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$1$ man can eat $1$ apple in $1$ day.How many apples can six men eat in six days?

My approach:

$1$ men can eat $1$ apple in $1$ day.

Now, 6 men can eat in 6 days 6 apples only assuming each men eat $1$ apple a day.

Am I right in my approach?

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    $\begingroup$ Nope. Each man has all of the six days to eat apples : ) $\endgroup$ – MonadBoy Oct 4 '15 at 11:21
  • $\begingroup$ @A.Sh Can you explain in detail what you are trying to say? $\endgroup$ – Jack Oct 4 '15 at 11:23
  • $\begingroup$ It depends on if they get fed up with apples at any point or keep the same apple eating speed. $\endgroup$ – mathreadler Oct 4 '15 at 11:24
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    $\begingroup$ @Jack Sure: 1 man can eat 1 apple in 1 day. In 6 days, that one man can eat 1 apple each day, so in total, 1 man can eat 6 apples in 6 days. Now multiply by the number of men... $\endgroup$ – MonadBoy Oct 4 '15 at 11:24
  • $\begingroup$ I can eat like twenty apples a day before I get sick. These puzzles are silently teaching you to view any relationship as (multi-)linear. Sure, work out the math that is intended for them, but also make sure to stay real, you know. If you can scramble five eggs in five minutes using one pan, how long does it take to scramble three eggs using four pans? Well, just use one of those pans to scramble them in five minutes as well! You know what I’m sayin’? $\endgroup$ – k.stm Oct 4 '15 at 11:31
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no. 1 man can eat 1 apple in 1 day therefor 1 man can eat 6 apples in 6 days. therefor 6 men can eat 36 apples in 6 days....

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    $\begingroup$ But we need to assume no one gets fed up with apples at any point. $\endgroup$ – mathreadler Oct 4 '15 at 11:29
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    $\begingroup$ right ;) :D :D :D and we have to assume that nobody becomes greedy and so on.... ;) $\endgroup$ – Börge Oct 4 '15 at 11:29
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    $\begingroup$ And that no one loses motivation. $\endgroup$ – mathreadler Oct 4 '15 at 11:36
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    $\begingroup$ haha :) :P Carry on $\endgroup$ – Jack Oct 4 '15 at 11:38
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    $\begingroup$ @mathreadler To overcome the obstacles you seem to foresee just keep in mind that 1 apple a day keeps 1 doctor away ;-) $\endgroup$ – Thomas Oct 4 '15 at 12:29
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The following piece by Lewis Carroll first appeared in ``The Monthly Packet'' of February 1880 and is reprinted in _The_Magic_of_Lewis_Carroll_, edited by John Fisher, Bramhall House, 1973.

                             Cats and Rats

If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100 rats in 50 minutes?

This is a good example of a phenomenon that often occurs in working problems in double proportion; the answer looks all right at first, but, when we come to test it, we find that, owing to peculiar circumstances in the case, the solution is either impossible or else indefinite, and needing further data. The 'peculiar circumstance' here is that fractional cats or rats are excluded from consideration, and in consequence of this the solution is, as we shall see, indefinite.

The solution, by the ordinary rules of Double Proportion, is as follows:

   6 rats   :  100 rats  \
                          >   :: 6 cats : ans.
  50 min.   :    6 min.  /

Then:  ans = (100)(6)(6)/(50)(6) = 12

But when we come to trace the history of this sanguinary scene through all its horrid details, we find that at the end of 48 minutes 96 rats are dead, and that there remain 4 live rats and 2 minutes to kill them in: the question is, can this be done?

Now there are at least four different ways in which the original feat, of 6 cats killing 6 rats in 6 minutes, may be achieved. For the sake of clearness let us tabulate them: A. All 6 cats are needed to kill a rat; and this they do in one minute, the other rats standing meekly by, waiting for their turn. B. 3 cats are needed to kill a rat, and they do it in 2 minutes. C. 2 cats are needed, and do it in 3 minutes. D. Each cat kills a rat all by itself, and take 6 minutes to do it.

In cases A and B it is clear that the 12 cats (who are assumed to come quite fresh from their 48 minutes of slaughter) can finish the affair in the required time; but, in case C, it can only be done by supposing that 2 cats could kill two-thirds of a rat in 2 minutes; and in case D, by supposing that a cat could kill one-third of a rat in two minutes. Neither supposition is warranted by the data; nor could the fractional rats (even if endowed with equal vitality) be fairly assigned to the different cats. For my part, if I were a cat in case D, and did not find my claws in good working order, I should certainly prefer to have my one-third-rat cut off from the tail end.

In cases C and D, then, it is clear that we must provide extra cat-power. In case C less than 2 extra cats would be of no use. If 2 were supplied, and if they began killing their 4 rats at the beginning of the time, they would finish them in 12 minutes, and have 36 minutes to spare, during which they might weep, like Alexander, because there were not 12 more rats to kill. In case D, one extra cat would suffice; it would kill its 4 rats in 24 minutes, and have 24 minutes to spare, during which it could have killed another 4. But in neither case could any use be made of the last 2 minutes, except to half-kill rats---a barbarity we need not take into consideration.

To sum up our results. If the 6 cats kill the 6 rats by method A or B, the answer is 12; if by method C, 14; if by method D, 13.

This, then, is an instance of a solution made indefinite by the circumstances of the case. If an instance of the impossible be desired, take the following: If a cat can kill a rat in a minute, how many would be needed to kill it in the thousandth part of a second? The mathematical answer, of course, is 60,000, and no doubt less than this would not suffice; but would 60,000 suffice? I doubt it very much. I fancy that at least 50,000 of the cats would never even see the rat, or have any idea of what was going on.

Or take this: `If a cat can kill a rat in a minute, how long would it be killing 60,000 rats?' Ah, how long, indeed! My private opinion is that the rats would kill the cat.

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