See attachment for diagramsolution.doc Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 15 are not having anything out of PP,bear,sd. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 I'm getting 15 by venn diagram. but why would anyone eat a spotted dick?! on second thought, I don't want to know.... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 34 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 34 We start with 90 total logicians. We know that 24 had pork pie (PP), 33 had beer (B) and 38 had spotted dick (SD). We also know that 5 had both PP and B, 10 had B and SD, 8 had PP and SD and finally 3 had all three PP, B and SD. Let us start by assuming that the 24 that had PP could also have had the B and/or SD, also that the 33 B could have had PP and/or SD and similarly that the 38 SD could have had PP and/or B. Since it looks like 95 people had at least the PP, B or SD (24 PP + 33 B + 38 SD) there must be some that are counted multiple times so let's start eliminating the extraneous counts. There are 10 people that are counted both in the 33 B and the 38 SD so we must eliminate them from one of the counts. (24 PP + 33 B + (38 SD - 10 B,SD)) = 85 There are also 8 people that are counted both in the 24 PP and the 38 SD so we must eliminate them from one of the counts while keeping the previous eliminations. (24 PP + 33 B + (38 SD - 10 B,SD - 8 PP,SD)) = 77 There are also another 5 people counted twice in each of the 24 PP and 33 B which must be eliminated from one of the counts while keeping the previous eliminations. (24 PP + (33 B - 5 PP,B) + (38 SD - 10 B,SD - 8 PP,SD)) = 72 Finally there are 3 people who are counted three times each in the 24 PP, 33 B and 38 SD which must be eliminated from two of the counts while again keeping the previous eliminations. (24 PP + (33 B - 5 PP,B - 3 PP,B,SD) + (38 SD - 10 B,SD - 8 PP,SD - 3 PP,B,SD)) = 66 We now have the true count of who had any of the pork pie (PP), beer (B) and spotted dick (SD) as being 66 of the 90 logicians. It is now a simple matter of subtraction to determine who did not eat any of the three foods. (90 logicians - 66 PP|B|SD) = 34 Therefore, 34 of the 90 did not eat any of the pork pie, beer and spotted dick. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 Edit to my earlier post: adding Venn Diagram solution PP circle contains 24 total which is split into 8 PP only, 5 PP and B, 8 PP and SD and 3 PP, B and SD SD circle contains 38 total which is split into 17 SD only, 8 PP and SD, 10 B and SD and 3 PP, B and SD B circle contains 33 total which is split into 15 B only, 5 PP and B, 10 B and SD and 3 PP, B and SD Add each segment together and you have (8 PP + 15 B + 17 SD + 5 PP,B + 8 PP,SD + 10 B,SD + 3 PP,B,SD) = 66 (90 - 66) = 34 not in either circle Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 31, 2010 Report Share Posted March 31, 2010 uh i used venn diagram too so i have no idea how they got 34... and i did my math wrong its actually 15 oops Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 1, 2010 Report Share Posted April 1, 2010 34 We start with 90 total logicians. We know that 24 had pork pie (PP), 33 had beer (B) and 38 had spotted dick (SD). We also know that 5 had both PP and B, 10 had B and SD, 8 had PP and SD and finally 3 had all three PP, B and SD. Let us start by assuming that the 24 that had PP could also have had the B and/or SD, also that the 33 B could have had PP and/or SD and similarly that the 38 SD could have had PP and/or B. Since it looks like 95 people had at least the PP, B or SD (24 PP + 33 B + 38 SD) there must be some that are counted multiple times so let's start eliminating the extraneous counts. There are 10 people that are counted both in the 33 B and the 38 SD so we must eliminate them from one of the counts. (24 PP + 33 B + (38 SD - 10 B,SD)) = 85 There are also 8 people that are counted both in the 24 PP and the 38 SD so we must eliminate them from one of the counts while keeping the previous eliminations. (24 PP + 33 B + (38 SD - 10 B,SD - 8 PP,SD)) = 77 There are also another 5 people counted twice in each of the 24 PP and 33 B which must be eliminated from one of the counts while keeping the previous eliminations. (24 PP + (33 B - 5 PP,B) + (38 SD - 10 B,SD - 8 PP,SD)) = 72 Finally there are 3 people who are counted three times each in the 24 PP, 33 B and 38 SD which must be eliminated from two of the counts while again keeping the previous eliminations. (24 PP + (33 B - 5 PP,B - 3 PP,B,SD) + (38 SD - 10 B,SD - 8 PP,SD - 3 PP,B,SD)) = 66 We now have the true count of who had any of the pork pie (PP), beer (B) and spotted dick (SD) as being 66 of the 90 logicians. It is now a simple matter of subtraction to determine who did not eat any of the three foods. (90 logicians - 66 PP|B|SD) = 34 Therefore, 34 of the 90 did not eat any of the pork pie, beer and spotted dick. You made a mistake friend. The three who have had all SD, PP, and B were counted 3 times when you added 24 PP + 33 B + 38 SD however they were also subtracted 3 times when you subtracted -10 B,SD -8 PP,SD - 5 PP,B Since they were added 3 times and then subtracted 3 times it is as if they were never counted at all. You must therefore add them back again as a lone group. + 3 SD,PP,B So you then have (24+33+38) -(10 + 8 + 5) + (3) = 75 Which gives an answer of 15. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 1, 2010 Report Share Posted April 1, 2010 34 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 1, 2010 Report Share Posted April 1, 2010 (edited) Venn diagrams are helpful in this but logic also enters the "equation". Edited April 1, 2010 by GPL88 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 1, 2010 Report Share Posted April 1, 2010 Venn diagrams are helpful in this but logic also enters the "equation". Care to explain how? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 1, 2010 Report Share Posted April 1, 2010 uh i used venn diagram too so i have no idea how they got 34... and i did my math wrong its actually 15 oops with the venn diagrams. I forgot to split the 3 PP, B & SD people from the PP & B, PP & SD, and B & SD groups which results in: PP total = 24 = 14 PP + 2 PP & B + 5 PP & SD + 3 PP, B & SD B total = 33 = 21 B + 2 PP & B + 7 B & SD + 3 PP, B & SD SD total = 38 = 23 SD + 5 PP & SD + 7 B & SD + 3 PP, B & SD 14 PP + 2 PP & B + 5 PP & SD + 3 PP, B & SD + 21 B + 7 B & SD + 23 SD = 75 PP|B|SD 90 L - 75 PP|B|SD = 15 ¬PP|B|SD You made a mistake friend. The three who have had all SD, PP, and B were counted 3 times when you added 24 PP + 33 B + 38 SD however they were also subtracted 3 times when you subtracted -10 B,SD -8 PP,SD - 5 PP,B Since they were added 3 times and then subtracted 3 times it is as if they were never counted at all. You must therefore add them back again as a lone group. + 3 SD,PP,B So you then have (24+33+38) -(10 + 8 + 5) + (3) = 75 Which gives an answer of 15. I see now where I messed up. I can't believe I missed the fact that the PP, B & SD group was counted among the PP, B, SD, PP & B, PP & SD and B & SD groups. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 6, 2010 Report Share Posted April 6, 2010 90 logicians went to a restaurant. 3 of them had a pork pie, beer, and spotted dick. 24 of them had a pork pie; 5 of them had a pork pie and beer; 33 had beer; 10 had a beer and spotted dick; 38 had spotted dick; 8 had spotted dick and a pork pie. How many of the logicians had nothing? 90-3=87. 87-2=85. 85-7=78. 78-9=70. 70-11=59. 59-18=41. 41-18=23. 23 had nothing. But- there is another solution! It is unknown! For others could have eaten other stuff! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 7, 2010 Report Share Posted April 7, 2010 (edited) 15 did not have pork pie, beer nor spotted dick. Though maybe they had something else. Edited April 7, 2010 by Eragon Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 7, 2010 Report Share Posted April 7, 2010 Is it 24??? Or did all of them have something to eat?? Quote Link to comment Share on other sites More sharing options...
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90 logicians went to a restaurant.
3 of them had a pork pie, beer, and spotted dick. 24 of them had a pork pie; 5 of them had a pork pie and beer; 33 had beer; 10 had a beer and spotted dick; 38 had spotted dick; 8 had spotted dick and a pork pie.
How many of the logicians had nothing?
Edited by K SenguptaLink to comment
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