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# Chronicles of the Redrum vs. Ecidius Prank War: The Purple Pool Party

## Question

(A comical aside to the B-kun/Y-san series...a snippet from the past...)

Having won their third consecutive state swim meet, Ecidius University was celebrating by throwing a pool party for the athletes, coaches, and their friends at their brand spanking new new pool (donated by one of their many alums who had more money than they knew what to do with courtesy of the .com era).

Ecidius's star athlete, as well as the indisputable top of his class in academics, B-kun, arrived early in order to get in his daily 100 laps (after having run his daily 10 miles, doing his daily 200 push-ups, and practicing his martial arts and kendo for his daily 2 hours...each).

As he approached, he noticed something strange...the pool water was...purple. He dipped his finger into it, and the hue left a stain on his skin.

He scowled as the lopsidedly smiling face popped into his mind. That woman...

After running some tests with the aid of a few (of many) of his admirers in the chemistry department, he determined that there was 5 kg of dye in the 60,000 gallons of water in the pool. He knew that the pool's filtering system could pump out water, remove the dye, and return the water to the pool at a rate of 200 gallons/min.

The doting chemistry majors had also informed him that the effects of the dye became imperceptible when the concentration came below 0.02 grams/gallon.

The pool party started in 4 hours, but he was slated to deliver a speech before anyone could jump in. How long would his speech need to last for to ensure no guests ended up with purple-tinted skin?

## 4 answers to this question

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Three hours, seven minutes and about 25 seconds.

Because every minute 200/60000 = 1/300 of the pool water is purified.

Thus (299/300)n is the concentration multiplier after n minutes.

Since we want 5000/60000 = 0.08333... to become 0.02 g/gal,

we want the concentration to be multiplied by 0.24.

Assuming perfect mixing of the pool water as it is being purified, this gives

(299/300)n = 0.24.

Logarithms are for nerds, so I tapped my old HP45 pocket calculator 427 times and estimated the fractional deficit to determine that n = 427.42 minutes. Subtract the four-hour lead time, and he has quite a speech to deliver.

But if you know what it is/was, you also know about penny post cards most likely.

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Leave it. And change the poem to read, When I am old I will become purple.

Working on an answer. You would have to invoke calculus...

The simple answer, which is wrong but permitted by an inaccurate reading

of the OP, is don't worry, the pool will be fine before the guests arrive.

Initial concentration is 0.08333 and final must be 0.02 g/gal.

Thus 76% [45,600 gallons] of the water must have its dye removed.

45,600/200 = 152 minutes = 2 hours and 32 minutes.

Plenty of time.

Of course that's not how it works.

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Eh, close enough...off from the calc solution by about a minute...

...of the amount (in mass) of dye at any time is q(t)=5000e^(-t/300), which gives ~7.136 hrs til the pool's 'safe'

Of course, when B-kun woke up the next morning with a sore threat and had lost his voice, he realized that his opponent, being a chemical engineer and a calculus lover, had been two steps ahead of him all along ;P

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The difference coming from [my] using finite 1-minute steps.

You allowed time to prefix "Unaccustomed as I am to public speaking ..."

Nice puzzle.

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