Help with a word Problem-Brain teaser
Robin J Kent
extra credit brain teaser once every so often. We got back the last
one And I would really like someone to explain it to me.
Farmer John has many hens. He knows that 20 of them , housed in three
coops, will hatch 30 eggs in 18 days.
How long will it take for 30 hens , housed in four coops, to hatch
the same number of Eggs?
Now how I worked it was to find the rate per hen per day (of the 20)
and plug that into a formula for thirty hens,
.0833(30 hens) = 30eggs/X days to get 12 days.
But of course that is not the answer, the answer is 18 days- It
doesn't matter how many hens.
This makes no sense to me at all, What am I not seeing in the word
problem to make the number of egg producing units be irrelative. The
way I see it if it was two billion hens they would produce 30 eggs way
before 18 days went by..
David Brownridge
> Farmer John has many hens. He knows that 20 of them , housed in three
> coops, will hatch 30 eggs in 18 days.
> How long will it take for 30 hens , housed in four coops, to hatch
> the same number of Eggs?
>
> Now how I worked it was to find the rate per hen per day (of the 20)
> and plug that into a formula for thirty hens,
> .0833(30 hens) = 30eggs/X days to get 12 days.
>
> But of course that is not the answer, the answer is 18 days- It
> doesn't matter how many hens.
> This makes no sense to me at all, What am I not seeing in the word
> problem to make the number of egg producing units be irrelative. The
> way I see it if it was two billion hens they would produce 30 eggs way
> before 18 days went by..
It's an old joke : if one woman can gestate one baby in nine months,
does that mean one could produce a baby in just one month by allocating
nine women to the task ?
--
rgds
DVD
(David Brownridge) <mailto:d...@melbpc.org.au>
Stan Brown
>Farmer John has many hens. He knows that 20 of them , housed in three
>coops, will hatch 30 eggs in 18 days.
> How long will it take for 30 hens , housed in four coops, to hatch
>the same number of Eggs?
>
> The
>way I see it if it was two billion hens they would produce 30 eggs way
>before 18 days went by..
Your fallacy is the same one used by a couple of generations of
programming managers, who assume that if you add more engineers to a
software project it will get done faster.
To see what is wrong, just think of this slogan:
Nine women, one month, one baby.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.concentric.net/%7eBrownsta/
My reply address is correct as is. The courtesy of providing a correct
reply address is more important to me than time spent deleting spam.
Aakash Mehendale
>
[snip]
> > The
> >way I see it if it was two billion hens they would produce 30 eggs way
> >before 18 days went by..
>
> Your fallacy is the same one used by a couple of generations of
> programming managers, who assume that if you add more engineers to a
> software project it will get done faster.
>
See also 'The Mythical Man-Month' by Frederick Brooks.
[snip]
--
Aakash Mehendale
aak...@hotmail.com
Stan Brown
>Stan Brown wrote:
>> Your fallacy is the same one used by a couple of generations of
>> programming managers, who assume that if you add more engineers to a
>> software project it will get done faster.
>
>See also 'The Mythical Man-Month' by Frederick Brooks.
Right. Brooks' Law should be tattooed on the heart of every programming
manager: "Adding people to a late software project makes it later."
Naz
>
> Robin J Kent wrote:
>
> > Farmer John has many hens. He knows that 20 of them , housed in three
> > coops, will hatch 30 eggs in 18 days.
> > How long will it take for 30 hens , housed in four coops, to hatch
> > the same number of Eggs?
> >
> > and plug that into a formula for thirty hens,
> > .0833(30 hens) = 30eggs/X days to get 12 days.
> >
> > But of course that is not the answer, the answer is 18 days- It
> > doesn't matter how many hens.
> > This makes no sense to me at all, What am I not seeing in the word
> > way I see it if it was two billion hens they would produce 30 eggs way
> > before 18 days went by..
>
> does that mean one could produce a baby in just one month by allocating
> nine women to the task ?
The key word in the chicken problem, of course, being _hatch_. Not lay,
or produce, but _hatch_.
-N-
Charlie Revie
>
> Stan Brown wrote:
> >
> [snip]
> > > The
> > >way I see it if it was two billion hens they would produce 30 eggs way
> > >before 18 days went by..
> >
> > programming managers, who assume that if you add more engineers to a
> > software project it will get done faster.
> >
>
> See also 'The Mythical Man-Month' by Frederick Brooks.
>
Hey! Where I work given a serial process similar to that of making one
human baby some of the genius level seem to think they they can throw
money at it, hire nine human females to create a parallel process and
create that one human baby in one month.
Charlie
Robin J Kent
>Robin J Kent wrote:
>
>> Farmer John has many hens. He knows that 20 of them , housed in three
>> coops, will hatch 30 eggs in 18 days.
>> How long will it take for 30 hens , housed in four coops, to hatch
>> the same number of Eggs?
>>
>> Now how I worked it was to find the rate per hen per day (of the 20)
>> and plug that into a formula for thirty hens,
>> .0833(30 hens) = 30eggs/X days to get 12 days.
>>
>> But of course that is not the answer, the answer is 18 days- It
>> doesn't matter how many hens.
>> This makes no sense to me at all, What am I not seeing in the word
>> problem to make the number of egg producing units be irrelative. The
>> way I see it if it was two billion hens they would produce 30 eggs way
>> before 18 days went by..
>
>It's an old joke : if one woman can gestate one baby in nine months,
>does that mean one could produce a baby in just one month by allocating
>nine women to the task ?
That would make sense except that I just confirmed that "real" hens
average an egg every two days.
But onto the next one...,
5 people get together to play poker once a week at a table with 6
chairs. How many times can they play without duplicating the seating
arrangement?
All I can start with is that it will be a rather large number. So
I'm thinking that it would be Six raised to the power of 5.
But that seems too easy and besides 5^6 sounds almost as good.
Is this the right track and if not is there a way to prove it without
in listing 4 friends and a table?
Steve Monson
>5 people get together to play poker once a week at a table with 6
>chairs. How many times can they play without duplicating the seating
>arrangement?
>
> All I can start with is that it will be a rather large number. So
>I'm thinking that it would be Six raised to the power of 5.
>But that seems too easy and besides 5^6 sounds almost as good.
> Is this the right track and if not is there a way to prove it without
>in listing 4 friends and a table?
Let's number the people from 1-5. The chairs could then be likened
to a 6-digit number, each of whose digits can be 1,2,3,4 or 5, to
indicate who's sitting in each chair. A zero indicates no one in that
chair.
Now, there are 5^6 6-digit numbers: 000000 ... 555555
But that's not what we want, since there's always a zero somewhere
in the number, as well as one each of the digits 1-5. Thus, our
usable numbers are
012345 ... 543210
Since each of these is basically a permutation of the six digits,
there are 6! = 720 different seating arrangements.
Or did I miss something critical?
Steve Monson
--
I said "no" to drugs, but they just woudn't listen.
Virgil Hancher
Monson) wrote:
>012345 ... 543210
>
>Since each of these is basically a permutation of the six digits,
>there are 6! = 720 different seating arrangements.
>
>Or did I miss something critical?
If they are sitting at a round table then there is no detectable łstarting
position" around the table, and all of
012345, 123450, 234501, 345012, 450123, and 501234
are essentialy the same arrangement, so you answer is 6!/6 = 5! = 120
--
vmh...@friix.comx | Un-x to reply
Theera
Steve Monson wrote in message <70aff6$4...@euphony.tri.sbc.com>...
>See ye here Robin J Kent's writings:
>>5 people get together to play poker once a week at a table with 6
>>chairs. How many times can they play without duplicating the seating
>>arrangement?
>>
ok, another way to look at this is to assume, with no loss of generality,
that the same person always chooses their chair first ... 6 possibilities.
and the next person always chooses second ................... 5
possibilities
and person #3 chooses third ..............................................4
psossib's
person #4
...........................................................................3
poss
person #5
.......................................................................... 2
poss
so the total number of "seatng arrangements" is 6*5*4*3*2 .. or 6 factorial.
(720)
and if you call it a "duplicate" to have everybody shift one seat to the
left....
diivide by 6 and call it 5 factorial (120) ... and if, in addition, the
empty chair's postion
didn't "change" the seating arrangement, ... it would be 4 factorial,, 24.
"some" ambiguity in the problem statement.
or so it seems to me.