Using the formula: (W + I) * C where W = the constant of wood, which is well known to be 61, as agreed in many scientific circles. I = the variable in this equation, and stands for the word "if" from the original problem. As there are three circumstances, with 0 equaling the chance that the woodchuck cannot chuck wood, 1 being the theory that the woodchuck can chuck wood but chooses not to, and 2 standing for the probability that the woodchuck can and will chuck wood, we clearly must choose 2 for use in this equation. C = the constant of Chuck Norris, whose presence in any problem involving the word chuck must there, is well known to equal 1.1 of any known being, therefore the final part of this calculation is 1.1. As is clear, this appears to give the answer of (61 + 2) * 1.1 = (63) * 1.1 = 69.3. However, Chuck Norris' awesome roundhouse kick declares that all decimal points cannot be used in formulas such as this, and so it must be rounded to the final solution of 69 units of wood.
Source(s):
http://wiki.answers.com/Q/How_much_wood_could_a_woodchuck_chuck_if_a_woodch...

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Work = Rate * Time
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Would he care enough to chuck wood, or would the forest to wood chucking woodchucks ratio mean that he wouldnt chuck much, myself i wouldnt know
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Here is my answer, as translated to five different languages and back:


If the beaver can install and persistent tree of the structure of the meat, it would have that to install and persistent tree of the structure of the meat. But, if the beaver not possibly must be installed, it would have, is not and with the tree of the structure of the persistent meat, the tree of the structure does not have the honorarium and the persistent meat. Mine, the tree of the structure installs the best beaver of the beaver for the MASER, the end to install the tree of the structure, with the mine installed with the tree of the structure, that will surround by the persistent meat of the persistent meat with the persistent meat.
Source(s):
http://tashian.com/perl/translate-babel.cgi

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