MollynDiesel Posted August 24, 2012 Report Share Posted August 24, 2012 (edited) This is a question @Sparks got asked at 1 of his interviews so lets see who can answer it... You have 7 metal balls, all LOOK the same size and weight. 1 of the balls is broken and slightly lighter than the others. You have some weighing scales. You have to weigh all the balls atleast once...you have 2 chances to weigh the balls to find the broken 1 GO! Edited August 25, 2012 by Molly'n'Diesel Quote Link to comment Share on other sites More sharing options...
Removed #5 Posted August 24, 2012 Report Share Posted August 24, 2012 That's so easy that it's obvious and I'm not going to spoil the fun of watching others .... I'll PM you my answer. Quote Link to comment Share on other sites More sharing options...
MollynDiesel Posted August 24, 2012 Author Report Share Posted August 24, 2012 That's so easy that it's obvious and I'm not going to spoil the fun of watching others .... I'll PM you my answer. Haha yh dont spoil it its good to make people use their brains every now and then Quote Link to comment Share on other sites More sharing options...
NicTurtle Posted August 24, 2012 Report Share Posted August 24, 2012 Ok, I'll play. You weigh 3 balls on each side. If the scale is even you know the remaining two contain the broken ball and you weigh them to find which. If the scale wasn't even, you take the lighter side and weigh two of those ball. Why do I feel so dirty talking about balls? Perverted mind.... Quote Link to comment Share on other sites More sharing options...
Mollys Dad Posted August 24, 2012 Report Share Posted August 24, 2012 Ok, I'll play. You weigh 3 balls on each side. If the scale is even you know the remaining two contain the broken ball and you weigh them to find which. If the scale wasn't even, you take the lighter side and weigh two of those ball. Why do I feel so dirty talking about balls? Perverted mind.... If you put 3 on each side that makes 6 so you can't put 3 on each side and have a remaining 2. Also, You have 7 metal balls, all the same size and weight. 1 of the balls is broken and slightly lighter than the others. If they are all the same weight, how is 1 slightly lighter than the others? Quote Link to comment Share on other sites More sharing options...
Elyse Posted August 25, 2012 Report Share Posted August 25, 2012 lol - math problem wording FAIL! Let's see [spoilers ahead; ye been warned.] : Get rid of the weighing scale. Take each ball in your hand, and measure them that way. Find out 2 that weigh the same; the weight of those balls are the 'normal not broken' ones. Now, keep on doing that until you pick up the one that is lighter than all the others. Boom. Quote Link to comment Share on other sites More sharing options...
MollynDiesel Posted August 25, 2012 Author Report Share Posted August 25, 2012 If you put 3 on each side that makes 6 so you can't put 3 on each side and have a remaining 2. Also' date=' [i']You have 7 metal balls, all the same size and weight. 1 of the balls is broken and slightly lighter than the others. If they are all the same weight, how is 1 slightly lighter than the others? Stop nit picking please! It is 7 balls with 1 slightly lighter than the rest, is that better! Quote Link to comment Share on other sites More sharing options...
BingBlaze n Skyla Posted August 25, 2012 Report Share Posted August 25, 2012 Can't ujust look for the broken one? Lol Sent from my ST18i using Forum Runner Quote Link to comment Share on other sites More sharing options...
brian brown Posted August 25, 2012 Report Share Posted August 25, 2012 (edited) Easy, place 1 ball on each side, if they're the same, discount them. you'd only need to weigh 6 (3 x 2) because if they all weighed the same, the 7th would be the broken one Then you could weigh the broken one against one of the discounted ones to confirm it's lighter, meaning all the balls have been weighed at least once Edited August 25, 2012 by brian brown! Quote Link to comment Share on other sites More sharing options...
Sparks Posted August 25, 2012 Report Share Posted August 25, 2012 Can't ujust look for the broken one? Lol Sent from my ST18i using Forum Runner They all look exactly the same, the defect is so small you can't tell by weighing them by hand.... You can only use the weighing scales twice Some of you have it right, when you think of it, it's simple... But in an interview its not so easy lol Sent from my GT-I9100 using Xparent Cyan Tapatalk 2 Quote Link to comment Share on other sites More sharing options...
Sparks Posted August 25, 2012 Report Share Posted August 25, 2012 lol - math problem wording FAIL! Let's see [spoilers ahead; ye been warned.] : Get rid of the weighing scale. Take each ball in your hand, and measure them that way. Find out 2 that weigh the same; the weight of those balls are the 'normal not broken' ones. Now, keep on doing that until you pick up the one that is lighter than all the others. Boom. No can do, you have to use the scales... Sent from my GT-I9100 using Xparent Cyan Tapatalk 2 Quote Link to comment Share on other sites More sharing options...
brian brown Posted August 25, 2012 Report Share Posted August 25, 2012 If you can only use the scale twice total, 1,place 3 balls in each side, if they're level, the one that's left out is broken. 2, If one side is lighter than the other, discount the 3 heavier ones plus the odd one, then put one ball in each side. If they're level the one that's left out is broken, if not it'll be apparent which one's lighter Quote Link to comment Share on other sites More sharing options...
Removed #5 Posted August 25, 2012 Report Share Posted August 25, 2012 Very good Brian, my answer ( which I PM'd in 5 minues after the question was posted) is almost word for word the same as yours. I enjoy math word puzzles - but I also enjoy them when they're a bit more complicated than this was. Quote Link to comment Share on other sites More sharing options...
brian brown Posted August 25, 2012 Report Share Posted August 25, 2012 Very good Brian, my answer ( which I PM'd in 5 minues after the question was posted) is almost word for word the same as yours. I enjoy math word puzzles - but I also enjoy them when they're a bit more complicated than this was. Cheers Al, Basically what I'd already put further up but using 3 balls at a time (each side) instead of one I read the question last night when I was knackered & couldn't think straight, it was only this morning that I read it properly that I 'got' the question Quote Link to comment Share on other sites More sharing options...
Elyse Posted August 25, 2012 Report Share Posted August 25, 2012 But how....okay. I get it now. I was never good at math puzzles... Quote Link to comment Share on other sites More sharing options...
SaraB Posted August 25, 2012 Report Share Posted August 25, 2012 I got it pretty quick, but, man, that's a mean one to ask in an interview!!! Quote Link to comment Share on other sites More sharing options...
Emma Posted August 25, 2012 Report Share Posted August 25, 2012 I'm still confused and I've read the answer! Mind u I only got a 'd' in maths (my worst subject) and I've hit the bottle tonight so not much of anything make sense right now! Only reason I'm not talking gibberish is coz my lovely phone is correcting my typos. I love u phone! Ok, will stop waffling now and leave all you mathematical geniuses in peace x Quote Link to comment Share on other sites More sharing options...
Mollys Dad Posted August 25, 2012 Report Share Posted August 25, 2012 Did the person posing this question understand Fermat's theorem / combinatorics / Euler / polar co-ordinates Quote Link to comment Share on other sites More sharing options...
persephona Posted August 25, 2012 Report Share Posted August 25, 2012 Did the person posing this question understand Fermat's theorem / combinatorics / Euler / polar co-ordinates Hahahaha... I do think it's a strange question to ask at an interview... I do understand all the concepts above but I don't think this question has a lot to do with maths, more like creative thinking! I love maths! All the concepts above are fairly simple things compared to some of the stuff that is out there. Also there is more than just one Fermat's theorem lol... Quote Link to comment Share on other sites More sharing options...
Mollys Dad Posted August 25, 2012 Report Share Posted August 25, 2012 Hahahaha... I do think it's a strange question to ask at an interview... I do understand all the concepts above but I don't think this question has a lot to do with maths, more like creative thinking! I love maths! All the concepts above are fairly simple things compared to some of the stuff that is out there. Also there is more than just one Fermat's theorem lol... You do understand those? Join the club! Kindred spirit or nerd? Quote Link to comment Share on other sites More sharing options...
persephona Posted August 25, 2012 Report Share Posted August 25, 2012 If you can only use the scale twice total, 1,place 3 balls in each side, if they're level, the one that's left out is broken. 2, If one side is lighter than the other, discount the 3 heavier ones plus the odd one, then put one ball in each side. If they're level the one that's left out is broken, if not it'll be apparent which one's lighter Actually I don't think this answer is right to be honest, because at least one ball doesn't get to be weighed. And you have to weight them all at least once. (you do get to determine which one is lighter, but not all the conditions in the question are fulfilled) Actually I don't think it is possible to have all the balls weighed in this challenge with only two attempts! You have to weigh all the balls atleast once...you have 2 chances to weigh the balls to find the broken 1 Quote Link to comment Share on other sites More sharing options...
Mollys Dad Posted August 25, 2012 Report Share Posted August 25, 2012 the original question is badly worded... Quote Link to comment Share on other sites More sharing options...
persephona Posted August 25, 2012 Report Share Posted August 25, 2012 You do understand those? Join the club! Kindred spirit or nerd? I don't like to think of myself as a nerd lol... but I do love maths, especially the theoretical stuff. It's kind of a big part of my degree too So yeah, maybe kindred spirit? Ironically, I do need a calculator for adding up figures and percentages (to my shame, I can't really do any arithmetics at all). Quote Link to comment Share on other sites More sharing options...
Removed #2 Posted August 26, 2012 Report Share Posted August 26, 2012 What kind of job is he interviewing for? Quote Link to comment Share on other sites More sharing options...
brian brown Posted August 26, 2012 Report Share Posted August 26, 2012 (edited) Actually I don't think this answer is right to be honest, because at least one ball doesn't get to be weighed. And you have to weight them all at least once. (you do get to determine which one is lighter, but not all the conditions in the question are fulfilled) Actually I don't think it is possible to have all the balls weighed in this challenge with only two attempts! Yes it is possible, think about it: 1,place 3 balls in each side, if they're level, the one that's left out is broken. You can then place the broken one & a heavier one in either side to show that one is lighter - they've all been weighed at least once & you've used 2 attempts to prove it 2, If one side is lighter than the other, discount the 3 heavier ones plus the odd one, then put one ball in each side. If they're level the one that's left out is broken, if not it'll be apparent which one's lighter - place the lighter one & the 'odd' one from the beginning in each side to show that one's lighter & again they've all been weighed at least once same answer but with a bit more added. It meets the criteria just didn't need saying Edited August 26, 2012 by brian brown! Quote Link to comment Share on other sites More sharing options...
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