Todays Brain Teaser #33 (09JUL2013)

vpnavy

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Today's Brain Teaser #33 (09JUL2013)

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A steer weighing 630 kilograms requires 13,500 calories a day to maintain its weight. That amount of food turns out to be proportional to its external surface.
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How many calories does a steer of 420 kilograms require?
 

Uncle Jeff

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Grrr VPNavy makin me due math and hurt my brain ...

10,300 calories
 

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vpnavy

vpnavy

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steer.gif
A steer weighing 630 kilograms requires 13,500 calories a day to maintain its weight. That amount of food turns out to be proportional to its external surface.
metal_detector.gif
How many calories does a steer of 420 kilograms require?
Grrr VPNavy makin me due math and hurt my brain ...10,300 calories
Correct!
yahoo.gif


Weight is proportional to linear dimension (length or girth of the steer) cubed. Surface area is proportional to linear dimension squared. Therefore - 13,500 x [ (420/630)1/3 ] 2 = 10,300 calories!

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GMD52

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VP, please correct me if I read everything wrong, but the math you did in the solution factered out to 10,330, making Uncle Jeff the winner, right?
 

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vpnavy

vpnavy

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VP, please correct me if I read everything wrong, but the math you did in the solution factered out to 10,330, making Uncle Jeff the winner, right?
metal_detector.gif
Wow - sorry gang - blew that one big time - it has now been corrected!
 

BryanM362

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Bryan,

Looks like we went to the same math class.:icon_thumright:

I think so, and it seems correct to me.

VP, can you elaborate on the solution, it doesn't make sense to me. Unless the weight and number of calories are not directly proportionate?

This is what I did:

X= number of calories for 420Kg steer needs.

630 420
----- X ----- = 630X = 420 x 13,500
13,500 X

630X = 5,670,000


5,670,000
x = ------------ = 9,000 calories
630


Thanks!


edit: Can't get the spaces to stay where I want them. Hopefully, you can figure out what I mean.
 

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