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# Re: SQL Fun Challenge #3

From: Noel <tbal_at_go2.pl>
Date: Fri, 12 Mar 2004 13:36:59 -0000
Message-ID: <c2salh\$sbu\$1@inews.gazeta.pl>

Uzytkownik "FlameDance" <flamedance_at_gmx.de> napisal w wiadomosci news:c2s3qo\$kcc\$03\$1_at_news.t-online.com...
> Daniel Morgan wrote:
>
> As the bird (constant 20mph) flies slower than the trains (constant
> 50mph) it won't fly back and forth at all, instead it will lag behind
> "Train 1". It will fly one-way until the trains collide, at 2/5th the
> trains speed. The trains collide after 114/2 miles, so the bird flies
> (114 miles *2)/(2*5) = 114/5 = 22.8 miles in the same time span.
>
> Or did I misunderstand the setup? It appears to me that to make sense
> the question requires that the bird flies faster than the trains.

The trains colide in (114/(50+50)) hours, so if bird fly with 20mph, then he flies (114/100) * 20 = 22.8 miles.
If bird speed would be 70 mph then it flies 79,8 miles.

Ah, SQL
SELECT (114/(50+50)) * 20 FROM DUAL;
;-)

```--
Noel
```
Received on Fri Mar 12 2004 - 07:36:59 CST

Original text of this message

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