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"Wanderer" <wanderer@ticnet.com> wrote in message
news:111gbq8hoalf388@corp.supernews.com...
> For those not on the board (and since this excellent math student managed
> to miss the Lenny... grrr), here's how it shakes out:
>
> There were five friends, a Kacheek, a Tuskaninny, a Lenny, an Acara, and a
> Bori. The Tuskaninny's age was that of the Acara plus the Kacheek. The
> Bori
> was twice the Acara's age, or the same age as the Kacheek and the
> Tuskaninny
> put together. The Lenny's age was three times the average of all five of
> their ages.
>
> If the Acara is four months older than the Kacheek, how many months old is
> the Lenny?
>
> Using the first letters to save space...
>
> T=A+K
> B=2A
> 2A=K+T
> L=3((K+T+L+A+B)/5)
>
> Now you define all of the variables by their relation to K. You can't do
> this with L, naturally, since it's represented by an average.
>
> 2A=K+(A+K)
> 2A=2K+A
> 2A-A=2K+A-A
> A=2K
>
> B=2A
> B=2(2K)
> B=4K
>
> 2A=K+T
> 2(2K)=K+T
> 4K=K+T
> T=3K
>
> Now you have:
>
> Acara=2K
> Bori=4K
> Kacheek=K
> Tuskaninny=3K
>
> Now, the Lenny tells you the Acara is just four months older than the
> Kacheek.
>
> A=2K=K+4
>
> 2K=K+4
>
> K=4
>
> Therefore:
>
> Acara=8 months
> Bori=16 months
> Kacheek=4 months
> Tuskaninny=12 months
>
> Now, it's quite true this doesn't give you the Lenny's age directly.
> Returning to the problem, though, notice that the average is the sum
> divided by five. It is then multiplied by three. Therefore, the final
> product has both three and five as factors. I just plugged in 15s until I
> hit 60:
>
> L=3((8+16+4+12+L)/5)
> L=3((8+16+4+12+60)/5)
> L=3(100/5)
> L=3(20)
> L=60
>
> I missed the Lenny entirely, so I might as well show off here... since I
> have no chance of finding out about all the various chocolate items for
> the current one...
>
I told ya all pretty much the same thing some time ago. (just didn't
spell the whole thing out) And I _still_ despise algebraic word problems.
Balyn