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Today's Brain Teaser - 02/15/2008 (61) Today's Brain Teaser - 02/15/2008 (61)
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Old 02-15-2008, 08:09 AM   #1 (permalink)
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Talking Today's Brain Teaser - 02/15/2008 (61)
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.

How many calories does a steer of 420 kilograms require?
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Old 02-15-2008, 08:16 AM   #2 (permalink)
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9,000?
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Old 02-15-2008, 08:20 AM   #3 (permalink)
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9001 (I actually agree with DJAXX)
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Old 02-15-2008, 08:22 AM   #4 (permalink)
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13,500[( 420/630)1/3]2 = 10,300 calories
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Old 02-15-2008, 08:27 AM   #5 (permalink)
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Quote:
Originally Posted by clevalley View Post
13,500[( 420/630)1/3]2 = 10,300 calories
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Old 02-15-2008, 08:32 AM   #6 (permalink)
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Originally Posted by clevalley View Post
13,500[( 420/630)1/3]2 = 10,300 calories
I like how there's no explanation for how he came up with this equation...
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Old 02-15-2008, 08:36 AM   #7 (permalink)
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Originally Posted by somd_newbie View Post
I like how there's no explanation for how he came up with this equation...
Surface area is proportional to linear dimension squared.

Therefore, 13,500 x [ ( 420/630)1/3 ] 2 = 10,300 calories
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Old 02-15-2008, 08:38 AM   #8 (permalink)
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I came up with 900.6 to be exact, but is .6 of a calorie like half a nibble of a snickers bar?
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Old 02-15-2008, 08:50 AM   #9 (permalink)
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Quote:
Originally Posted by clevalley View Post
13,500[( 420/630)1/3]2 = 10,300 calories
Correct!

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

Stay tuned for tomorrow's!

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Old 02-15-2008, 10:13 AM   #10 (permalink)
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Gawd, I feel like such a tard!
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