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Recurrence Relations - HSN forum

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Recurrence Relations


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#1 Karennm

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Posted 07 April 2009 - 03:28 PM

The monks of St. Columbas Priory distil their own whisky. The whisky is matured for 12 years in oaks casks, during which time it loses 3.5% of its volume each year due to evaporation.
The monks prepare 8000litres of spirit estimating that this will produce at least 4000litres of whisky in 12 years time. Are they right?
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#2 Philip

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Posted 08 April 2009 - 01:30 PM

QUOTE (Karennm @ Apr 7 2009, 03:28 PM) <{POST_SNAPBACK}>
The monks of St. Columbas Priory distil their own whisky. The whisky is matured for 12 years in oaks casks, during which time it loses 3.5% of its volume each year due to evaporation.
The monks prepare 8000litres of spirit estimating that this will produce at least 4000litres of whisky in 12 years time. Are they right?
huh.gif


Hope i get this right... wink.gif

Ok, so we know that they prepare 8000 litres of spirit. And each year it loses 3.5% of volume; or you could look at it as having 96.5% of its original volume (which is probably easier here). And the time is 12 years.

This is just a depreciation question.

Therefore the numbers we have taken out of the question can be put like this:

8000(0.965^12) (Don't know why but the 2 of '12' doesn't want to be powered, just imagine it is laugh.gif )

Firstly, we have the number being inputted - 8000. Then we just multiply it by the 96.5% or 0.965 - as 1.0 would be 100%; 0.5 would be 50%. This number is the 'powered' by the number of years in question - in this case it's 12.

So all that is left to do is put the number into the equation, to get:

8000(0.965^12)=5217

Once you have written that, you write a conclusion saying: "After 12 years there is 5217 litres left. As 5217>4000, the monks are correct."

Hope this helped!

Yeah!





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