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If 3 cats eat 3 mice in 3 minutes; how many cats for 100 mice in 100 minutes?

  1. Nov 22, 2004 #1
    i know this seems simple, and perhaps it's a trick, but i'm confused! is the answer 100?
  2. jcsd
  3. Nov 22, 2004 #2

    Math Is Hard

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    well, let's see, it seems it takes each cat three minutes to eat a mouse...

    must be a trick, my cat can't finish a mouse in less than 5 minutes... :rofl:
  4. Nov 22, 2004 #3


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    No it's not. The cats are all eating at the same time, aren't they ? Just picture a row of cats eating :yuck: mice. They all start at the same time, and finish at the same time. How much time has elapsed ?
  5. Nov 22, 2004 #4
    3 minutes.
  6. Nov 22, 2004 #5
    how many mice can one cat eat in 100 minutes? 1 cat eats 100/3 mice. 100mice/(100/3micepercat)=3cats
  7. Nov 22, 2004 #6


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    Well, that depends on if it still counts if a cat vomits up previous mice and continues eating, or if you have to factor in cat stomach volume, or how much time is lost to vomiting up previously eaten mice.
  8. Nov 22, 2004 #7
    I didn't think of the cat vomiting.
    I know I could eat you all day and have no problem whatsoever.
  9. Nov 23, 2004 #8


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    I doubt the author of the original question considered it either. It's a good proof of how out of touch mathematicians are with reality. :biggrin: They think just because 3 cats can eat 3 mice in 3 minutes, those same cats can just keep eating at the same rate and finish off 100 mice in 100 minutes. But, that's 33 1/3 mice per cat in 100 minutes, so it raises a lot of problems in reality. 33 1/3 mice is a lot of mice for a cat. You'll definitely be finding lots of vomited mouse bits on the carpet if the cat tries to eat that many mice. And, how do they work out that 1/3 mouse each? If they are fighting over the last mouse, it might take a long longer to eat that one. Plus, once the cat is full, they probably won't be very fast at chasing mice, they will go take a nap for a hour or so, during which time the mice could be running across their tails and they still wouldn't get up and chase them. And even if they kept eating (more typical of a dog than a cat though), they'd get slower and slower as they got fuller, playing with their food more before eating it. You need to factor in the changing rate of eating with fullness. See, this is an amazingly complex problem in reality, but I'm sure some math teaching writing the question wasn't aware of that :biggrin:

    (This is why I used to score really horribly on multiple choice exams. :tongue2:)
  10. Nov 24, 2004 #9
    i think the problem was more focussed on finding out whether a person can spot the mathematical answer for it, and not if it is applicable in real life or not.

    anycase, if u try to apply it in reality, then u will have to specify,
    the cat,
    the exact volume of its stomach,
    the no. of days for which it was starving to prepare for such an event,
    the amount of body fat stored in the cat,
    general inclinatio0n of the cat towards eating, and mice particularly.
    apart from that, the general conditions of temprature, humidity, and other factors, which will play a role in the cat getting tired while eating, and thus affecting its speed.

    and a very important thing which mised out, is the mice, u will have to specify the mice which are being eaten by the cats, the smaller the mice, the faster and more in no. can be eaten by the cats.

    after considering all this, i came up with an equation, but it was way too complicated for me to understand, forget about explaining it to you all.
    [ just kidding :) ]

    but still i think that it possible for three extremely large cats,( maybe, lions) to eat 33 1/3 mice without vomiting.

    so, it was not such an unrealistic problem after all.
  11. Nov 24, 2004 #10


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    Ah, you've beat me at my own game...but of course, cat is a very generic term and I was foolishly assuming a house cat. 33 1/3 mice in 100 minutes would be no problem at all for a large cat like a lion or tiger, at least assuming they aren't ob/ob mice (those are the ones with a genetic mutation that allows them to get VERY fat).
  12. Nov 24, 2004 #11
    The question is How many cats?
  13. Nov 24, 2004 #12
    It takes a cat three minutes to eat one mouse, 100 mice is equivlent to 300 minutes. So, we must divide 300 by a number to get 100 minutes, that number is three. Three cats.
  14. Nov 25, 2004 #13
    well, if you are trying to solve this problem in purely mathematical terms, then the answer is simply 3 cats.

    although i don't think that this problem can be easily applied to real life situations as they would involve far too many parameters.
  15. Dec 7, 2004 #14
    33.333..... cats
  16. Dec 8, 2004 #15
    Four or more Cats
    3 cats would eat 99 mice in 99 minutes.
    Only one cat would be able to get the the last one and would not be able to eat it within the remaining one minute.

    So it would take 4 or more cats to eat 100 mice "in" (within) 100 minutes.

    RB :wink:
  17. Dec 8, 2004 #16
    Based on the data it appears that it takes a single cat 3 minutes to eat a mouse

    At that rate a single cat could eat 33.333… mice in 100 minutes

    So 3 cats could eat 3 times as much as one cat or 100 mice in 100 minutes.

    So I don't see why the answer wouldn't be just 3 cats.

    This assumes they can eat continuously at that rate and share at least one mouse as a meal. It also assumes that the don't spend any time hunting for, or playing with, their food. Very unrealistic. :biggrin:

    The brain teaser is from the way the question is worded. If it takes 3 cats 3 minutes to eat 3 mice, the first instinct is to say that it will take 100 cats 100 minutes to eat 100 mice. This brain teaser is probably meant to be told in a bar as a joke around drunkards. :wink:
  18. Dec 8, 2004 #17


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    Hmm, I have 3 cats handy. Does anyone have 100 mice I can 'borrow'? My cats are rather incompetent mousers though, I may have to tie up the mice (blindfold them as well for humane reasons).
  19. Dec 8, 2004 #18
    Well, after 99 mice, the cats won't be so hungry, and will peacefully share the last one.
    So, its gonna take only 3 cats to do the job :-)
  20. Dec 8, 2004 #19


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    And it helps if the last mouse is already cut into thirds. :biggrin:
  21. Dec 9, 2004 #20
    Well I figure it will take 3 cats about 5 minutes just to agree to share and even more time to agree on making a fair split. So the best chance is if the big cat get the last mouse. And after 33 mice they will all be large kittys. And even with there best chance it will still take 102 minutes to do the job.
    So with a 100 minute dead line I'd need 4 cats.

    But I agree using 4 cats to do the job in 75 minutes - that would be "within" 100 minutes - but not really "IN" 100 minutes. .... What to DO to solve??

    ---- Lets review ---
    In our special or general relativity view of the problem 3 cats would do.
    BUT down on the quantum level where Cats only come in whole cats
    And mice are eaten only in whole mice.
    We only see a statistical probability of how many Mice are eaten at a given defined time. Or a statistical probability of the time when exactly 100 mice will be eaten.

    So far sending cats around a Cat Accelerator into a Cat Collider has not generated any ¾ Cat or 1/4 Cat sized elements. And even if it did - they probable would not BE cats and may not be able to eat mice.

    Just a part of the problem and conflict between GR and Quantum Theory.

    There is only one thing to do - we need we need higher level math here.
    Any body know some people on String Theory (String prefered over M).
    They may be able to help us towards a unified theory and solution.!!

    Ya Think. :biggrin:

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