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Maths Help please

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quizbabe | 19:11 Mon 02nd Aug 2010 | Science
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A flour mill wheat is to be adjusted to moisture content 15% on a dry basis. If the whole grain receivd at the mill is found to contain 11.4% water initially how much water must be added per 100kilo of input grain as received to produce the required water content. The answer is 1.8 kilo to 100 kilo but I cannot work it out. Help please.
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You cant just take 3.6% and use that, because every time you add y kg of water water you are also increasing the overall weight by y kg.
I said suppose the original composition of 100kg is 88.6 kg dry grain and 11.4 kg water.

We want to add y kg of water such that water now accounts for 15% of the overall weight.

New total weight = 100+y
New amount of water =...
07:13 Tue 03rd Aug 2010
I’m with factor 30 on this, here’s how it goes;
The original 100kg contains 88.6kg grain & 11.4kg water
You want a ratio of 85kg grain to 15kg water
But you have 88.6kg of grain so you have to have proportionately more water
ie 88.6/85x15which equals 15.635kg of water
You already have 11.4kg of water in the grain so you only have to add the difference between what you have and what you want ie. 15.635 - 11.4 = 4.235
So you now have 104.235 Kg of grain containing 15.635 Kg of water which gives a water content of 14.9997 %
The question clearly states "how much water must be added per 100kilo of input grain as received to produce the required water content" so the final weight is not important. As this is a math question I really don't think an answer of even 1.81 or 1.79 is good enough.
Quizman55...the grain as received is found to contain 11.4% water (relative to its dry weight) not 11.4kg of water. Your ratio of 88.6:11.4 works out as 12.8%.

Think of this as a VAT problem, if a product costs £100 inc VAT at 17.5% then to get the ex VAT cost you divide 100 by 1.175, you don't subtract 17.5 from 100. Substitute 11.4 for 17.5, kg for £ and ex-VAT for DRYWEIGHT and you have the same calculation.
The question is poorly defined, although it may be written using a convention with which professionals in the field are familiar.
We think we now agree on what is meant by 15% on a dry basis but the question doesn't say whether the 11.4% was also calculated on a dry weight basis or whether it means water represents 11.4% of the grain as received. I think the answer that is being sought refers to "input grain as received" which I assume is the gross weight including water but before you add the extra water.

The consensus seems to be that the answer is not 1.8.
I now prefer davethedog's solution of 1.89 which assumes that the 11.4% and 15% figures are being calculated on different bases.
Having had another look at this problem I agree with Factor 30 the amount of water that needs to be added is 4.235 Kg This problem obviously was not set by a mathematician in the first place and whoever did set it assumed that it would be half of the 3.6 Kg "difference" in the 11.4% and 15% water content without calculating the percentage of the new weight of grain. The correct answer cannot possibly be 1.8 Kg. So to sum up (pardon the pun) the consist of the modified grain would be 88.6 Kg Dry grain (original value) 15.635 Kg water therefore the new percentage of water in the modified grain is 15.635/104.235 *100 = 14.999760157.......%
It doesn't seem an unreasonable assumption that the 11.4% has been calculated on the same basis as the 15%

The mistake in the question as posed would appear to be, either 11.4% should have read 11.4kg (in which case I still have an issue with 1.89 being rounded to 1.8) or 1.8kg is wildly wrong and the answer should 3.23kg.

Go here for actual practice and note the warning at the end which cautions against adding water directly to grain and not seeking to increase content by this method by more than 2%!! http://www.ag.ndsu.ed...tsci/crops/ae905w.htm
Factor30...Table 1 & 2 in the link from my last post gives a weight increase of 100kg in going from 11.5% to 15% (calculated on a wet basis) in the order of 4kg which is in line with your original calculation

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